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Math 6310. Algebra
Taught by David Zywina
Notes by Linus Setiabrata
This course is the first semester of the graduate algebra sequence at Cornell University, as it was taughtin the Fall semester of 2018. Please let me know if you spot any mistakes! These notes are probably quitefar from being typo-free. Most things in [blue font square brackets] are personal comments. Most thingsin [red font square brackets] are (important) announcements. David Mehrle’s wonderful Algebraic NumberTheory notes inspired me to do this.
Last edited May 20, 2020. Theorem numbering unchanged since Dec 3, 2018.
Contents
1 Groups 31.1 Aug 24, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Aug 27, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Aug 29, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Aug 31, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Sep 5, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Sep 7, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Sep 10, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8 Sep 12, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.9 Sep 14, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.10 Sep 17, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.11 Sep 19, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.12 Sep 21, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.13 Sep 24, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2 Rings 312.13 Sep 24, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.14 Sep 26, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.15 Sep 28, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.16 Oct 1, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.17 Oct 3, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.18 Oct 5, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.19 Oct 10, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.20 Oct 12, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.21 Oct 15, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.22 Oct 17, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3 Modules 513.23 Oct 19, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.24 Oct 22, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.25 Oct 24, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.26 Oct 29, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.27 Oct 31, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
1
3.28 Nov 2, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.29 Nov 5, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.30 Nov 7, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.31 Nov 9, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4 Fields 714.31 Nov 9, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.32 Nov 12, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724.33 Nov 14, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 744.34 Nov 16, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.35 Nov 19, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
5 Algebraic Geometry 795.36 Nov 26, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.37 Nov 28, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.38 Nov 30, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.39 Dec 3, 2018 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
2
1 Groups
1.1 Aug 24, 2018
This course has a blackboard page with some info on it. You can get some course details if you emailProf Zywina. [There is some homework due every Friday.] The first one will be until September 7, so 14days from now, and there will be a takehome final (namely, a big homework). There are some “interesting”questions to justify not-having-a-midterm.
The main book is Dummit/Foote, but we won’t follow it exactly. There are also a bunch of other booksat blackboard that you can get.
This course is “very serious”. In particular, it’s not aimed at undergraduate students. This is a 2ndcourse in abstract algebra. [You know: groups, rings, modules, fields.]
We have a TA who has office hours and does other TA things (the TA is Avery St. Dizier).
Group Actions
Let G be a group (multiplicative with 1). A (left) group action of G on a set X is a map G ×X → X,that is, (g, x) 7→ g · x = gx, such that
a) g1(g2x) = (g1g2)x for g1, g2 ∈ G and x ∈ X.
b) 1 · x = x for x ∈ X.
For g ∈ G, define the map ϕg : X → X given by x 7→ gx. Using (a) and (b), we have
• ϕg1g2 = ϕg1 ϕg2
• ϕ1 = idX
This defines a map ϕ : G → SX by g 7→ ϕg. Here SX is the group of permutations of X, which wewill sometimes denote SX or SymX . One finds that ϕ is a homomorphism. Conversely, a homomorphismG→ SX defines a group action (by g x := ϕg(x)). Sometimes this is useful for succinctness reasons.
Next semester we’ll replace SX with Aut(V ) (automorphisms of a vector space) when we do representa-tions.
Remark 1.1.1. There are right actions (we defined left actions). You can guess what they’re going to be:
a) (xg1)g2 = x(g1g2)
b) x1 = x.
But the theories are the same because a right action gives rise to a left action by g ∗ x := xg−1.
Group actions are intimately linked with the structure of G (we’ll get to that on Monday).
The action of G on X (sometimes denoted by G X) breaks x up into orbits. Say x ∼ y if x = gy forg ∈ G. Observe that ∼ is an equivalence relation:
• x ∼ x (reflexive)
• x ∼ y ⇐⇒ y ∼ x (symmetric)
• x ∼ y and y ∼ z implies x ∼ z (transitive)
This equivalence relation breaks X into pieces (these are our orbits):
3
Definition 1.1.2. The G-orbit (or just orbit, when the group is clear) of G on X containing x ∈ X isy ∈ X : x ∼ y = gx : g ∈ G = G · x (sometimes called the equivalence class of X).
Definition 1.1.3. Let G\X be the set of G-orbits in X.
We in fact haveX =
⊔O∈G\X
O,
that is, X is partitioned into disjoint orbits, and each O has a G-action.
Definition 1.1.4. We say G acts transitively on X if for any x, y ∈ X, we have y = gx for some g ∈ G.Equivalently, G\X has only one orbit.
Each G-orbit has a transitive G-action.
Example 1.1.5. Let X = G. Define the group action g ∗ x = gxg−1 (conjugation by g). Then ϕ : G →Aut(G) ⊆ SG. We will define Inn(G) := ϕ(G) to be the inner automorphisms of G; it turns out Inn(G) EAut(G). We will define Out(G) := Aut(G)/Inn(G), and it turns out that
Out(Sn) ∼=
1 if n 6= 6
Z/2Z if n = 6.
In homework 2 we’ll construct an automorphism of S6 that is not given by conjugation.
Example 1.1.6. Let X = G. Define the group action g ·x := g ·x, ie. let G act on itself by left multiplication.This gives an injective homomorphism G → SG (since it sends 1 to the identity in SG). This is a theoremof Cayley: every finite group is isomorphic to a subgroup of Sn for some n.
Example 1.1.7. Let G = z ∈ C× : |z| = 1 = S1 and X = z ∈ C : |z| ≤ 1. Then G
X by rotation.
Example 1.1.8. Sn
[n] := 1, 2, . . . , n which induces the homomorphism Sn → S[n] = Sn (given byidSn).
Example 1.1.9. Take a subgroup H of G. Then G acts by left multiplication on the cosets G/H = gH : g ∈G. This will be a key example (and in his mind, the “only” example).
Groups act on many things (at least, they act on themselves, and in practice, they generally naturallyact on many other things). Group actions sometimes gives information about the group itself.
Definition 1.1.10. Take x ∈ X. Define
Gx := g ∈ G : gx = x;
it is the stabilizer of x.
In fact, Gx is a subgroup of G. Define a (surjective) map f : G → G · x via g 7→ g · x. Take g, h ∈ G.Then
f(g) = f(h) ⇐⇒ gx = hx ⇐⇒ (h−1g)x = x ⇐⇒ h−1g ∈ Gx ⇐⇒ g ∈ hGx ⇐⇒ gGx = hGx
so the map f : G/Gx → Gx given by gGx 7→ gx is a bijection. Also, it respects G-actions: G acts on bothG/Gx and Gx, and f(ga) = gf(a)[such maps are sometimes called “G-equivariant”]. So group actions andsubgroups are intimately linked. Next time we’ll explore the following:
There is a correspondence
transitive sets with a G-action/isomorphism←→ subgroups of G/conjugacy
and if we can find interesting actions we can find interesting subgroups, and vice versa. So actions give youa way of attacking group theory.
4
1.2 Aug 27, 2018
Last time:
Let G be a group acting on a set X (so think ϕ : G→ SX).
For x ∈ X, we defined
• G · x = gx : g ∈ G[⊆ X] “a G-orbit”
• Gx = g ∈ G : gx = x ≤ G “stabilizer of x”
We have a bijection
G/Gx∼−→ G · x
gGx 7→ gx
that respects the G-actions.The key example is G/H with H ≤ G, and G acting by left multiplication. Then
GaH = g ∈ G : g · aH = aH= g ∈ G : a−1ga ∈ H= aHa−1.
Assume X is finite and x1, . . . , xh are representatives of the G-orbits. Then
|X| =h∑i=1
|G · xi|
because X is the disjoint union of the G · xi, and also
|X| =h∑i=1
[G : Gxi ]
(and notice that the terms on the right side all divide |G|).
Example 1.2.1. Fix integers 0 ≤ k ≤ n. Say that
X =
([n]
k
):= A ⊆ [n] := 1, 2, . . . , n : |A| = k.
Then G = Sn acts on X transitively. Pick a representative (any element works, since the action is transitive),say A0 = [k], then GA0
= S[k] × S[n]\[k]. Then
|X| = [G : GA0] =
|G||GA0
|=
n!
k!(n− k)!.
“Not groundbreaking.”
Proposition 1.2.2. Let H be a subgroup of a finite group G. Suppose that [G : H] is equal to the smallestprime p dividing |G|. Then H is a normal subgroup.
Proof. We have an action G
G/H, and ϕ : G → SX ∼= Sp (since |X| = [G : H] = p). Thus the firstisomorphism theorem gives
G/ ker(ϕ) → Sp
where the cardinality of G/ ker(ϕ) is not divisible by any q < p [because |G/ ker(ϕ)| divides |G|, and |G|is not divisible by any q < p], and |Sp| = p! = p(p − 1)!. This implies that |G/ ker(ϕ)| ∈ 1, p. But
5
|G/ ker(ϕ)| 6= 1 because G acts on X transitively. So |G/ ker(ϕ)| = p. Notice that ker(ϕ) consists of theelements of G that act trivially on G/H, so ker(ϕ) ⊆ G1·H = H. Thus
ker(ϕ) ⊆index p︷ ︸︸ ︷H ⊆ G︸ ︷︷ ︸
index p
so H = ker(ϕ), which is normal in G.
Proposition 1.2.3 (Burnside’s Lemma). Let G be a finite group acting on a finite set X. For g ∈ G, defineXg := x ∈ X : gx = x. Then
|G\X| = 1
|G|∑g∈G|Xg|.
Observe that the left side counts the number of G-orbits of X, whereas the right side counts the averagenumber of elements fixed by g ∈ G.
Proof. It suffices to prove this for each orbit of X. So G acts transitively, and without loss of generalityX = G/H. ∑
g∈G|Xg| =
∑g∈G
∑x∈Xgx=x
1
=∑x∈X
∑g∈Ggx=x
1
=∑x∈X|Gx|
=∑x∈X|H|
= |X||H| = |G/H| · |H| = |G|
where the fourth equality is true since GaH = aHa−1, and in particular |GaH | = |H|.
Corollary 1.2.4 (Jordan’s Lemma). Let G be a finite group acting transitively on a set X with |X| ≥ 2.Then there exists a g ∈ G that fixes no points in X.
(Such g ∈ G are sometimes called “derangements”)
Proof. Suppose not, so |Xg| ≥ 1 for all g ∈ G. By Burnside,
1 =1
|G|∑g∈G|Xg| ≥ 1
|G|∑g∈G
1 = 1
but we have |X1| = |X| ≥ 2.
Corollary 1.2.5 ([also] Jordan’s Lemma). Let G be a finite group. If H is a proper subgroup of G, thenthen there is a conjugacy class C of G such that
H ∩ C = ∅.
Remark 1.2.6. This is false for infinite groups. Also, when Prof Zywina does Galois theory, this is whathe uses to show that a subgroup H ≤ G is actually equal to G: he shows it hits every conjugacy class.
Proof of Corollary 1.2.5. There exists g ∈ G that does not fix any points in X = G/H. This says that
gaH 6= aH for a ∈ G=⇒ a−1ga 6∈ H for a ∈ G=⇒ H ∩ C = ∅,
where C is the conjugacy class of g in G.
6
p-groups
Fix a prime p.
Definition 1.2.7. A finite group is a p-group if its cardinality is a power of p.
Some examples include Z/pe1Z× · · · × Z/perZ, and the matrix group1 ∗ ∗0 1 ∗0 0 1
⊆ GL3(Fp),
which has order p3.
Remark 1.2.8. There are 49,487,365,422 groups of order 210 up to isomorphism. Also, there are 11,759,892groups of order < 210. It’s a folklore conjecture that “most” finite groups are 2-groups.
Proposition 1.2.9. Let G be a p-group acting on a finite set X. Define XG := x ∈ X : gx = x, ∀g ∈ G.Then
|X| ≡ |XG| (mod p).
Remark 1.2.10. Usually |XG| is mysterious and you use the proposition to show it’s not empty by counting|X| (mod p).
Proof of Proposition 1.2.9. Take any G-orbit O ⊆ X\XG (notice that G X\XG). Then
|O| = |G · x| = [G : Gx] = pk (G is a p-group)
Notice that |O| > 1, and |O| ≡ 0 (mod p). We have
X\XG =⊔
O⊆X\XGO,
so|X| − |XG| = |X\XG| =
∑O⊆X\XG
|O| ≡ 0 (mod p)
Proposition 1.2.11 (Cauchy). Let G be a finite group and let p be a prime dividing |G|. Then G has anelement of order p.
Proof. DefineX = (g1, . . . , gp) ∈ Gp : g1 . . . gp = 1
and observe that Z/pZ X by [1] ∗ (g1, . . . , gp) = (gp, g1, . . . , gp−1) since
gpg1 . . . gp−1 = gp (g1 . . . gp)︸ ︷︷ ︸=1
g−1p = 1.
Then |X| = |Z/pZ|p ≡ 0 (mod p), and XZ/pZ = (g, . . . , g) : g ∈ G, gp = 1. Then
g ∈ G : gp = 1 = |XZ/pZ| ≡ |X| ≡ 0 (mod p).
Since XZ/pZ is nonempty (it contains (1, 1, . . . , 1)), there is g ∈ G\1 such that gp = 1.
7
1.3 Aug 29, 2018
[There are Office Hours, on Thursdays at 12:30-2:30 in his office.]
Last time, we discussed:
Theorem 1.3.1 (Cauchy). For a finite group G and a prime p dividing |G|, there is a g ∈ G of order p.
A major tool to prove this theorem is the following
Proposition 1.3.2. Let G be a p-group acting on a finite set X. Then |X| ≡ |XG| (mod p).
There are some generalizations of these, called the three Sylow Theorems.
Sylow Theorems
We begin with
Proposition 1.3.3 (Lagrange). Let H ≤ G, where G is a finite group. Then |H| divides |G|.
The converse is false: given n ≥ 1 dividing |G|, there need not be a subgroup H with |H| = n.
Anyways, fix a prime p. Let G be a finite group. Define
Definition 1.3.4. A p-Sylow subgroup (sometimes called a Sylow p-subgroup) of G is a subgroup H ≤ Gsuch that H is a p-group and p - [G : H] (that is, if |G| = pαm with p - m, then |H| = pα).
We fix the notation Sylp(G) to denote the set of p-Sylow subgroups.
Example 1.3.5. Say G = Z/NZ, where N = pαm with p - m. It has a p-Sylow subgroup (the one generatedby [m]). Alternatively, you can recall the Chinese Remainder Theorem, which asserts
Z/NZ ∼= Z/pαZ× Z/mZ.
Example 1.3.6. Say G = GLn(Fp), the n× n invertible matrices with entries in Fp. Consider the p-group
H =
1 ∗ . . . ∗0 1 . . . ∗...
.... . .
...0 0 . . . 1
≤ G
and observe that|H| = p1+2+···+(n−1) = pn(n−1)/2
We can also compute #GLn(Fp): observe that this is the number of bases of the Fp vector space Fnp , sinceA ∈Mn(Fp) is invertible if and only if its columns form a basis of Fnp . But this is just
(pn − 1)(pn − p)(pn − p2) . . . (pn − pn−1) = p1+2+···+(p−1)n∏i=1
(pi − 1)︸ ︷︷ ︸6≡0 (mod p)
so H is actually a p-Sylow subgroup.
There are three Sylow theorems; the first shows p-Sylow subgroups exist, the second tells you how to findother ones given one, and the third gives a count on how many there are. We have the following:
Proposition 1.3.7. Let H be a subgroup of G. Suppose that G has a p-Sylow subgroup S. Then H∩(gSg−1)for some g ∈ G is a p-Sylow subgroup of H. In particular, Sylp(G) 6= ∅ =⇒ Sylp(H) 6= ∅
8
Proof. Define X = G/S. The group G, and hence the subgroup H, acts on X by left multiplication. Nowtake x := gS ∈ X. What is Hx? Well,
Hx = H ∩Gx= H ∩ (gSg−1)︸ ︷︷ ︸
p-group
.
We want to show that for some x ∈ X, we have p - [H : Hx]. So suppose not, ie. p÷ [H : Hx] for all x ∈ X.Then
|X| =h∑i=1
|Hxi| =h∑i=1
[H : Hxi ]︸ ︷︷ ︸≡0 (mod p)
where the xi represent the H-orbits in X. But now p divides |X| = |G/S| = [G : S], which can’t happensince S ∈ Sylp(G).
Theorem 1.3.8 (Sylow I). Every finite group has a p-Sylow subgroup, ie. Sylp(G) 6= ∅.
Proof. We want to embed G into a bigger group, which we know has p-Sylow subgroups (and then applyProposition 1.3.7). Cayley’s theorem says that G → Sn for some n ≥ 1. Furthermore, there is an injectivehomomorphism
Sn → AutFp(Fnp ) ∼= GLn(Fp)
given by Sn
Fnp : fix a basis e1, . . . , en of Fnp and σ(ei) = eσ(i). Now we are done, because we showedGLn(Fp) has a p-Sylow subgroup.
Remark 1.3.9. This is a nonstandard proof; it’s not a bad idea to look at the book for the standard onetoo. Prof Zywina mentioned that he got this proof from a class with Serre.
Now observe that if S ∈ Sylp(G), then gSg−1 ∈ Sylp(G) for g ∈ G, since conjugation doesn’t change thecardinality of a group. The second Sylow theorem says that all p-Sylow subgroups arise in this manner:
Theorem 1.3.10 (Sylow II). The p-Sylow subgroups of G are conjugate to each other. Moreover, everyp-subgroup of G is contained in a p-Sylow subgroup.
(So the p-Sylow subgroups are those which are maximal with respect to cardinality, but Sylow II guar-antees that they are those which are maximal with respect to inclusion.)
Proof. Fix S ∈ Sylp(G) (this uses Sylow I (Theorem 1.3.8)). Take any p-subgroup H ⊆ G. By Proposi-tion 1.3.7 there is a g ∈ G such that H ∩ gSg−1 is a p-Sylow subgroup of H. But H = H ∩ gSg−1 since His already a p-subgroup, so the unique p-Sylow subgroup is itself. Thus, H ⊆ gSg−1 ∈ Sylp(G).
Also, take any H ∈ Sylp(G). Then H ⊆ gSg−1 for some g ∈ G. But they have the same cardinality forsome g ∈ G, so H = gSg−1.
Define np(G) to be the number of p-Sylow subgroups.
Theorem 1.3.11 (Sylow III). We have np(G) ≡ 1 (mod p) and np(G) divides m, where |G| = pαm withp - m.
We’ll prove this next lecture. For now, here’s an example of the power of this theorem:
Example 1.3.12. There is no simple group of order 200 = 23 · 52. Here’s a proof:
Sylow III (Theorem 1.3.11) says that n5(G) ≡ 1 (mod 5) and n5(G) divides 8. Then n5(G) = 1 and|S| = 25. So when you conjugate S with any element, you still have S, that is, gSg−1 = S for all g ∈ G, soS is a nontrivial normal subgroup of G. Thus, G is not simple.
9
1.4 Aug 31, 2018
Last time:
Let G be a finite group and p be a prime. Denote by Sylp(G) to be the set of p-Sylow subgroups of G,that is, a p-group H ≤ G such that p - [G : H]. We had three big theorems:
Theorem 1.4.1 (Sylow I). We have Sylp(G) 6= ∅.
Theorem 1.4.2 (Sylow II). The action of G on Sylp(G) by conjugation is transitive.
Theorem 1.4.3 (Sylow III, to be proven). Let np(G) be the number of p-Sylow subgroups. Then np(G) ≡ 1(mod p) and np(G)|m, where n = pαm and p - m.
As an application of the theorems, we have:
Example 1.4.4. Let G be a group of order pq with p, q primes so that p < q and q 6≡ 1 (mod p). Then Gis cyclic.
Proof. Recall (from Example 1.3.12) that if S ∈ Sylp(G), we have np(G) = 1 ⇐⇒ S EG. Let
np := np(G) ≡ 1 (mod p)
and note that np|q. Since q is prime we have np = 1 or np = q. But q 6≡ 1 (mod p) so np = 1. Let P be theunique p-Sylow subgroup of G, so P EG.
Choose Q ∈ Sylq(G). Notice that Q acts on P by conjugation, that is, there’s a map
ϕ : Q→ Aut(P ) ∼= (Z/pZ)×
If kerϕ = 1, then |ϕ(Q)| = q|p− 1, which contradicts p < q. So ϕ = 1, ie. P and Q commute. This impliesthat G is abelian of order pq. It follows that G is cyclic.
Let’s now prove Sylow III.
Proof of Theorem 1.3.11. Recall that G acts transitively on Sylp(G) via conjugation. Fix S ∈ Sylp(G). Then
np(G) = #Sylp(G) = [G : GS ],
where GS = g ∈ G : gSg−1 = S =: NG(S) is the normalizer of S in G. In other words, np(G) = [G :NG(S)]. Note that S ⊆ NG(S). This implies that
np(G) divides [G : NG(S)][NG(S) : S] = [G : S] = m.
Also, observe that S acts on Sylp(G) by conjugation. But np(G) = #Sylp(G) ≡ #Sylp(G)S (mod p), by
Proposition 1.2.9 (recall that Sylp(G)S = S′ ∈ Sylp(G) : gS′g−1 = S′,∀g ∈ S).
Fix an S′ ∈ Sylp(G) such that gS′g−1 = S′ for all g ∈ S. This is the same as saying S ⊆ NG(S′). Alsonote that S′ ENG(S′) (so that np(NG(S′)) = 1). So S, S′ ∈ Sylp(NG(S′)), that is, S = S′. It follows that
the unique element in Sylp(G)S is S itself (we fixed S′ and showed it was S). In other words,
np(G) ≡ Sylp(G)S = 1 (mod p).
Now we can do something a little more serious.
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Example 1.4.5. We will show that A5 is the only simple group of order 60.
Define X = Syl5(G), where G is simple of order 60. Then G acts by conjugation on X; it is transitiveby Sylow II (Theorem 1.3.10). This gives a map ϕ : G → SX . Let’s compute |X| = n5(G). By Sylow III(Theorem 1.3.11), we have n5(G) ≡ 1 (mod 5) and n5(G) divides 12. So n5(G) = 1 or n5(G) = 6, but G issimple so n5(G) = 6 (since n5(G) = 1 would imply that G has a normal subgroup of order 5).
Since G is simple, kerϕ = 1 or G. But the action is transitive, so kerϕ 6= G. So kerϕ = 1, and G ⊆ S6.In fact, G ⊆ A6, since G ⊆ S6 ±1 is given by the sign map, and if G contains an element not in A6,the kernel of the sign map restricted to G would yield an index 2 normal subgroup.
Now recall that G acts on A6/G by left multiplication (observe that |A6/G| = 6). Note that G fixes 1 ·G.So G acts on Y := (A6/G)\1 ·G, where |Y | = 5. Now we have
ψ : G→ SY
and G simple implies that the kernel is trivial or G. We claim that the kernel is trivial:
Suppose not; let ker(ψ) = G. So g · aG = aG for all a ∈ A6, g ∈ G. So a−1ga ∈ G for all a ∈ A6. Thismeans that GEA6. But A6 is simple! So
ψ : G → SY
So G ⊆ S5. As before, G ⊆ A5, that is, G ∼= A5.
Remark 1.4.6. We don’t actually need that A6 is simple; we just need that A6 has no order 6 subgroups(we won’t prove this so we can end lecture on time). Later in the course we’ll meet the alternating groupsagain, anyways.
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1.5 Sep 5, 2018
[A reminder that homeworks are due on Friday]
Recall that G acts on itself by conjugation, which gives a homomorphism ϕ : G → Aut(G) ⊆ SG. Theimage is denoted Inn(G), called the “inner automorphisms”. One can check Inn(G)EAut(G):
Proof. Let f ∈ Aut(G), and g ∈ G. We want to show that f ϕ(g) f−1 is an inner automorphism: indeed,
(f ϕ(g) f−1)(x) = (f ϕ(g))(f−1(x))
= f(gf−1(x)g−1)
= f(g)xf(g)−1
= ϕ(f(g)).
Thus we can define Out(G) = Aut(G)/Inn(G).
Observe that G/Z(G) → Aut(G), where Z(G) = ker(ϕ) = x : xg = gx∀g ∈ G is the center of G. Wehave Z(G)EG. In fact, Z(G) is a characteristic subgroup of G:
Definition 1.5.1. Let H ≤ G. We say H is characteristic if f(H) ⊆ H for all f ∈ Aut(G).
Of course, H ≤ G is normal if f(H) ⊆ H for all f ∈ Inn(G). Examples of characteristic subgroups includeH = 1, H = G, and any H ≤ Z/mZ. Maybe a more nontrivial characteristic subgroup is the commutatorsubgroup of G, that is
[G,G] := 〈g−1h−1gh︸ ︷︷ ︸:=[g,h]
: g, h ∈ G〉
Lemma 1.5.2. Let N EG. Then G/N is abelian if and only if N ⊇ [G,G]. In particular, G/[G,G] is thelargest abelian quotient of G.
Proof. The quotient G/N is abelian if and only if
gNhN = hNgN ⇐⇒ g−1h−1gh ∈ N ⇐⇒ [G,G] ⊆ N.
Proposition 1.5.3. Let G 6= 1 be a p-group. Then Z(G) 6= 1.
Proof. We have an action G
X = G by conjugation. So |X| ≡ |XG| (mod p) (this is Proposition 1.2.9).In particular,
0 ≡ |G| ≡ |Z(G)| (mod p)
so p divides |Z(G)|.
Recall that a group G is simple if G 6= 1 and its only normal subgroups are 1 and G.
For example, Z/pZ is simple for p prime (Proposition 1.5.3 says that these are the only finite simplep-groups). Also, An is simple for n ≥ 5, and so is
PSLn(K) := SLn(K)/Z(SLn(K)) = SLn(K)/aI : a ∈ K×, an = 1
for n ≥ 2, and K a field with |K| > 3 if n = 2 (and of course, SLn(K) denotes n × n matrices with entriesin K and det = 1). Also, there are the sporadic groups (Monster, Baby Monster, ...).
Here’s some archaic terminology, (Prof Zywina prefers “filtrations”, or something):
Definition 1.5.4. A series of a group G is a sequence of subgroups
G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = 1
with Gi+1 normal in Gi. The integer n is called the length of the series. Since the Gi are normal, we canconsider the quotients G0/G1, . . . , Gn−1/Gn. Such a series is a composition series of G if all these quotientsare simple.
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Series always exist (take G0 = G,G1 = 1...), but it’s not clear that composition series always exist. Infact, for infinite groups, this is not always true. But:
Lemma 1.5.5. Every finite group G has a composition series.
Proof. We induct on |G|. If |G| = 1, then G = 1. Then it is true with n = 0.
Take G with |G| > 1 and assume the lemma is true for smaller groups. If G is simple then then we aredone. So assume G is not simple. Then there is a normal subgroup N E G, and can assume there is nolarger normal subgroup. By our choice of N , we have G/N is simple (otherwise, there would be a largernormal subgroup of G). But |N | < |G|, so N has a composition series N = N0 ⊃ N1 · · · ⊃ Nr = 1. ThenG ⊃ N ⊃ N1 ⊃ · · · ⊃ Nr = 1 is a composition series of G.
The integers Z do not have a composition series, so this lemma is false for infinite groups. This is becausenontrivial subgroups of Z are isomorphic to Z itself, and Z is not simple, so you could never end a chain.
Theorem 1.5.6 (Jordan-Holder). Let G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = 1 be a composition series of a group G.Then the simple quotients G0/G1, G1/G2, . . . , Gn−1/Gn are unique up to isomorphism and reordering (i.e.,regardless of the choice of composition series), counted with multiplicities. The length does not depend onthe composition series either.
This is really powerful. As an example, consider G = Z/6Z and observe that
Z/6Z = G0 ⊃ G1 = 〈2〉 ⊃ G2 = 1
induces G0/G1∼= Z/2Z and G1/G2
∼= Z/3Z, whereas
Z/6Z = G0 ⊃ G1 = 〈3〉 ⊃ G2 = 1
induces G0/G1∼= Z/3Z and G1/G2
∼= Z/2Z.
Example 1.5.7. Consider G = S5. Notice that Sn ⊃ An ⊃ 1 (recall that An is simple so this is a compositionseries). But there is no NESn such that Sn/N ∼= An and N ∼= Z/2Z. This is because An acts by conjugationon N ∼= Z/2Z so An commutes with N , which implies Z(An) 6= 1 (which is false, apparently!).
Example 1.5.8. Let |G| = 1024. The quotients that appear in the composition series are just Z/2Z, sinceall quotients that would appear are 2-groups, so if they were simple they would have to be Z/2Z.
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1.6 Sep 7, 2018
[The new homework is up.]
Last time, we stated a very miraculous theorem. Recall:
A composition series for a group G is a series
G = G0 ⊃ G1 ⊃ · · · ⊃ Gn = 1
where Gi+1 is normal in Gi, and Gi/Gi+1 are simple. In general, groups don’t have composition series, butfinite groups do. There may be multiple composition series; last class we saw the following (miraculous)theorem:
Theorem 1.6.1 (Jordan-Holder). Take any composition series as above. Then the simple groups
G0/G1, . . . , Gn−1/Gn
depend only on G up to isomorphism and reordering.
As a remark, this theorem also holds for infinite groups which have a composition series. We defined thelength of G, denoted by `(G) := n, and define `(G) =∞ if G does not have a composition series.
Example 1.6.2. Fix an integer m ≥ 2. Consider G = Z/mZ. Factor m = p1 . . . pr into primes. We havethe composition series
Z/mZ ⊃ p1Z/mZ ⊃ p1p2Z/mZ ⊃ · · · ⊃ p1 . . . pr︸ ︷︷ ︸=m
Z/mZ = 0
which gives quotients Z/p1Z,Z/p2Z, . . . ,Z/prZ with cardinalities p1, . . . , pr. Jordan-Holder says that theseprimes are unique up to reordering; the uniqueness amounts to the Fundamental Theorem of Arithmetic.
Let’s prove Jordan-H(Theorem 1.5.6)
Proof. Take any simple group S. Define the number e(G,G•, S) to be the number of factors Gi/Gi+1 with0 ≤ i < n that are isomorphic to S. We need to show that e(G,G•, S) does not depend on G•. We provethis by induction on n.
The n = 0 case is easy: we have G = G0 = 1 and hence e(G,G•, S) = 0. So assume n ≥ 1, and that thetheorem is known for composition series of length strictly less than n. If G is simple, then G = G0 ⊃ G1 = 1and e(G,G•, S) = 1 if S ∼= G and 0 otherwise. If G is not simple, then it has a normal subgroup N . For0 ≤ i < n, define Ni := N ∩Gi and observe that we have the chain of inclusions
N = N0 ⊇ N1 ⊇ · · · ⊇ Nn = 1
which is not necessarily a composition series. Similarly, for 0 ≤ i < n, define (G/N)i := GiN/N , andsimilarly observe that we have the chain of inclusions
G/N = (G/N)0 ⊇ (G/N)1 ⊇ · · · ⊇ (G/N)n = 1.
This gives short exact sequences
1 Ni Gi GiN/N 1,
which in turn give short exact sequences (“this is a good exercise to check if you know your isomorphismtheorems well”)
1 Ni/Ni+1 Gi/Gi+1 (G/N)i/(G/N)i+1 1.
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Since Gi/Gi+1 is simple, exactly one of Ni/Ni+1, and (G/N)i/(G/N)i+1 is simple and isomorphic toGi/Gi+1, and the other is 1. So we have
e(G,G•, S) = e(N,N•, S) + e(G/N, (G/N)•, S),
where we abuse notation a little because N• and (G/N)• are not composition series, but they still meanwhat they should mean (number of quotients which are isomorphic to S). We want to show that both thesenumbers are strictly less than e(G,G•, S), so that we can apply the induction hypothesis.
The quotient Ni/Ni+1 is simple or equal to 1 for all 0 ≤ i < n. In fact, Ni/Ni+1 = 1 for some 0 ≤ i < n(otherwise, we have (G/N)i = (G/N)i+1 and we get G/N = (G/N)0 = · · · = (G/N)n = 1, which impliesG = N). So we can obtain a composition series of length strictly less than n by removing some Ni. Theinductive hypothesis implies e(N,N•, S) depends only on N and S. Similarly, (G/N)i = (G/N)i+1 for somei (otherwise, Ni = Ni+1 for all 0 ≤ i < n, so N = N0 = · · · = Nn = 1 implies N = 1), so e(G/N, (G/N)•, S)depends only on G/N and S. So since
e(G,G•, S) = e(N,N•, S) + e(G/N, (G/N)•, S),
we conclude that e(G,G•, S) depends on N,G/N , and S (and not G•).
Remark 1.6.3. In the proof, we actually showed that for N EG, we have `(G) = `(N) + `(G/N). This istrue when G doesn’t even have a composition series (so if G doesn’t have a composition series, then eitherN or G/N doesn’t have one!)
Remark 1.6.4. There is a Holder program for classifying finite groups. It has two ridiculously hard steps.The idea is:
1. Classify finite simple groups
2. Find all ways to “put simple groups together”.
Composition series are related to the second part, i.e. the study of how can we put simple groups together.
Part 1 was solved∗ in the ’80s. Maybe not everything was published and some people are still reprovingthings. There are:
• Infinite families: Z/pZ, An, for n ≥ 5
• 16 families of Lie types (such as PSLn(Fq) = SLn(Fq)/Z(SLn(Fq)), for n ≥ 2 and q > 3 if n = 2).
• 26 sporadic groups (Monster, Baby Monster, Thompson, Janko,...)
There’s some overlap in these families. For example PSL2(F4) and PSL2(F5) are simple of order 60, sothey’re both isomorphic to A5.
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1.7 Sep 10, 2018
Today, we’re going to show An is simple for n ≥ 5. We’re going to prove this by induction; the base casewas the homework. There are a lot of proofs of this; this will be one that is hopefully not “magical” (thatis, we won’t have to appeal to any random facts about cycles)
Definition 1.7.1. Fix n ≥ 1. A group G acts n-transitively on a set X if |X| ≥ n and G acts transitivelyon (x1, . . . , xn) : xi ∈ X distinct, where g · (x1, . . . , xn) = (gx1, . . . , gxn).
For example, Sn acts n-transitively on [n] = 1, 2, . . . , n. A slightly less trivial example is the following:
Example 1.7.2. The gorup An acts (n− 2)-transitively on [n]. Indeed, pick distinct a1, . . . , an−2 ∈ [n] anddistinct b1, . . . , bn−2 ∈ [n]. There are two σ ∈ Sn such that σ(ai) = bi for 1 ≤ i ≤ n − 2; they differ by atransposition. So one of the σ is in An.
Note that if σ is n-transitive, then it’s k-transitive for 1 ≤ k ≤ n.
Lemma 1.7.3. Let G be a group acting 2-transitively on X. Then Gx (the stabilizer of x ∈ X) is a maximalsubgroup of G for any x.
This will be a nice way of generating large subgroups.
Proof. If not, then Gx < M < G for some M . Recall that there is a natural bijection G/Gx → X.
G/Gx =⊔
C∈G/M
κ : κ ∈ G/Gx, κ ⊆ C
so thatX =
⊔C∈G/M
XC
where [G : M ] > 1, and |XC | = [M : Gx] > 1. Furthermore, G permutes the XC amongst themselves. HereXC denotes the image of κ : κ ∈ G/Gx, κ ⊆ C under the bijection G/Gx → X.
This will contradict 2-transitivity on G X: take distinct x1, x2 in the same XC and distinct y1, y2 notin the same set XC . There is no g such that gx1 = y1 and gx2 = y2. Otherwise, G does not permuteXC.
We now prove An is simple. “It’s not an easy proof”.
Theorem 1.7.4. The group An is simple for n ≥ 5.
Proof. Let n ≥ 6 and assume that An−1 is simple. By abuse of notation, we identify An−1 with a subgroupof An in the following way: for a σ ∈ An−1, define τ ∈ An by τ(i) = σ(i) for i ≤ n − 1, and τ(n) = n.In his notation, An−1 = (An)n = σ ∈ An : σ(n) = n, that is, it’s the stabilizer of n ∈ [n] under the ac-tion An
[n]. Note that An acts 2-transitively on [n], so An−1 ⊆ An is a maximal subgroup, by Lemma 1.7.3.
Suppose N 6= 1 is a normal subgroup of An. We need to show that N = An. Consider N ∩An−1EAn−1.Since An−1 is simple we have N ∩An−1 = 1 or An−1. We have two cases:
Case 1. We have N∩An−1 = An−1. Since An−1 is maximal either N = An (and we’re done) or N = An−1.But An−1 is not normal in An: conjugate any element by the transposition (1, n) and we no longer fix n, sowe’re not in An−1.
Case 2. We have N ∩ An−1 = 1. So the only element of N that fixes n is the identity. Thus, observeN · An−1 ⊃ An−1, since N 6= 1 and N ∩ An−1 = 1; since An−1 is maximal we have N · An−1 = An. Weconclude |N | = n, e.g. because there is a surjective map An−1 An−1 · N/N = An/N , whose kernel isN ∩ An = 1. Consider the map ϕ : N → [n] which sends g 7→ g · n. We claim that ϕ is bijective: Thereis σ ∈ An so that σ(n) = i, and since An = N · An−1 we can decompose σ = gh for g ∈ N,h ∈ An−1, so
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i = σ(n) = g(h(n)) = g(n) so that ϕ is surjective.
We claim that ϕ is An−1-equivariant, where An−1
N by conjugation and An−1
[n] as usual (i.e., “theseactions are compatible”). That is, we should show that for any g ∈ N,h ∈ An−1, we have ϕ(hgh−1) = h·ϕ(g).But
ϕ(hgh−1) = hgh−1 · n = hg · n,
since h−1 ∈ An−1. Notice also that h · ϕ(g) = h(g · n), so that ϕ(hgh−1) = h · ϕ(g).
By factoring through ϕ, we get an action An−1
N − 1 by conjugation that is 3-transitive (using thefact that An−1
[n − 1] is (n − 3)-transitive). So Aut(N) acts 3-transitively on N − 1 and |N | = n ≥ 6.This will prove to be too good to be true:
Lemma 1.7.5. There is no finite group G with |G| ≥ 5 such that Aut(G) acts 3-transitively on G− 1.
Proof. Suppose G exists. Then all g ∈ G − 1 have the same order, since Aut(G) preserves order. Theorder must be a prime p, by Proposition 1.2.11.
As always, suppose first that p > 2. Since |G| ≥ 5, there is x ∈ G − 1, and y ∈ G − 1, x, x−1.We have two pairs (x, x−1) and (x, y). There is no f ∈ Aut(G) such that f(x) = x and f(x−1) = y, since1 = f(xx−1) = xy but y 6= x−1. So G
X is not 2-transitive. On the other hand, if p = 2, then G is abelian(for a, b ∈ G, we have ab = a−1b−1 = (ba)−1 = ba). So G = Fr2 with r ≥ 3. Then Aut(G) = GLr(F2). Onecan check that GLr(F2) Fr2 − 0 is not 3-transitive.
This proves that An is simple for n ≥ 5.
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1.8 Sep 12, 2018
Let K be a field, and n ≥ 2. Recall that
PSLn(K) := SLn(K)/Z(SLn(K)) = SLn(K)/aI : an = 1.
We’re going to prove today that PSLn(K) is simple for all n ≥ 2 and fields K, where |K| > 2 if n = 2. Thetechniques we use for this group will apply to the groups of Lie type, so except for the sporadic groups we’llhave everything. Note that PSL2(F2) ∼= S3 and PSL2(F3) ∼= A4.
Here’s a theorem, due to Iwasawa:
Theorem 1.8.1. Let G be a group acting 2-transitively on a set X. Suppose that:
a) G is “perfect”, that is, G has no nontrivial abelian quotients (equivalently, we have G = [G,G]).
b) for some x ∈ X, the stabilizer Gx has a normal subgroup A that is abelian and ∪g∈GgAg−1 generate G.
Then G/H is simple, where H is the kernel of the action of G
X
(Recall that G X gives ϕ : G→ SX ; the kernel of the action is H.)The conditions seem very restrictive, but it apparently applies to most of the groups of Lie type. In our
case, G is SLn; this is easier to work with.
Proof. Take a normal subgroup N of G, with N ) H. We need to show that N = G, since G/H is simple ifand only if G has no normal subgroup containing H except H and G.
Take x ∈ X. Define M := Gx; it is a maximal subgroup of G (this was proven in Lemma 1.7.3; weused G
X is 2-transitive). So M = Gx ⊇ H, because elements of H fix all elements (it’s the kernel of theaction), whereas elements of Gx fix only x. Since M is maximal, N ·M is equal to M or G.
Suppose that N ·M = M , i.e. that N ⊆M . Take g ∈ G and n ∈ N ; since NEG we have g−1ng ∈ N ⊆M .So n · gM = gM , that is, n fixes G/M = G/Gx = X, so n ∈ H. But this says N ⊆ H, even though weassumed N ) H.
So G = N ·M . Define G := G/N . Let A be the image of A in G. The homomorphism
M → G G
is surjective because G = N ·M . Observe that we’re trying to show that G is trivial; we showed that it’s aquotient of M . Note that AE G, since AEM and M G. Because ∪g∈GgAg−1 generates G, we conclude
that ∪g∈GgAg−1 generate G.
So A = G, because ∪g∈GgAg−1 ⊆ A (we have A is normal), and G = 〈∪g∈GgAg−1〉 ⊆ A. This means
that G is abelian. But G is perfect, so if G = G/N is abelian we must have G/N = 1, that is, N = G.
Example 1.8.2. Let G = SLn(K), where K is a field, n ≥ 2, and |K| > 3 when n = 2. Observe that Gacts on X = 1-dimensional subspaces of Kn; we chose this action so that the kernel is H = aI : an =I = Z(SLn(K)). Sometimes X is denoted Pn−1(K), called projective space. Indeed, G/H = PSLn(K). Wewant to apply Theorem 1.8.1.
We should verify that G X is 2-transitive. The idea is the following: given a basis v1, . . . , vn of Kn andw1, . . . , wn of Kn, there is a B ∈ SLn(K) such that Bv1 = w1, . . . , Bvn = cwn, where c ∈ K×.
We should also verify that condition b) holds. Indeed, take x := K · e1 ∈ X, so that x = (∗, 0, . . . , 0).Now
Gx =
a ∗ . . . ∗0 b11 . . . b1(n−1)
......
. . ....
0 b(n−1)1 . . . b(n−1)(n−1)
: a ∈ K×, b ∈ GLn−1(K), adetB = 1
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has a normal subgroup
A =
a ∗ . . . ∗0 1 . . . 0...
.... . .
...0 0 . . . 1
∼= Kn−1
which is abelian. Now for 1 ≤ i, j ≤ n distinct, let Ei,j ∈ Mn(K) with 1 at the (i, j)th coordinate and 0elsewhere. Then note that I + cEi,j ∈ SLn(K) for any c ∈ K, and it is conjugate in SLn(K) to an elementin A.
Observe that SLn(K) is generated by such I + cEi,j , for 1 ≤ i, j ≤ n distinct, and c ∈ K. The idea isthe following: Multiplying by I + cEi,j is doing an elementary row operation, that is, adding c times row jand leaving other rows the same. Any B ∈ SLn(K) can be put in the form I using these operations.
We should also verify that condition a) holds. Indeed, it suffices to show that I + cEi,j are commutators:since they generate SLn(K), this will imply that G = [G,G]. The idea is as follows. If n ≥ 3, then1 c 0
0 1 00 0 1
=
[1 0 c0 1 00 0 1
,1 0 0
0 1 00 1 1
]
and if n = 2 then [1 −b(a2 − 1)0 1
]=
[ [1 b0 1
],
[a 00 a−1
] ]where we note that −b(a2 − 1) can be anything in K if a 6∈ 0, 1,−1, which holds since |K| > 3.
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1.9 Sep 14, 2018
We’ll begin by talking about a concept that’s falling out of fashion; they’re not really in the newer books.We’ll talk about groups with operators. The idea is to generalize group theory in the following way. Beginby fixing a set Ω.
Definition 1.9.1. An Ω-group (or a group with operator set Ω) is a group (G, ·) such that for all x ∈ Ωand g ∈ G, we have a gx ∈ G satisfying (gh)x = gxhx for all g, h ∈ G.
So if Ω = ∅, then we just have a usual group, and we could redo group theory!
Definition 1.9.2. An Ω-subgroup H of G is a subgroup stable under the Ω-action.
For a normal Ω-subgroup N EG, we have that G/N is an Ω-group, given by (gN)x = gxN .
Definition 1.9.3. G 6= 1 is a simple Ω-group if it has no normal Ω-subgroups.
And you say “hey, most of this still works”. For eample, the Jordan-Holder theorem works with Ω-groups.Let’s look at an example that’s not Ω 6= ∅.
Example 1.9.4. Let R be a commutative ring. Then an R-module M is an “R-group”: for a ∈ R,m ∈M ,define ma := am. An R-subgroup is just a submodule, and so on. We say an R-module M 6= 0 is simple ifthe only submodules are 0 and itself. With this, we get a Jordan-Holder theorem. In particular:
A composition series for an R-module M is a chain of submodules
M = M0 ⊂M1 ⊂ · · · ⊂Mn = 0
such that Mi/Mi+1 are simple for all 0 ≤ i < n. So Jordan-Holder says that up to reordering and isomor-phism, the simple R-modules
M0/Mi, . . . ,Mn−1/Mn
are independent of composition series. The length of M is defined to be `(M) := n.
This is somewhat important in algebraic geometry, when one tries to figure out intersection properties;it’s sometimes a good generalization of dimension (when R = K is a field then it is exactly the dimension ofthe vector space).
Okay, let’s move on.
Solvable Groups
Let G be a group, and recall [G,G], called the commutator subgroup of G, denotes the subgroup of Ggenerated by commutators [g, h] = g−1h−1gh. We showed that [G,G] is characteristic subgroup of G. Wealso saw that G/[G,G] is the (unique) largest abelian quotient of G, which shouldn’t be too surprising.
Let G(0) = G, and G(1) = [G,G] = [G(0), G(0)]. Inductively, define G(i+1) = [G(i), G(i)], so that
G(0) ⊇ G(1) ⊇ G(2) ⊇ . . . .
We say that G is solvable if G(n) = 1 for some n ≥ 0. The British call these soluble groups. The smallestsuch n ≥ 0 is called the derived length.
For example, when G is abelian, then [G,G] = G(1) = 1, and in fact, G is abelian if and only if its derivedlength is 1.
20
Example 1.9.5. Let
G =
[a b0 1
]: a ∈ R×, b ∈ R
and observe that
[G,G] =
[1 b0 1
]: b ∈ R
∼= R
so that G/[G,G]∼−→ R×, given by [
a b0 1
]7→ a.
Observe that [G,G] is abelian and so is G/[G,G], but G itself is not abelian. This will come up again later.The group G is solvable, with derived length 2.
Some nonexamples include A5 = [A5, A5], and more generally perfect groups, and also S5. Sometimesinfinite groups G are also not solvable even though G(i+1) ⊂ G(i) for all i. We have the following lemma:
Lemma 1.9.6. The group G is solvable if and only if G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = 1 is a series with Gi/Gi+1
abelian.
Proof. The forwards direction is easy: define Gi := G(i); recall that
G0 ⊇ G1 ⊇ · · · ⊇ Gn︸︷︷︸=G(n)
= 1
for n large enough; we have G(i)/G(i+1) = G(i)/[G(i), G(i)] is abelian, as desired.
For the backwards direction, we claim that G(i) ⊆ Gi; roughly, at each step Gi removes something sothat the quotient is abelian, whereas G(i) removes the largest thing possible so that the quotient is abelian.We will prove this claim by induction. For i = 0, of course we have G(0) = G = G0, we are done. Supposethat G(i) ⊆ Gi for i ≥ 0. Then
G(i+1) = [G(i), G(i)] ⊆ [Gi, Gi] ⊆ Gi+1.
Lemma 1.9.7. Suppose G is solvable.
1. Any subgroup of G is solvable.
2. Any quotient of G is solvable.
Proof. For part i), we take H ≤ G. Note that H(i) ⊆ G(i) for all i, so eventually H(i) is trivial. This alsoimplies that the derived length of H is at most the derived length of G.
For part ii), we take N EG. Note that
(G/N)(i) = G(i)N/N
for i ≥ 0. Then G(n) = 1 implies (G/N)(n) = 1, so G/N is solvable.
Proposition 1.9.8. Fix a group G. Let N be a normal subgroup. Then G is solvable if and only if N andG/N are solvable.
Remark 1.9.9. Recall Example 1.9.5, which says that Proposition 1.9.8 is not true if we replace “solvable”with “abelian”; this makes proving things about solvable groups with induction much easier. Solvable groupsare the smallest class of groups with this property.
Proof. The forward direction is just Lemma 1.9.7. For the backwards direction recall that (G/N)(i) =G(i)N/N . So for n large enough, (G/N)(n) = 1, so G(n) ⊆ N , and for m large enough, N (m) = 1. ThenG(m+n) = (G(n))(m) ⊆ N (m) = 1, that is, G is solvable.
21
1.10 Sep 17, 2018
Last time, we defined solvable groups:
The group G is solvable if and only if there exists a series
G = G0 ⊇ G1 ⊇ · · · ⊇ Gn = 0
such that Gi+1 EGi and Gi/Gi+1 is abelian.
It suffices to consider G(i), where G(0) := G and G(k+1) := [G(k), G(k)], for k ≥ 0. Solvable groups hadsome nice properties: subgroups and quotients of solvable groups are solvable. Also, if NEG, then G is solv-able if and only if N and G/N are solvable. This should be thought of as the key property of solvable groups.
Historical Aside
Consider an irreducible polynomial f ∈ Q[x]. What are its roots?
For example, when f = ax2 + bx+ c with a 6= 0, then
x =−b±
√b2 − 4ac
2a.
Can we find similar solutions to other polynomials, where “similar” means “in terms of radicals”? Thismotivates the following definition:
Definition 1.10.1. We say f is solvable in radicals if its roots are in a finite extension L/Q, with L ⊆ C,and L = Q(α1, . . . , αn) such that αnii ∈ Q(α1, . . . , αi−1) for i = 1, . . . , n for some ni ≥ 1.
As an example, Q(√
2,3√
3 +√
5) is a field that satisfies the above conditions.
Example 1.10.2. Let f = x4 + x3 + x2 + x+ 1 = x5−1x−1 . One solution is x = e2πi/5, that is,
x =
(√5
2− 1
4
)+
1
2
(√5
4+
√5
2+
√5
4−√
5
2
)i.
So f is solvable in radicals.
Let K ⊆ C be the extension of Q generated by the roots of f .
Definition 1.10.3. The Galois group of f is G := Aut(K), that is, the group of field automorphisms of K.That is, if α1, . . . , αd ∈ C are the roots of f , then we get a homomorphism
ϕ : G → Sd
given by σ(αi) = αϕ(σ)·i.
We have
Theorem 1.10.4. The polynomial f is solvable in radicals if and only if G = Aut(K) is solvable.
For example, if f = x4 + x3 + x2 + x + 1 then G ∼= (Z/5Z)× (because elements σ of G correspond tointegers a ∈ (Z/5Z)×, where the correspondence comes from σ(e2πi/5) = e2πia/5).
As another example, if d = deg f ≤ 4, then G → Sd, where Sd is solvable (for d ≤ 4). So G is solvablebecause it is a subgroup, and so f is solvable in radicals.
Finally, if f = x5 − x− 1, then G ∼= S5, which is not solvable, so f is not solvable in radicals.
This is foreshadowing; we’ll see more of this in 6320, and more of field theory in 6310.
By the way, we have two major results on solvable groups:
22
Theorem 1.10.5 (Burnside). If |G| = paqb, for p, q prime, then G is solvable.
The proof isn’t so bad, it’s in Dummit/Foote, but one needs some representation theory to do it.
Theorem 1.10.6 (Feit-Thompson). If |G| is odd, then G is solvable.
The proof of this one lasts more than 250 pages. It was a pivotal first step in the classification of finitesimple groups.
Let’s reset.
Nilpotent Groups
We begin with a definition:
Definition 1.10.7. The lower central series of G is the series defined by
C0G := G,Ci+1G = [G,CiG]
for i ≥ 0.
We should observe thatG = C0G ⊇ C1G ⊇ · · · ⊇ CnG ⊇ . . .
The book apparently uses Gi, and the previous professor of this class used G[i]; we’ll use Ci, where C is the“central” in “lower central series”.
Definition 1.10.8. We say G is nilpotent if CnG = 1 for some n ≥ 0. The number n is the nilpotency class.
As an example, if G is abelian, then C1G = [G,G] = 1.
Proposition 1.10.9. If G is nilpotent, then G is solvable.
Proof. Just observe thatCiG/Ci+1G = CiG/[G,CiG],
but also notice that [G,CiG] ⊇ [CiG,CiG], so the quotient is abelian (see Lemma 1.5.2).
Definition 1.10.10. We say G is a central extension of a group Γ by an abelian group A if there is an exactsequence
1 A G Γ 1
with A ⊆ Z(G).
Proposition 1.10.11. The group G is nilpotent of nilpotency class at most n if and only if G is a centralextension of a nilpotent group Γ of nilpotency class at most n− 1.
The idea is that this proposition gives us a way to induct (similar to the case of solvable groups). Let’sprove this:
Proof. Let’s first prove the forward direction: we have CnG = 1, so Cn−1G ⊆ Z(G) since [G,Cn−1G] =CnG = 1: recall that by definition [G,Cn−1G] = 〈g−1h−1gh〉 for g ∈ G, h ∈ Cn−1G and g−1h−1gh = 1; thissays gh = hg and so h ∈ Z(G).
So we defineA = Cn−1G, Γ = G/A
and since
1 A G Γ 1
23
is a short exact sequence, we have Cn−1Γ = A/A = 1 ⊆ Γ, so Γ is nilpotent, with nilpotency class at mostn− 1.
Conversely, if we have a short exact sequence
1 A G Γ 1
such that Cn−1Γ = 1, then Cn−1G ⊆ A and CnG = [G,Cn−1G] ⊆ [G,A] = 1 since A ⊆ Z(G). So G isnilpotent of nilpotency class at most n.
We get a different way of building up groups: if we considered short exact sequences
1 A G Γ 1
with AEG and solvable, then G itself was solvable; if now we insist that A ≤ G is abelian, then the G weconstruct will be nilpotent.
Example 1.10.12. Let G be a p-group. We claim that G is nilpotent.
Proof. We’ll induct on |G|. For the base case, we have |G| = 1, which is nilpotent. We want to find subgroupsof the center, but we showed that p-groups have nontrivial center (in fact, we showed that p divides Z(G),see Proposition 1.5.3).
Suppose G 6= 1 and smaller p-groups are nilpotent. Also, Z(G)EG, so we have that G/Z(G) is nilpotent(by the inductive hypothesis).
1 Z(G) G G/Z(G) 1
Next time we’ll talk about some of the key properties of nilpotent groups and begin finishing up grouptheory.
24
1.11 Sep 19, 2018
Today, we’ll talk just a little bit more about nilpotent groups and then we’ll move on.
Recall that we had an exact sequence
1 N G G/N 1
and that if N and G/N are solvable, then G itself was also solvable. Also, all abelian groups are solvable.“These are basically what solvable groups are”: you start with abelian groups and build up, that is, you candeconstruct solvable groups into abelian groups.
We also had nilpotent groups, which are what you get when the exact sequences above are more restricted.That is, we talked about central extensions
1 A G Γ 1
where A ⊆ Z(G). We have that abelian groups are nilpotent, and if Γ is nilpotent, then so is G. In thissetup it’s clear that nilpotent groups are solvable.
These families of groups are nice for inductive proofs. Here’s an example:
Proposition 1.11.1. Let G be nilpotent, and H < G a proper subgroup. Then H ( NG(H).
As an example of the power of this, this implies that maximal subgroups are normal.
Proof. We induct on the nilpotency class n of G (recall we had the sequence C0G = G and Ci+1G = [G,CiG],and n is the smallest integer such that CnG = 1). If n = 0 then G = 1, so it’s vacuously true. For n = 1,then G is abelian, which means NG(H) = G, as desired. Now we can do the inductive step.
Let A = Z(G). Last time we showed that G/Z(G) = G/A is nilpotent with nilpotency class at mostn− 1. We have two cases:
If A ⊆ H then H/A < G/A; the inductive hypothesis says that
H/A ( NG/A(H/A) = NG(H)/A
where the equality follows because A ⊆ H. So this implies H ( NG(H), as desired.
If A 6⊆ H, then A = Z(G) ⊆ NG(H), and so NG(H) ) H.
There is the following result, which is not so nice to present:
Theorem 1.11.2. Let G be a finite group. The following are equivalent:
1. G is nilpotent
2. G is a product of p-groups for some primes p (you can have different primes appearing)
3. For every prime p, G has a unique p-Sylow subgroup
4. Any two elements of G of relatively prime order commute.
Remark 1.11.3. It’s not hard to show that the product of two nilpotent groups is nilpotent, and last timewe showed that p-groups are nilpotent. So 2 is not so hard, and 3 is not so hard either (the p-Sylow subgroupis the product of the p-groups for a fixed p).
Remark 1.11.4. Number theory is full of theorems where one “fights their way up” from abelian groupsto nilpotent/solvable groups. Often the abelian case is well understood and the nonabelian case is harder,but you can scrape the abelian things together and say something about the nonabelian things.
25
Free Groups
Fix a set S. We will define a group FS and a map i : S → FS . Intuitively, S is a set of generators for FSwith no relations except those imposed by the group axioms (so for example, ac = abb−1c is imposed by thegroup axiom, but ab = ba is not). We will prove:
Theorem 1.11.5 (Universal property of free groups). For any map ϕ : S → G, where G is a group, thenthere is a unique group homomorphism Φ: FS → G such that Φ i = ϕ, that is,
S FS
G
ϕ
i
∃!Φ
commutes.
Remark 1.11.6. Suppose there is i′ : S → F ′S has the same property. Then the universal property for FSand again for FS′ gives
FS
S
FS′
f
i′
i
g
We claim that f and g are inverses. Indeed,
S FS
FS
i
igf
commutes, but uniqueness of the homomorphism says that g f = idFS . For exactly the same reason,f g = idFS′ . So the moral is that FS is determined (with i : S → FS up to natural isomorphism).
Let us construct FS . We first choose a set S−1 such that it is that is disjoint from S and that there isa bijection S → S−1, given by s 7→ s−1, and s−1 ← [ s. (this are just some symbols, this isn’t a group yet).Let T = S ∪ S−1, which you can think of T = (s, i) : s ∈ S, i ∈ ±1.
A word in T is a finite sequence w = (a1, . . . , an) with an ∈ T . Alternatively, we can concatenate theseto form some string a1 . . . an. The word is reduced if ai+1 6= a−1
i for all i.
Let FS be the set of all reduced words in T , with the following group structure (a1 . . . an) · (b1 . . . bm) isthe reduced word of the concatenation a1 . . . anb1 . . . bm. There are things to check: associativity is pedantic,the identity is the empty word (which we still denote by 1), and the inverse of a1 . . . an is a−1
n . . . a−11 .
Also, i : S → FS is the inclusion s 7→ s. Now the universal property holds if we define Φ to be the mapΦ(ae11 . . . ae1n ) = ϕ(a1)e1 · ϕ(an)en . Here are some results about free groups:
1. Up to isomorphism, FS depends only on the cardinality of S (will be on the next homework).
2. (Nielsen-Schreier) A subgroup of a free group is free.
The second result is not as easy as one might expect: for |S| = 2, we have [FS , FS ] ∼= FS′ with S′ infinite.
26
1.12 Sep 21, 2018
Last time, we fixed a set S and constructed a free group FS with S ⊆ FS generating the group. Themain property was that any map ϕ : S → G extends to a unique homomorphism Φ: FS → G.
Consider the wedge of n circles, as a topological space its fundamental group is π1(X,P ) ∼= F[n] [here[n] = 1, 2, . . . , n], so sometimes one can use algebraic topology to understand free groups (see Nielsen-Schreier; the proof uses algebraic topology).
Presentations
[Prof Zywina thought that the previous 6310 had group presentations, and then realized that they weretalking about presentations of groups.]
Let S be a set, and consider again FS the free group on S generators. Let R be a set of words in S∪S−1,and define the group
〈S|R〉 := FS/N,
where N is the smallest normal subgroup of FS containing R. This is called a presentation of the group〈S|R〉.
Remark 1.12.1. Any group is of the form 〈S|R〉. Indeed, choose generators S of G, and get a ϕ : S G,so it induces a unique Φ: FS G, and the kernel of this map can be chosen to be N .
We say G is finitely presented if G ∼= 〈S|R〉 with both S and R finite.
Example 1.12.2. Consider 〈a|an〉 = 〈a|an〉 (forget about the brackets); this is isomorphic to Z/nZ(via the isomorphism a 7→ 1).
Example 1.12.3. Consider 〈a, b|b3, a−1b−1ab〉 = 〈a, b|b3 = 1, ab = ba〉. We let a, b be the image of a, b inG. Then G ∼= Z× Z/3Z, given by the map a 7→ (1, 0) and b 7→ (0, 1).
Example 1.12.4. We have D2n∼= 〈r, s|rn = 1, s2 = 1, s−1rs = r−1〉. Since rs = sr−1 = srn−1 we can write
elements of 〈r, s|rn = 1, s2 = 1, s−1rs = r−1〉 =: G as sirj : i = 0, 1, 0 ≤ j < n, so |G| ≤ 2n. This gives theisomorphism to D2n.
Example 1.12.5. Let G := 〈a, b|a2 = b3 = 1〉. It’s a fact that
G ∼= SL2(Z)/±I.
The isomorphism is given by
a 7→[0 −11 0
], and b 7→
[1 −1−1 0
]and there is some work to show that the images of a, b generate SL2(Z)/±I and even more work to showthat this really does give an isomoprhism.
In general, it is undecidable to determine if 〈S|R〉 is finite (or trivial, or abelian) when S and R is finite(where “undecidable” is in the precise computer science sense). So just given a presentation it can be verymysterious what the group is. More generally, given a word g, there is no algorithm to determine if it istrivial in 〈S|R〉.
In general, it is a challenge to show that there is a homomorphism out of the presentation into the groupyou think it should be; thankfully usually there is some extra information (like in the case of D2n, we knewits size). On the other hand a homomorphism from the group you think it should be into the presentationisn’t so hard.
27
Direct Products
We’ll talk about direct products and next lecture we’ll also talk about semidirect products and that’llbasically be the end of our group theory.
Fix groups H and K. This gives a group H ×K (endowed with component-wise multiplication). Therelevant question to ask is: Given a group G, how does one detect whether or not it is a direct product?
By abuse of notation, we let H := H × 1 ⊆ G and K := 1×K ⊆ G (where G = H ×K). There are someproperties we knew about H and K: we have
• H EG,K EG
• HK = hk : h ∈ H, k ∈ K = G
• H ∩K = 1.
The claim is that these properties are sufficient, in the following sense:
Proposition 1.12.6. Fix a group G. Suppose H and K are normal subgroups of G such that HK = G andH ∩K = 1. Then
H ×K → G
(h, k) 7→ hk
is an isomorphism.
Proof. We first show that H and K commute. Let h ∈ H and k ∈ K, and consider the commutatorh−1k−1hk = (kh)−1hk. Now we have
h−1k−1hk = h−1 (k−1hk)︸ ︷︷ ︸∈H
∈ H, and h−1k−1hk = (h−1k−1h)︸ ︷︷ ︸∈K
k ∈ K
so h−1k−1hk ∈ H ∩K = 1, and hence hk = kh. Knowing this it is easier to check that ϕ : H ×K → G givenby (h, k) 7→ hk is a homomorphism. Notice that ϕ is surjective since HK = G. To see that ϕ is injective,notice that if ϕ(h, k) = 1 we have hk = 1, and hence h = k−1 ∈ H ∩K = 1. So (h, k) = (1, 1), as desired.
Semidirect Products
Now we want to loosen the assumptions:
Let H and K be subgroups of G such that H E G and H ∩ K and G = HK. Note that without theassumption G = HK, we still have that HK is always a subgroup of G.
As before, ϕ : H × K → HK = G given by (h, k) 7→ hk is still a bijection, though it need not be ahomomorphism. To define H oK (the semidirect product), we’ll understand how multiplication works inHK and pull the multiplication over to H ×K (as a set) to give it a group structure.
So let h1, h2 ∈ H, and k1, k2 ∈ K. Then
(h1k1)(h2k2) = h1k1h2(k−11 k1)k2
= (h1 k1h1k−11︸ ︷︷ ︸
∈H (HEG)
) · k1k2︸︷︷︸∈K
and since ϕ was a bijection this is the unique way of writing (h1k1)(h2k2) as an element of HK. Note thatK acts on H by conjugation. So this will define a multiplication on H ×K.
28
1.13 Sep 24, 2018
We’re going to finish up semi-direct products and then begin rings.
Last time, we had a group G with two subgroups H and K such that
• H EG
• HK = G
• H ∩K = 1.
We saw that there was a map H ×K → G given by (h, k) 7→ hk is a bijection, but not a homomorphism.In particular, we had, for h1, h2 ∈ H, k1, k2 ∈ K,
(h1k1)(h2k2) = (h1k1h2k−11︸ ︷︷ ︸
∈H
) · (k1k2︸︷︷︸∈K
)
and since H×K → G is a bijection this is the unique way of writing (h1k1)(h2k2) as a product of two elementsh′k′ for some h′ ∈ H, k′ ∈ K. Note that K acts on H by conjugation, that is, there is a ϕ : K → Aut(H).
Let’s do a reset. Let H and K be groups, and let K be an action on H such that ϕ : K → Aut(H). Wedefine
H oϕ K = H oK
to be the group consisting of the set H ×K with the multiplication
(h1, k1) · (h2, k2) := (h1(k1 · h2), k1k2) = (h1 · ϕk1(h2), k1k2).
We say that H oK is the semidirect product of H and K. This is a group.
The triangle in o points in the direction it would in the above setup (we had H E G). The identity inthis group is (1, 1). We have ϕ : K → Aut(H) is a homomorphism into the automorphism group, ϕ1 = idH ,so
(1, 1) · (h, k) = (1 · (1 · h), 1 · k) = (h, k).
Also since ϕk is an automorphism of H, we have ϕk(1) = 1, and so
(h, k) · (1, 1) = (h · (k · 1), k · 1) = (h, k).
We have (h, k)−1 = (k−1 · h−1, k−1), since
(h, k) · (k−1 · h−1, k−1) = (h · k · (k−1 · h−1), k · k−1) = (1, 1).
We can identify H and K with H × 1 ⊆ H oϕ K and 1×K ⊆ H oϕ K. We can also check that H and Kare subgroups of H oϕ K, with HK = H oϕ K and H ∩K = 1 and H EHoϕ.
Furthermore, for h ∈ H and k ∈ K, we have k · h = khk−1 (where on the left side k ∈ K is acting onh ∈ H, giving an element of H → H × 1, whereas on the right side we are multiplying in H oϕK [the claimis that this will land inside H × 1 and in particular is the element k · h ∈ H × 1]).
Example 1.13.1. For ϕ = 1, we have H oϕ K = H ×K as groups!
Example 1.13.2. For H = R, and K = R×, we have K H by k ·h = kh (multiplication as a real number).From this, we get a group H oK. It turns out that
H oK ∼= G =
[a b0 1
]: a ∈ R×, b ∈ R
.
29
Inside G we have H1 EG given by
H1 =
[1 b0 1
]: b ∈ R
∼= R
and
K1 =
[a 00 1
]: a ∈ R×
∼= R×.
We can also verify that [a 00 1
]︸ ︷︷ ︸
:=k
[1 b0 1
]︸ ︷︷ ︸
:=h
[a 00 1
]−1
︸ ︷︷ ︸:=k−1
=
[a ab0 1
] [a−1 00 1
]=
[1 ab0 1
]so that khk−1 = k · h, as claimed earlier.
Example 1.13.3. Let H = 〈r〉 and K = 〈s〉, where H is the cyclic group of order n, and K is the cyclicgroup of order 2. We have ϕ : K → Aut(H) given by ϕs : H → H sending x 7→ x−1. We have HoϕK ' D2n.
Remark 1.13.4. If G = H oK then we actually have a split exact sequence
1 H G K 1
since G has a copy of K sitting inside it.
30
2 Rings
2.13 Sep 24, 2018
[We should read chapter 7 of Dummit/Foote]. Everybody should know what a ring is.
Definition 2.13.1. A ring is a set R with two binary operations + and · (ie. addition and multiplication)such that the following hold:
a) (R,+) is an abelian group (with additive identity denoted by 0)
b) Multiplication is associative, that is, a(bc) = (ab)c for a, b, c ∈ R.
c) There is a distributive law, that is, a(b+ c) = ab+ ac and (a+ b)c = ac+ bc for a, b, c ∈ R.
d) There is an element 1 ∈ R such that 1a = a = a1 for all a ∈ R.
As a warning, Dummit and Foote do not include d). But to Prof Zywina, these are called rngs (there’sno identity).
We say that R is commutative if ab = ba for all a, b ∈ R. It’s noncommutative otherwise. Some exam-ples include Z, Z/nZ, R[x], Mn(R), R[[x]], C(R), and R = 0 (though some people assume an axiom 0 6= 1).
We also have the ring RG (called a group ring) for G a finite group and R a ring, defined to be the setof formal sums
∑g∈G ag · g with ag ∈ R, given by∑
g∈Gagg +
∑g∈G
bgg =∑g∈G
(ag + bg)g
and ∑g∈G
agg ·∑g∈G
bgg =∑g,h∈G
agbhgh =∑g∈G
( ∑h1,h2∈Gh1h2=g
ah1bh2
)g.
It turns out that the group ring CG ∼= Πri=1Mni(C), and this is linked to the representation theory of G
(take 6320!).
Under his definition, 2Z is not a ring, though it is a rng (and is hence a ring under Dummit/Footeconvention).
Definition 2.13.2. A subring of a ring R is a subset S ⊆ R such that
• a+ b ∈ S for a, b ∈ S
• −a ∈ S for a ∈ S
• ab ∈ S for a, b ∈ S
• 1 ∈ S.
We have S is a ring with +, ·, 1 inherited from R.
We say an element a ∈ R is a unit if ab = 1 = ba for some b ∈ R. The set of units in R is denoted byR×. This can be given a group structure via multiplication from elements of R.
For example, Z× = ±1 and R× = R− 0 and Mn(R)× =: GLn(R) and R[x]× = R×.
For a wild example, for a commutative ring R we have
R[x]× = a0 + a1x+ · · ·+ anxn : a0 ∈ R×, ai nilpotent , i ≥ 1,
where nilpotent elements a ∈ R are those such that ak = 0 for some k ≥ 1. This would need some proof,which is omitted.
Of course, this means that for R a field, we have R[x]× = R×.
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2.14 Sep 26, 2018
Let R be a ring. Recall that a (left) ideal of R is a nonempty I ⊆ R such that
• a+ b ∈ I for a, b ∈ I
• ra ∈ I for r ∈ R, a ∈ I.
In other words, I is an R-submodule of R (where R acts on itself by left multiplication). Right idealsalso exist; when R is commutative this distinction does not matter.
Example 2.14.1. Let f : R→ S be a ring homomorphism, that is, a map f such that
• f(a+ b) = f(a) + f(b)
• f(ab) = f(a)f(b)
• f(1) = 1
Then ker f = r ∈ R : f(r) = 0 is an ideal of R.
Example 2.14.2. Let a1, . . . , an ∈ R be elements of R. One can consider the ideal generated by a1, . . . , an,given by
〈a1, . . . , an〉 = r1a1 + · · ·+ rnan : ri ∈ R.
You can have infinitely many ai’s but the ideal generated by them will be the set of finite linear combinationsof the ai.
Let I ⊆ R be an ideal of a commutative ring R. We can look at the cosets
R/I := r + I : r ∈ R
where r+ I = r+ a : a ∈ I. We have that R/I is a ring with operations (a+ I) + (b+ I) = (a+ b) + I and(a+ I) · (b+ I) = ab+ I. There is a homomorphism R→ R/I given by r 7→ r + I; its kernel is precisely I.
Example 2.14.3. Let R = Z and fix n ≥ 1 consider the ideal 〈n〉 = nZ. The homomorphism Z/nZ is thefamiliar ring of integers mod n.
Example 2.14.4. Consider R = R[x]/〈x2 + 1〉. We have R ∼= C because there is a ring homomorphismR[x]→ C given by f(x) 7→ f(i) that is surjective and has kernel 〈x2 + 1〉.
Let R be a commutative ring.
Definition 2.14.5. An element a ∈ R is a zero divisor if ab = 0 for some nonzero b ∈ R.
As an example, [2] ∈ Z/6Z is a zero divisor.
Definition 2.14.6. We say R 6= 0 is an integral domain if it has no nonzero zero divisors.
Definition 2.14.7. We say R 6= 0 is a field if every a ∈ R−0 has a multiplicative inverse (that is, ab = 1for b ∈ R). Equivalently, R is a field if it has exactly two ideals.
To see the equivalence, notice that 〈a〉 ⊇ 〈1〉 = R, so 〈a〉 = R.
Fix an ideal I ⊆ R.
Definition 2.14.8. We say I is a prime ideal if ab ∈ I with a, b ∈ R, then a ∈ I or b ∈ I. Equivalently, I isprime if and only if R/I is an integral domain.
Definition 2.14.9. We say I is a maximal ideal if I ( R and there are no ideals between I and R (wrtinclusion). Equivalently, I is maximal if and only if R/I is a field.
Observe that maximal ideals are prime (since fields are integral domains).
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Theorem 2.14.10. Let R 6= 0 be a commutative ring (with 1). Then R has a maximal ideal.
Proof. Use Zorn’s lemma. Let P be the set of proper ideals of R, and say that P 6= ∅ (and 0 ∈ P). Then Pwith ⊆ is a partially ordered set. Take any nonempty chain C ⊆ P, that is, for A,B ∈ C, we have A ⊆ B orB ⊆ A. We want to find an upper bound for all the elements in this chain, but
J :=⋃I⊆C
I ⊆ R.
We claim that J is an ideal. It’s nonempty because:
• It’s nonempty, since 0 ∈ J
• Take a, b ∈ J . Then a ∈ A or b ∈ B; either A ⊆ B or B ⊆ B so a+ b ∈ A or a+ b ∈ B. In either case,a+ b ∈ J
• Take a ∈ J, r ∈ R. Then a ∈ A for some A ∈ C, so ra ∈ A, and so ra ∈ J .
We also need to show that J is a proper ideal of P. But notice that if J = R, then 1 ∈ J , and hence1 ∈ A for some A ∈ C, which is a contradiction because then A = R and so A 6∈ P. So J ( R and J ∈ P.In particular, J is an upper bound of C.
By Zorn’s lemma, P has a maximal element with respect to inclusion, that is, R has a maximal ideal.
Remark 2.14.11. There are some crazy rings out there (especially the ones from analysis). This theoremin its full generality is equivalent to Zorn’s lemma. On the other hand, for Noetherian rings, one can avoidusing any big hammers.
Remark 2.14.12. This theorem fails for commutative rings without an identity.
Localization
This is in Dummit and Foote, in section 15.4. It’s only a little fancier than the field of fractions con-struction, which is in Dummit and Foote 7.5.
Let R be a commutative ring with 1, and let S ⊆ R be a multiplicatively closed set, that is, if a, b ∈ Sthen ab ∈ S (we don’t assume any other structure). Assume that 1 ∈ S (you can replace S by S ∪ 1).
Theorem 2.14.13. There is a commutative ring S−1R with a ring homomorphism π : R → S−1R suchthat π(b) ∈ (S−1R)× for b ∈ S and for any ring homomorphism ϕ : R → A with ϕ(b) ∈ A× for all b ∈ S,then there is a unique ring homomorphism Φ: S−1R → A such that Φ π = ϕ. This is expressed in thecommutative diagram
R S−1R
A
ϕ
π
∃!Φ
Loosely, S−1R is the ring you obtain from R when you “invert” all elements of S. As an example, letR = Z, and S = Z− 0, then the construction S−1R should produce Q.
We call S−1R the localization of P at S, or sometimes the ring of fractions of R with respect to S.
Remark 2.14.14. The idea is that you want to invert some elements of R which may not be units, but wecan homomorph into an A such that elements of S get sent to units. The construction says that “S−1R” isthe “best ring” to which we should send S to units (it’s a fact that π(b) ∈ (S−1R)× for all b ∈ S).
Remark 2.14.15. The theorem gives a universal property that determines S−1R and π : R→ S−1R up toa uniquely determined isomorphism.
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2.15 Sep 28, 2018
Let R be a commutative ring S ⊆ R multiplicatively closed, with 1 ∈ S. We had this theorem last timestated without proof:
Theorem 2.15.1. There is a commutative ring S−1R with a homomorphism π : R → S−1R such thatπ(b) ∈ (S−1R)× for all b ∈ S. Moreover, for any ring homomorphism ϕ : R→ A with ϕ(b) ∈ A× for b ∈ S,there is a unique homomorphism Φ: S−1R→ A such that ϕ = Φ π, that is,
R S−1R
A
ϕ
π
∃!Φ
commutes.
We say S−1R is the localization of R at S.
Proof. Define a relation on R × S given by (a, b) ∼ (c, d) if and only if x(ad − bc) = 0 for some x ∈ S. Ofcourse, if S has no zero divisors then this is true if and only if ad− bc = 0.
We claim that this is an equivalence relation. Obviously we are reflexive and symmetric. We are transitive:if (a, b) ∼ (c, d), so that x(ad − bc) = 0 for some x ∈ S, and (c, d) ∼ (e, f), so that there is y so thaty(cf − ed) = 0 for some y ∈ S, then
0 = fyx(ad− bc) + bxy(cf − de) = xyd(af − be)
so that (a, b) ∼ (e, f). We let ab be the equivalence class of (a, b); we can now define
S−1R :=
a
b: a ∈ R, b ∈ S
.
As an exercise, show that S−1R is a ring with
a
b+c
d=ad+ bc
bdand
a
b· cd
=ac
bd.
This not worth doing in lecture.
We have identities in S−1R; the multiplicative one is 1 = 11 and the additive one is 0 = 0
1 . We also havea canonical homomorphism
π : R→ S−1R
that maps r 7→ r1 . If S has no zero divisors (of R), then π is injective. Also, for b ∈ S, we have ( b1 )−1 = 1
b ,so elements in S have inverses.
Finally, take any map ϕ : R → A such that ϕ(b) ∈ A× for b ∈ S. Then Φ: S−1R → A given byab 7→ ϕ(a)ϕ(b)−1 is a homomorphism such that Φ π = ϕ.
The key example is the following:
Example 2.15.2. Let R be an integral domain and S = R−0. We have that S is multiplicatively closed(because R is integral domain) and contains 1. So we can consider S−1R: it is called the fraction field orquotient field of R. Since S has no zero divisors R → S−1R by r 7→ r
1 .
We understand these well: this construction maps Z→ Q and R[x]→ R(x).
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Example 2.15.3 (Localization at a prime). Let R be a commutative ring, and p ⊆ R prime ideal, andS := R\p; notice that S is multiplicatively closed since if a, b ∈ S such that ab ∈ p (that is, ab 6∈ S), but thismeans that either a or b is also in p, which is not true (they were both in S). So ab ∈ S.
The localization S−1R is sometimes denoted Rp. For example, we have Z〈p〉 = ab : p - b. You can alsoconsider C[x, y]〈x,y〉, where 〈x, y〉 = f ∈ C[x, y] : f(0, 0) = 0. Thus,
C[x, y]〈x,y〉 =
f
g: f, g ∈ C[x, y], g(0, 0) 6= 0
;
that is, we have functions that make sense in a neighborhood of (0, 0), so in this sense things are “local”.
Example 2.15.4. Let R = Z/6Z and S = 1, 2, 4 (so we are trying to invert zero divisors). Then one canshow that S−1R ∼= Z/3Z, andS−1R is smaller than R, and π : R→ S−1R is not injective.
Theorem 2.15.5. There is an inclusion preserving bijectionprime ideals p of R with p ∩ S = ∅
←→
prime ideals of S−1R
p 7−→ S−1p :=
a
b: a ∈ p, b ∈ S
π−1(q)←− [ q
Let R be a commutative ring with 1. We can do some operations: for I, J ⊆ R ideals, we have
• I ∩ J is an ideal (largest ideal in I and J)
• I + J = a+ b : a ∈ I, b ∈ J is an ideal (smallest ideal containing I and J)
• IJ = 〈ab : a ∈ I, b ∈ J〉 = ∑ni=1 aibi : ai ∈ I, bi ∈ J
We say I and J are coprime (some people say comaximal) if I + J = 〈1〉 = R. Observe that if R = Z,and a, b ∈ Z not both 0, then 〈a〉+ 〈b〉 = 〈gcd(a, b)〉, so 〈a〉 and 〈b〉 are coprime if and only if gcd(a, b) = 1.
Fix ideals I1, . . . , In of R. Define a map by
ϕ : R→n∏i=1
R/Ii
r 7→ (r + I1, . . . , r + In).
It is a homomorphism.
Theorem 2.15.6 (Chinese Remainder Theorem). Suppose the ideals Ii, Ij are coprime for i 6= j. Then ϕis surjective, and the kernel is I1 ∩ · · · ∩ In = I1 . . . In.
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2.16 Oct 1, 2018
Let R be a commutative ring. For ideals I, J ⊆ R, we have new ideals I + J , I ∩ J , and IJ ; they satisfythe inclusions
I + J
I J
I ∩ J
IJ
We say that I and J are coprime if I + J = R. For R = Z, the above picture looks like
〈gcd(a, b)〉
〈a〉 〈b〉
〈lcm(a, b)〉
〈ab〉
Lemma 2.16.1. If I and J are coprime, then I ∩ J = IJ .
Proof. We just need to show I ∩ J ⊆ IJ .
Take a ∈ I ∩ J . Recall I and J are coprime, so that x + y = 1 for some x ∈ I, y ∈ J . So a = ax + ay,but a ∈ J and x ∈ I, whereas a ∈ I and y ∈ J , so ax+ ay ∈ IJ .
Let I1, . . . , In be ideals of R such that Ii and Ij are coprime for i 6= j (that is, they’re pairwise coprime).We define the usual ring homomorphism
ϕ : R→n∏i=1
R/Ii,
that is, ϕ(r) = (r + I1, . . . , r + In).
Theorem 2.16.2 (Chinese Remainder Theorem). With assumptions as above, ϕ is surjective. Furthermore,
ker(ϕ) = I1 ∩ · · · ∩ In = I1 . . . In.
Remark 2.16.3. There is content in the second equality. It also gives the immediate consequence that
R/(I1 . . . In)∼−→ R/I1 × · · · ×R/In.
So you can use the Chinese Remainder Theorem in two ways: you can break up a ring into simpler rings(by the above isomorphism), or you can generate some interesting elements (because ϕ above is surjective)which are useful in proofs.
Example 2.16.4. Let n = pe11 . . . perr . Then
Z/nZ ∼−→r∏i=1
Z/peii Z.
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Proof. Let’s first show that I1 ∩ · · · ∩ In = I1 . . . In. The proof is by induction; for n = 1 this is obvious, andn = 2 is Lemma 2.16.1. So assume n > 2 and that the result is known for smaller n.
We have
I1 ∩ · · · ∩ In = I1 ∩ (I2 ∩ · · · ∩ In)
= I1 ∩ (I2 . . . In)
by induction; we need only show that I1 and I2 . . . In are coprime (this will imply that I1 ∩ (I2 . . . In) =I1 . . . In, as desired).
For each 2 ≤ i ≤ n there is xi ∈ I1, yi ∈ Ii so that 1 = xi + yi. So
1 =
n∏i=2
(xi + yi) ≡n∏i=2
yi (mod I1),
where we define a ≡ b (mod I1) if and only if a− b ∈ I1. This means
1 =
( n∏i=2
(xi + yi)−n∏i=2
yi︸ ︷︷ ︸∈I1
)+
n∏i=2
yi︸ ︷︷ ︸I2...In
and so 1 ∈ I1 + I2 . . . In.
Now we show that ϕ is surjective. It suffices to show that there is r ∈ R so that ϕ(r) = (1, 0, . . . , 0) (andevery other standard basis element of R/I1 × · · · ×R/In), because ϕ is linear.
Recall that I1 + Ii = R. Then 1 = xi + yi with xi ∈ I1 and yi ∈ Ii. Take r := y2 . . . yn =∏ni=2 yi. This
says that r ∈ Ii for i ≥ 2, since yi ∈ Ii. On the other hand,
r =
n∏i=2
yi =
n∏i=2
(1− xi) ≡n∏i=2
(1− 0) = 1 (mod I1)
since xi ∈ I1. This completes the proof.
Definition 2.16.5. We say that a commutative ring R is Noetherian if one of the following (equivalent)conditions hold:
(a) R satisfies the Ascending Chain Condition (ACC): If we have an ascending chain
I1 ⊆ I2 ⊆ I3 ⊆ . . .
for ideals In of R, then for some n, it stabilizes, that is,
In = In+1 = In+2 = . . . .
(b) Every nonempty set of ideals in R has a maximum (with respect to inclusion).
(c) All ideals are finitely generated.
Remark 2.16.6. Condition (b) allows us to avoid Zorn’s lemma in some proofs (unfortunately, we’d haveto restrict to Noetherian rings...)
Proof. We first show (a) implies (b).
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Let Σ be a nonempty set of ideals with no maximum. Choose I1 ∈ Σ. Inductively define I2, I3, . . . byobserving that for n ≥ 2, In−1 is not maximal in Σ, so there is In ∈ Σ such that In−1 ( In. Thus
I1 ( I2 ( I3 ( . . . ,
contradicting (a).
Now we prove (b) implies (c).
Take any ideal I ⊆ R. We define the set
Σ := J : J ⊆ I, J finitely generated ideal of R.
Take a maximal set J ⊆ Σ. If J = I we’re done because I is finitely generated. If J ( I, then take a ∈ I−J ,and observe that the finitely generated ideal J + 〈a〉 ) J , so J could not have been maximal.
Finally we should show (c) implies (a). Given an ascending chain
I1 ⊆ I2 ⊆ . . .
we consider the ideal
I :=
∞⋃i=1
Ii ⊆ R,
which is finitely generated, so I = 〈a1, . . . , ar〉. But there is an N such that ai ∈ In for all n ≥ N ; this saysthat
〈a1, . . . , ar〉 ⊆ In ⊆ I
so that In = I for all n ≥ N .
Remark 2.16.7. It’s quite terrifying when one encounters a non-Noetherian ring, because they’re so ubiq-uitous.
Theorem 2.16.8 (Hilbert’s Basis Theorem). If R is Noetherian, then R[x] is Noetherian.
We’ll prove this next time.
Corollary 2.16.9. If R is Noetherian, then R[x1, . . . , xn] is Noetherian.
Of course, this follows from repeatedly applying Hilbert’s Basis Theorem. This was a controversialtheorem at that time, because the proof was very nonconstructive. Apparently, this is false if we replace“Noetherian” with PID and other good adjectives.
Lemma 2.16.10. If R is Noetherian and I is an ideal, then R/I is Noetherian.
Proof. Recall that there is an inclusion preserving bijection:
ideals of R containing I ↔
ideals of R/I
J 7→ J/I
Now the Lemma is clear, because if R satisfies ACC, then R/I now satisfies ACC.
Remark 2.16.11. LetR be a Noetherian ring. Any finitely generatedR-algebra is isomorphic toR[x1, . . . , xn]/I,which is Noetherian. It’s also useful to note that I = 〈f1, . . . , fr〉 is itself finitely generated.
We’ll talk about Noetherian rings a lot throughout the second half of the course.
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2.17 Oct 3, 2018
We’re going to prove the Hilbert Basis Theorem.
Theorem 2.17.1 (Hilbert’s Basis Theorem). If R is Noetherian, then R[x] is Noetherian.
Proof. Take any ideal I ⊆ R[x]. Suppose that I is not finitely generated. We have I 6= 0. Choose a nonzerof0 ∈ I of minimal degree. We inductively define f1, f2, f3, · · · ∈ I with fn ∈ I − 〈f1, . . . , fn−1〉; this set isnonempty since I is not finitely generated. In particular, pick an fn of minimal degree in I − 〈f1, . . . , fn−1〉.
Define dn = deg(fn). Note that d0 ≤ d1 ≤ d2 ≤ . . . . Also define an to be the leading coefficient of fn,that is,
fn = anxdn + . . .
so that each ai ∈ R− 0. Consider the ascending chain of ideals of R given by
〈a0〉 ⊆ 〈a0, a1〉 ⊆ 〈a0, a1, a2〉 ⊆ . . .
and observe that this chain eventually stabilizes. This says that for some N ≥ 1, we have
〈a0, . . . , an〉 = 〈a0, . . . , aN 〉
for all n ≥ N .
Now take any n > N , and observe that an ∈ 〈a0, . . . , aN 〉. Thus
an = r0a0 + · · ·+ rNaN
with ri ∈ R. We can define
g := fn −(r0x
dn−d0f0 + · · ·+ rNxdn−dN fN
)=
(an − (r0a0 + · · ·+ rNaN )︸ ︷︷ ︸
=0
)xdn + lower order terms.
We claim that g ∈ I − 〈f0, . . . , fn−1〉, even though the above computation says that deg g < dn = deg fn.To see that this is true, g − fn ∈ 〈f0, . . . , fn−1〉 but fn 6∈ 〈f0, . . . , fn−1〉, so g 6∈ 〈f0, . . . , fn−1〉 either.
This gives a contradiction to the assumption that I is not finitely generated. So R[x] is Noetherian.
Example 2.17.2. There are non Noetherian rings: rings of continuous (or holomorphic) functions, the ringR[x1, x2, . . . ], and
∏∞i=1 F2.
Let R be an integral domain (a commutative ring with no zero divisors).
We say a divides b if b = ac for some c ∈ R, or if equivalently 〈b〉 ⊆ 〈a〉. This is denoted a|b.
We say a and b are associates if a = ub for some u ∈ R×, or if equivalently 〈b〉 = 〈a〉. This is anequivalence class, so sometimes this is denoted a ∼ b if there is no confusion.
We say p ∈ R is irreducible if p 6= 0 and p is not a unit, and if p = ab , then a or b is a unit.
We say p ∈ R is prime if p 6= 0 and p is not a unit, and if p|ab then p|a or p|b, or if equivalently 〈p〉 is aprime ideal of R.
Lemma 2.17.3. If p ∈ R is a prime, then it is irreducible.
Of course, the point is that the converse does not hold in general.
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Proof. If p = ab, then since p is prime p|a or p|b. Without loss of generality, suppose p|a, so that a = pc.Then p = (pc)b = pcb; since R is an integral domain we have 1 = cb and so b is a unit.
Proposition 2.17.4. Let R be a Noetherian integral domain. Then for any a 6= 0 in R, we have
a = up1 . . . pr
with u ∈ R× and pi ∈ R irreducible.
Proof. Define the set
Σ := 〈x〉 : x ∈ R, x 6= 0, x not a product of irreducibles.
Suppose that Σ 6= ∅. Since R is Noetherian, there is a maximal 〈x〉 ∈ Σ. So x cannot be factored intoirreducibles, i.e., x = ab with a, b not units where either a or b cannot be factored into irredicubles.
Without loss of generality, suppose a cannot be factored into irreducibles. Then
〈x〉 ⊆ 〈a〉 =⇒ 〈x〉 = 〈a〉
and so a = xy with y ∈ R, but as before we now have x = (xy)b = xyb and so 1 = yb and b is a unit.
Remark 2.17.5. The ring R[x1, x2, . . . ] is non-Noetherian but has fatorization into irreducibles.
We didn’t need the full power of Noetherian rings; we just needed that principal ideal domains satisfyACC – that weaker assumption is equivalent to the conclusion of the above lemma.
Definition 2.17.6. A Unique Factorization Domain (UFD) is an integral domain R such that
• every non-zero element a ∈ R satisfiesa = up1 . . . pr
for some unit u ∈ R× and irreducible pi ∈ R
• such a factorization is unique up to factoring and associates, that is, if
a = u′q1 . . . qs
for U ′ and unit and qi irreducible, we have r = s and σ ∈ Sr such that pi ∼ qσ(i).
Nonexamples of UFD include: Z[√−5] (classic example: 6 = 2 ·3 = (1+
√−5) · (1−
√−5)), R[cosx, sinx]
(classic example: cos2 x = (1 + sinx)(1− sinx)).
Even worse is H the ring of holomorphic functions C → C is such that elements don’t even factor intoirreducibles! The classic example is that sin(πz) = z(z − 1)(z − 2) . . . (z − n)g(z) for some holomorphic g;it’s not hard to show that these are all irreducible.
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2.18 Oct 5, 2018
Last time, we defined a UFD: an integral domain R such that every a ∈ R− 0 has a factorization
a = up1 . . . pr
with u ∈ R× and pi irreducible, where the p1, . . . , pr are unique up to reordering and scaling by units (asso-ciates).
Let a = upe11 . . . perr and b = u′pf11 . . . pfrr for irreducible pi (not associates of each other) and ei ≥ 0 andfi ≥ 0. Then
Definition 2.18.1. A greatest common divisor of a and b is a common divisor that divides all other commondivisors. In other words,
gcd(a, b) = pmin(e1,f1)1 . . . pmin(er,fr)
r .
Note that this is only defined up to a unit, or by the ideal 〈gcd(a, b)〉.
Lemma 2.18.2. Let R be a UFD. Take p ∈ R. Then p is prime if and only if p is irreducible
Proof. The forward direction was done in Lemma 2.17.3. For the backwards direction, take p irreducible,and suppose p|ab with a, b ∈ R nonzero, that is,
p(u′′p′′1 . . . p′′t ) = pc = ab = (up1 . . . pr)(p
′1 . . . p
′s)
where u, u′, u′′ ∈ R× and pi, p′i, p′′ irreducible. Then by uniqueness of factorizations both sides are the same
up to reordering and units, so p is an associate of pi or p′i for some i. In the first case p|a and in the secondp|b.
Lemma 2.18.3. Let R be an integral domain. Suppose all a 6= 0 in R can be factored into irreducibles.Then, R is a UFD if and only if all irreducibles of R are prime.
Proof. The forward direction is the previous lemma. For the backwards direction, take two different factor-izations
up1 . . . pr = q1 . . . qs
for pi, qj irreducible in R×. Look at p1: we have p1|q1 . . . qs, and since p1 is now prime (by assumption),we have p1|qj for some j, and hence p1 and qj are associates. So reorder and scale by units to get p1 = q1.Because we are in an integral domain we can cancel, that is,
u′p2 . . . pr = q2 . . . qs
for possibly a different u′. So we can do [lazy man’s] induction: keep repeating so that u = qs−r . . . qs (wehave s ≥ r because for each pi there must be at least one qi on the right side). Since
1 = u−1qs−r . . . qs
we have s = r and pi = qi for all i.
Definition 2.18.4. An integral domain is a principal ideal domain (abbreviated PID) if it is a domain andif all ideals of R are principal.
Standard examples are Z, F [x], F for F a field. Nonexamples include F [x, y] and the ideal 〈x, y〉, andZ[x] and the ideal 〈5, x〉. But these are still UFDs (we’ll prove this next time: if R is a UFD then so is R[x]).
Proposition 2.18.5. All PIDs are UFDs.
Proof. Let R be a PID. Any a 6= 0 has a factorization into irreducibles (since R is Noetherian!). Take anyirreducible p ∈ R; we need to show it is prime.
Observe that 〈p〉 ( R since p is not a unit. It follows that there is a maximal ideal m ⊇ 〈p〉 (and wedo not need Zorn here, because R is Noetherian). Then m = 〈π〉 because R is a PID. So 〈p〉 ⊆ 〈π〉 andπ|p. Note that π is not a unit, and so p = uπ for some u ∈ R× (because p is irreducible). It follows that〈p〉 = 〈π〉 = m, so 〈p〉 is a maximal ideal and hence a prime ideal. So p is now prime.
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This proof also gives the corollary
Corollary 2.18.6. All nonzero prime ideals of a PID are maximal.
Definition 2.18.7. A Euclidean domain is an integral domain R such that there is a function N : R−0 →Z such that N(a) ≥ 0 for a 6= 0 ∈ R and for any a, b ∈ R, with b 6= 0, we have a = qb+ r with q, r ∈ R suchthat r = 0 or N(r) < N(b).
Secretly N above stands for norm. Examples of Euclidean domains include Z with N(b) = |b|, and F [x]with N(f) = deg f . More nontrivial is
Example 2.18.8. Let R = Z[i] ⊆ C. We have N(a + bi) = |a + bi|2 = a2 + b2. To see that we can divide,take any α, β ∈ Z[i] with β 6= 0. Note that α/β = x + yi is not necessarily in Z[i]; we can only concludex, y ∈ R. But we can take a, b ∈ Z such that |x− a| ≤ 1/2 and |y − b| ≤ 1/2; define q := a+ bi ∈ Z[i], andobserve that ∣∣∣∣αβ − q
∣∣∣∣ ≤√(
1
2
)2
+
(1
2
)2
so that
|α− qβ︸ ︷︷ ︸:=r
|2 ≤ 1
2|β|2
so that α = qβ + r, with N(r) < N(β).
As an exercise, what goes wrong with R = Z[√−5]?
Proposition 2.18.9. A Euclidean domain R is a PID.
Proof. Let I be any ideal of R; we wish to show it’s principal. We can assume I 6= 〈0〉. Take any b ∈ I −0with N(b) minimal (recall that N(·) is an integer, so we can actually take this minimum). Note that 〈b〉 ∈ I;we claim that the other inclusion is true. Indeed, take any a ∈ I so that a = qb+ r with q, r ∈ R and r = 0or N(r) < N(b). But we have r = a − qb ∈ I so r = 0 (otherwise N(r) < N(b), contrary to assumption onb). So a = qb ∈ 〈b〉 so I = 〈b〉.
Remark 2.18.10. Euclidean domains have a nice Euclidean algorithm, that is, if a = qb+ r you can provegcd(a, b) = gcd(b, r). So these are important computationally. They’re also useful for proving certain ringsare PIDs.
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2.19 Oct 10, 2018
Last week, we considered several classes of rings. In particular, we have the chain of inclusions
Integral domains ) UFDs ) PIDs ) Euclidean Domains ) Fields.
Fix an integral domain R. Take a, b ∈ R, not both 0.
• If R is a UFD, then there is a gcd(a, b) ∈ R, defined up to a unit (or equivalently defined up to theideal 〈gcd(a, b)〉).
• If R is a PID, then this ideal 〈gcd(a, b)〉 is just the (principal) ideal 〈a, b〉. Thus there are x, y ∈ Rsuch that ax+ by = gcd(a, b).
• If R is Euclidean, with norm map N : R−0 → Z≥0, then if b ∈ R−0, for a ∈ R we have a = qb+rwith q, r ∈ R and r = 0 or N(r) < N(b). Then any d divides both a and b if and only if it divides bothb and r. So 〈gcd(b, r)〉 = 〈gcd(a, b)〉.
Let’s do an example of the Euclidean algorithm, which says that we should use a = qb+ r over and overuntil we get 0 as the remainder. Let R = Z with N(b) = |b|; say that a = 46391, b = 4301. Then
gcd(46391, 4301)
46391 = 10 · 4301 + 3381 = gcd(4301, 3381)
3381 = 3 · 920 + 621 = gcd(3381, 920)
920 = 1 · 621 + 299 = gcd(621, 299)
621 = 2 · 299 + 23 = gcd(299, 23)
299 = 13 · 23 + 0 = gcd(23, 0)
= 23
Today we’ll focus on the Gaussian integers R = Z[i] and do some computations with it. We showed that itwas a Euclidean domain with N(a+bi) = a2+b2; this was Example 2.18.8. Observe that N(αβ) = N(α)N(β)respects multiplication.
Lemma 2.19.1. We have Z[i]× = ±1,±i with (±1)−1 = ±1 and (±i)−1 = ∓i. Furthermore, each primeπ of Z[i] divides a unique prime p ∈ Z, with N(π) = p or N(π) = p2.
Proof. Let α ∈ Z[i]×, so that there is β ∈ Z[i] with αβ = 1. Thus
N(α)N(β) = N(αβ) = N(1) = 1
so N(α) = 1. Now α = a + bi where a2 + b2 = 1 and a, b are integers. So (a, b) ∈ (±1, 0), (0,±1), andα = ±1 or α = ±i.
For the second part, let N(π) = p1 . . . pr. Then since ππ = N(π), then since π is prime we have π dividessome pi =: p. We can factor p into primes in Z[i], say p = ππ2 . . . πs, and hence
p2 = N(p) = N(π)N(π2) . . . N(πs)
so that N(π) = p or N(π) = p2.
In fact, this proof says that if N(π) = p2 then π = p up to a unit, and if N(π) = p then p = ππ. In otherwords, we saw that any prime π ∈ Z[i] divides a unique p. When we factor p in Z[i], we either have p primein Z[i] or p = ππ for primes π, π conjugate to each other, both of norm p. We want to understand whichcase happens.
We should deal with the case p = 2 separately: we have 2 = (1 + i)(1 − i) = −i(1 + i)2, and so up tounits, 1 + i is the unique prime dividing 2.
Now consider p odd. We have isomorphisms
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Z[i]/〈p〉 Z[x]/〈x2 + 1, p〉 Fp[x]/〈x2 + 1〉∼ ∼
Notice that x2 + 1 ∈ Fp[x] is separable since p 6= 2. If x2 + 1 ∈ Fp[x] is irreducible, then Fp[x]/〈x2 + 1〉 is afield, which means that Z[i]/〈p〉 is a field, and hence p is prime in Z[i]. On the other hand, if x2 + 1 ∈ Fp[x]factors, say x2 + 1 = (x+ α)(x− α), then
Fp[x]/〈x2 + 1〉 ∼−→CRT Fp[x]/〈x+ α〉 × Fp[x]/〈x− α〉 ∼= Fp × Fp.
This is not an integral domain, so p is not prime in Z[i] and p = ππ.
It remains to answer when −1 is a square in F×p .
Proposition 2.19.2. The group F×p is cyclic. Moreover, any finite subgroup G of F×, with F a field, iscyclic.
Proof. The second part implies the first part, so we’ll prove that. We have that G is finite and cyclic, so it isisomorphic to the direct product of its p-Sylow subgroups [we technically aren’t allowed to use the structuretheorem for finite abelian groups yet]. So without loss of generality assume G is a p-group, and suppose
that G is not cyclic. Since |G| = pe, we have gpe−1
= 1 for all g ∈ G (otherwise we’d have a generator). So
xpe−1 − 1 has at least |G| = pe roots in F . However, deg(xp
e−1 − 1) < pe, which is a contradiction.
So when p is odd, we have −1 a square in F×p , that is, x2 = −1 for some x ∈ F×p , if and only if we havean element of order 4 in F×p , which means that p− 1 = |F×p | is divisible by 4.
So to summarize, if p = 2, then 1 + i is the unique prime dividing 2. If p ≡ 3 (mod 4), then p is prime inZ[i] and it is the unique prime dividing itself. If p ≡ 1 (mod 4) then p = ππ, where if we write p as a sumof two squares p = a2 + b2 then π = a+ bi. Hence a+ bi, a− bi are the primes that divide p.
Theorem 2.19.3 (Fermat). Let p be an odd prime. Then p = a2 + b2 with a, b ∈ Z if and only if p ≡ 1(mod 4).
Proof. We just did the backwards direction, which is the hard one. To get the forward direction, notice thatx2 ≡ 0, 1 (mod 4) for every x, and so p ≡ 0, 1, 2 (mod 4) but p is odd.
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2.20 Oct 12, 2018
We’ll begin with an example:
Example 2.20.1. Find all solutions of y2 = x3 − 1 with x, y ∈ Z.
Proof. We fix a solution (x, y) ∈ Z2. Notice that
x3 = y2 + 1 = (y + i)(y − i) in Z[i].
We claim that y + i and y − i are relatively prime: suppose not, and take a prime π ∈ Z[i] dividing bothy+ i and y− i. So π divides their difference, and up to a unit, π|2 so π = 1 + i (we characterized the primesdividing 2 last class). Also, π divides x3 = (y + i)(y − i), but 2 = N(π)|N(x3) = x6 so x is even. But theny2 = x3 − 1 ≡ 7 (mod 8), a contradiction. So y + i and y − i are relatively prime.
So x3 = (y + i)(y − i) for some relatively prime y + i and y − i. It follows that y + i and y − i arethemselves cubes: take π a prime dividing y + i, and let e be such that πe|y + i and πe+1 - y + i. It followsthat πe|x3 and πe+1 - x3, and so e must be divisible by 3. So
y + i = uπ3e11 . . . π3er
r
where u is a unit and πi are nonassociate primes. Then
y + i = (a+ ib)3 = (a3 − 3ab2) + (3a2b− b3)i
with a, b ∈ Z. So we have integers a, b ∈ Z satisfying
y = a3 − 3ab2
1 = 3ab2 − b3
The second equation says b = ±1. When b = 1 then 3a2 − 1 = 1, which is impossible, and when b = −1then 3a2 + 1 = 1, which has the unique solution a = 0. Then the first equation says y = 0 and hence x = 1.
The only solution (x, y) is (1, 0).
Last time, we showed that if p is an odd time, then p can be written as a sum of two squares if and onlyif p ≡ 1 (mod 4). So take such a p ≡ 1 (mod 4), and define
N := #(a, b) ∈ Z2 : p = a2 + b2.
What is N? We have at least 8 solutions: (±a,±b) and (±b,±a). We claim that these are precisely all ofthe solutions: note that
N = #π ∈ Z[i] : N(π) = pbut then p = ππ is the prime factorization of p, so the primes that divide p are uπ, uπ with u ∈ Z[i]×.
Consider n = 232 + 1 = 4294967297. Observe that n = (216)2 + 12 = 622642 + 204492 is a sum of twosquares in at least two ways. Hence n is not prime!
We can find integer factors of n in the following way: write
(216 + i)(216 − i) = n = (62264 + 20449i)(62264− 20449i).
We can compute gcd(216 + i, 62264− 20449i) via the Euclidean Algorithm (steps are suppressed, the pointis that it’s doable). The gcd happens to be −4− 25i. So in particular
N(−4− 25i)|N(216 + 1)
and since N(−4− 25i) = 641 and N(216 + 1) = n, we see that 641|n.
And more specifically n = 641 · 6700417 (this factorization was originally due to Euler).
Let R be an integral domain. We defined F ⊇ R the field of fractions of R.
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Definition 2.20.2. We say that R is integrally closed if all α ∈ F that is a root of a monic polynomial inR[x] is actually in R.
Proposition 2.20.3. A UFD is integrally closed.
For example, Z is integrally closed, since Z is a UFD. We can conclude that x3 − 2 is irreducible in Q[x](if not it has a root α ∈ Q, but by the proposition we have α ∈ Z, and obviously there is no α ∈ Z so thatα3 = 2). Let’s prove the above proposition:
Proof. Takef = xn + an−1x
n−1 + · · ·+ a1x+ a0 ∈ R[x]
such that f(α) = 0 with α ∈ F . Then α = cd with c, d ∈ R, d 6= 0. Since R is a UFD, we can assume that c
and d are relatively prime (we can “reduce to lowest terms”). Then
0 = dnf(α) = cn + an−1dcn−1 + · · ·+ a1d
n−1c+ a0dn
which says that−cn = d(an−1c
n−1 + · · ·+ a1dn−2c+ a0d
n−1)
and since gcd(c, d) = 1 we have d ∈ R×, and now α = cd−1 ∈ R.
Example 2.20.4. Fix d ∈ Z squarefree, d 6= 1 so that d ≡ 1 (mod 4). Then Z[√d] is not a UFD. This is
because we can consider the monic polynomial
x2 − x+1− d
4=
(x−
(1 +√d
2
))(x+
(1−√d
2
))with roots in the fraction field (1±
√d)/2 6∈ Z[
√d].
For d 6= 1 squarefree, define
Rd :=
Z[√d] if d 6≡ 1 (mod 4)
Z[ 1+√d
2 ] if d ≡ 1 (mod 4)
The Rd are integrally closed; there is a formal process we’ll get to later that gives us integral closures of rings.
Some facts:
• Rd is a UFD if and only if Rd is a PID
• For d < 0, we have Rd is a UFD if and only if
d ∈ −1,−2,−3,−7,−11,−19,−43,−67,−163
• (Conjecture by Gauss): There are infinitely many d > 1 such that Rd is a UFD. We expect about 76%of these to work, by some Cohen-Lenstra heuristics.
The second fact was conjectured by Gauss, and is a theorem due to Stark and Heegner.
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2.21 Oct 15, 2018
Last time we defined, for squarefree d 6= 1, the ring
Rd :=
Z[√d] if d 6≡ 1 (mod 4)
Z[ 1+√d
2 ] if d ≡ 1 (mod 4)
and Rd need not be UFD, but it is a Dedekind domain, which is a domain in which every nonzero idealfactors uniquely into prime ideals. These Rd are actually the ring of integers of Q(
√d).
Example 2.21.1. Consider R := R−5 = Z[√−5]; one can show that R× = ±1 (by looking at norm
maps). We have failure of unique factorization in R, since
2 · 3 = 6 = (1 +√−5)(1−
√−5)
with every element irreducible in R. We consider the ideals
p = 〈2, 1 +√−5〉
q1 = 〈3, 1 +√−5〉
q2 = 〈3, 1−√−5〉.
We claim that these are maximal ideals, for example,
R/q1∼= Z[x]/〈3, x2 + 5, 1 + x〉 ∼−→ Z/〈3〉
where the isomorphism is x 7→ −1.Observe that
p2 = 〈4, 2(1 +√−5), (1 +
√−5)2〉 = 〈4, 2 + 2
√−5, 2
√−5〉 = 〈4, 2, 2
√−5〉 = 〈2〉.
So 〈2〉 = p2, and similarly one can show that 〈3〉 = q1q2. One can also do pq1 = 〈1 +√−5〉 and pq2 = 〈1−√
−5〉. Sop2 · q1q2 = 〈2〉〈3〉 = 〈6〉 = 〈1 +
√−5〉〈1−
√−5〉 = pq1 · pq2.
Thus, although unique factorization fails in R, the nonzero ideal 〈6〉 factors uniquely into a product of primeideals.
Remark 2.21.2. PIDs are both UFDs and Dedekind domains, but not all UFDs are Dedekind domains(and as we saw earlier, not all Dedekind domains are UFDs).
Example 2.21.3. Let’s factor 4x3 + 11x− 6 in Q[x]. Because it’s a cubic we just need to know if it has aroot, say α = r
s with r, s ∈ Z and gcd(r, s) = 1 and s ≥ 1.
Thus 4r3 + 11rs2−6s3 = 0. In other words, r(4r2 + 11s2) = 6s3 so r divides 6 (because r, s are coprime).So r = ±1,±2,±3. Similarly, 4r3 = s(−11rs + 6s2), so s divides 4, and s = 1, 2, 4 (we assumed s ≥ 1, bysymmetry).
So we can check all 18 possible α. It turns out that α = 12 is the only root. It follows that
4x3 + 11x− 6 = 4
(x− 1
2
)(x2 +
1
2x+ 3
).
In Z[x], this says 4x3 + 11x− 6 = (2x− 1)(2x2 + x+ 6).
Gauss’ Lemma tells us how factorizations in Z[x] and Q[x] compare. More generally, let R be a UFDand F be the field of fractions of R; we want to compare factorizations in R[x] with those in R and F [x].
The irreducible elements of degree 0 in R[x] are just the irreducible elements of R. So you can’t escapeunderstanding factorization of R if you want to understand factorizations of R[x].
47
Proposition 2.21.4 (Rational root theorem). Let R be a UFD. Fix f = anxn + · · ·+ a1x+ a0 ∈ R[x], with
an 6= 0. If α ∈ F is a root, take α = rs with r, s ∈ R relatively prime. Then r|a0 and s|an.
The proof was Example 2.21.3.
Definition 2.21.5. We say that a nonzero f ∈ R[x] is primitive if its coefficients are relatively prime (inR). So the only elements of R that divide all coefficients are units.
Take any nonzero f ∈ F [x]. We claim that there is a value c(f) ∈ F× such that f = c(f) · f1 withf1 ∈ R[x] primitive.
Proof. We can clear denominators, that is, pick d so that df ∈ R[x], so without loss of generality f ∈ R[x].Let g be a gcd of the coefficients of f . Now f = g · fg , and f
g ∈ R[x] is primitive.
We also have that c(f) ∈ F× and f1 are unique up to multiplication by a unit in R:
Proof. Say that αf1 = βf2 with α, β ∈ F× and fi ∈ R[x] primitive. So αβ−1f1 = f2, and if αβ−1 = rs with
r, s ∈ R relatively prime, we have rf1 = sf2, and the gcd of the coefficients is r and is s. So r = su foru ∈ R×. Now αβ−1 := u is a unit, so f2 = uf1.
The element c(f) is called the content of f . Note that if f ∈ R[x] then c(f) ∈ R.
Lemma 2.21.6 (Gauss’ lemma). Let R be a UFD, and F its fraction field. If f, g ∈ F [x] are nonzero,then c(fg) ≈ c(f)c(g), that is, c(fg) and c(f)c(g) are equal up to a unit. In particular, if f, g ∈ R[x] areprimitive, then fg is also primitive.
Of course, a polynomial is primitive if and only if its content is a unit.
Let’s prove the special case (if f, g ∈ R[x] are primitive, then so is fg). Suppose fg is not primitive, sothere is a prime π ∈ R dividing the coefficients of fg. Now consider
fg = 0 ∈ (R/〈π〉)[x],
where R/〈π〉 is an integral domain (since π is prime). It follows that (R/〈π〉)[x] is an integral domain, too. Soeither f = 0 or g = 0 in (R/〈π〉)[x]. Then either f or g are not primitive (because π divides their coefficients).
We will prove the rest of the lemma next lecture.
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2.22 Oct 17, 2018
Last time, we let R be a UFD, and R ⊆ F be its fraction field. We say f ∈ R[x] is primitive if itscoefficients are relatively prime. For nonzero f ∈ R[x], we have f = c(f) ·f1 for some c(f) ∈ F× (the contentof f) and primitive f1 ∈ R[x]. We have that c(f) is unique up to multiplication by a unit u ∈ R×. We hadthe following lemma:
Lemma 2.22.1 (Gauss’ Lemma). If f, g ∈ R[x] are primitive, then fg is primitive.
The proof wasn’t so bad, we just took fg modulo a prime and got a contradiction. Here’s another Gauss’Lemma.
Lemma 2.22.2 (Gauss’ Lemma). If f, g ∈ F [x] are nonzero, then c(fg) = u · c(f)c(g) for some u ∈ R×.
Proof. We can write f = c(f)f1 and g = c(g)g1 with f1, g1 primitive; we now have
fg = c(f)c(g)f1g1
where f1g1 is primitive due to the other Gauss’ Lemma. Because of uniqueness of c up to multiplication bya unit, we have c(f)c(g) = c(fg) up to a unit.
Recall that if f ∈ R[x] is of degree 0, then f is irreducible in R[x] if and only if f is irreducible in R. Forexample, in Z[x], the usual primes p ∈ Z are irreducible.
Proposition 2.22.3 (Gauss’ Lemma?). Take f ∈ R[x] of degree ≥ 1. Then f is irreducible in R[x] if andonly if f is primitive and f is irreducible in F [x].
Proof. The backwards direction is easier. Suppose that f is irreducible in F [x] and primitive. Suppose thatf = gh for g, h ∈ R[x] nonunits. Note that deg g = 0 or deg h = 0, otherwise this would be a factorizationin F [x]. Say that g ∈ R− 0. Now g|f and f is primitive, so g is a unit of R.
To prove the forwards direction, we assume that f is irreducible. Clearly f is primitive. Say that f = ghwith g, h ∈ F [x] of degree at least 1. Then f = c(g)c(h)g1h1, where c(g)c(h) ∈ F× and g1, h1 ∈ R[x] areprimitive. But f is primitive, so c(f) = 1 up to a u ∈ R×, so in fact c(g)c(h) ∈ R×.
It follows that f = ug1h1 with u ∈ R× and g1, h1 ∈ R[x], but since f is irreducible in R[x] means thateither g1 or h1 is in R[x]× = R×, which means that g or h is in F×. This means that f is irreducible inF [x].
Lemma 2.22.4 (Gauss’ Lemma). Take f ∈ R[x] nonzero. Suppose f = gh with g, h ∈ F [x]. Then there isa c ∈ F× such that f = (cg)(c−1h) with cg, c−1h ∈ R[x].
The proof is exactly the same as the proof above.
Theorem 2.22.5. Let R be a UFD. Then R[x] is a UFD.
Hilbert’s Basis Theorem is similar (replace UFD with Noetherian), but this is not true for PIDs.
Proof. Take a nonzero f ∈ R[x]. Then f = df1 with f1 primitive, and d ∈ R. Since R is a UFD we canfactor d into irreducibles in R (and hence in R[x]). We also have
f1 = p1 . . . pr
with pi ∈ F [x] irreducible, and there are pi = cipi ∈ R[x] with ci ∈ F×. We have
f1 = p1 . . . pr.
Since f1 is primitive we have pi are all primitive. Now pi are primitive and irreducible in F [x], so by Gausspi ∈ R[x] are irreducible.This gives existence of factorizations.
49
To get uniqueness, we let f = uπ1 . . . πrp1 . . . ps where u ∈ R×, and πi are degree 0 irreducibles in R[x],and pi degree ≥ 1 irreducibles in R[x]. Since c(f) ∈ R is, up to a unit, is exactly π1 . . . πr, the πi are uniqueup to reordering and units since R is a UFD.
So assume f = p1 . . . ps. Now the pi are unique up to reordering and units, since pi ∈ F [x] is in a PID(and hence a UFD). But also c(pi) = 1 so pi is unique in R[x].
Theorem 2.22.6 (Eisenstein’s Criterion). Let R be a UFD, and let p ∈ R be a prime. Consider
f = anxn + an−1x
n−1 + · · ·+ a1x+ a0 ∈ R[x]
with the condition that p|a0, a1, . . . , an−1, and p - an, and p2 - a0. Then f is irreducible in F [x]. If f isprimitive (for example, f is monic), it is also irreducible in R[x].
Proof. Suppose f = gh with g, h ∈ R[x]. Reduce this mod p, that is,
f = gh ∈ (R/〈p〉)[x]
with f = anxn, and an 6= 0. Since R/〈p〉 is a field, so g = c1x
d and h = c2xn−d with ci ∈ (R/〈p〉)× and
0 ≤ d ≤ n. But if 0 < d < n then p divides constant terms of g and h, which is contradictory to assumption(that p2 - a0). So deg g or deg h is 0.
This is the classic example (and the one Eisenstein proved his criterion for):
Example 2.22.7. Let p be a prime, and f = xp−1 + · · ·+ x+ 1 ∈ Z[x]. Then f(x) = xp−1x−1 , and
f(x+ 1) =(x+ 1)p − 1
x=
p∑j=1
(p
j
)xj−1
and since p|(pj
)for j = 1, . . . , p− 1, p2 -
(p1
)= p, and p -
(pp
)= 1, Eisenstein says that f is irreducible in Z[x]
and hence it’s irreducible in Q[x].
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3 Modules
3.23 Oct 19, 2018
We saw that if R is a UFD, then R[x] is a UFD. The primes of R[x] are the primes of R and the primitivepolynomials in R[x] irreducible in F [x], where F is the fraction field. The story is pretty much the same whenone asks for R[[x]]; more things become units. In general power series are more complicated but algebraicallyR[[x]] is easier to work with.
Today we’re going to talk about modules; it’ll be mostly definitions. Let R be a ring with 1.
Definition 3.23.1. A (left) module over R (or R-module) is a set with with operations
M ×M +−→M
(m,n) 7→ m+ n
and
R×M ·−→M
(r,m) 7→ r ·m
called “addition” and “scalar multiplication” such that the following hold:
• (M,+) is an abelian group, with identity 0;
• 1 ·m = m for m ∈M
• r · (s ·m) = (rs) ·m for r, s ∈ R and m ∈M
• (r + s) ·m = r ·m+ s ·m for r, s ∈ R and m ∈M
• r · (m+ n) = rm+ rn ro r ∈ R and m,n ∈M
Remark 3.23.2. There is a similar definition for right modules, where the multiplication is M × R ·−→ M .But if R is commutative, and M is a right R-module, we can define a left module with r · m := m · r.Commutativity is required to keep the r · (s ·m) = (rs) ·m axiom.
Example 3.23.3. If R is a field, then R-modules are precisely vector spaces over F .
Example 3.23.4. Fix an integer n ≥ 0 and define Rn := (a1, . . . , an) : ai ∈ R with the operations(a1, . . . , an) + (b1, . . . , bn) = (a1 + b1, . . . , an + bn) and r · (a1, . . . , an) = (ra1, . . . , ran). This is called thefree R-module of rank n.
Example 3.23.5. Let R = Z. Then Z-modules are precisely abelian groups, with the multiplication
a ·m =
m+ · · ·+m︸ ︷︷ ︸a times
if a > 0
0 if a = 0
−m− · · · −m︸ ︷︷ ︸|a| times
if a < 0
Example 3.23.6. (Left) ideals I ⊆ R are an R-module: they’re closed under addition, and closed undermultiplication by R. Similarly, quotients are R-modules with the usual addition and r(a+ I) = ra+ I
Example 3.23.7. Polynomial rings R[x] with the usual addition and multiplication are also an R-module.
Recall that a homomorphism of R-modules is a map f : M → N such that f(m+ n) = f(m) + f(n) andf(rm) = rf(m) for m,n ∈M and r ∈ R. We also say f is an isomorphism if it is also bijective. The inverseis automatically an isomorphism.
51
Example 3.23.8. If f : M → N is a homomorphism, then f(M) is an R-module, and ker f = m ∈M : f(m) = 0 is also an R-module.
Definition 3.23.9. A submodule of M is a subset N ⊆ M that is an R-module using the operations fromthe ambient module M .
Observe that if N ⊆M , then N is a submodule of M if and only if 0 ∈ N , m+ n ∈ N for all m,n ∈ N ,and rn ∈ N for all r ∈ R, n ∈ N . The other axioms follow in N because they are inherited from M .
Example 3.23.10. Submodules of R are precisely ideals.
Example 3.23.11. We can take quotients: Let M be a module and N ⊆M be a submodule. The quotientM/N is an abelian group with scalar multiplication defined by r(m+N) = rm+N for r ∈ R and m ∈M .This turns M/N into an R-module. We have a natural projection M →M/N given by m 7→ m+N ; it is ahomomorphism of R-modules.
There are the four isomorphism theorems as always.
Example 3.23.12. For a set S ⊆M , we have the submodule generated by S:
R · S :=
∑s∈S′
rs · s : S′ ⊆ S finite, rs ∈ R,
that is, we have finite linear combinations of elements in S. This is the smallest submodule containing S.
Definition 3.23.13. We say M is finitely generated if M = R · S for some finite S ⊆M .
Definition 3.23.14. Let M be an R-module. We say M is Noetherian if any chain
N1 ⊆ N2 ⊆ N3 ⊆ . . .
of submodules stabilizes, ie., Nn = Nn+1 for n large enough. This agrees with the previous notation, thatis, if M = R as an R-modules, then R is Noetherian in the usual sense if and only if it is Noetherian in thissense.
Proposition 3.23.15. We say that M is Noetherian if and only if all submodules of M are finitely generated.
The proof is the same as the proof of (the equivalence of) Definition 2.16.5.
Proposition 3.23.16. Let M be an R-module, and N a submodule of M . Then M is Noetherian if andonly if N and M/N are Noetherian.
Proof. The forward direction is easy: if M is Noetherian then so is N because submodules of N are inparticular submodules of M and are hence finitely generated. Also, a submodule of M/N is of the form L/Nwith N ⊆ L ⊆ M (there’s an isomorphism theorem being used here). Since M is Noetherian L is finitelygenerated as an R module, so L/N is also finitely generated. So M/N is Noetherian.
For the backward direction, we recall that M/N is finitely generated, say with generators x1, . . . , xn ∈M ,and N is finitely generated, say with generators y1, . . . , ym ∈ N . Take any m ∈ M . Then m + N =∑ni=1 ri(xi +N) for some ri ∈ R. Now
m−n∑i=1
rixi ∈ N
so we have n−∑ni=1 rixi =
∑mj=1 sjyj with sj ∈ R. Now we win; M is generated by x1, . . . , xn, y1, . . . , ym.
Proposition 3.23.17. Let R be Noetherian. Then if M is an R-module, we have M is Noetherian if andonly if M is finitely generated.
As a special case, we get that Rn is Noetherian, by induction on n (the base case is our assumption on R).
If M is finitely generated, there is a surjective homomorphism Rn M , so M is a quotient of Rn andM is Noetherian. Of course, if M is Noetherian, then M is finitely generated.
52
3.24 Oct 22, 2018
Last time, for R a commutative ring, we saw that if R is Noetherian, then an R-module M is Noetherianif and only if M is finitely generated.
Today, we let R be a PID. We’ll describe modules over a PID.
Let M be a finitely generated R-module, say with generators m1, . . . ,mn ∈M . We have a unique homo-morphism of R-modules ϕ : Rn M mapping ei (the “standard basis” of Rn) to mi. It’s clear what this mapshould be, since ϕ(r1e1 + · · ·+rnen) = r1m1 + · · ·+rnmn. Also, ϕ is surjective since m1, . . . ,mn generate M .
The first isomorphism theorem says Rn/ kerϕ∼−→ M . Notice that kerϕ is finitely generated, since Rn
is Noetherian. So there is another surjective homomorphism ψ : Rm kerϕ, so we get the short exactsequence
Rm Rn M 0ψ ϕ
Take any 1 ≤ j ≤ m, and observe that
ψ(ej) =
n∑i=1
Ai,jei for unique Ai,j ∈ R.
If you view Rm and Rn as “columns”, then ψ is given by left multiplication by a matrix A ∈ Mn,m(R). SoM ∼= Rn/ψ(Rm) = Rn/A(Rm).
Some remarks: Let P ∈ GLn(R) and Q ∈ GLm(R). Then because P,Q are isomorphisms
Rn/(PAQ)(Rm) = Rn/(PA(Rm)) = Rn/P (A(Rm)) = Rn/A(Rm)
given by P v ← [ v.
So to understand modules M over a PID, we want to understand Rn/A(Rm); an idea is to find P and Qso that A is “nice”.
In linear algebra, for R = F a field, and a matrix A ∈ Mm,n(F ), there are invertible P ∈ GLm(F ) andQ ∈ GLn(F ) such that
D = PAQ =
11
. . .
10
. . .
0
where D is not necessarily a square and there are r 1’s, where r is the rank of A (the point is that we’ll givea generalization of this result later, for PIDs). Then
Rm/D(Rn) ∼= R× · · · ×R︸ ︷︷ ︸r times
×0× · · · × 0 ∼= Rm−r = Fm−r
which says that finitely generated vector spaces are free, ie. have a finite basis.
53
The generalization is called Smith Normal Form: Let R be a PID, and take any matrix A ∈ Mm,n(R).Then there are invertible P ∈ GLm(R) and Q ∈ GLn(R) such that
D = PAQ =
a1
a2
. . .
ar0
. . .
0
with a1, . . . , ar ∈ R nonzero and a1|a2, a2|a3, . . . , ar−1|ar. Moreover, r is unique and the ai are unique upto multiplication by u ∈ R×.
Theorem 3.24.1 (Structure theorem for finitely generated modules over PIDs). Let R be a PID, and letM be a finitely generated R-module. Then there are r, s ≥ 0 and nonzero a1, . . . , as ∈ R that are not unitssuch that
M ∼= R/〈a1〉 × · · · ×R/〈as〉 ×Rr
and a1|a2, a2|a3, . . . , as−1|as. Moreover, r and s are unique, and the ais are unique up to units.
Sometimes people write a1|a2, a2|a3, . . . , as−1|as as “a1|a2| . . . |as”. The ai are called invariant factorsand r the free rank.
Proof. We first prove existence assuming Smith Normal Form. We saw (with m and n switched) thatM ∼= Rm/A(Rn) for some A ∈ Mm,n(R), so M ∼= Rm/D(Rn) (with the isomorphism coming from SmithNormal Form), where as before
D = PAQ =
a1
a2
. . .
ar0
. . .
0
with ai ∈ R− 0 and a1|a2| . . . |ar. Then
Rm/D(Rn) ∼= Rm/(a1R× · · · × arR× 0× · · · × 0) ∼= R/〈a1〉 × · · · ×R/〈ar〉 ×R× · · · ×R
where we remove the early ais that are units (since R/〈ai〉 = 0).
We’ll prove Smith Normal Form next time; the proof will be an algorithm, that is, one could actuallycompute the ai.
Theorem 3.24.2 (Structure theorem for finitely generated abelian groups). Let M be a finitely generatedabelian group. Then
M ∼= Z/n1Z× · · · × Z/nsZ× Zr
for unique r ≥ 0 and unique positive integers n1, . . . , ns ≥ 2 satisfying n1|n2| . . . |ns.
Observe that if M is well understood, one could compute r and ni; every step of the proof is constructive.
If ai = uπe11 . . . πett with u ∈ R× and π1, . . . , πt ∈ R primes (not associates) and ei ≥ 1, then by CRT wehave R/〈ai〉 ∼= R/〈πei1 〉 × · · · ×R/〈πetr 〉. This gives another description of the modules over a PID.
Next time we’ll give the proof of Smith Normal Form, and specialize to R = F [x].
54
3.25 Oct 24, 2018
Last time, we talked about Smith Normal Form: Let R be a PID, and take a matrix A ∈Mm,n(R). Thenthere are invertible matrices P ∈ GLm(R) and Q ∈ GLn(R) such that
D = PAQ =
a1
a2
. . .
ar0
. . .
0
with ai ∈ R nonzero satisfying a1|a2| . . . |ar. Moreover, r is unique and ai is unique up to multiplication byu ∈ R×. The matrix D is called the Smith Normal Form of A.
We’ll prove the existence of the Smith Normal Form of A. We say A,B ∈ Mm,n(R) are equivalent ifPAQ = B for some P ∈ GLm(R) and Q ∈ GLn(R). Of course, this is an equivalence relation on Mm,n(R).
Proof. Fix A ∈Mm,n(R). Our goal is to find an equivalent matrix of the form[a1 00 B
]with B ∈ Mm−1,n−1(R), where a1 divides all entries of B. Doing this is sufficient; we can use induction onthe size of the matrix. Notice that a1 divides all entires equivalent to B, too.
We can assume A 6= 0 (otherwise, we are done). We can perform elementary row operations: we canswap rows, add a multiple of one row to another, and scale a row by u ∈ R×.
They are reversible by another elementary row operation. Also, each corresponds to left multiplicationby a matrix P ∈ GLm(R). We can also perform column operations, corresponding to right multiplication bya matrix P ∈ GLm(R). By swapping rows and columns, we can assume A1,1 6= 0.
For 2 ≤ i ≤ n, we can replace A by an equivalent matrix A′ such that A′i,1 = gcd(A1,1, Ai,1). To seethis, say that (without loss of generality, because we can swap rows) that i = 2; for a = A1,1, b = A2,1, andg = gcd(a, b), then if R is a PID we have 〈a, b〉 = 〈g〉,that is, ax+ by = g for some x, y ∈ R.
In step 1, we split into cases. The good case is when A1,1|Ai,1 for all 2 ≤ i ≤ m, say with Ai,1 = qiA1,1
with rowi := rowi − qirow1. This turns the first column into the forma ∗ . . . ∗0 ∗ . . . ∗...
.... . .
...0 ∗ . . . ∗
and go to Step 2.
In the bad case, A1,1 - Ai,1 for some 2 ≤ i ≤ n. Then replace A by an equivalent matrix A′ such thatA′1,1 = gcd(A1,1, A1,i−1) and restart Step 1 (we’ll discuss why this must terminate later).
In step 2, we also split into cases. The good case is when A1,1|A1,j for all 2 ≤ j ≤ n, say with A1,j = qjA1,1
with colj := colj − qjcol1. This turns A into the forma 0 . . . 00 ∗ . . . ∗...
.... . .
...0 ∗ . . . ∗
55
and go to Step 3.
In the bad case, A1,1 - A1,j for some j. Then replace A by an equivalent matrix A′ such that A′1,1 =gcd(A1,1, A1,j) and go to step 1.
In step 3, we also split into cases. The good case is when A1,1 divides all entries of B. Then we’d be done.
If there is an 2 ≤ i ≤ m such that A1,1 does not divide all elements in i-th row, we add the i-th row tothe first row, and go to step 2.
To show this is a proof/algorithm, we need to know why it halts. Whenever we were in a bad case, weobtained an equivalent matrix A′, such that A′1,1|A1,1 and A1,1 - A′1,1. Let α1, α2, . . . be the A1,1 we getover the algorithm. Then
〈α1〉 ⊆ 〈α2〉 ⊆ . . .
and a bad case corresponds to 〈αn〉 ( 〈αn+1〉. Eventually, we never hit a bad case, so it stops!
Example 3.25.1. Let G = Z3/〈(3, 3, 6), (8, 4, 0), (0, 12, 12)〉. Consider
A =
3 8 03 4 126 0 12
,and observe that G = Z3/A(Z3). We want to reduce this matrix. Set r2 := r2 − r1 and r3 := r3 − 2r1, sothat
A′ =
3 0 00 −4 120 −16 12
and then set c2 := c2 − 3c1, and set c2 := −1c2 and swap c1 and c2, so that
A′ =
1 3 04 0 1216 0 12
and set r2 := r2 − 4r1 and r3 := r3 − 16r1, so that
A′ =
1 3 00 −12 120 −48 12
and set c2 := c2 − 3c1 and r3 := r3 − 4r2, so that
A′ =
1 0 00 −12 120 0 −36
and set c3 := c3 + c2, c2 := −c2, and c3 := −c3, so that
A′ =
1 0 00 12 00 0 36
.Thus, A is equivalent to A′ as above; we have
G ∼= Z3/
1 0 00 12 00 0 36
Z3 ∼= (Z× Z× Z)/(Z× 12Z× 36Z) ∼= Z/12Z× Z/36Z.
56
3.26 Oct 29, 2018
[Homeworks are due Monday for a few weeks, since class was cancelled Friday and Thanksgiving willhappen.]
Last time, we talked about Smith Normal Form:
Let R be a PID, and take any A ∈Mm,n(R). Then there are P ∈ GLm(R) and Q ∈ GLn(R) so that
D = PAQ =
a1
a2
. . .
ar0
. . .
0
with nonzero a1, . . . , ar ∈ R satisfying a1|a2| . . . |ar. Moreover, the ai are unique up to units.
We proved the existence of P and Q.
Today we’ll set R = F [x], where F is a field. Let V be a finite dimensional vector space over F . Considera linear operator T : V → V . We can turn V into an F [x]-module in the following way: elements c ∈ F acton V as usual, and x ∈ F acts on V via T , that is, xv := T (v).
The converse is true; if M is an F [x] module then it is a (maybe not necessarily finite dimensional) vectorspace with a linear operator given by T (v) := xv.
Choose a basis v1, . . . , vn of V . We set
T (vj) =
n∑i=1
Ai,j · vi
with Ai,j ∈ F . This gives rise to A ∈ Mn(F ). We can now view Fn as an F [x] module by letting x act asleft multiplication by A, and V ∼= F [x].
Define B := xI −A ∈Mn(F [x]). By homework 8, problem 7, we can say
V ∼= Fn ∼= F [x]n/B(F [x]n).
This is supposed to be a good thing; in the classification of finitely generated modules over a PID we tooka finitely generated module and understood F [x]n/B(F [x]n).
Consider the Smith Normal Form of B, so
B = P
11
. . .
1a1
. . .
as0
. . .
Q
57
for P,Q ∈ GLn(F [x]) and a1, . . . , as ∈ F [x] monic of degree at least 1 and a1|a2| . . . |as. Moreover, a1, . . . , asare unique polynomials; they are the invariant factors of B or A.
Define the characteristic polynomial of A to be cA(x) = det(xI − A) ∈ F [x], so that cA(x) is monic ofdegree n. Because detB 6= 0, we in fact have
B = P
11
. . .
1a1
. . .
as
Q
that is, we have no zeros. Notice that
cA(x) = detB = detP · a1 . . . as · detQ
and furthermore that detP,detQ ∈ F× (since P ·P−1 = I, so det(P−1) is an inverse to detP in F [x]× = F×).Now
cA(x) = (detP detQ)a1 . . . as
for monic ai and monic cA. Since detP detQ ∈ F we have detP detQ = 1. So cA(x) = a1 . . . as. This couldbe a determinant-free definition of characteristic polynomials!
Again, by homework 8, we see that (for D the Smith Normal Form of B)
V ∼= F [x]/B(F [x]n)∼= F [x]n/DF [x]n
∼=F [x]× · · · × F [x]× F [x]× · · · × F [x]
F [x]× · · · × F [x]× 〈a1〉 × · · · × 〈as〉∼= F [x]/F [x]× · · · × F [x]/F [x]× F [x]/〈a1〉 × · · · × F [x]/〈as〉= F [x]/〈a1〉 × · · · × F [x]/〈as〉
As an aside, consider a monic polynomial
f = xd + bd−1xd−1 + · · ·+ b1x+ b0 ∈ F [x]
Then V := F [x]/〈f〉 has a linear operator T given by multiplication by x: there is a basis 1, x, x2, . . . , xd−1and T (1) = x, and T (x) = x2, and so on; also, T (xd−1) = xd = −b01− b1x− · · · − bd−1x
d−1.
With respect to this basis, T is given by the matrix
Cf :=
0 0 . . . −b01 0 . . . −b1...
. . ....
...0 . . . 1 −bd−1
This is called the companion matrix of f . We have that A is similar to
Ca1Ca2
. . .
Cas
.
58
This is painful to prove with classical linear algebra, but in terms of modules and PIDs this is just sayingyou can put a module structure on vector spaces.
In any case, we had V ∼= F [x]/〈a1〉 × · · · × F [x]/〈as〉. Fix f ∈ F [x] so that f(T )v = 0 for all v ∈ V .This is equivalent to saying that f(x)v = 0 for all v ∈ V , and by the isomorphism f(x)m = 0 for allm ∈ F [x]/〈a1〉 × · · · × F [x]/〈as〉, which happens if and only if as|f (because a1|a2| . . . |as), that is, as isthe minimal polynomial of T (classically, the minimal polynomial is a monic polynomial in F [x] of minimaldegree such that as(T ) = 0).
Since as|cA, we have cA(A) = 0. This is classically called the Cayley Hamilton theorem.
Example 3.26.1. We fix
A =
13 24 50−6 −11 −250 0 1
∈Mn(Q)
so that
B = xI −A =
x− 13 −24 −506 x+ 11 250 0 x− 1
To get to Smith Normal Form we can reduce:x− 13 −24 −50
6 x+ 11 250 0 x− 1
r1:=r2,r2:=r1−−−−−−−−−→
6 x+ 11 25x− 13 −24 −50
0 0 x− 1
r2:=6r2−−−−−→
6 x+ 11 256(x− 13) −144 −300
0 0 x− 1
r2:=r2−(x−13)r1−−−−−−−−−−−→
6 x+ 11 250 −x2 + 2x− 1 −25x+ 250 0 x− 1
column operations−−−−−−−−−−−→
1 0 00 −x2 + 2x− 1 −25x+ 250 0 x− 1
simplify−−−−−→
1 0 00 x− 1 00 25(x− 1) −(x− 1)2
−→
1 0 00 x− 1 00 0 (x− 1)2
And so A is similar to 1 0 0
0 0 −10 1 2
59
3.27 Oct 31, 2018
[I was out of town for this lecture. Cosmo Viola and Yichi Zhang took notes that I borrowed from.]
Recall
Theorem 3.27.1 (Theorem A: Structure Theorem for finitely generated modules over a PID). Let M be afinitely generated module over a PID R. Then
M ∼= Rr ×R/〈a1〉 × · · · ×R/〈as〉
with ai ∈ R nonzero and nonunits such that a1|a2| . . . |as. Moreover, r, s ≥ 0 are unique and a1, . . . , as,called the invariant factors of M , are unique up to units.
We will talk about
Theorem 3.27.2 (Theorem B). Let M be a finitely generated module over a PID R. Then
M ∼= Rr ×R/〈pe11 〉 × · · · ×R/〈pett 〉
with pi ∈ R irreducible and ei ≥ 1. Moreover, r and t are unique, and pe11 , . . . , pett , called the elementary
divisors of M , are unique up to scaling by units and reordering.
Proof. The existence just follows from Theorem A (Theorem 3.27.1) and the Chinese Remainder Theorem:if a = upe11 . . . pett for irreducible and mutually nonassociate then
R/〈a〉 ∼= R/〈pe11 〉 × · · · ×R/〈pett 〉
as an isomorphism of rings and R-modules.
To prove uniqueness, we fix an irreducible p ∈ R dividing a. Without loss of generality, assume p = pifor some i, where p and pi are associates. Notice that if j ≥ 1 then we have an isomorphism
pj−1R/pjR← R/pR
pj−1x←[ x.
Define the field F := R/〈p〉. Notice that
pj ·R/〈pe〉 =
〈pj〉/〈pe〉 if j ≤ e0 if j ≥ e
andpj−1 ·R/〈pe〉pj ·R/〈pe〉
∼=
〈pj−1〉/〈pj〉 ∼= R/〈p〉 = F if j ≤ e0 if j ≥ e
Thus, if q ∈ R is irreducible and not an associate of p, then there are a, b ∈ R so that ap+ bqe = 1, that is,ap = 1 in R/〈qe〉; this means that p has an inverse in R/〈qe〉, and hence
prR/〈qe〉 ∼= R/〈qe〉.
Now we understand
pj−1M
pjM∼=pj−1(Rr ×R/〈pe11 〉 × · · · ×R/〈p
ett 〉)
pj(Rr ×R/〈pe11 〉 × · · · ×R/〈pett 〉)
∼= Fr ×∏
1≤i≤tpi=pj≤ei
F
and in particular the quanitity
dimF pj−1M/pjM = r + #i : 1 ≤ i ≤ t, pi = t, and j ≤ ei
60
depends only on m, e, and j. By letting j grow arbitrarily large, we find that r depends only on M .
Furthermore, we see that #i : 1 ≤ i ≤ t, pi = t, and j ≤ ei depends only on M,p and j. From this,for all j we can recover the sequence of peii with p = pi (up to units), and by considering all p we acquireuniqueness.
Remark 3.27.3. We can use uniqueness in Theorem B (Theorem 3.27.2) to prove uniqueness in TheoremA(Theorem 3.27.1), and uniqueness of the Smith Normal Form.
One application of Theorem B (Theorem 3.27.2) is the following:
Corollary 3.27.4 (Jordan Canonical Form). Fix a matrix A ∈Mn(F ), where F is a field. Assume that thecharacteristic polynomial cA(x) = det(xI −A) ∈ F [x] splits (ie., factors into degree 1 terms).
For λ ∈ F and e ≥ 1, we have a Jordan block
Jλ,e =
λ 1
. . .. . .
. . . 1λ
.Then the matrix A is similar to Jλ1,e1
. . .
Jλr,er
called the Jordan canonical form of A, with λi ∈ F and ei ≥ 1. The pairs (λ1, e1), . . . , (λr, er) are unique upto reordering.
Let’s prove this!
Proof. Let V := Fn be the F [x] module for which multiplication by x is left multiplication by A. We have
V ∼= F [x]/〈a1〉 × · · · × F [x]/〈as〉
for a1| . . . |as and ai monic of degree ≥ 1. Last time we showed that cA = a1 . . . as, so ai ∈ F [x] splits. Asbefore, the Chinese Remainder Theorem says that
V ∼= F [x]/〈(x− λ1)e1〉 × · · · × F [x]/〈(x− λr)er 〉
with cA = (x − λ1)e1 . . . (x − λr)er . Now consider W := F [x]/〈(x − λ)e〉 for λ ∈ F and e ≥ 1. We
have a linear map T : W → W given by multiplication by x. Then a basis of W over F is given by
(x− λ)e−1
, . . . , (x− λ), 1. We have
T ((x− λ)j) = x(x− λ)j = (x− λ)(x− λ)
j+ λ(x− λ)
j,
so with respect to this basis the matrix T is given by
T =
λ 1
. . .. . .
. . . 1λ
=: Jλ,e
so there is a basis of V such that multiplication by A is given byJλ1,e1
. . .
Jλr,er
.
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3.28 Nov 2, 2018
Last time, we proved Jordan canonical form:Take A ∈ Mn(F ) such that its characteristic polynomial is cA(x) = det(xI − A) ∈ F [s] splits, that is,
factors into degree 1 terms. Then A is similar toJλ1,e1
. . .
Jλr,er
with λi ∈ F and ei ≥ 1. Moreover, (λ1, e1), . . . , (λr, er) are unique up to reordering.
The idea is that V := Fn can be given an F [x]-module structure with x acting as multiplication by A.Then as F [x]-modules, we have
V ∼= F [x]/〈a1〉 × · · · × F [x]/〈as〉
with ai ∈ F [x] monic of degree at least 1, and a1| . . . |as. In this setting, we have cA = a1 . . . as with as theminimal polynomial of A, which we sometimes denote mA. This implies that cA and mA = as have the sameirreducible factors (because of the divisibility conditions a1| . . . |as). We proved Jordan Canonical Form byusing Chinese Remainder Theorem.
Example 3.28.1. Consider A ∈ GLn(C). Let’s prove that B2 = A for some B ∈ Mn(C). We’ll prove thisby reducing this to the case where A is a Jordan block, and in this case it’s not so hard to prove this.
So let’s say J is the Jordan Canonical Form of A, so that A = PJP−1 where J is now invertible (sinceA is invertible). If B2 = J for some B, then (PBP−1)2 = PB2P−1 = A, so A is a square.
So without loss of generality, assume A is in Jordan canonical form, so
A =
J1
. . .
Jr
.Notice that if B2
i = Ji, for some Bi ∈Mn(C) thenB1
. . .
Br
2
=
B21
. . .
B2r
=
J1
. . .
Jr
= A
so without loss of generality
A =
λ 1
. . .. . .
. . . 1λ
.Recall the (power series) binomial theorem, which states that as formal polynomials( ∞∑
k=0
(1/2
k
)xk)2
= 1 + x, where
(n
k
)=n(n− 1) . . . (n− k + 1)
k!for n ∈ Q.
We have A = λ(I + λ−1N) with
N =
0 1
. . .. . .
. . . 10
62
and in particular Nn = 0. So
B0 :=
∞∑k=0
(1/2
k
)(λ−1N)k =
n−1∑k=0
(1/2
k
)λ−kNk
and B20 = I + λ−1N . So (
√λB0)2 = λ(I + λ−1N) = A, where
√λ exists because λ ∈ C.
Let’s do another example:
Example 3.28.2. Let
A =
1 1 . . . 11 1 . . . 1...
.... . .
...1 1 . . . 1
∈Mn(Q).
What is its Jordan Canonical Form?
The rank of A is 1, so its nullity is n − 1, that is, we have eigenvalues 0, . . . , 0, n. Also, A2 = nA. SomA has to divide x2 − nx = x(x − n). Because degmA > 1 we must have mA = x2 − nx. So the JordanCanonical Form is
n0
. . .
0
.Alternatively, you could have seen that A was diagonalizable because mA splits into linear factors.
What about A as a matrix in Mn(F )? If charF - n, then the same proof works and we have the sameanswer. On the other hand, if charF |n, then mA = x2. Then the Jordan Canonical Form is
0 10 0
0. . .
0
.
Example 3.28.3. Let
A =
−2 2 1−7 4 25 0 0
∈M3(Q).
so that
xI −A =
x+ 2 −2 −17 x− 4 −2−5 0 x
and swapping c1 and c2, and then scaling c1 by −1, gives 1 −2 x+ 2
2 x− 4 7−x 0 −5
so that swapping r2 := r2 − 2r1 and r3 := r3 + xr1 gives1 −2 x+ 2
0 x −2x+ 30 −2x x2 + 2x− 5
63
and cleaning up gives 1 0 00 x −2x+ 30 −2x x2 + 2x− 5
so that setting c3 : c3 + 2c2 gives 1 0 0
0 x 30 −2x x2 − 2x− 5
and swapping c2 and c3, and scaling r2 by 1/3, gives1 0 0
0 1 x/30 x2 − 2x− 5 −2x
so that with r3 := r3 − (x2 − 2x− 5)r2 gives1 0 0
0 1 x/30 0 (−x3 + 2x2 − x)/3
and with c3 := c3 − (x/3)c3 and scaling r3 by -3 gives1
1x3 − 2x+ x
which is the Smith Normal Form of B. So we have cA = mA = a1 = x3−2x2 +x = x(x−1)2. So the JordanCanonical Form of A is given by 0
1 10 1
Example 3.28.4. Suppose that A ∈ M13(Q), with a1 = (x − 1)2(x − 2)(x − 3)2, and a2 = x(x − 1)2(x −2)2(x− 3)3. Then the Jordan Canonical Form of A is given by
1 11
23 1
30
1 11
2 12
3 13 1
3
.
[Today, I learnt that if you want to make a matrix with more than 10 columns, you need to set the counterMaxMatrixCols to however many columns you need]
64
3.29 Nov 5, 2018
Today we’ll talk about free modules and tensor products.
Let R be a commutative ring (we’ll talk about noncommutative rings some time in the future; it’s usefulfor next semester) with 1. Let A be a set. We can define
FA := f : A→ R with f(a) = 0 for all but finitely many a ∈ A.
This is clearly an R-module. For a ∈ A, we have the element
δa(b) =
1 if b = a
0 otherwise
that is, δa ∈ FA. In fact, FA is a free module over R with basis δa : a ∈ A. Indeed, take f ∈ FA, andobserve that
f =∑a∈Af(a)6=0
f(a)δa
which is a finite sum. This shows that the δa span. To show that they are linearly independent, say that∑a∈A′
caδa = 0
for A′ ⊆ A finite and ca ∈ R. Plugging in a ∈ A′ says that ca · 1 = 0, that is, ca = 0 for all a ∈ A′, so they’relinearly independent. This just says that every element of FA is a unique linear combination (with scalarsin R) of the set δa : a ∈ A. We can identify a ∈ A with δa. We can view FA as a free R-module with basis A.
So this is much less painful than the construction of a free group. Like in that case, we have the universalproperty that for any ϕ : A→M there is a unique Φ: FA →M so that ϕ = Φ ι, where ι : A → FA that is,the following diagram commutes:
A FA
M
ϕ ∃!Φ
If A is a finite, set that, is A = 1, . . . , n=: [n], then FA = Rn. Let’s move on.
Tensor Products
Let R be a commutative ring. For R-modules M,N,P , we say that a map ϕ : M ×N → P is R-bilinear(or just bilinear, if R is clear) if
M → P N → P
m 7→ ϕ(m,n) n 7→ ϕ(m,n)
are homomorphisms of R modules for all m ∈M and n ∈ N . Some examples include
Example 3.29.1. For R = R, the dot product Rn × Rn → R,
Example 3.29.2. The map Mn(R)×Mn(R)→Mn(R) given by (A,B) 7→ AB −BA,
Example 3.29.3. The map R×R→ R given by (a, b) 7→ ab,
Example 3.29.4. The map R3 × R3 → R3 given by (v, w) 7→ v × w.
65
We don’t want to redo the whole theory of modules and homomorphisms to capture bilinear maps.Bilinear maps are nice: fix a ϕ : M × N → P bilinear; observe that for any R-module homomorphismf : P → P ′, the map f ϕ : M ×N → P ′ is bilinear. So the idea is to find one universal “best” bilinear mapand then compose with homomorphisms to get all other bilinear maps. This is captured by
Theorem 3.29.5. Take R-modules M and N . Then there is an R module M ⊗R N and a bilinear mapι : M ×N → M ⊗R N such that for any bilinear map ϕ : M ×N → P , there is a unique homomorphism ofR-modules Φ: M ⊗R N → P such that ϕ = Φ ι, that is, the following diagram commutes:
M ×N M ⊗R N
P
ϕ
ι
∃!Φ
Remark 3.29.6. This universal property characterizes M⊗RN and ι : M×N →M⊗RN up to (canonical)isomorphism.
Remark 3.29.7. The map ι : M × N → M ⊗R N is usually written as ι(m,n) = m ⊗ n; first of all, thenotation m⊗ n is ambiguous, unless we know what M,R,N are, and secondly, ι need not be surjective.
Proof. Consider FM×N , the free module on the set M ×N . Its basis can be identified with M ×N . So wehave the inclusion map M × N → FM×N . This inclusion map is most definitely not bilinear. So we willdefine
M ⊗R N := FM×N/K
where K is the submodule of M ×N generated by
• (m+m′, n)− (m,n)− (m′, n)
• (m,n+ n′)− (m,n)− (m,n′)
• (rm, n)− r(m,n)
• (m, rn)− r(m,n)
for all r ∈ R,m,m′ ∈M,n, n′ ∈ N .
Observe thatι : M ×N → FM×N M ⊗R N
is R-bilinear. Now take any bilinear ϕ : M ×N → P . Then because FM×N is a free group we have
M ×N FM×N
P
ϕ∃!f
and observe that f(K) = 0, where K is the same K as above (that is, we defined M ⊗R N := FM×N/K).For example,
f((m+m′, n)−(m,n)−(m′, n)) = f(m+m′, n)−f(m,n)−f(m′, n) = ϕ(m+m′, n)−ϕ(m,n)−ϕ(m′, n) = 0
because ϕ is bilinear. For similar reasons all other generators are annihilated by f . So we can factor theinclusion and the f through the projection by K:
M ×N M ⊗R N
P
ϕ∃!Φ
66
as desired.
Let’s do an example:
Example 3.29.8. We have Z/2Z⊗Z Z/3Z = 0. Indeed, take a ∈ Z/2Z and b ∈ Z/3Z. Then we have
a⊗ b = (3 · a)⊗ b = 3 · (a⊗ b) = a⊗ (3b) = a⊗ 0 = a⊗ (0 · 0) = 0 · (a⊗ 0) = 0.
In general, we’ll see at some point that if a, b ≥ 1 that
Z/〈a〉 ⊗[Z?] Z/〈b〉 ∼= Z/〈gcd(a, b)〉.
Example 3.29.9. We have R⊗RM ∼= M . Indeed, we have j : R×M →M given by (r,m) 7→ rm. Then ifϕ : R×M → P is bilinear, we have
R×M M
P
j
ϕm7→ϕ(1,m)
and uniqueness gives R⊗RM ∼= M .
The tensor product distributes:
Proposition 3.29.10. We have (M ⊕M ′)⊗R N ∼= (M ⊗R N)⊕ (M ′ ⊗R N).
Proof sketch. We have a bilinear map (M ⊕M ′) ×N → (M ⊗R N) ⊕ (M ′ ⊗R N) given by ((m,m′), n) 7→(m⊗ n,m′ ⊗ n) which gives a homomorphism (M ⊕M ′)⊗R N → (M ⊗R N)⊕ (M ′ ⊗R N).
We also have a bilinear map M×N → (M⊕M ′)⊗RN by (m,n) 7→ (m, 0), n), so we get a homomorphismM⊗RN → (M⊕M ′)⊗RN . Do the same thing with M replaced by M ′, and check that they are inverses.
So we get
Corollary 3.29.11. We have Rm ⊗R Rn ∼= (R ⊗R Rn)m ∼= (Rn)m ∼= Rnm. n general, we have Rm ⊗R Rnis a free module with basis ei ⊗ ej : 1 ≤ i ≤ m and 1 ≤ j ≤ m.
67
3.30 Nov 7, 2018
Last time, for R a commutative ring and M,N two R-modules, we defined the tensor product M ⊗R N ;it is an R-module with an R-bilinear map ι : M × N → M ⊗R N , given by (m,n) 7→ m ⊗ n. We have thefollowing universal property: for any R-bilinear map ϕ : M × N → P , there is a unique homomorphismΦ: M ⊗R N → P of R-modules such that ϕ = Φ ι. This was expressed in the commutative diagram
M ×N M ⊗R N
Pϕ
ι
∃!Φ
To appreciate this construction we probably need to prove some theorems, but at this point at least thetensor product is turning bilinear maps (mysterious) to homomorphisms (less so).
The elements of M ⊗R N are called tensors; the element m⊗ n is called a pure, or sometimes simple, orsometimes elementary, tensor. We saw that R⊗RM ∼= M , and (M ⊕M ′)⊗RN ∼= (M ⊗RN)⊕ (M ′⊗RN).This isomorphism was given by ((m,m′), n) → (m ⊗ n,m′ ⊗ n) and checking that the universal propertygives an inverse. There is also (⊕
i∈IMi
)⊗R N ∼=
⊕i∈I
(Mi ⊗R N).
Example 3.30.1. We have M ⊗RN ∼= N ⊗RN . This will use that R is commutative somehow. Obviously,the map will be given by m⊗ n↔ n⊗ n but there are details to check:
Consider the map M ×N → N ×M given by (m,n) 7→ (n,m); this map is bilinear so it factors to a mapϕ : M ×N → N ⊗RM , that is, we have the diagram:
M ⊗R N
M ×N N ×M N ⊗R N∃!Φι
ϕ
and now we have Φ(m⊗ n) = Φ(ι(m,n)) = ϕ(m,n) = n⊗m.
Example 3.30.2. We have for I ⊆ R an ideal, the R-module R/I. If N is an R-module, we have theR-module IN =
∑ni=1 aini : ai ∈ I, ni ∈ N, which is the smallest submodule of n containing all the an
with a ∈ I, n ∈ N . We claim thatR/I ⊗R N ∼= N/IN.
Indeed, we have a bilinear map
R/I ×N j−→ N/IN
(r, n) 7→ rn
and one can check that j is indeed well defined and bilinear. We claim that the universal property holds forj, that is we should verify
R/I ×N N/IN
Pϕ
j
∃!Φ
But if it exists, then this map sends Φ(n) = Φ(j(1, n)) = ϕ(1, n). So this is a possible definition of Φ; if Φexists then it is unique. One should check that Φ is well defined (it is clearly a homomorphism if it’s well
68
defined). To see this we compute
Φ
(n+
m∑i=1
aini
)= ϕ(1, n) +
m∑i=1
aiϕ(1, ni) = ϕ(1, n) +
m∑i=1
ϕ(ai, ni) = ϕ(1, n) +
m∑i=1
ϕ(0, ni) = ϕ(1, n).
So j satisfies the universal property and proves the claim. Let’s do an example of this example:
Set R = Z and I = 〈a〉. Furthermore let N = Z/〈b〉 (where a, b ≥ 1 are integers). We have
Z/〈a〉 ⊗Z Z/〈b〉 ∼=Z/〈b〉a · Z/〈b〉
=Z/〈b〉
〈gcd(a, b)〉/〈b〉∼= Z/〈gcd(a, b)〉
Let’s now consider a (not necessarily commutative) ring R (still with a multiplicative identity). Let M bea right R-module and N a left R-module. We will define an abelian group M ⊗RN (one needs the ⊗R to bein the middle, as if it was “connecting two trains”). In this setting, bilinear maps are not quite the right thing:
Definition 3.30.3. Let L be an abelian group (written additively). A map ϕ : M ×N → L is R-balanced if
• ϕ(m+m′, n) = ϕ(m,n) + ϕ(m′, n)
• ϕ(m,n+ n′) = ϕ(m,n) + ϕ(m,n′)
• ϕ(m, rn) = ϕ(rm, n)
for m,m′ ∈ M,n, n′ ∈ N, r ∈ R. Notice that rϕ(m,n) doesn’t necessarily make sense because L is not(necessarily) an R-module.
One obvious example of an R-balanced map was given last lecture, with R×R→ R given by (a, b) 7→ ab.
There is a universal R-balanced map M×N ι−→M⊗RN such that for any R-balanced map ϕ : M×N → L,there is a unique group homomorphism Φ: M ⊗RN → L such that Φ ι = ϕ. The proof is exactly the sameas last time; the commutative diagram is also exactly the same (with P replaced by L).
Definition 3.30.4. Let R,S be rings. An abelian group M is an (S,R)-bimodule if M is a left S-module,and a right R-module such that s(mr) = (sm)r for all s ∈ S, r ∈ R,m ∈M .
Example 3.30.5. The ring R is an (R,R)-bimodule. Dummit/Foote calls this “the standard R-module”.
Example 3.30.6. Let R be a commutative ring, and M a left R-module. Then m · r := rm turns M intoa (R,R)-module.
Let M be an (S,R)-bimodule and N a left R-module. Then M⊗RN is an abelian group, but the S-actionon M gives some extra structure: M ⊗R N is actually an S-module, given by
s
( e∑i=0
mi ⊗ ni)
:=
e∑i=0
(smi)⊗ ni.
So if R is commutative and M,N are R-modules, then they’re (R,R)-bimodules and M⊗RN is an R-module,so it agrees with before.
Example 3.30.7. Let G be a group, and H ⊆ G a subgroup. Say that H acts on a vector space V over F .So this is a map ϕ : H → AutF (V ) = GL(V ). We have the induced representation FG ⊗FH V , where FHand FG are the group rings. This is an FG-module, that is, a vector space over F with a G-action.
More generally, let M be a left R-module and S an R-algebra (that is, a ring homomorphism f : R→ Swith f(R) contained in the center of S). Then S is an (S,R)-bimodule, that is, s · r := sf(r). Then we getthe left S-module S ⊗RM , called the extension of scalars.
69
3.31 Nov 9, 2018
[I was out of town for this lecture. Cosmo Viola and Yichi Zhang took notes that I borrowed from.]
Let R,S, T be rings, and M an (S,R)-bimodule and N an (R, T )-bimodule. We defined the Z-module(abelian group) denoted M ⊗R N , and can in fact turn M ⊗R N into an (S, T )-bimodule.
Take an R-algebra S (that is, a ring homomorphism f : R→ S such that f(R) is contained in the centerof S; for r ∈ R, s ∈ S, we have r · s = f(r)s = sf(r) = s · r, where the second equality follows because f(r)is in the center). Then, if M is a left R-module, we have S ⊗RM a left S-module.
Now suppose R is commutative. Then, for R-algebras A and B, we have that A⊗RB is also an R-algebra.Multiplication is defined by (a⊗ b) · (a′ ⊗ b′) = (aa′)⊗ (bb′).
Example 3.31.1. Let S be a commutative R-algebra, with R → S. We claim that S ⊗R R[x] ∼= S[x].Indeed, consider the (R-bilinear) map
S ×R[x]j−→ S[x]
(a, f) 7→ af
which induces a unique R-module homomorphism
S ⊗R R[x]→ S[x]
a⊗ f 7→ af
that is in fact an isomorphism with inverse
S[x]→ S ⊗R R[x]m∑i=0
aixi 7→
m∑i=0
ai ⊗ xi
Now let
0 A B C 0f g
be an exact sequence of left R-modules. Then, for a right R-module M , we have an exact sequence ofZ-modules
0 M ⊗R A M ⊗R B M ⊗R C 0f∗ g∗
defined by f∗(m⊗a) = m⊗f(a). The failure of surjectivity of g∗ leads to functors TorRn (M,−) : R−mod→Z−mod, where TorR0 (M,−) = M ⊗R −.
More in 6320.
70
4 Fields
4.31 Nov 9, 2018
Let F be a field. There is a unique homomorphism of rings ϕ : Z→ F , where ϕ(1) = 1 and hence
ϕ(n) =
1 + · · ·+ 1︸ ︷︷ ︸n times
if n ≥ 0
−1− · · · − 1︸ ︷︷ ︸n times
if n < 0
Then observe thatZ/ kerϕ → F,
where F is a field, so Z/ kerϕ better not have any zero-divisors; we have Z/ kerϕ is an integral domain, andkerϕ is a prime ideal of Z. So kerϕ = 〈n〉 for n = 0 or n = p a prime.
The characteristic of F is charF = n. Now if charF = 0 then
ϕ : Z → F extends to Q → F
which we view as Q ⊆ F . Otherwise if charF = p then Fp = Z/〈p〉 → F , which we view as Fp ⊆ F . We saythat Q/Fp is the prime field of F .
Consider a subfield F ⊆ K, where both F and K are fields. We call K a field extension of F , and denotethis K/F . The degree of an extension K/F is [K : F ] = dimF K. For extensions F ⊆ K ⊆ L, we have[L : F ] = [L : K][K : F ]. The proof idea is as follows: if either [L : K] or [K : F ] are infinite, then this is easy;if both are finite then take a basis a1, . . . , am of L over K and a basis b1, . . . , bn of K over F . As an exercise,one can show that aibj for i ∈ [m], j ∈ [n] is a basis of L over K. Then [L : F ] = mn = [L : K][K : F ].
Consider an extension K/F and fix α ∈ K. Let F [α] and F (α) be the smallest subring and subfieldrespectively of K containing F and α. We have a homomorphism of F -algebras
ϕ : F [x]→ K
f(x) 7→ f(α)
and again this induces F [x]/〈kerϕ〉 ∼−→ F [α] ⊆ K, and since K is a field we have that kerϕ is a prime idealof F [x]; we have kerϕ = 〈0〉 or kerϕ = 〈p〉 with p ∈ F [x] monic and irreducible.
As before, if kerϕ = 〈p〉 then we getF [x]/〈p〉 ∼−→ F [α]
and note that since F [x]/〈p〉 is a field then so is F [α], and hence F [α] = F (α). We have
[F (α) : F ] = dimF F [x]/〈p〉 = deg p,
and we say p ∈ F [x] is the minimal polynomial of α over F (where the minimal polynomial is the monicpolynomial in F [x] of minimal degree such that p(α) = 0).
On the other hand, if kerϕ = 0, that is, f(α) 6= 0 for all nonzero f ∈ F [x], then F [α] is not a field, and ϕextends to a map F (x)
∼−→ F (α). In this case we say α is transcendental over F ; if kerϕ = 〈p〉 as discussedabove we say α is algebraic over F .
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4.32 Nov 12, 2018
Let F be a field, and K/F a field extension (that is, a field K such that F ⊆ K).
Take α ∈ K. Then we might have that α is transcendental over F , that is, there is no non-zero f ∈ F [x]with f(α) = 0. Then we have
F [x]∼−→ F [α] ⊆ F (α)
f(x) 7→ f(α)
that is, transcendental elements are basically like free variables (from the point of view of algebra). Since F [x]is an infinite dimensional vector space over F (it has F -basis 1, x, x2, . . . ), we also have [F (α) : F ] =∞.
Otherwise, we have that α is algebraic over F . Then there is a unique monic mα,F ∈ F [x] of minimaldegree such that mα,F (α) = 0 [there is at least one such monic polynomial, and if there are two, one canapply the division algorithm]. We call mα,F the minimal polynomial of α over F . We have, like last time,
F [x] F [α] ⊆ Kf 7→ f(α),
inducing F [x]/〈mα,F 〉∼−→ F [α]
where we also have F [x]/〈mα,F 〉 is a field because mα,F is irreducible. Also, we have
[F (α) : F ] = [F [α] : F ] = degmα,F .
Example 4.32.1. Fix α = 3√
2 ∈ R and we have mα,Q = x3 − 2. We have the inclusion of fields
Q(α)
Q
and the isomorphism
Q[x]/〈x3 − 2〉 ∼−→ Q[α]
f 7→ f(α).
Then, Q[α] has Q-basis 1, α, α2, and Q[x]/〈x3−2〉 has Q-basis 1, x, x2. Here’s a question: what is (2+α)−1?
We can work in Q[α]: polynomial division gives
x3 − 2 = (x2 − 2x+ 4)(x+ 2)− 10
so for x = α we have (α2 − 2α+ 4)(α+ 2) = 10 and
(2 + α)−1 =4
10− 2
10α+
1
10α2.
Now let β = 3√
2e2πi/3 ∈ C. We have
Q[β] Q[x]/〈x3 − 2〉 Q[α]∼∼
sending c0 + c1β + c2β2 to c0 + c1α+ c2α
2.
Consider an isomorphism of fields ϕ : F∼−→ F ′, and fix M ∈ F [x] irreducible. Set m′ = ϕ(m) ∈ F ′[x]
(that is, apply ϕ to the coefficients of m). Then m′ is irreducible. Take a root α of m (in some extension ofF ) and a root β of m′ (in some extension of F ′). Then we have isomorphisms
72
F (α) F [x]/〈m〉 F ′[x]/〈m′〉 F ′(β)
f(α) f f f(β)
f ϕ(f)
∼ ∼ ∼
and we get an isomorphism ϕnew : F (α)∼−→ F ′(β) such that ϕnew|F = ϕ and ϕnew(α) = β. Note that ϕnew
is unique!
Definition 4.32.2. An extension K/F is a splitting field for f ∈ F [x] if f factors into degree 1 terms inK[x] (sometimes called “splits completely”) and no field F ⊆ K ′ ( K has this property.
Theorem 4.32.3. Fix f ∈ F [x] and set n = deg f . Then a splitting field K for f exists, and [K : F ] ≤ n!.
Proof. Induct on n ≥ 1. For n = 1 this is easy. So suppose this is true for polynomials of degree at mostn− 1, and pick f ∈ F [x]. Take an irreducible factor M ∈ F [x] of f . Then L := F [x]/〈f〉 is a field extensionof F of degree degm ≤ n. There is an α ∈ L such that f(α) = 0. Then f(x) = (x−α)g(x) with g(x) ∈ L[x],and deg g = n− 1. By the inductive hypothesis, there is an extension K/L with [K : L] ≤ (n− 1)! such thatg splits completely in K[x]. This says that f splits completely in K[x], and
[K : F ] = [K : L]︸ ︷︷ ︸≤(n−1)!
[L : F ]︸ ︷︷ ︸≤n
≤ n!
Now we can replace K by the smallest dimensional F ⊆ K ′ ⊆ K for which f splits completely in K ′[x].
Take f ∈ F [x]. A splitting field K/F of f is unique up to isomorphism fixing f :
K K ′
F
∼
The idea is to build up inductively. Take m ∈ F [x] irreducible dividing f and degm > 1 if possible (and ifnot, then we’d be done), and choose roots α1 ∈ K and β1 ∈ K ′, so that
F1 := F0(α1) F0(β1) =: F ′1
F0 = F F ′0 = F
∃!ϕ1
id
As before, take m ∈ F1[x] irreducible dividing f and degm > 1 (again, if possible; if not, we’d be done), andchoose roots α2 ∈ K and β2 ∈ K ′ of m′; we can repeat
K = Fm(αm+1) F ′m(βm+1) = K ′
......
F2 := F1(α2) F ′1(β2) =: F ′2
F1 := F0(α1) F0(β1) =: F ′1
F0 = F F ′0 = F
∃!ϕ2
∃!ϕ1
id
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4.33 Nov 14, 2018
Fix a nonzero f ∈ F [x] for some field F . We had
Definition 4.33.1. An extension K/F is a splitting field of f over F if f splits completely in K[x] and doesnot split completely in K ′[x] for any field F ⊆ K ′ ( K. So
f(x) = c
n∏i=1
(x− αi)
with c ∈ F× and ai ∈ K, and F (α1, . . . , αn) = K.
Last time, we showed the existence of a splitting field K/F of f over F , and moreover showed that[K : F ] ≤ n!. We also showed that it is essentially unique: for two splitting fields of f over F , say K and K ′,there is an isomorphism ϕ : K → K ′ such that ϕ|F = idF . This is not unique, and the failure of uniquenessis measured in Galois theory (which is a 6320 thing). In any case, up to isomorphism, we can talk about“the” splitting field.
Note that if f(x) = c(x− α1) . . . (x− αn) is the factorization of f in K, then f(x) factors as
f(x) = c
n∏i=1
(x− ϕ(αi))
in K ′.
Example 4.33.2. Let f = x4− 2 ∈ Q[x]. This is irreducible (by Eisenstein at p = 2). We will work over C,that is,
f = (x− 4√
2)(x+4√
2)(x− 4√
2i)(x+4√
2i)
and hence its splitting field is K = Q(± 4√
2,± 4√
2i) = Q( 4√
2, i). We have
[K : Q] = [K : Q(4√
2)]︸ ︷︷ ︸=2
[Q(4√
2) : Q]︸ ︷︷ ︸=4
= 8
and by the way 8 ≤ 4!.
Example 4.33.3. Let F = Fp(t) and consider f = xp + t. We claim that f is irreducible in (Fp(t))[x]: byGauss’ lemma, it suffices to check it is irreducible in (Fp[t]︸︷︷︸
UFD
)[x]. But clearly f is irreducible in
(Fp[x])[t] ∼= Fp[t][x] ∼= (Fp[t])[x]
because it is degree 1 in t.
Let α be a root of f (in some extension of F ). We have
(x− α)p =
p∑n=0
(p
n
)xn(−α)p−n = xp + (−α)p ∈ Fp[x].
But since α is a root we have 0 = αp + t and hence
(x− α)p = xp − αp = xp + t
so that xp + t = (x − α)p. It follows that K = F (α) and [K : F ] = p. We’ll see later that this sort ofphenomenon doesn’t happen in characteristic 0, that is, an irreducible polynomial will give distinct roots.
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Consider nonzero f ∈ F [x], and let K/F be a splitting field of f . Then
f(x) = c
r∏i=1
(x− αi)ei
with c ∈ F×, αi ∈ K are distinct, and ei ≥ 1. We say f is separable if it has no multiple roots, that is,e1 = · · · = er = 1.
We’ll give a simple criterion to check if f is separable.
Let f = anxn + an−1x
n−1 + · · · + a1x + a0. The derivative of f is f ′ = nanxn−1 + (n − 1)an−1x
n−2 +· · · + a1. We get some of the usual properties (when we restrict to polynomials): it’s linear, and there arethe chain/product rules.
Lemma 4.33.4. Let α be a root of f (in some field). Then α is a multiple root of f if and only if f ′(α) = 0.
Proof. If f(x) = (x − α)eg(x) for e ≥ 1 and g is such that g(α) 6= 0, then f ′(x) = e(x − α)e−1g(x) + (x −α)eg′(x). So
f ′(α) = e(x− α)e−1g(α) =
eg(α) 6= 0 if e = 1
0 if e > 1
Theorem 4.33.5. For f ∈ F [x] nonzero, we have f is separable if and only if f and f ′ are relatively primein F [x].
The point is that we don’t have to understand splitting fields, we can just work over the base field.
Proof. The backwards direction follows from the fact that 〈f, f ′〉 = 〈1〉 = F [x], and so af + bf ′ = 1 fora, b ∈ F [x]. Now if α (in some extension of F ) satisfies f(α) = f ′(α) = 0 then a(α)f(α) + b(α)f ′(α) = 0, acontradiction. So there are no multiple roots, by the previous lemma. Hence f is separable.
Conversely, if f and f ′ are not relatively prime. So there is an irreducible m ∈ F [x] dividing f and f ′.Take α to be a root of m in some extension of F . Then f = mg1 and f ′ = mg2 implies f(α) = m(α)g1(α) = 0and f ′(α) = m(α)g2(α) = 0, so that f has a repeated root at α by the previous lemma.
Example 4.33.6. Consider f = xp + t ∈ Fp(t)[x], and observe that f ′ = pxp−1 = 0. This means that anyroot of f is a multiple root of f , as we saw. In particular, f is not separable (which we already knew).
Example 4.33.7. Consider f = xn− 1 ∈ F [x], and observe that f ′ = nxn−1. If charF |n, then we have thesame situation in Example 4.33.6, and f is not separable. In the other case, if charF - n then xn − 1 andnxn−1 are relatively prime (because the only irreducible dividing nxn−1 is x); it follows that f is separable.
Proposition 4.33.8. We have
a) If f ∈ F [x] is irreducible and f ′ 6= 0, then f is separable.
b) When charF = 0, we always have f ′ 6= 0 when deg f ≥ 1. In particular, f is separable if and only if f isa product of nonassociate irreducible polynomials.
Proof. Everything is clear except for part (a), so we’ll prove that:
If f is irreducible then the only factors of f up to units are 1 and f . This shows that gcd f, f ′ = 1 or f .Note that f - f ′ since deg f ′ < deg f and f ′ 6= 0. This implies that f is separable.
Remark 4.33.9. If f ∈ F [x] is irreducible and charF = p > 0, then f(x) = fsep(xph
) for a unique h ≥ 0and a unique separable fsep ∈ F [x].
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4.34 Nov 16, 2018
Today we’ll talk about cyclotomic fields.
Example 4.34.1. (Cyclotomic Fields) Fix n ≥ 1. Consider f = xn − 1 ∈ Q[x]. The roots of f ,say in C, are the n-th roots of unity. They form a cyclic group (under multiplication); call this group
µn = 1, ζn, ζ2n, . . . , ζ
n−1n . Usually we pick ζn = e2πi/n. The splitting field of f is Q(ζn).
We say ζ ∈ µn is primitive if it has order n in µn. Define the n-th cyclotomic polynomial to be
Φn(x) =∏ζ∈µn
primitive
(x− ζ) ∈ Q(ζn)[x].
Then we haveΦn(x) =
∏a∈(Z/nZ)×
(x− ζan)
which is well defined in the sense that if a ∼= b (mod n) then x− ζan = x− ζbn. If you like, you can define
Φn(x) =∏a≤n
gcd(a,n)=1
(x− ζan)
and it is obvious that deg Φn = #(Z/nZ)× =: φ(n) (where φ is the Euler totient function).
We have ∏d|n
Φd(x) = xn − 1.
This is a factorization (a priori in Q(ζn)); we claim that Z[x].
We will induct on n ≥ 1. The claim is true for n = 1, that is, Φ1(x) = x− 1. Now assume Φd ∈ Z[x] ford < n. Then ∏
d|n
Φd︸ ︷︷ ︸∈Z[x],
primitive
·Φn = xn − 1︸ ︷︷ ︸∈Z[x],
primitive
so by Gauss’ Lemma Φn ∈ Z[x].
Proposition 4.34.2. The polynomial Φn ∈ Z[x] is irreducible.
Proof. Suppose Φn = fg with f, g ∈ Z[x] and f irreducible. Take any root ζ of f . Take any prime p - n. Wehave Φn(ζp) = 0. So ζp is a root of f or g.
Suppose first that g(ζp) = 0. Then f(x) and g(xp) have a common root (x = ζ). Since f is irreducibleand primitive we have f |g(xp) in Z[x]. Taking mod p shows that f |g(xp) = g(x)p in Fp[x]. So in particularf and g are not relatively prime (in Fp[x]). So xn − 1 = Φn = f g ∈ Fp[x] is not separable. This is acontradiction, because h := xn − 1 and h′ = nxn−1 are relatively prime in Fp[x] (remember we assumedp - n), so by Theorem 4.33.5 we see that h is in fact separable.
So f(ζp) = 0 for all primes p - n. So for any root ζ ′ of Φn, we have ζ ′ = ζa for some a ∈ Z withgcd(a, p) = 1. If a = p1 . . . pr with pi - n, then ζ ′ = (. . . ((ζp1)p2) . . . )pr and so ζ ′ is a root of f . Then f hasall the roots of Φn. Furthermore both Φn and f are monic, and Φn is separable. Then we are done since fis irreducible.
Corollary 4.34.3. We have [Q(ζn) : Q] = deg Φn = φ(n).
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This corollary is equivalent to the previous proposition.
Remark 4.34.4. The polynomials Φn are important for number theory. For p - n, we have Φn splitscompletely in Fp[x] if and only if p ≡ 1 (mod n).
Example 4.34.5. (Finite fields) Fix a finite field F and let p = charF > 0. We have Fp ⊆ F. We have|F| = p[F:Fp], where recall that [F : Fp] = dimFp F.
Fix a prime p and an integer n ≥ 1. Let’s construct a field with pn elements. Consider the group F×,which has order pn − 1. Then if a ∈ F×, we have ap
n−1 = 1; this says that for any a ∈ F, a is a root ofxp
n − x. It follows thatxp
n
− x =∏a∈F
(x− a)
and this is supposed to be suggestive (this is not a proof; we assumed F exists).
Indeed, fix p and n ≥ 1. Let K be a splitting field of xpn − x over Fp. Then xp
n − x ∈ Fp[x] is separablesince xp
n − x and pnxpn−1 − 1 = −1 are relatively prime. So F := α ∈ K : αp
n − α = 0. Note thatseparability means that |F| = pn.
We claim that F is a subfield of K. Obviously, 0 and 1 are in F; if α, β ∈ F we have (α+β)pn
= αpn
+βpn
so α+ β ∈ F, and (αβ)pn
= αpn
βpn
so αβ ∈ F.
Finally, the field F with pn elements is unique up to isomorphism, so it is often denoted Fpn . Uniquenessfollows from the fact that K is unique up to isomorphism.
Definition 4.34.6. A field K is algebraically closed if every non-constant f ∈ K[x] has a root in K(equivalently, if all nonzero f ∈ K[x] split completely in K).
An example is C.
Next time, we’ll prove
Theorem 4.34.7. Every field F is contained in an algebraically closed field.
The proof will be a bit painful.
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4.35 Nov 19, 2018
We’ll finish fields today. [There’s a final homework that’s on blackboard; it’ll be due next Friday, Nov30. There will be a takehome final that’ll go up after the last class; it’ll be due on Dec 13 at 4:30 pm.]
Recall that a field K is algebraically closed if every non-constant f ∈ K[x] has a root in K (equivalently,all nonzero f ∈ K[x] split completely). Today we will prove
Theorem 4.35.1. Every field F is contained in an algebraically closed field.
Proof. Sometimes fields are uncountably large, so we have to do something more drastic (Zorn’s lemma).Here’s a proof due to Artin:
For each non-constant f ∈ F [x], denote by xf an indeterminate variable. Consider the polynomialring R = F [xf : f ∈ F [x]]. Let I be the ideal of R generated by all the f(xf ) with f ∈ F [x], that is,
I = 〈f(xf ) : f ∈ F [x]〉. Observe that for xf ∈ R/I, we have f(x) = f(xf ) = 0.
The idea is to choose a maximal ideal I ⊆ m ( R, and so R/m is a field extension of F such that everyf ∈ F [x] has a root (and it has a root of the same reason as above).
For this idea to work we better show that I 6= R. Suppose that 1 ∈ I. Then g1f1(xf1)+· · ·+gnfn(xfn) = 1for some fi ∈ F [x] and gi ∈ R. For ease of notation set xi = xfi for all i. We have other variablesxn+1, . . . , xm such that g1(x1, . . . , xm)f1(x1) + · · ·+ gn(x1, . . . , xm)fn(xn) = 1. Let K/F be a splitting fieldof f1(x) . . . fn(x), that is, a field where all of the fi simultaneously split. Let αi ∈ K be a root of fi(x). Nowset xi = αi for i ∈ [n]. Then
1 = g1(α1, . . . , αn, xn+1, . . . , xn) f1(α1)︸ ︷︷ ︸=0
+ · · ·+ gn(α1, . . . , αn, xn+1, . . . , xn) fn(αn)︸ ︷︷ ︸=0
= 0.
Now we can use our idea: by Zorn’s Lemma, there is a maximal ideal I ⊆ m ( R [since R/I has a maximalideal]. We have a K1 := R/m ⊇ F = K0 so that every f ∈ K0[x] has a root in K1. We can repeat thisconstruction to get K2 ⊆ K3 ⊆ . . . with the property that every f ∈ Kn[x] has a root in Kn+1. Now define
K =⋃n≥1
Kn,
an extension of F . One should check that K is a field (it’s important that the Ki are nested); for f ∈ Ke[x]with degree d ≥ 1. Then f splits completely in Kd+e ⊆ K: in each extension Kn+1/Kn we are guaranteeda new root of f . So K is algebraically closed since
K[x] =⋃e≥1
Ke[x].
Definition 4.35.2. Fix an extension K/F . The algebraic closure of F in K is
L = α ∈ K : α algebraic over F.
Observe that F ⊆ L ⊆ K is a subfield: we have F ⊆ L and 0, 1 ∈ F ⊆ L; if α, β ∈ L we have
[F (α, β) : F ] = [F (α)(β) : F (α)][F (α) : F ] ≤ [F (β) : F ][F (α) : F ] <∞
because α, β are algebraic over F . So all γ ∈ F (α, β) are algebraic over F ; in particular, α ± β, αβ, α−1 (ifα 6= 0) are all algebraic.
Let K be an algebraically closed extension of F . Let F be the algebraic closure of F in K. It’s a factthat F is unique up to an isomorphism that fixes F . So we call F the algebraic closure of F .
Remark 4.35.3. One can show that Q exists without Zorn’s lemma, because every α ∈ Q is a root of somef ∈ Q[x]; enumerate the irreducibles of Q[x] and repeatedly take splitting fields (there are only countablymany irreducibles).
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5 Algebraic Geometry
5.36 Nov 26, 2018
There won’t be so much geometry today.
Theorem 5.36.1 (Hilbert’s Nullstellensatz – weak form). If F is algebraically closed, then the maximalideals of the ring F [x1, . . . , xn] are precisely 〈x1 − a1, . . . , xn − an〉 with ai ∈ F .
Note that 〈x1 − a1, . . . , xn − an〉 are clearly maximal ideals; the content of the theorem is that all themaximal ideals are of this form. Observe that this gives a bijection between maximal ideals of F [x1, . . . , xn]and tuples in Fn.
Also, this theorem is not necessarily true when F is not algebraically closed (even in one variable, andF = Q).
Corollary 5.36.2. For F algebraically closed, and fixed f1, . . . , fr ∈ F [x1, . . . , xn], then
〈f1, . . . , fr〉 = 〈1〉 ⇐⇒ there is no a ∈ Fn such that f1(a) = · · · = fr(a) = 0
Proof. The forward direction is easy, because if there are g1, . . . , gr so that g1f1 + · · · + grfr = 1, and thefi(a) = 0 for all i ∈ [r], then 0 = 1.
For the backward direction, suppose that 〈f1, . . . , fr〉 ( 〈1〉, then 〈f1, . . . , fr〉 ⊆ m. But the weakNullstellensatz tells us that m = 〈x1 − a1, . . . , xn − an〉. In particular, we have
fi =
n∑j=1
gi,j(xj − aj)
so fi(a) = 0 for all i.
We’ll prove the theorem after some lemmas:
Lemma 5.36.3 (Artin-Tate). Let A ⊆ B ⊆ C be commutative rings. Assume
a) A is Noetherian
b) C is a finitely generated A-algebra
c) C is a finitely generated B-module.
Then B is a finitely generated A-algebra.
Proof. Take x1, . . . , xm that generate C as an A-algebra, and y1, . . . , yn that generate C as a B-module. Wehave
xi =
n∑j=1
bi,jyj
with bi,j ∈ B, and
yiyj =
n∑k=1
bi,j,kyk
with bi,j,k ∈ B. Let B0 ⊆ B be the A-algebra generated by the bi,j and bi,j,k. Note that B0 is Noetherian,because B0 = A[bi,j , bi,j,k], which is a quotient of A[xi,j , xi,j,k], which is Noetherian by the Hilbert basistheorem (Theorem 2.16.8), since A is also Noetherian.
We have that C is generated as an A-algebra by x1, . . . , xm, and hence it is generated as a B0 moduleby yi, . . . , yn. Because B ⊆ C and B0 is Noetherian, then B is a finitely generated B0-module, so B is afinitely generated A-algebra.
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Theorem 5.36.4 (Zariski’s Lemma). Let K/F be a field extension such that K is a finitely generatedF -algebra. Then K is a finite extension of F .
Remark 5.36.5. It is important that K is a field. Observe that F [x] is not a field; it’s a finitely generatedF -algebra, but it has infinite dimension as an F -vector space.
Being a finitely generated F -algebra is in general very different from being a finitely generated F -vectorspace. But when K is a field then it’s the same.
Proof of Theorem 5.36.4. We have K = F [x1, . . . , xn] for some xi ∈ K. Observe that if the xi are all alge-braic over F , then K = F [x1, . . . , xn] is a finite extension of F .
Otherwise, there is an r such that (perhaps after reordering):
• x1, . . . , xr are algebraically independent over F (that is, F (x1, . . . , xi−1)(xi) is transcendental overF (x1, . . . , xi−1) for all i ∈ [r]).
• xr+1, . . . , xn are algebraic over F (x1, . . . , xr).
This number r is well defined (it is called the transcendence degree of F ).
Secretly, we should be looking for a contradiction (otherwise K would not be finite dimensional). Sodefine L = F (x1, . . . , xr). Observe that K is a finite extension of L, and K is a finitely generated F -algebra(by assumption). Thus Artin-Tate says that L = F (x1, . . . , xr) is a finitely generated algebra over F , saywith generators f1/g1, . . . , fm/gm with fi, gi ∈ F [x1, . . . , xr] relatively prime.
If π ∈ F [x1, . . . , xr] is an irreducible dividing a denominator of some α ∈ L, then π|g1 . . . gm. Thiscontradicts the fact that F [x1, . . . , xr] has infinitely many irreducibles up to units.
We are now able to prove the weak Nullstellensatz.
Proof of Theorem 5.36.1. Fix a maximal ideal m ⊆ F [x1, . . . , xn], and consider F ′ = F [x1, . . . , xn]/m as anextension of F . Here F ′ is a field and is finitely generated as an F -algebra. By Zariski’s lemma F/F ′ is a finiteextension and is hence algebraic. Because F is algebraically closed, we have F ′ = F . So F [x1, . . . , xn]/m = F ,and we have the quotient map
ϕ : F [x1, . . . , xn]→ F [x1, . . . , xn]/m = F
which is a homomorphism of F -algebras. If ai := ϕ(xi) ∈ F for all i, we have ϕ(xi − ai) = 0, and som = kerϕ ⊇ 〈x1 − a1, . . . , xn − an〉, which is a maximal ideal. So m = 〈x1 − a1, . . . , xn − an〉.
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5.37 Nov 28, 2018
Last time we talked about Hilbert’s Nullstellensatz (at least, its weak form):
Theorem 5.37.1. If F is algebraically closed, then the maximal ideals of F [x1, . . . , xn] are 〈x1−a1, . . . , xn−an〉 with ai ∈ F .
What if F is not algeraically closed? Let F be an algebraic closure of F . For a ∈ Fn, we get a homomor-phism ϕa : F [x1, . . . , xn]→ F given by f 7→ f(a). This homomorphism is an homomorphism of F -algebras.The image of ϕa is F [a1, . . . , an] ⊆ F . Because the ai are all algebraic (they’re in F ), the kernel of ϕa is amaximal ideal of F [x1, . . . , xn].
We won’t prove it, but it’s a fact that all maximal ideals are of this form. The proof uses Zariski’s lemma;it’s the same idea as last time. But note that different a ∈ Fn may give the same ideal kerϕa (it’s not abijection anymore). For example, a = i or a = −i give homomorphisms R[x] → C given by f(x) 7→ f(a);they both have kernel 〈x2 + 1〉.
Let F be a field. We’ll give two key definitions:
Definition 5.37.2. For a set S ⊆ F [x1, . . . , xn], its vanishing locus is
V(S) := a ∈ Fn : f(a) = 0∀f ∈ S.
If I is the ideal of F [x1, . . . , xn] generated by S, then V(S) = V(I). Also, observe that it usually sufficesto consider finite S, because the ideal generated by S is finitely generated (Hilbert basis theorem; see Theo-rem 2.16.8). Dummit and Foote uses different notation; they call this set Z(S) := V(S).
A subset of Fn is an algebraic set if it is V(S) for some S.
Definition 5.37.3. For a set X ⊆ Fn, define the ideal
I(X) := f ∈ F [x1, . . . , xn] : f(a) = 0∀a ∈ X.
Let R be a commutative ring, and let I ⊆ R be an ideal. The radical of I is
rad(I) := f ∈ R : fn ∈ I for some n ≥ 1.
You can check it’s an ideal. Also I ⊆ rad(I). We say I is radical if rad(I) = I. For example, prime idealsare radical, and intersections of radical ideals are radical.
Observe that I(X) is radical, since if fn ∈ I(X), then (f(a))n = 0 for all a ∈ X; because we are in a fieldf(a) = 0 and f ∈ I(X).
Let I be an ideal. Then consider I(V(I)). It’s easy to show that I ⊆ I(V(I)). Also, I(V(I)) is a radicalideal so we can say more; I(V(I)) ⊇ rad(I).
Theorem 5.37.4 (Hilbert’s Nullstellensatz). If F is algebraically closed and I ⊆ F [x1, . . . , xn] is an ideal,then
I(V(I)) = rad(I).
Proof. This is called the Rabinowitsch trick. We only need to prove I(V(I)) ⊆ rad(I). Take any non-zerof ∈ I(V(I)).
Define g := 1 + xn+1f ∈ F [x1, . . . , xn+1]. Let J ⊆ F [x1, . . . , xn+1] be the ideal generated by g and I.Observe that V(J) is empty, because if (a1, . . . , an, an+1) ∈ V(J) ⊆ Fn+1, then f(a1, . . . , an) = 0 becausef ∈ I ⊆ J , also g = 1 + an+1f(a1, . . . , an) = 0, which is a contradiction.
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Weak Nullstellensatz says that J = 〈1〉 = F [x1, . . . , xn+1] (otherwise, J ⊆ m = 〈x1−a1, . . . , xn+1−an+1〉and (a1, . . . , an+1) ∈ V(J)). In particular, this says that
1 = hg +
m∑i=1
hibi
with h, hi ∈ F [x1, . . . , xn+1] and bi ∈ I. In particular, for
xn+1 =−1
f
we have
1 = 0 +
m∑i=1
hi
(x1, . . . , xn,
−1
f
)bi(x1, . . . , xn).
Clear denominators: multiplying by a large enough power of f gives
fr =m∑i=1
frhi
(x1, . . . , xn,
−1
f
)︸ ︷︷ ︸
∈F [x1,...,xn]
bi
and bi ∈ I. Thus fr ∈ I, and f ∈ radI.
Corollary 5.37.5. Let F be algebraically closed. We have inclusion reversing bijections
algebraic sets in Fn radical ideals of F [x1, . . . , xn]
I
V
Proof. We should show that V is surjective. This follows from the observation that V(I) = V(rad(I))(which isn’t hard to check). The Nullstellensatz says that I V = id, and these two give the corollary.
As an example of this Corollary, observe that
points in Fn ←→ maximal ideals of F [x1, . . . , xn]
This is the statement of the weak Nullstellensatz.
Let F be algebraically closed. On Fn, we can define the Zariski topology: closed sets are precisely thealgebraic sets in Fn. This defines a topology. If X ⊆ Fn is an algebraic set, it gets a topology from Fn.From here we can define a notion of irreducible spaces (we’ll talk about this more next time), and there willbe a correspondence
irreducible algebraic sets in Fn ←→ prime ideals of F [x1, . . . , xn]
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5.38 Nov 30, 2018
Last time, we fixed an algebraically closed field F and considered an ideal S ⊆ F [x1, . . . , xn]; we definedthe algebraic set
V(S) = a ∈ Fn : f(a) = 0 for all f ∈ S,and for a subset X ⊆ Fn, we defined the ideal
I(X) = f ∈ F [x1, . . . , xn] : f(a) = 0 for all a ∈ X
of F [x1, . . . , xn]. We proved the Nullstellensatz, which gave a correspondence
algebraic sets in Fn ↔ radical ideals of F [x1, . . . , xn]X 7→ I(X)
V(I)←[ I
which we proved using the weak Nullstellensatz, which gave a correspondence
points in Fn ↔ maximal ideals of F [x1, . . . , xn]
which then extended to a correspondence between all the algebraic sets and all the radical ideals.
We kept giving Fn more and more properties (topologies, morphisms), so we will (“pretentiously”) denoteit by AnF so we don’t confuse it with the boring vector space Fn.
We want to define morphisms of AnF . More specifically, let X ⊆ AnF and Y ⊆ AmF be algebraic sets.
Definition 5.38.1. A morphism from X to Y is a function ϕ : X → Y such that
ϕ(a) = (f1(a), . . . , fm(a))
for all a ∈ X, with f1, . . . , fm ∈ F [x1, . . . , xn].
This is the category of algebraic sets over F . The main thing is that composing two morphisms is amorphism (because when you compose two polynomials it’s a polynomial).
Example 5.38.2. Let X ⊆ AnF = Fn. Take any f ∈ F [x1, . . . , xn]; the map ϕ : X → A1F = F given by
a 7→ f(a) is a morphism. Note that different f might give the same morphism: note that f(a) = g(a) for alla ∈ X precisely when f − g ∈ I(X), so two functions will give the same morphism precisely when they differby a function in I(X).
Motivated by this we define
Definition 5.38.3. The coordinate ring of X ⊆ AnF is
F [X] := F [x1, . . . , xn]/I(X),
which we can view as the ring of morphisms X → A1F .
Observe that F [AnF ] = F [x1, . . . , xn], since I(AnF ) = 0. Also, the ring F [X] is reduced, that is, ifa ∈ F [X] with an = 0 and n ≥ 1, then a = 0; this is because F [X] is a quotient by a radical ideal. Indeed,if an = 0, then a = f for f ∈ F [x1, . . . , xn], so 0 = an = fn; we have fn = 0 ∈ I(X), which is a radical idealso f ∈ I(X), so a = f = 0.
Example 5.38.4. Let X = A1F , and Y ⊆ V(y − x3) ⊆ A2
F . Consider the map
ϕ : X → Y
t 7→ (t, t3)
which is actually a morphism (in particular because ϕ(t) ∈ V(y − x3)); it has an inverse ψ : Y → X givenby (x, y) 7→ x so X and Y are isomorphic algebraic sets. We remark that we might want to stop thinking ofY as living inside its ambient space A2
F (like with manifolds, where after we have the charts we might wantto forget about the ambient space).
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Example 5.38.5. Let X = A1F , and Y = V(y2 − x3) ⊆ A2
F . The morphism
ϕ : X → Y
t 7→ (t2, t3)
is bijective, but one can show that ϕ has no inverse morphism (intuitively, you’d want the map to be(x, y) 7→ y/x, but this is not a polynomial). Geometrically we also see this: for F = R we can plot the setof points Y ⊆ R2:
x
y
and (0, 0) ∈ Y is the problem because it’s not smooth there.
Suppose you have a morphism ϕ : X → Y of algebraic sets. This induces a homomorphism of F -algebras
ϕ∗ : F [Y ]→ F [X]
f 7→ f ϕ
which we call the pullback.
Here’s a fact: there is a bijection
morphisms X → Y ←→ homomorphism of F -algebras F [Y ]→ F [X]
where ϕ 7→ ϕ∗ is the pullback described earlier, and the content is that one can go backwards. In particularit becomes clear that morphisms from X → Y only depend on these rings F [X] and F [Y ] rather than theambient spaces.
We have a contravariant functor
algebraic sets over F → finitely generated reduced F -algebras
so that on the objects X 7→ F [X] and on the morphisms (ϕ : X → Y ) 7→ (ϕ∗ : F [Y ]→ F [X]).
Here’s a FACT: This functor is an equivalence of categories (we’ll talk about this more next time). Inmany ways this says that the category of algebraic sets is the same as the category of finitely generatedreduced F -algebras, though algebraic sets are very geometric and the F -algebras are very algebraic.
With this equivalence of categories in mind, we can ask for a way to “go back”: given a finitely generatedreduced F -algebra R, can you construct an algebraic set (in a reasonable way)? (Yes; come to lecture nexttime to see Specm(R))
What if you start with a general commutative ring? Can we make some sort of geometric object (evenif it’s not quite an algebraic set)? (Yes; come to lecture next time to see affine schemes and Spec(R))
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5.39 Dec 3, 2018
We’re trying to get a hint of schemes.
We fix a commutative ring R. We can define
Spec(R) := p : p prime ideal of R
andSpecm(R) := m : m maximal ideal of R.
Of course, Specm(R) ⊆ Spec(R).
Prior to this we were thinking about ideals I ⊆ F [x1, . . . , xn] for algebraically closed F , and algebraicsets
X = V(I) = a ∈ Fn : f(a) = 0∀f ∈ I.
We also had the coordinate ring F [X] := F [x1, . . . , xn]/I(X), where
I(X) = f ∈ F [x1, . . . , xn] : f(a) = 0 ∀a ∈ X.
This coordinate ring is a finitely generated reduced F -algebra.
What is Specm(F [X])? Recall that we had a correspondence: maximal ideals of F [X] correspond tomaximal ideals of F [x1, . . . , xn] that contain I(X) by some isomorphism law (the fourth one, apparently).But now the maximal ideals of F [x1, . . . , xn] are precisely ma = 〈x1−a1, . . . , xn−an〉 with a ∈ Fn, by weakNullstellensatz. Observe that ma ⊇ I(X) precisely when a ∈ X. In this sense,
Specm(F [X]) = X
ma ← [ a
For algebraic sets X ⊆ Fn and Y ⊆ Fm, recall that a morphism ϕ : X → Y is a function such thatϕ(a) = (f1(a), . . . , fm(a)), with fi ∈ F [x1, . . . , xn]. This induces a pullback homomorphism of F -algebras
ϕ∗ : F [Y ]→ F [X]
g︸︷︷︸Y→A1
F
7→ g ϕ
We can recover ϕ from ϕ∗.
For a ∈ X, we claim that(ϕ∗)−1(ma) = mϕ(a);
to see this fix a ∈ X and take g ∈ F [Y ]. Then
g ∈ mϕ(a) ⇐⇒ g(ϕ(a)) = 0
⇐⇒ (g ϕ)(a) = 0
⇐⇒ ϕ∗(g)(a) = 0
⇐⇒ ϕ∗(g) ∈ ma
⇐⇒ g ∈ (ϕ∗)−1(ma).
Remark 5.39.1. In general, for a homomorphism ψ : R→ R′ of rings and m ⊆ R′ a maximal ideal, ψ−1(m)need not be a maximal ideal of R; we have a map
R/ψ−1(m) R′/mψ
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but rings could certainly embed into fields. On the other hand, the inverse image of a prime ideal must bea prime ideal.
If X ⊆ Fn and f ∈ F [X], one can observe that
X −V(f) = Specm(F [x]f )
where the right side is the spectrum of the localization of the ring F [X] with respect to f .
For R a commutative ring, we can endow the set Spec(R) with a topology; define for S ⊆ R the set
V(S) = p ∈ Spec(R) : f ∈ p for all f ∈ S.
The Zariski topology on Spec(R) is defined by letting the closed sets be the V(S) = V(I) (for I = 〈S〉).Let’s quickly check this:
We have V(0) = Spec(R) and V(1) = ∅;
If I and J are ideals then V(I) ∪V(J) = V(IJ) because p ∈ V(IJ) ⇐⇒ IJ ⊆ p, and since p is primethis happens precisely when I ⊆ p or J ⊆ p. This means that p ∈ V(I) ∪V(J);
We also have ⋂a∈A
V(Ia) = V
(∑a∈A
Ia
)since p ∈ ∩a∈AV(Ia) if and only if Ia ⊆ p for all a precisely when
∑a∈A Ia ⊆ p, which happens precisely
when p ∈ V(∑a∈A Ia).
Example 5.39.2. Let R be a local integral domain such that Spec(R) = 0,m and observe that the closureof 0 is 0,m.
We can endow X := Spec(R) with more structure that makes it more useful. In particular, we want togive it a structure sheaf OX : we want
• A ring OX(U) for every open U ⊆ Spec(R), with the additional assumption that OX(∅) = 0
• For U ⊆ V an inclusion of open sets, we have a restriction homomorphism resV,U : OX(V ) → OX(U)with the additional property that resU,U = id, and if U ⊆ V ⊆W we have resW,U = resV,U resW,V
• If we have open sets U,Uα ⊆ X so that
U =⋃α∈A
Uα
and we have fα ∈ OX(Uα) such that
fα|Uα∩Uβ = fβ |Uα∩Uβfor all α, β ∈ A, then there exists a unique f ∈ OX(U) such that f |Uα = fα for all α ∈ A.
• For f ∈ R, we also wantOX(X −V(f)) = Rf
where Rf is the localization of R with respect to f .
The first two items define a presheaf. Also, one can check that there is a unique such OX up to a naturalnotion of isomorphism. We say that Spec(R) with its topology and OX is an affine scheme. Recall earlier
algebraic sets over F ←→ finitely generated reduced F -algebras
sending X 7→ F [X]. The inverse is given by Specm(R)← [ R. It turns out there is an equivalence of categories
affine schemes ←→ commutative rings
with Spec(R)← [ R and X 7→ OX(X).
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