NOTES FROM MATH 494, WINTER 2013

CALEB SPRINGER (FROM LECTURES OF MITYA BOYARCHENKO)

Class, 1-9-13

Reminders: Commutative rings and fields.

Definition 1. A commutative ring is a triple (A,+, ·) where A is a set and + and · are binaryoperations on A satisfying:

(1) Both + and · are associative and commutative(2) There exists 0 2 A such that a+ 0 = a for all a 2 A(3) There exists 1 2 A such that a · 1 = a for all a 2 A(4) For all a 2 A there exists b 2 A such that a+ b = 0. Write b = �a.(5) a · (b+ c) = ac+ ac for all a, b, c 2 A

In this situation, we say that A is a field if 0 6= 1 and every element has a multiplicative inverse.

Theorem 1. If K is a field, there exists a field extension L � K such that L is algebraically closed.(Note: algebraically closed means that every f(x) 2 L[x] of degree � 1 has a root in L.)

Proof Idea: Recall that if f(x) 2 K[x] and deg(f) � 1, then K has a field extension in whichf(x) has a root. Namely, let g(x) be any irreducible factor of f(x), then form K[x]/(g(x)). Then,x 2 K[x]/(g(x)) is a root of f (where x = x+ (g(x)).)

Strategy of proof:Step 1: Repeat the above for all irreducible polynomials in K[x] at once.

Lemma 1. (proved later): If K is a field, there exists a field extension K1 � K such that for anyf(x) 2 K[x] with deg(f) � 1 there exists ↵ 2 K1 with f(↵) = 0 (i.e. f has a root in K1.)

Step 2: Iterate this procedure. We started with K and obtained K1. Apply the same procedureto K1 and get K2 � K1. Keep going. Now we have 8n � 1, every f(x) 2 Kn[x] of degree � 1 has aroot in Kn+1[x]. Form L = [1n=1Kn. There is a unique way to make L into a field such that eachKn becomes a subfield.

By construction, L is a field extension of K. Let’s show that L is algebraically closed. Takef(x) 2 L[x] with deg f � 1. Then there exists n 2 N such that all the coe�cients of f(x) are inKn (i.e. f(x) 2 Kn[x].) By construction, f(x) has a root in Kn+1, hence also in L. (QED)

Now back to the Lemma. Write P = {f 2 K[x]| deg(f) � 1}. First, form A = the ring ofpolynomials with coe�cients in K, where the variables correspond to elements of P.

Notation: for each f 2 P , introduce variable Xf . Then, A = K[{Xf}f2P ]. The elements of

A are finite K-linear combinations of monomials in the variables of {Xf}f2P . A monomial is aproduct of finitely many of the variables Xf , possibly with repetitions. Define · and + on A in theusual way.

Next, we want to create something (namely a quotient of A by a certain ideal) in which everyf(x) 2 P has a root. Begin by considering I = ideal of A generated by {f(Xf )}f2P . That is tosay that

I =

(nX

i=1

gi · fi(Xfi) : n � 1, f1, . . . , fn 2 P and g1, . . . , gn 2 A

).

1

Example: K = Q and f1(x) = x+ 2, f2(x) = x2 � 5 and f3(x) = x2 + x+ 1. In A, we have, inparticular, Xf1 , Xf2 , Xf3 . Example element of I:

(Xf4Xf5 +X2f6) · (Xf1 + 2) + (Xf2 �

7

11Xf1) · (X2

f2 � 5) + (Xf3 +Xf4 �Xf6) · (X2f3 +Xf3 + 1).

Form R = A/I. In R, we have the elements Xf = Xf + I. Because f(Xf ) 2 I by construction,we get f(Xf ) = 0 in R for every f 2 P. Note this is because ⇡(f(Xf )) = f(⇡(Xf )) = f(Xf ) and⇡(f(Xf )) = 0 in R. (Slight abuse of notation: f is first as a polynomial with coe�cients in K,then coe�cients in R.)

K ,! A! A/I = R. So we get a map K[x]! R[x].Remaining Steps;

• check that I 6= A.• Proposition: If B is any commutative ring and J ⇢ B is any ideal with J 6= B, then thereexists maximal ideal M ⇢ B with J ⇢M . (Note maximal ideal () B/M is a field.)

Class, 1-11-13: Zorn’s Lemma

Setup: Let (X,) be a partially ordered set:

• X is a set• is a binary relation on X such that:

– x x for all x 2 X– If x y and y x, then x = y– x y and y z, then x z

Suppose Y ⇢ X is a subset. An element z 2 X is an upper bound for Y if y z for all y 2 Y .

Zorn’s Lemma: Let (X,) be a nonempty poset (short for “partially ordered set”.) Assumethat every totally ordered subset Y ⇢ X has an upper bound in X. Then X has a maximal element,i.e., an element x 2 X such that if y 2 X and x y, then x = y.

Notes:

(1) There can be many y 2 X such that y 6 x and x 6 y.(2) X could have many di↵erent maximal elements.(3) Y ⇢ X is called Totally Ordered if given y1, y2 2 Y , either y1 y2 or y2 y1. (For short,

write “toset”.)

Applications:

(1) Let A be a (commutative) ring and let I ⇢ A be an ideal such that I 6= A. Then thereexists a maximal ideal M ⇢ A such that I ⇢M .

Proof. Let X be the set of all ideals J ⇢ A such that J 6= A and I ⇢ J . Define J1 J2if and only if J1 ⇢ J2. This is a partial order on X. We know X 6= ; because I 2 X.Let’s verify the hypothesis of Zorn’s Lemma. Take a totally ordered subset Y ⇢ X. If weshow that Y has an upper bound in X, then we will be done. Define J0 to be the union (assubsets of A) of all J 2 Y . If we check that J0 2 X, we win.

Why is J0 an ideal? Choose a, b 2 J0 and � 2 A. It is enough to show that � · a+ b 2 J0.By construction, we can find J1, J2 2 Y such that a 2 J1 and b 2 J2. But either J1 ⇢ J2 orJ2 ⇢ J1 because Y is totally ordered. So �a+ b is in the larger of J1, J2. Hence, �a+ b 2 J0.Clearly, I ⇢ J0.

Why is J0 6= A? If J0 = A, then 1 2 J0. But this means that there is some ideal J 2 Ysuch that 1 2 J , but this means that J = A, which is a contradiction. ⇤

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(2) Let K be a field, let V,W be vector spaces over K, let U ⇢ V be a subspace and letT : U !W be a linear map. Then, there is a linear map T 0 : V !W such that T 0|U = T .

Proof. (Secretly a more algebraic version of Hahn-Banach.) LetX be the set of pairs (U 0, T 0)where U 0 ⇢ V is a subspace with U ⇢ U 0 and T 0 : U 0 ! W is a linear map such T 0|U = T .Declare (U 0, T 0) (U 00, T 00) when U 0 ⇢ U 00 and T 00|U 0 = T 0. It is clear that this is a partialordering. Then, (X,) is a nonempty poset.

Claim 1: X has a maximal element (U, T ).Claim 2: U = V .Suppose Y ⇢ X is a (nonempty) toset. For each y 2 Y , denote by Uy ⇢ V the corre-

sponding subspace and by Ty : Uy ! W the corresponding linear map. (So, y = (Uy, Ty).)Form U0 = [y2Y Uy. Since Y is totally ordered, U0 ⇢ V is a subspace. Define T0 : U0 !Was follows. If v 2 U0, then v 2 Uy for some y 2 Y , and we set T0(v) := Ty(v).

Exercise 1: If v 2 Uy1 and v 2 Uy2 , then Ty1(v) = Ty2(v). (So, T0 is well defined)Exercise 2: Prove T0 is linear.

So, (U0, T0) 2 X and i s an upper bound for X. By Zorn’s Lemma, X has a maximalelement, call it (U, T ).

Claim: U = V . Assume that U 6= V . Choose any v0 2 V such that v0 /2 U . ConsiderU 0 = U + (K · v0). By construction, U ⇢ U 0 and U 6= U 0, and since (K · v0) \ U = {0}(since K is a field), every element of U 0 can be written as u + � · v0 for unique u 2 U and� 2 K. Hence, we can define T 0 : U 0 !W by T 0(u+ �v0) = T (u). Easy. T 0 is linear, so geta contradiction with the maximality of (U, T ). ⇤

Class, 1-14-13

Definition 2. A poset (X,) is inductively ordered if every nonempty totally ordered Y ⇢ X hasan upper bound in X.

Zorn’s Lemma: Every nonempty inductively ordered poset has a maximal element.

Application 3: If D is a division ring and M is a D-module, then M has a basis, i.e. there existsa subset B ⇢M such that 8y 2M there exists a unique choice of ab 2 D for every b 2 B such thatab = 0 for all but finitely many b 2 B and y =

Pb2B ab · b.

Proof. Note: A subset S ⇢M is linearly independent if given as 2 D for all s 2 S such that as = 0for all but finitely many s and

Ps2S as · s = 0, it necessarily follows that as = 0 for all s 2 S.

Define X = {all linearly independent subsets of M} and order X via inclusion. Note that X isa nonempty poset by construction. Let Y ⇢ X be totally ordered and define T = [S2Y S (union ofsubsets of M .)

Claim: T 2 X (and hence T is an upper bound for Y .)Note: By definition, T is linearly independent () any finite subset of T is linearly independent.

But if F ⇢ T is a finite subset, then F ⇢ S for some S 2 Y because Y is totally ordered.So T is linearly independent.By Zorn’s Lemma, there exists a maximal element S0 2 X. We claim that S0 is a basis of M .

Given m 2M , we need to show that m 2 spanD(S0).If m 2 S0, then this is clear. Otherwise, consider S0 [ {m} /2 X by maximality, so there exists

c 2 D and there exists as 2 D (for each s 2 S0), all but finitely many of which are 0, such thatc ·m+

Ps2S0

as · s = 0 and either c 6= 0 or as 6= 0 for some s 2 S0. But c = 0 would contradict thelinear independence of S0. Hence m =

Ps2S0

(�c�1 · as)s 2 spanD(S0) ⇤3

More Definitions.

Definition 3. If L � K is a field extension, then we can view L as a vector space over K. If thisvector space is finite dimensional, we call the extension L � K finite and write [L : K] = dimK(L) =the degree of L over K.

Let L � K be a field extension and let ↵ 2 L.

Definition 4. We say that ↵ is algebraic over K if there exists f(x) 2 K[x] such that deg(f(x)) � 1and f(↵) = 0. Otherwise, ↵ is transcendental over K.

Definition 5. K[↵] = the subring of L generated by K and ↵ (or {f(↵)|f(x) 2 K[x]}.)

Definition 6. K(↵) = the subfield of L generated by K and ↵ (or {f(↵)g(↵) |f, g 2 K[x] and g(↵) 6= 0}.)

Terminology: ↵ 2 C is called algebraic/transcendental if it is so over Q.

E.g. e and ⇡ are transcendental. Non-stupid theorem proving this:

Theorem 2. If 0 6= ↵ 2 C is algebraic, then e↵ is transcendental.

Note: i =p�1 is algebraic, so if ⇡ were algebraic, then 2i · ⇡ would be algebraic. But e2⇡i = 1

is obviously not transcendental.

Proposition 1. Let L � K be any field extension and let ↵ 2 L The following are equivalent:

(1) ↵ is algebraic over K.(2) ↵ 2 E for some finite field extension E � K such that E ⇢ L.(3) K(↵) is finite over K.(4) K(↵) = K[↵].

Proof. Note that (3) =) (2) is obvious. Also, (2) =) (1) was proved in Math 493 (reminder:look at the sequence 1,↵,↵2, . . . .) And (2) =) (4) by Math 493 because if (2) is satisfied, thenK[↵] ⇢ E, so K[↵] has finite dimension over K and K[↵] is an integral domain, so it is a field. ⇤

Class 1-16-13

Continued proof of Proposition. Remember that we have proven that (3) =) (2) =) (4).

Proof. Proof of (4) =) (1): If ↵ = 0, we win. Otherwise, ↵�1 2 K(↵) = K[↵], so ↵�1 = f(↵) forsome 0 6= f(x) 2 K[x]. Then, ↵ is a root of x · f(x)� 1, so (1) holds.

Proof of (1) =) (3): Since ↵ is algebraic over K, there exists n 2 N such that ↵n 2spanK(1,↵, . . . ,↵n�1). Then by induction, ↵m 2 spanK(1,↵, . . . ,↵n�1) for all m � n. Hence,K[↵] = spanK(1,↵, . . . ,↵n�1) is finite dim over K. It is a domain, so by Math 493, K[↵] is a field.So K[↵] = K(↵) =) dimK(↵) <1, proving (3). ⇤

(Note: It is proved in the bible that 0 /2 N.)Reminders:

(1) Given field extensions F ⇢ E ⇢ K, K � F is finite () E � F and K � E are finite, andif this holds, [K : F ] = [K : E] · [E : F ]

(2) Suppose L � K is a field extension and let ↵ 2 L be algebraic over K. The map ' : K[x]!L, f(x) 7! f(↵), is a ring homomorphism with Im(') = K[↵] = K(↵) and ker(') 6= 0 =)ker(') = (p↵(x)) for a uniquely determined monic polynomial p↵(x) 2 K[x]. This is theminimal polynomial of ↵ over K. Note: p↵(x) is irreducible over K and deg(p↵) = [K(↵) :K].

4

Corollary 1. Let L � K be any field extension. Then X = {↵ 2 L |↵ is algebraic over K} is asubfield of L.

Proof. Let ↵,� 2 X. Then K(↵) is finite over K. Also, � is algebraic over K(↵). So, [K(↵)(�) :K(↵)] <1. Hence, [K(↵)(�) : K] <1. But ↵±�,↵·� 2 K(↵)(�) and ↵�1 2 K(↵) if ↵ 6= 0. (Thisimplies these are all elements of X because K(↵)(�) is a finite extension, so it is algebraic.) ⇤

What is Galois theory about? (roots of polynomials)“Theorem”: There is no general formula for finding a root of a degree 5 equation that uses

only radicals and arithmetic operations.(This has to do with the fact that S5 is not solvable.)

Definition 7. Say that a 2 C is expressible in terms of radicals if there exists a chain of subfieldsQ = F0 ⇢ F1 ⇢ F2 ⇢ · · · ⇢ FN ⇢ C such that a 2 FN and for each 1 j N , there exists �j 2 Fj

such that Fj = Fj�1(�j) and �mj

j 2 Fj�1 for some integer mj � 2.

Proposition 2. Suppose that f(x) 2 [x] is irreducible of degree 5 that has exactly 3 real roots.Then no complex (i.e. both real and non-real) root of f(x) can be expressed in terms of radicals.

Let’s define the Galois group:

Definition 8. Let f(x) 2 Q[x] be irreducible of degree n � 1. Then f(x) has exactly n distinct rootsin C, call them ↵1, . . . ,↵n. The Galois group of f(x) is the subgroup of Perm({↵1, . . . ,↵n}) (⇠= Sn)defined as follows.

Consider L = Q(↵1, . . . ,↵n) (a finite field extension of Q.) Define G = {field automorphisms of L}(where the group law is composition.) If g 2 G and 1 i n, g(↵i) is also a root of f(x) becausef has rational coe�cients:

0 = f(↵i) =) 0 = g(f(↵i)) = f(g(↵i))

So to g 2 G we associate a permutation of {↵1, . . . ,↵n}, so we get a homomorphism G! Sn. It isclearly injective because if g 2 G and g(↵i) = ↵i for all 1 i n, then g(�) = � for every � 2 Lbecause L = Q(↵1, . . . ,↵n).

Facts:

(1) f(x) has a complex root expressible in terms of radicals () G is solvable.(2) If deg(f(x)) = 5 and f has exactly 3 real roots, then G = S5. (And S5 is not solvable

because it contains A5.)

Class 1-18-13

Recall: f 2 Q[x], deg f � 1. Over C, we can factor f(x) = a(x� ↵1)(x� ↵2) . . . (x� ↵n).

Definition 9. The Galois Group of f(x) is G = {automorphisms of the field Q(↵1, . . . ,↵n)}.

Last time: get an injective homomorphism G! Perm({↵1, . . . ,↵n}).Question: What is the image of this homomorphism?Example 1: f(x) has degree 2. Hereafter, we will always assume that f(x) is monic and irreducible

over Q. Then, f(x) = (x � ↵1)(x � ↵2) where ↵1,↵2 2 C \ Q. If f(x) = x2 + ax + b, then

↵1,2 =�a±

pa2�4b2 .

Easy case: if a2�4b < 0, then ↵1,↵2 /2 R, then ↵2 = ↵1. The map z 7! z is a field automorphismof Q(↵1,↵2) that interchanges ↵1 and ↵2.

Claim: In general (with f(x) monic, quadratic and irreducible), there exists g 2 G = Gal(f) thatswitches ↵1 and ↵2.

5

Note that ↵1 + ↵2 = �a, so ↵2 = �a � ↵1 and a 2 Q, so Q(↵1) = Q(↵1,↵2) = Q(↵2).({1,↵1} is a Q-vector basis of Q(↵1,↵2).) So the g 2 Gal(f) that we want has to be given byg(�+ µ · ↵1) = �+ µ↵1 for every �, µ 2 Q. In fact, this formula does define a field automorphism,though this needs to be checked.

Better approach:

' : Q[x]! Q(↵1,↵2) by '(h(x)) = h(↵1)

: Q[x]! Q(↵1,↵2) by (h(x)) = h(↵2)

Note: ', are surjective ring homomorphisms and ker(') = ker( ) = (f(x)) ⇢ Q[x].

Aside [Math 493]: A = any ring and I ⇢ A two sided ideal, then form the quotient ring A/I. Ifany homomorphism � : A! B is a ring homomorphism and I ⇢ ker(�), then we get a well definedring homomorphism � : A/I ! B given by �(a+ I) = �(a).

See the pictures we drew.The same argument actually proves:

Lemma 2. Let L � K be a field extension and let ↵,� 2 L be such that ↵,� are algebraic overK and have the same minimal polynomial f(x). Then, there exists a unique field isomorphismg : K(↵)! K(�) such that g(a) = a for all a 2 K and g(↵) = �.

Proof. (Used picture proof.) ⇤Example 2: f(x) = x3� 2 = (x� 3

p2)(x� ⇣ 3

p2)(x� ⇣2 3

p2) where ⇣ = exp(2⇡i3 ) (label these three

roots ↵1,↵2,↵3 in that order.) Then L = Q(↵1,↵2,↵3) = Q( 3p2, ⇣). This has degree 6 over Q.

Claim: Every permutation of ↵1,↵2,↵3 can be realized by a field automorphism of L, so Gal(x3�2) ⇠= S3.

Note: z 7! z is an automorphism of L and interchanges ↵2 with ↵3.Question: How to get the other permutations?Attempt: Lemma =) there is a unique isomorphism between Q(↵1) and Q(↵2) that sends

↵1 7! ↵2. We would like to know: this can automatically be extended to an automorphism of L.

Class January 28, 2013

Motivation: Let L � K be a field extension such that L = K(↵) for some ↵ 2 L which is algebraicover K. Let f(x) be the minimal polynomial of ↵ over K, so L ⇠= K[x]/(f(x)). Suppose E is analgebraically closed field and ' : K ! E is a field homomorphism. How many homomorphisms : L! E such that |K = ' are there?

Answer: There is a bijection between the set {all such } and {roots of g(x) in E} where g(x) 2E[x] is obtained from f(x) by applying ' to the coe�cients. The bijection is 7! (↵). Inparticular, the number of such is deg f and equality holds () g(x) has no repeated roots inE () f is separable () f 0(x) 6= 0.

Definition 10. Let L � K be a finite extension. The separable degree of L over K is

[L : K]S = number of extensions of a given ' : K ! E to L! E where E is some algebraically closed field.

Lemma 3. The Boring Lemma: This definition of separable degree is well defined.

Proof. Suppose '1 : K ! E1 and '2 : K ! E2 are field homomorphisms where E1, E2 arealgebraically closed. We need to show that the number of extensions of '1 to L is equal to thenumber of extensions of '2 to L.

Define E0i = {� 2 Ei : � is algebraic over'i(K)}

Easy exercise: show that E01, E

02 are algebraically closed.

6

Earlier result =) there exists a field homomorphism � : E01 ! E0

2 such that '2 = � � '1.Moreover, �(E0

1) � '2(K). As fields, �(E01)⇠= E0

1, so �(E01) is algebraically closed. Hence, �(E0

1) =E0

2.Thus, � is surjective and therefore an isomorphism. We get a bijection

{ 1 : L! E1 extending '1}! { 2 : L! E2 extending '2}by 1 7! � � 1. ⇤

L � K finite extension of fields.

Lemma 4. |AutK(L)| [L : K]S, with equality () L � K is normal.

Lemma 5. [L : K]S [L : K], and equality holds () L � K is separable.

Corollary 2. L � K is Galois if and only if it is both normal and separable.

Lemma 6. If F ⇢ K ⇢ L are finite extensions, then [L : F ]S = [L : K]S · [K : F ]S.

Proof. Choose a homomorphism ' : F ! E, where E is algebraically closed. Note that an extensionof ' to L is the same thing as a pair consisting of an extension of ' to : K ! E and an extensionof to L. ⇤

January 30, 2013

Separability.

Definition 11. Algebraic extension L � K

(1) ↵ 2 K is called separable over K if min↵,K(x) = minimal polynomial of ↵ over K isseparable.

(2) L is separable if every ↵ 2 L is separable over K.

Proposition 3. Let L � K be a finite extension.

(1) [L : K]S [L : K](2) The following are equivalent:

(a) L is separable over K(b) [L : K]S = [L : K](c) There exists ↵1, . . . ,↵n 2 L, separable over K, such that L = K(↵1, . . . ,↵n).

Corollary 3. If F ⇢ K ⇢ L are finite extensions, then L � F is separable if and only if K � Fand L � K are both separable.

Proof. (of Proposition): Since L � K, we can write L = K(↵1, . . . ,↵n) for some ↵1, . . . ,↵n. Weget

K ⇢ K(↵1) ⇢ K(↵1,↵2) ⇢ · · · ⇢ K(↵1, . . . ,↵n�1) ⇢ L.

Each intermediate extension is generated by 1 element. So,

[K(↵1, . . . ,↵i+1), [K(↵1, . . . ,↵i)]S [K(↵1, . . . ,↵i+1), [K(↵1, . . . ,↵i)]

for all i by last time. Both sides of this inequality are multiplicative, this implies (1).For (2), (a) =) (c) is obvious. For (c) =) (b), use the same notation as in the prof of (1)

and assume that each ↵i is separable over K.Exercise: ↵i is also separable over K(↵1, . . . ,↵i�1). (Hint: The minimal polynomial of ↵i over

K(↵1, . . . ,↵i�1) divides the minimal polynomial of ↵i over K.)By what we proved last time, [K(↵1, . . . ,↵i), [K(↵1, . . . ,↵i�1)]S = [K(↵1, . . . ,↵i), [K(↵1, . . . ,↵i�1)]

for all i. Both sides multiply, so [L : K]S = [L : K].7

Prove (b) =) (a): If L � K is not separable, then there exists ↵ 2 L which is not separableover K. By last time, [K(↵) : K]S < [K(↵) : K]. By part (1), [L : K(↵)]S [L : K(↵)]. So,[L : K]S < [L : K], which is a contradiction. ⇤Normality.

Definition 12. Let L � K be an algebraic extension and choose an extension E � K such that E isalgebraically closed. We say that L is normal over K if every field homomorphisms '1,'2 : L! Esuch that '1|K = idK = '2|K , we have '1(L) = '2(L).

Boring Lemma: Whether or not L is normal over K is independent of the choice of an alge-braically closed field E � K.

Proposition 4. Let L � K be a finite extension.

(1) |AutK(L)| [L : K]S ( [L : K].)(2) The following are equivalent:

(a) L is normal over K.(b) |AutK(L)| = [L : K]S.(c) L is a splitting field of some f(x) 2 K[x], which means that f(x) = c(x � �1) · (x �

�2) . . . (x � �m) for some c 6= 0,�i 2 L and L = K(�1, . . . ,�m). This is the same assaying that f has all of its roots in L. Important: f(x) is not assumed to be irreducibleover K.

(d) if g(x) 2 K[x] is irreducible and g(x) has a root in L, then g(x) is a product of linearfactors in L.

Note: Under (2), (a) is a definition, (b) is the Galois connection, (c) is useful in practice and(d) provides nontrivial results. So all of these are important results.

Proof. Proof of (1): Choose one field homomorphism ' : L! E with '|K = idK . If g 2 AutK(L),then ' � g : L! E is also a field homomorphism with ' � g|K = idK . If g, h 2 AutK(L) and g 6= h,' � g 6= ' � h. We immediately get |AutK(L)| [L : K]S . Moreover, if equality holds, then every : L! E with |K = idK has the form 'K � g for some g 2 AutK(L). Hence, '(L) = (L).

This proves (1) and also (b) =) (a) in (2). Conversely, assume (a). If : L ! E is anyhomomorphism with |K = idK , then (L) = '(L) by normality. Since ', are both injective,there exists a unique map of sets g : L ! L with = ' � g. Exercise: g 2 AutK(L). Hence, (b)holds. ⇤

Note that we have a corollary that a field extension is Galois if and only if it is both separableand normal.

February 1, 2013

Already know for finite extension L � K:

• Separable () [L : K]S = [L : K].• Normal () [L : K]S = |AutK(L)|.• Galois () [L : K] = |AutK(L)| () Both normal and separable.

Remember These Two Important Main Ideas:

• Definition of [L : K]S (useful for proofs, etc.)• Idea explained today.

Com back to main theorem: L � K is a finite Galois extension. G = Gal(L/K) = AutK(L).We want to prove that we have mutually inverse bijections

{Subgroups of G} ! {Intermediate fields L � F � K}8

From left to right, we will use the map H 7! LH and from right to left, we will use the mapF 7! AutF (L).

Sanity Checks: Under these bijections, we should have {1} ! L and G ! K, so we’d betterhave LG = K.

Proof. We have K ⇢ LG ⇢ L and we can view G as a subgroup of AutK(L). Hence,

[L : K] = |G| |AutK(L)| [L : LG] [L : K].

So, all of these must be equalities and [L : K] = [L : LG]. Hence, LG = K. ⇤Lemma 7. Given any intermediate field K ⇢ F ⇢ L, let H = AutK(L) ⇢ G. Then, L is Galoisover F and F = LH .

Proof. L � K is Galois () normal and separable =) L � F is normal and separable (byProblem Set 3) =) L � F is Galois =) F = LH by the previous argument (i.e. the SanityCheck proof.) ⇤Second Main Idea: Take any subgroup H ⇢ G and define F = LH . We need to check thatAutF (L) = H. A priori, H ⇢ AutF (L), so this is equivalent to checking that [L : F ] = |H|.

Annoying Part: No obvious reason why we couldn’t have [L : LH ] > |H|.(But let’s just ignore this annoying part for now.)

Nice Idea (Remember This Idea!) Choose ↵ 2 L. Consider the polynomial

f↵(x) =Y

h2H(x� h(↵)) 2 L[x].

Define action of H on L[x] by

h(a0 + a1x+ · · ·+ anxn) = h(a0) + h(a1)x+ · · ·+ h(an)x

n

It is easy to show that this is an action of H on L[x] by ring automorphisms. Hence,

h(f↵(x)) =Y

h02H(h(x� h0(↵))) =

Y

h02H(x� hh0(↵)) = f↵(x)

Consequence: if ↵ 2 L, then [LH(↵) : L] deg(f↵(x)) = |H|.

Lemma 8. (Annoying Lemma) There exists ↵ 2 L with LH(↵) = L.

(If we prove this Lemma, we get |H| � [L : LH ] � |H|. Thus, the proof of the bijection will becomplete.)

Proof. If L is finite, we know this from Math 493. So, assume that L (and hence K and LH) isinfinite. From what we already proved, the map

{Intermediate fields L � F � K} �! {Subgroups of G} by F 7! AutF (L)

is injective since F = LAutF (L). But G has finitely many subgroups because G is finite, so thereare only finitely many intermediate field K ⇢ F ⇢ L. Each such F can be viewed as a K-vectorsubspace of L.

Linear Algebra Lemma =) there exists ↵ 2 L such that ↵ /2 F for any intermediate fieldsK ⇢ F $ L. So, K(↵) = L. ⇤Summary: [L : LH ] |H|, but H ⇢ AutLH (H) (tautology) and |AutLH (L)| [L : LH ]. So,H = AutLH (L). Also, K ⇢ F ⇢ L and H ⇢ G, so we find:

F = LAutF (L), H = AutLH (L)

which proves our main theorem.9

February 4, 2013

Theorem 3. Let L � K be a finite field extension. The following are equivalent:

(1) L � K is Galois, i.e. |AutK(L)| = [L : K](2) K = LAutK(L)

(3) K = L� for some finite subgroup � 2 Aut(L)(4) L � K is normal and separable(5) L is a splitting field of some separable polynomial over K.(6) L is a splitting field of a separable irreducible polynomial over K.

We already know that (1) =) (2) by the main theorem and (2) =) (3) is obvious. We alsoproved (1) =) (4). Note that (6) =) (5) is a tautology.

We need (3) =) (4), (5) =) (4) and (1) =) (6)

Proof of (3) =) (4). : We know L � K is finite and K = L� for a subgroup � ⇢ Aut(L). Then,� ⇢ AutK(L), so |�| [L : K] <1. Choose ↵ 2 L. Need: ↵ is separable over K.

Consider the stabilizer H = {� 2 � | �(↵) = ↵}. Let �1, �2, . . . , �r be a set of representatives ofall the cosets �H of H in �. Then, {�1(↵), �2(↵), . . . , �r(↵)} is precisely the �-orbit of ↵. Now,

f(x) =rY

j=1

(x� �j(↵))

is a separable polynomial by construction. Like last time, f(x) is �-invariant, so f(x) 2 K[x].Hence, ↵ is separable over K. Moreover, each root of f(x) is contained in L, so L contains asplitting field for f(x) over K. Thus, L is normal and separable (by the homework.)

Proof of (5) =) (4). Assume that there exists a separable f(x) 2 K[x] such that in L[x],

f(x) = c(x� ↵1)(x� ↵2) . . . (x� ↵n)

with ↵i 2 L pairwise distinct and L = K(↵1, . . . ,↵n). In particular, the minimal polynomial ofeach ↵i over K divides f(x), whence it is separable. Thus L is separable over K.

For normality, choose any algebraically closed extension E � K and let �1, . . . ,�n be all theroots of f(x) in E. If ' : L ! E is any homomorphism with 'K = idK , then {'(↵1), . . . ,'(↵n)}is some permutation of {�1, . . . ,�n}. So, '(L) = K(�1, . . . ,�n) is independent of '.

Proof of (1) =) (6): Assume that L � K is Galois. Last time =) L = K(↵) for some↵ 2 L. Let f(x) be the minimal polynomial of ↵ over K. Then f(x) is irreducible over K, itis separable because L is separable over K, and f(x) splits completely in L: f(x) must divideQ

g2Gal(L/K)(x� g(↵)) (in fact, they are equal.)

Assume L � K is a finite Galois extension, G = Gal(L/K), H ⇢ G is a subgroup and K ⇢ F :=LH ⇢ L (so H = AutF (L).) Recall: F is separable over K.

Lemma 9. F is Galois over K if and only if H is normal in G.

Proof. Assume that H is normal. If g 2 G and ↵ 2 F , then h(g(↵)) = g(g�1hg)(↵) = g(↵)for all h 2 H. Hence, g(↵) 2 F . Therefore, restriction to F defines a group homomorphism : G! AutK(F ), where ker( ) = H. This implies that

[F : K] � |AutK(F )| � |Im( )| = |G||H| =

[L : K]

[L : F ]= [F : K].

So these must be equalities. In particular, F � K is Galois, is surjective and hence induces anisomorphism G/H ! Gal(F/K). ⇤

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Proof. (other direction) Finally, suppose F � K is Galois. Written proof in note paper. ⇤

February 8, 2013

Recall: �n(x) 2 Z[x] is a monic polynomial defined inductively for all n 2 N.• �1(x) = x� 1

• �n(x) =xn � 1Q

1d<n,d|n �d(x)for all n � 2.

Equivalently, using Z ,! C,

�n(x) =Y

!2C⇥ is primitive n-th root of 1

(x� !)

Owed from last time: If p is prime and p does not divide n, then �n(x) 2 Fp[x] is a separablepolynomial.

Proof. �n(x) divides xn�1 in Fp[x]. But xn�1 is already separable because gcd(n·xn�1, xn�1) = 1

for all n 6= 1. ⇤Applications.

(1) �n(x) 2 Z[x] is irreducible, hence 8primitive n-th root of unity ! 2 C⇥, �n(x) is the minimalpolynomial of ! over Q. Hence, there is an isomorphism Gal(Q(!)/Q)! (Z/nZ)⇥.

(2) Let K be a field such that char(K) does not divide n. If a 2 K, then a is a primitive n-throot of 1 in K if and only if �n(a) = 0. (Does not require the irreducibility of �n.)

(3) If n 2 N, then there are infinitely many primes p ⌘ 1 mod n. This relies on : p ⌘ 1mod n () F⇥

p has an element of order n () p does not divide n and �n has a root inFp.

(4) IfA = finite abelian group, then there exists a finite Galois extension L � Q withGal(L/Q) ⇠=A. (Idea: using (3), check that there exists N 2 N such that A ⇠= a quotient group of(Z/nZ)⇥.)

Note: If Z/nZ is cyclic, then n = pk or n = 2pk for odd prime p or n = 2 or n = 4. But (Z/nZ)⇥is always cyclic for all n 2 N.

Open problem: Describe all finite groups that are isomorphic to the Galois group Gal(L/Q) forsome finite Galois extension L � Q.

Claim 1. : Given any finite group � and given any number p 2 {0} [ {primes}, there exists afinite Galois extension L � K where K has characteristic p and Gal(L/K) ⇠= �.

Let F = any field of characteristic p > 0. Form L = F (x1, x2, . . . , xn) (field of rational functions).Sn acts on L by permuting the xi, which gives an embedding Sn ⇢ Aut(L). We’ll describe LSn andsee that [L : LSn

] = n!. In particular, L � LSn is a finite Galois extension with Gal(L/LSn) ⇠= Sn.Now, if � ⇢ Sn is a subgroup, take K = L�. By the Main Theorem, L � K is Galois andGal(L/K) ⇠= �.

Description of LSn:

Definition 13. For 1 k n, the k-th elementary symmetric polynomial in F [x1, . . . , xn] is

�k =X

1i1···ikn

xi1 . . . xik homogeneous of degree k.

E.g.: n = 3 =) �1 = x1 + x2 + x3, �2 = x1x2 + x1x3 + x2x3, �3 = x1x2x3.

Lemma 10. Let F be any field. If L = F (x1, . . . , xn) and K := F (�1, . . . ,�n) (i.e. the extensionof F generated by the �i’s in L), then LSn = K.

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Proof. By construction, K ⇢ LSn , so it su�ces to show that [L : K] n!. We have L =K(x1, . . . , xn) (This means that L is generated over K by the elements x1, . . . , xn.) ConsiderL = K0 � K1 � · · · � Kn = K where Kj = K(xj+1, . . . , xn). We will show that [Kj�1 : Kj ] jfor any 1 j n. This will imply that [L : K] n!.

Form f(t) :=Qn

i=1(t�xi) 2 L[t]. But we see f(t) = tn��1tn�1+�2tn�2� · · ·+(�1)n�n 2 K[t].By construction, Kn�1 = K(xn) and f(xn) = 0, so [K(xn) : K] deg(f) = n.

More generally, consider 1 j n. Then, Kj�1 = Kj(xj) and xj is a root of

fj(t) := (t� x1)(t� x2) . . . (t� xj) =f(t)Q

i = j + 1n(t� xi)

whereQ

i = j + 1n(t� xi) has coe�cients in Kj .So, fj 2 Kj [t], which implies that [Kj�1 : Kj ] deg(fj) = j. ⇤

February 11, 2013

Galois Groups. K = field, f(x) 2 K[x] separable. The Galois group of f(x) over K is Gal(L/K)where L = any splitting field for f(x) over K. (i.e. L � K is a finite extension such thatf(x) = c · (x� ↵1)(x� ↵2) . . . (x� ↵n) and L = K(↵1, . . . ,↵n).

How to find these? E.g. f(x) = x4 � 1. If K = Q, we can take L = Q(i). Thus, Gal(L/K) =

Z/2Z.But suppose Q ⇢ K ⇢ C. What’s the Galois group of f(x) relative to K? We can still take

L = K(i). But what does this imply? If i /2 K, then we still get Gal(L/K) = Z/2Z. But if i 2 K,then Gal(L/K) is trivial.

Now suppose that K is finite of characteristic p > 2. Always have f(x) = (x� 1)(x+1)(x2 +1),so we only need x2+1 to be separable. By the earlier argument: x2+1 is irreducible in K[x] =)the Galois group is Z/2Z and x2 + 1 is reducible =) the Galois group is trivial. Also, x2 + 1 isreducible in K[x] () it has a root in K () �1 is a square in K⇥ () |K| ⌘ 1 mod 4.

Ex: f(x) = x4 + 1. This is irreducible in Q[x]. Reason:

x4 + 1 = x4 + 2x2 + 1� 2x2 = (x2 + 1)2 � (p2x)2 = (x2 +

p2x+ 1)(x2 �

p2x+ 1)

is the prime factorization in R[x].Claim: x4 + 1 is reducible over any finite field K.

Proof. If char(K) = 2, then x4 + 1 = (x+ 1)4. Assume char(K) 6= 2. If �1 = a2 for some a 2 K,then x4 + 1 = (x2 � a)(x2 + a). If 2 = b2 for some b 2 K, then x4 + 1 = (x2 + bx+ 1)(x2 � bx+ 1).If �2 = c2 for some c 2 K, then x4 + 1 = x4 � 2x2 + 1� c2x2 = (x2 + cx� 1)(x2 � cx� 1).

But the set of squares in K⇥ has index 2 because K⇥ is cyclic of even order, so if two elementsare not squares, then their product is. Therefore, one of these three elements 1, 2,�2 must be asquare. ⇤Solvability By Radicals. Consider a field extension K ⇢ L. Choose an element ↵ 2 L.

Definition 14. ↵ is expressible in terms of radicals over K if there exists a sequence of fields K =K0 ⇢ K1 ⇢ K2 ⇢ · · · ⇢ KN�1 ⇢ KN ⇢ L such that KN 3 ↵ and for each 1 j N there exists�j 2 Kj with Kj = Kj�1(�j) and �

mj

j 2 Kj�1 for some mj 2 N.

For example:7p

4p5� i 6

p2p

3 +p11

is expressible in terms of radicals over Q.

Theorem 4. Assume that char(K) = 0, L � K is a field extension and ↵ 2 L. Then ↵ satisfiesthe definition above () if f(x) = minimal polynomial of ↵ over K, then the Galois group off(x) over K is solvable.

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Corollary 4. There is no general formula for a root of a degree 5 polynomial that uses only thearithmetic operations and uses radicals.

Proof. In fact, there is no such formula for f(x) = 2x5 � 10x+ 5 because the Galois group of f(x)over Q is ⇠= S5, which is not solvable. ⇤

Reminders.

(1) Subgroups and quotients groups of solvable groups are solvable.(2) If G is a group, N ⇢ G is a normal subgroup and N , G/N are both solvable, then G is

solvable.(3) The class of solvable groups is the smallest class that contains all abelian groups and satisfies

(2).(4) Ad hoc (recursive) definition of solvability for finite groups:

Definition 15. Let G be a finite group. Then, G is solvable () either G = {1} or thereexists nontrivial cyclic group H with a surjective group homomorphism f : G ! H suchthat ker(f) is also solvable.

Hilbert’s Theorem 90 =) proved next time!

February 13, 2013

Solvability by radicals: Note that char 0 assumption is important! Keep in mind: K = perfectfield of char p > 0, a 2 K such that xp � x� a is irreducible over K. L = K[x]/(xp � x� a) � Kis Galois with Gal(L/K) ⇠= Z/pZ.

Regarding the theorem: important ingredients:

(1) Adjoining roots of unity to the base field makes life better.(2) Increasing the base field makes Galois groups easier to understand.(3) Hilbert’s Theorem 90.(4) Characterization of finite solvable groups from last time.

For the rest of today, all fields have characteristic 0.Simplest situation: L � K finite Galois extension such that Gal(L/K) ⇠= Z/nZ and K contains

a primitive n-th root of unity ! 2 K⇥.

Claim 2. There is an element ↵ 2 L such that L = K(↵) and ↵n 2 K. (So, informally, L isobtained from K by adjoining n

p� for some � 2 K.)

Proof. Let � 2 Gal(L/K) be a generator. Then, �0 = id,�,�2, . . . ,�n�1 : L⇥ ! L⇥ are distincthomomorphisms. The linear independence lemma =)

Pn�1j=0 !

j�j(x) 6= 0 for at least one x 2 L⇥.

Choose such an x and set ↵ =Pn�1

j=0 !j�j(x). Then, �(↵) = !�1 · ↵ because

! · �(↵) =n�1X

j=0

!j+1�j+1(x) = ↵

Now,

�(↵n) = �(↵)n = (!�1↵)n = !�n↵n = ↵n.

This implies that ↵n 2 LGal(L/K =) ↵n 2 K. Also,

Gal(L/K(↵)) = {g 2 Gal(L/K)|g(↵) = ↵} = {�k|0 k n� 1 and �k(↵) = ↵} = {1}.

But �k(↵) = �k�1(�(↵)) = �k�1(!�1↵) = !�1�k�1(↵) = · · · = !�k↵. Thus, L = K(↵). ⇤13

Theorem 5. (Hilbert’s Theorem 90): Let L � K be a Galois extension with Gal(L/K) ⇠= Z/nZ.Choose a generator � 2 Gal(L/K). Question: Which elements of L can be written as �(↵)/↵ forsome ↵ 2 L⇥. Consider the norm homomorphism

NL/K : L⇥ ! K⇥ given by � 7! � · �(�)�2(�) . . .�n(�) = �(�)�2(�) . . .�n+1(�)

Note: If ↵ 2 L⇥, then

NL//K

✓�(↵)

↵

◆=�(↵)

↵· �

2(↵)

�(↵)· · · · · �

n(↵)

�n�1↵=�n(↵)

↵=↵

↵= 1.

Hilbert’s Theorem 90: If � 2 L⇥ and NL/K(�) = 1, then � = �(↵)↵ for some ↵ 2 L⇥.

Example: If � 2 K⇥ and �n = 1, then NL/K(�) = 1 because NL/K(x) = xn for all x 2 K⇥.

Proof. Assume that � 2 L⇥ and � · �(�) · �2(�) · · · · · �n�1(�) = 1. Linear independence lemma

=) x+ � · �(x) + ��(�)�2(x) + · · ·+ � · �(�) . . .�n�1(�)�n�1(x) 6= 0 for some x 2 L⇥.

Choose such an x and call this ↵0. Then, �(↵0) = ��1↵0. Take ↵ = 1↵0 . ⇤

Roots of cubics, revisited: f(x) 2 Q[x] monic, irreducible and of degree 3. Over C, we canfactor f(x) = (x�↵1)(x�↵2)(x�↵3). Then, L = Q(↵1,↵2,↵3) is a splitting field of f(x) over Q.Let G = Gal(L/Q). Then, 3 | |G| and G ,! S3. So, G ⇠= S3 or G ⇠= A3.

Structure of G: A3 ⇢ G and G/A3 ,! S3/A3⇠= Z/2Z. Galois Theory =) first step: try to

compute LA3 . Guess: D = (↵1�↵2)(↵1�↵3)(↵2�↵3) 2 L. By inspection, D 2 LA3 and D is notfixed by any g 2 G, g /2 A3. Hence, LA3 = Q(D). Moreover, D2 2 LG = Q.

Second step: L � A3 = Q(D) is a Galois extension with Galois group A3⇠= Z/3Z. Pick a

generator � 2 A3 given by ↵1 7! ↵2,↵2 7! ↵3,↵3 7! ↵1.Imitate the proof of the claim from the start of the lecture: consider ⇣3 = exp(2⇡i/3) and form

↵1 + ⇣3�(↵1) + ⇣23�2(↵) = ↵1 + ⇣3↵2 + ⇣23↵3.

And the cube of this is in LA3 = Q(D).

February 15th, 2013

Terminology:

(1) Radical tower of fields: K0 ⇢ K1 ⇢ · · · ⇢ KN such that 81 i N , there exists ↵i 2 Ki

and mi 2 N with ↵mii 2 Ki�1 and Ki = Ki�1(↵i).

(2) A radical extension is a field extension K 0 � K such that there exists a radical tower withK = K0, K 0 ⇢ KN . (Using the notation from (1).)

Obvious/boring properties:

• Transitivity for both radical towers and radical extensions. (E.g. if K ⇢ K 0 and K 0 ⇢ K 00

are radical extensions, so is K ⇢ K 00.)• If K ⇢ K 00 is a radical extension and K ⇢ K 0 ⇢ K 00, then K ⇢ K 0 and K 0 ⇢ K 00 are radicalextensions.

• Given fields K ⇢ F ⇢ E and K ⇢ K 0 ⇢ E, if K 0 is radical over K, then FK 0 is radical overF .

Proof. Choose a radical tower K0 ⇢ K1 ⇢ · · · ⇢ KN such that K 0 ⇢ KN . From thedefinition: F = FK0 ⇢ FK1 ⇢ · · · ⇢ FKN is also a radical tower, and K 0F ⇢ KNF . ⇤

Theorem 6. char(K) = 0, K ⇢ K 0 finite extension. Then, K 0 � K is a radical extension ()there exists a finite extension L � K 0 such that L � K is Galois and Gal(L/K) is solvable.

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Proof. Proof of (: It’s enough to show that L is a radical extension of K. First reduction: letn = [L : K].

Claim: It’s enough to consider the case where K 3 primitive n-th root of 1.Proof: Let E � L be an algebraically closed extension. Let ⇣n 2 L⇥ be a primitive n-th root ofunity. Consider K ⇢ L ⇢ L(⇣n) and K ⇢ K(⇣n) ⇢ L(⇣n) (imagine picture). We know L is Galoisover K and K(⇣n) is a radical extension over K. Also, both L and K(⇣n) are normal over K =)L(⇣n) is normal over K =) L(⇣n) is Galois over K because char(K) = 0 =) separable. Hence,L(⇣n) � K(⇣n) is also Galois. Note: if g 2 Gal(L(⇣n)/K(⇣n)), then g(L) = L because L � K isnormal, so g|L 2 Gal(L/K). Moreover, g is determined uniquely by g|L, so we get an injectivegroup homomorphism Gal(L(⇣n)/K(⇣n)) ,! Gal(L/K) by g 7! g|L. Hence, Gal(L(⇣n)/K(⇣n)).

From now on, replace K,L with K(⇣n), L(⇣n) if necessary, so assume that ⇣n 2 K.Now, we know that Gal(L/K) is solvable. To prove that L is a radical extension of K, use

transitivity and induction on n = [L : K] = |Gal(L/K)|. If n = 1, we are done. By the claimfrom last time, if Gal(L/K) is cyclic, then L � K is radical. If Gal(L/K) is not cyclic, then thereexists normal subgroup H ⇢ G such that H 6= G, G/H is cyclic, and H is solvable. Look atK ⇢ LH ⇢ L. These are Galois extensions, Gal(LH/K) ⇠= G/H and Gal(L/LH) ⇠= H, so we aredone by induction. ⇤

Proof. Proof of =) : Special case: K0 ⇢ K1 of length 1. We want: a finite extension K1 � L suchthat L � K0 is Galois and Gal(L/K0) is solvable.

By assumption, K1 = K0(↵) for some ↵ 2 K1 such that � = ↵m 2 K0 for some m 2 N. Choosean algebraically closed extension E ⇢ K1. Let L = K0 (all roots of xm��) = K0(↵, ⇣m) = K1(⇣m).We have char(K0) = 0 =) L � K0 is Galois because it is normal.

Know: K0(⇣m) is Galois over K0 and Gal(K0(⇣m)/K0) ,! (Z/nZ)⇥ is abelian. We get the tower:

K0 ⇢ K0(⇣m) ⇢ L = K0(⇣m)(↵)

where K0(⇣m) � K0 and L � K0(⇣m) are Galois by HW.Have: Gal(L/K0(⇣m)) ⇢ Gal(L/K0) is a normal subgroup with quotient Gal(K0(⇣m)/K0), which

is abelian.Claim: Gal(L/K0(⇣m)) ⇠= a subgroup of Z/mZ (So Gal(L/K0(⇣m)) is solvable.)Proof: If g 2 Gal(L/K0(⇣m)), then g is determined by g(↵) and g(↵) = ⇣am ·↵ for some a 2 Z/mZ.

Then, g 7! a is an injective group homomorphism Gal(L/K0(⇣m)) ,! Z/mZ. ⇤

February 18th, 2013

Elementary Symmetric Polynomials. Recall:

�k(x1, . . . , xn) =X

1i1<i2<···<ikn

xi1 . . . xxk2 Z[x1, . . . , xn]

for every 1 k n. Equivalently:

(T + x1)(T + x2 . . . (T + xn) = Tn +nX

k=1

�k · Tn�k

If A is any ring, we can also view �1, . . . ,�n as elements of A[x1, . . . , xn].Consider the action of Sn on A[x1, . . . , xn] given by permuting x1, . . . , xn.

Theorem 7. A[x1, . . . , xn]Sn = A[�1, . . . ,�n].15

Proof:Step 1: The case where A = k is a field. What do we know?

(?) k(x1, x2, . . . , xn)Sn = k(�1,�2, . . . ,�n)

(looking at degrees and fixed fields, easy application of Galois theory.) So, we need to deduce thesame about polynomials: k[x1, x2, . . . , xn]Sn = k[�1,�2, . . . ,�n].

Nontrivial part: (?) only implies that k[�1, . . . ,�n] ⇢ k[x1, . . . , xn]Sn ⇢ k(�1, . . . ,�n).Digression: Suppose that R ⇢ S are commutative rings. An element ↵ 2 S is integral over R is

there exists a monic f(x) 2 R[x] with f(↵) = 0.Note:

(1) If R,S are fields, then “integral” () “algebraic”.(2) Take R = Z, S = Q =) 1

2 is not integral over Z, and in fact, if ↵ 2 Q, then ↵ is integralover Z () ↵ 2 Z.

Lemma 11. If R ⇢ S are commutative rings, then {↵ 2 S : ↵ is integral over R} is a subring ofS. (proof later.)

Lemma 12. If R is a UFD and K = field of fractions of R, an element ↵ 2 K is integral overR () ↵ 2 R.

Proof. Suppose ↵ 2 K and f(x) = xn + a1xn�1 + · · · + an�1x + an 2 R[x] is such that f(↵) = 0.

Write ↵ = bc in reduced form: gcd(b, c) = 1. If c is not a unit, there exist a prime element p 2 R

with p|c. Since f(↵) = 0, we get bn + a1cbn�1+ · · ·+ an�1c

n�1b+ cn = 0 =) p|bn =) p|b, whichis a contradiction. ⇤

Remark 1. Let R = C[x, y]/(y2 � x3). This is a domain because y2 � x3 is irreducible in C[x, y](exercise!) Consider K = field of fractions of R. Then, y

x 2 K is integral over R because ( yx)2�x =

0 in R. But yx /2 R (exercise).

Back to proof of theorem:

R = k[�1, . . . ,�n] ⇢ k[x1, . . . , xn]Sn ⇢ k(�1, . . . ,�n) = field of fractions of R

by Galois theory we know.Consider R = k[�1, . . . ,�n] ⇢ S = k[x1, . . . , xn]. For each 1 i n xi 2 S is integral over R

because it is a root of the polynomial:

(T � x1)(T � x2) . . . (T � xn) = Tn + Tn +nX

k=1

�k · Tn�k 2 R[T ].

By the first Lemma we prove today, S is integral over R. A fortiori, SSn is also integral over R.But R is a UFD, and SSn ⇢ field of fractions of R, so by the second Lemma, SSn = R. This provesthe theorem in the case where A is a field.

Small gap in proof: Why is k[�1, . . . ,�n] a UFD?Issue: �1, . . . ,�n are defined as certain elements of k[x1, . . . , xn] = S where the xi’s are “freevariables”. Now, k[�1, . . . ,�n] = (by def) = the subring of S generated by k,�1, . . . ,�n.

Need: R ⇠= an abstract polynomial ring in n variables over k.Warning : Consider k[t] and the subring k[t2, t3] ⇢ k[t]. I.e. k[t2, t3] = {a0 + a2t

2 + a3t3 + · · ·+

antn}. This is not isomorphic to k[x, y], and in fact is not a UFD because k[t2, t3] ⇠= k[x, y]/(y2�x3).

Reason: We have a surjective homomorphism k[x, y]! k[t2, t3] given by f(x, y) 7! f(t2, t3). Checkthat the kernel is generated by y2 � x3.

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February 20, 2013

Most Important Theorem Of The Year.

Theorem 8. (Remember setup with A = any ring) =) Main conclusion:

A[x1, . . . , xn]Sn = A[�1, . . . ,�n].

Proof. Step 1: Prove the theorem for A = Z (or in fact for any UFD.)Step 2: Fix the gap in Step 1. ⇤

Proposition 5. Assume A is a commutative integral domain. The ring homomorphism

A[y1, . . . , yn]! A[x1, . . . , xn] by P 7! P (�1, . . . ,�n)

(where �k =P

xi1 . . . xik) is injective and hence gives a ring isomorphism

A[y1, . . . , yn]! A[�1, . . . ,�n].

In particular, if A is a UFD, so is A[�1, . . . ,�n].

Proof. Assume not and let P (y1, . . . , yn) 6= 0 be in the kernel. LetK be the field of fractions of A andlet E � K be an algebraically closed field extension. Then P (�1(x1, . . . , xn), . . . ,�n(x1, . . . , xn)) = 0as an element of A[x1, . . . , xn].

The same equality holds for any choice of (x1, . . . , xn) 2 En. (Proof continues after Lemma.) ⇤Lemma 13. The map En ! En given by (a1, . . . , an) 7! (�1(a1, . . . , an), . . . ,�n(a1, . . . , an)) issurjective.

Proof. Choose any (b1, . . . , bn) 2 En and form the polynomial f(T ) = Tn +Pn

k=1(�1)kbk · Tn�k 2E[T ]. Since E is algebraically closed, f(T ) = (T � a1) . . . (T � an) for some ai 2 E. Then,bk = �k(a1, . . . , an) for all 1 k n. ⇤

Back to the proposition: The lemma implies P (b1, . . . , bn) = 0 for every (b1, . . . , bn) 2 En.Exercise: This implies that P = 0 as an element of E[y1, . . . , yn] (Hint: use induction on n and thefact that E is infinite.) This concludes the proof of the proposition.

Compare the proposition to:Fact 1: k = any field, f1, . . . , fn 2 k[x1, . . . , xn] such that k[x1, . . . , xn] is integral over k[f1, . . . , fn].

Then, k[y1, . . . , yn]! k[f1, . . . , fn] by P 7! P (f1, . . . , fn) is a ring isomorphism. Think about: Con-sider k

n 7! kngiven by (a1, . . . , an) 7! (f1(a1, . . . , an), . . . , fn(a1, . . . , an)).

Fact 2: k = any field, f1, . . . , fn 2 k(x1, . . . , xn) such that the extension k(x1, . . . , xn) � k(f1, . . . , fn)is algebraic =) k(f1, . . . , fn) ⇠= k(y1, . . . , yn).

Step 1: R = Z[�1, . . . ,�n] ⇢ B = Z[x1, . . . , xn]. Want: BSn = R. Consider the field of fractionsof B and R: L = Q(x1, . . . , xn) � K = Q(�1, . . . ,�n). Know: LSn = K by Galois theory argumentfrom before. This implies that R ⇢ BSn ⇢ K.

Now each xi 2 B is integral over R (because it’s a root of: Tn +Pn

k=1(�1)k�kTn�k 2 R[T ].)=) B is integral over R =) BSn is integral over R =) BSn = R by a lemma from Mondaybecause R is a UFD by the previous proposition.

So we have finished the theorem when A = Z (or when A is a UFD.)Step 3: Generalize. New theorem: (V,+) = any abelian group. Then, V [x1, . . . , xn]Sn =

V [�1, . . . ,�n]. Note:

V [x1, . . . , xn] =

8<

:X

i1,...,in�0

ai1,...,inxi11 . . . xinn : ai1,...,in 2 V and all but finitely many ai1,...,in are 0

9=

; .

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V [�1, . . . ,�n] =

8<

:X

i1,...,in�0

bi1,...,in�i11 . . .�inn : bi1,...,in 2 V and all but finitely many bi1,...,in are 0

9=

; .

and�k =

X

1j1···jnn

xj1 . . . xjn

Note that because (V,+) is a group, we have no multiplication. Thus, we cannot view thesepolynomials as functions because evaluation requires multiplication. We can only view them asformal expressions.

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