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MATH5253M: Commutative algebra and algebraic geometry
Oliver King
www.maths.leeds.ac.uk/∼pmtok
adapted from K. Houston
2016
Contents
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
I Commutative Algebra 51 Revision of rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Revision of ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Maximal ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Localisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 The radical, nilradical and Jacobson radical . . . . . . . . . . . . . . . . . . . . . . 147 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 Nakayama’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Exact sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1910 Free modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2211 Noetherian rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2412 Primary decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2513 Hilbert’s Basis Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2814 Noether normalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
II Algebraic Geometry 3215 Algebraic sets and varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3216 Intersection multiplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3717 Projective space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1
Introduction
Books:• Miles Reid - Undergraduate algebraic geometry, LMS Student Texts 12, CUP, 1988.
• Miles Reid - Undergraduate commutative algebra, LMS Student Texts 29, CUP, 1995.
• Rodney Sharp - Steps in commutative algebra 2nd Ed, LMS Student Texts 51, CUP,2000.
• Robin Hartshorne - Algebraic Geometry, Springer Verlag, 1997. (First chapter only)
• M.F. Atiyah and I.G. MacDonald - Introduction to commutative algebra, WestviewPress, 1994
• W. Fulton - Algebraic Curves.
Given an equation or set of equations, we can ask various questions. For instance:
(i) Are there any solutions? Any integer, real, complex etc. solutions?
(ii) How many solutions are there? What are their multiplicities? For example, x3 = 0 hassolution x = 0 but with multiplicity 3.
(iii) Can we find them?
Example 0.1. What are the integer solutions of X2 + Y 2 = Z2? Solutions certainly exist, forexample
(X,Y, Z) = (3, 4, 5),
but are there others? Can we find all of them?Note first that if Z = 0 then both X and Y must also be zero. Assuming henceforth then that
Z 6= 0 we substitute x = XZ and y = Y
Z we can re-frame the question as
What are the rational solutions of x2 + y2 = 1?
Think of the solution set as a circle in R2:
x
y
−1 1
−1
1
Consider a line of slope t through (0, 1) that rotates about (0, 1). We can then find all solutions of
2
our new equation by using t as a new parameter.
x
y
solution
So we want rational solutions ofy − tx = 1x2 + y2 = 1
}Substituting the first of these into the second we see
x2 + (tx+ 1)2 = 1 =⇒ x2 + t2x2 + 2tx+ 1 = 1
=⇒ x2(t2 + 1) + 2tx = 0
=⇒ x(x(t2 + 1) + 2t
)= 0.
This gives two solutions, x = 0 and x =−2tt2 + 1
. The first solution for x gives y = 1, and the second
gives
y =−2t2
t2 + 1+ 1 =
1− t2
1 + t2.
The solutions are therefore
(x, y) = (0, 1) and (x, y) =(−2t
1 + t2,
1− t2
1 + t2
)(t ∈ R).
We need to see which values of t give rational values of x and y. A bit of checking shows thatx, y ∈ Q ⇐⇒ t ∈ Q. So let t = m
n , where m and n are coprime integers. Then
x =−2mnm2 + n2
and y =n2 −m2
m2 + n2.
Returning to our original variables X, Y and Z we see that integer solutions to X2 +Y 2 = Z2 canbe given by
Y = 2mn, Y = n2 −m2, Z = m2 + n2, m, n ∈ Z, m, n coprime, or
X = mn, Y =n2 −m2
2, Z =
m2 + n2
2if both m and n are odd.
For instance, m = 1, n = 3 gives X = 3, Y = 4, Z = 5.
The core idea here is to swap between algebra and geometry, and use the tools we have in bothareas to solve problems. One family of equations, conics, have been covered extensively (e.g. bythe ancient Greeks, Omar Khayyam, Rene Descartes. . . ). Another family with many open (andinteresting/difficult!) questions is cubics, in particular those of the form y2 = x3+ax+b. These are
3
known as elliptic curves, and are found throughout number theory (cf. Fermat’s Last Theorem),cryptography etc. They are also the focus of the Birch and Swinnerton-Dyer conjecture, one ofthe Clay Institute Millennium problems.
Unfortunately our parameterisation trick in Example 0.1 doesn’t always work. For instance,see [2, Section 2.2] for a curve y2 = x(x − 1)(x − λ) which has no rational parameterisation. Wewill therefore have to build a bigger bridge between the two fields.
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Part I
Commutative Algebra
1 Revision of rings
Definition 1.1. A ring is a triple (R,+, ·) of a set R and two binary operations
+ : R×R −→ R (addition)· : R×R −→ R (multiplication)
such that the following hold:
(i) (R,+) is an abelian group, with identity 0 = 0R;
(ii) there is an element 1 = 1R such that 1 · r = r · 1 = r for all r ∈ R;
(iii) · is associative, i.e. (r · s) · t = r · (s · t) for all r, s, t ∈ R;
(iv) · distributes over +, i.e. r · (s+ t) = r · s+ r · t and (s+ t) · r = s · r + t · r for all r, s, t ∈ R.
We will often abbreviate the triple (R,+, ·) to just R with the operations implicit, and moreoverthe multiplication r · s to just rs.
Definition 1.2. A ring R is called commutative if rs = sr for all r, s ∈ R.
Remark. In this course all rings will be commutative rings, and so hereafter we will take “ring”to mean “commutative ring”.
Example 1.3. (i) Z, the set of integers.
(ii) Zn = Z/nZ, the integers modulo n.
(iii) R, the set of real numbers.
(iv) C, the set of complex numbers.
(v) C[0, 1], the set of continuous functions on [0, 1].
(vi) Gaussian integers Z[i] = {a+ bi : a, b ∈ Z}.
(vii) Let X be any set, and define FX = RX = {functions f : X −→ R}. Define +, · : FX×FX −→FX by
(f + g) : X → Rx 7→ f(x) + g(x),
(f · g) : X → Rx 7→ f(x)g(x).
Then FX is a commutative ring, with additive identity 0FX: x 7→ 0 and multiplicative
identity 1FX: x 7→ 1.
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(viii) We can also construct new rings from old ones. Let R be any commutative ring, and define
R[x] = {polynomials in x with coefficients in R} =
{n∑i=0
rixi : n ∈ N and ri ∈ R ∀i
}.
This is also a commutative ring. We can then define R[x1, . . . , xn] inductively by
R[x1, . . . , xn] = R[x1, . . . , xn−1][xn].
This is just polynomials in the variables x1, . . . , xn with coefficients in R.
(ix) R[[x]] = {formal power series in x with coefficients in R} =
{ ∞∑i=0
rixi : ri ∈ R ∀i
}. Note
that these are formal objects, not necessarily functions from R to R. For instance,∑∞i=0 x
i
is an element of R[[x]], but we cannot evaluate this at x = 1 so it does not define a functionR→ R.
Definition 1.4. A field is a ring K where every element other than 0K has a multiplicative inverse.Formally, for each r ∈ K\{0} there exists an r−1 ∈ K\{0} such that rr−1 = r−1r = 1K .
Example 1.5. (i) Familiar fields are C,R,Q. Another example is Zp = Z/pZ for any prime p.
(ii) Z itself is not a field, nor is the set Z[i] of Gaussian integers. For instance, 2 + 0i has noinverse. In fact the units of Z[i] are ±1,±i.
We will now see another way of constructing rings and fields from old ones:
Example 1.6. Let R,S be rings. The Cartesian product R× S = (R× S,+, ·) of R and S is alsoa ring, where we define
(r1, s1) + (r2, s2) = (r1 + r2, s1 + s2)(r1, s1) · (r2, s2) = (r1r2, s1s2).
for all r1, r2 ∈ R, s1, s2 ∈ S. We have 0R×S = (0R, 0S) and 1R×S = (1R, 1S). Note that if K andL are fields then K × L is not a field, for instance (0, 1) has no multiplicative inverse.
Definition 1.7. A subset S ⊆ R of a ring R is called a subring if (S,+) is a subgroup of (R,+),1R ∈ S and S is closed under multiplication. Similarly, if K is a field then a subset L ⊆ K is calleda subfield if it is a subring of K and r−1 ∈ L for all non-zero r ∈ L.
Example 1.8. Let R = R and S = {a+ b√
5 : a, b ∈ Z}. Clearly 0 = 0 +√
5, 1 = 1 + 0√
5 ∈ S, sowe will check that it is additively and multiplicatively closed. For all a, b, c, d ∈ R, we have
(a+ b√
5) + (c+ d√
5) = (a+ c) + (c+ d)√
5 ∈ S,
(a+ b√
5)(c+ d√
5) = ac+ ad√
5 + bc√
5 + 5bd
= (ac+ 5bd) + (ad+ bc)√
5 ∈ S.
Similarly if R = C, then S = {a+b√−5 : a, b ∈ Z} is a subring. Rings like these play an important
role in areas of number theory.
Definition 1.9. Let R,S be rings. A ring homomorphism from R to S is a map ϕ : R→ S suchthat for all r1, r2 ∈ R:
(i) ϕ(r1 + r2) = ϕ(r1) + ϕ(r2);
(ii) ϕ(r1r2) = ϕ(r1)ϕ(r2);
(iii) ϕ(1R) = 1S .
6
If ϕ is bijective then we say ϕ is an isomorphism.
Exercise (Exercise sheet 0). If ϕ : R → S is a ring isomorphism, prove that ϕ−1 : S → R is aring homomorphism (and hence also an isomorphism).
Definition 1.10. Let ϕ : R → S be a ring homomorphism. The kernel of ϕ, denoted Kerϕ, isthe set
Kerϕ = {r ∈ R : ϕ(r) = 0S}.
The image of ϕ, denoted Imϕ, is the set
Imϕ = {ϕ(r) : r ∈ R}.
The proof of the following proposition is left as an easy exercise:
Proposition 1.11. (i) Imϕ is a subring of S.
(ii) Kerϕ is not necessarily a subring of R.
Proof. Exercise.
2 Revision of ideals
That Kerϕ is not a subring of R causes us problems if we wish to introduce quotient rings likewe introduced quotient groups. Note that if H is a subgroup of G then G/H does not necessarilyexist. Note also that dealing with commutative groups circumvents this problem, but that is notthe case when dealing with rings. The “correct” notion of a substructure that allows us to takequotients is that of an ideal.
Definition 2.1. Let R be a ring. A subset I ⊆ R is called an ideal if:
(i) I 6= ∅;
(ii) for all x, y ∈ I, x− y ∈ I;
(iii) for all x ∈ I and r ∈ R, rx ∈ I.
We write I CR to mean I is an ideal of the ring R.If I 6= R, then we say that I is a proper ideal of R.
Example 2.2. (i) Let R be a ring. Then {0R} and R are both ideals of R, usually referred toas trivial ideals.
(ii) For any n ∈ Z, nZ is an ideal of Z.
(iii) For a ring homomorphism ϕ : R → S, Kerϕ is an ideal of R. Indeed let x, y ∈ Kerϕ andr ∈ R, then
ϕ(0) = 0 so 0 ∈ Kerϕ (Kerϕ 6= ∅),ϕ(x+ y) = ϕ(x) + ϕ(y) = 0 + 0 = 0 so x+ y ∈ Kerϕ,
ϕ(rx) = ϕ(r)ϕ(x) = ϕ(r)0 = 0 so rx ∈ Kerϕ.
(iv) A crucial example for algebraic geometry, and one we will encounter many times later inthe course, is the following. Let K be a field (usually R or C), V ⊆ Kn be a set andR = K[X1, . . . , Xn]. Then
I(V ) = {f ∈ R : f(v) = 0 for all v ∈ V }
is an ideal of R.
7
Definition 2.3. Let A be a non-empty subset of a ring R. The ideal generated by A, denoted 〈A〉,is the set of all elements
〈A〉 =
{n∑i=1
riai : n ∈ N, r1, . . . , rn ∈ R, a1, . . . , an ∈ A
}.
We say an ideal I is finitely generated if there exists a finite subset A ⊆ R such that I = 〈A〉.
We can also perform operations on ideals as per the following proposition.
Proposition 2.4. Let I, J be ideals of a ring R. The following are then also ideals of R:
(i) I ∩ J = {x : x ∈ I and x ∈ J};
(ii) IJ = 〈{xy : x ∈ I, y ∈ J}〉;
(iii) I + J = 〈I ∪ J〉;
(iv) (I : J) = {r ∈ R : rJ ⊆ I} (the ideal quotient of I and J).
Proof. Exercise. For part (iv) see Exercise Sheet 0, and note that the properties exhibited justifyits name.
We will now move on to quotient rings.
Definition 2.5. Let I be an ideal of a ring R. A coset of I in R is a set
r + I = {r + x : x ∈ I}
for some r ∈ R. This may also be denoted by r, and we denote by R/I the set of cosets of I in R.
The following proposition is straightforward:
Proposition 2.6. (i) Two cosets are either equal or disjoint, and the union of all cosets is R.We say that the cosets partition R.
(ii) Cosets r + I and s+ I are equal if and only if r − s ∈ I.
(iii) We can define multiplication and addition on R/I by setting (r + I) + (s+ I) = (r + s) + Iand (r + I)(s+ I) = rs+ I.
(iv) The additive and multiplicative identities of R/I are 0 + I = I and 1 + I respectively.
This proposition shows that we have a ring structure on R/I, with much of the structureinherited from the ring structure on R.
Proposition 2.7. Let I be an ideal of a ring R. Define ϕ : R→ R/I by ϕ(r) = r + I. Then:
(i) ϕ is a ring homomorphism (called the quotient homomorphism);
(ii) Kerϕ = I;
(iii) there is a bijection between ideals of R/I and the ideals of R which contain I, given by
J CR/I 7−→ ϕ−1(J) = {r ∈ R : r + I ∈ J}I ⊆ K CR 7−→ ϕ(K) = {r + I : r ∈ K}.
Proof. (i) See Exercise Sheet 0.
(ii) See Exercise Sheet 0.
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(iii) For an ideal K such that I ⊆ K C R, we first show that ϕ(K) is an ideal of R/I (notethat this may not be true for any ϕ). Clearly ϕ(K) 6= ∅, as ϕ(I) = I ∈ ϕ(K). For anytwo cosets r + I, s + I ∈ ϕ(K) we have r, s ∈ K, and since K is an ideal then r − s ∈ K.Hence (r + I) − (s + I) = (r − s) + I ∈ ϕ(K). If now we also choose any t + I ∈ R/I then(t+ I)(r + I) = tr + I ∈ ϕ(K), since tr ∈ K again due to K being an ideal of R.
We now show that the assignment K 7−→ ϕ(K) is injective. Suppose K 6= K ′ are both idealsof R containing I, then without loss of generality there is some r ∈ K such that r /∈ K ′. Weclearly have r + I ∈ ϕ(K). We will show that r + I /∈ ϕ(K ′), thus ϕ(K) 6= ϕ(K ′). Assumefor a contradiction that r + I ∈ ϕ(K ′), then r + I = s+ I for some s ∈ K ′. By the equalityrule for cosets, we have r − s ∈ I ⊆ K ′, and hence (r − s) + s = r ∈ K ′, a contradiction.
Finally, we show the map K 7−→ ϕ(K) is surjective. Given an ideal J C R/I we clearlyhave ϕ(ϕ−1(J)) = J , so we must show that ϕ−1(J) is an ideal of R containing I. Thecontainment is easy, since I = ϕ−1(0) ⊆ ϕ−1(J). If now r, s ∈ ϕ−1(J), then r + I, s+ I ∈ Jand hence (r − s) + I ∈ J . Therefore r − s ∈ ϕ−1(J). Similarly if t ∈ R then t + I ∈ R/Iand (t+ I)(r + I) = tr + I ∈ J , hence tr ∈ ϕ−1(J).
Theorem 2.8. Let ϕ : R → S be a ring homomorphism. Then ϕ : R/Kerϕ → Imϕ given byϕ(r + Kerϕ) = ϕ(r) is an isomorphism.
Proof. See Exercise Sheet 0 (remember to check that this is well defined!).
3 Prime ideals
Definition 3.1. An ideal P of R is called a prime ideal if;
(i) P 6= R;
(ii) xy ∈ P =⇒ x ∈ P or y ∈ P .
The first example below explains the name of these ideals.
Example 3.2. (i) The ideal nZ of Z is prime if and only if n is prime (Exercise).
(ii) The ideal 〈f〉 of C[x] is prime if and only if f is irreducible, i.e. f cannot be written as theproduct of two polynomials of positive degree.
Proposition 3.3. Let ϕ : R → S be a ring homomorphism. If P C S is a prime ideal, thenϕ−1(P ) CR is a prime ideal.
Proof. Let x, y ∈ R be such that xy ∈ ϕ−1(P ), i.e. ϕ(xy) ∈ P . Now ϕ(xy) = ϕ(x)ϕ(y), andsince P is prime we therefore have either ϕ(x) ∈ P or ϕ(y) ∈ P . Hence either x ∈ ϕ−1(P ) ory ∈ ϕ−1(P ).
Proposition 3.4. Let I be an ideal of a ring R. If P is a prime ideal of R containing I, then theimage of P in R/I is also prime.
Proof. Denote by P the image of P in R/I. Suppose x+I, y+I ∈ R/I are such that (x+I)(y+I) ∈P . Then xy + I ∈ P , so there is some p ∈ P such that xy − p ∈ I ⊆ P . Therefore xy ∈ P , soeither x ∈ P or y ∈ P as P is prime, thus either x+ I ∈ P or y + I ∈ P .
Remark 3.5. These two propositions show that the bijection between ideals of R/I and ideals ofR containing I restricts to a bijection between prime ideals of R/I and prime ideals of R containingI.
Definition 3.6. A ring R is an integral domain if:
(i) R 6= {0};
9
(ii) for all r, s ∈ R, rs = 0 =⇒ r = 0 or s = 0, i.e. there are no non-zero zero divisors.
Example 3.7. (i) Z is an integral domain.
(ii) Z4 is not an integral domain, as (2 + 4Z)(2 + 4Z) = 4 + 4Z = 0.
(iii) The ring M2(R) of 2× 2 real matrices is not an integral domain, as(1 00 0
)(0 00 1
)=(
0 00 0
).
Theorem 3.8. Let I �C R be an ideal. Then I is prime iff R/I is an integral domain.
Proof. Suppose I is prime. Then since I 6= R we have R/I 6= {0}. Now suppose a+ I is non-zeroin R/I and there is some b+ I ∈ R/I such that (a+ I)(b+ I) = I. Then ab+ I = I and ab ∈ I.Since I is prime we have either a ∈ I or b ∈ I, but since a+I 6= I this forces b ∈ I. Hence b+I = 0in R/I, and R/I is an integral domain.
Suppose now that R/I is an integral domain. Since R/I 6= {0} we must have I 6= R. Now letab ∈ I for some a, b ∈ R, then ab + I = (a + I)(b + I) = I. Since R/I is an integral domain, wemust have either a+ I = I or b+ I = I, and hence either a ∈ I or b ∈ I. Therefore I is prime.
Theorem 3.9. Let R be a ring, I1, . . . , In C R be ideals, and P C R be a prime ideal. Then thefollowing are equivalent:
(i) Ij ⊆ P for some 1 6 j 6 n;
(ii) I1 ∩ · · · ∩ In ⊆ P ;
(iii) I1 . . . In ⊆ P .
Proof. (i) =⇒ (ii) =⇒ (iii) are trivial.(iii) =⇒ (i): Assume that I1 . . . In ⊆ P but for all 1 6 j 6 n we can choose aj ∈ Ij\P . Then
a1 . . . an ∈ I1 . . . In\P as P is prime, a contradiction.
4 Maximal ideals
Definition 4.1. An ideal I of a ring R is called a maximal ideal if:
(i) I 6= R;
(ii) there is no ideal J of R such that I ( J �C R.
Example 4.2. (i) pZC Z is a maximal ideal for p prime (we will see a proof of this soon).
(ii) 〈X〉CR[X,Y ] is not maximal, as 〈X〉 ( 〈X,Y 〉 �C R[X,Y ].
Theorem 4.3. Maximal ideals are prime.
Proof. Let m be a maximal ideal of a ring R and suppose ab ∈ m for some a, b ∈ R. If neither anor b are in m then both 〈a〉+m and 〈b〉+m are strictly bigger than m. As m is maximal, we mustthen have 〈a〉+ m = 〈b〉+ m = R. But now
R = RR
= (〈a〉+ m)(〈b〉+ m)
= m2 + 〈a〉m + 〈b〉m + 〈ab〉⊆ m 6= R,
which is a contradiction.
Proposition 4.4. Let R be a ring. Then:
10
(i) R is a field iff {0} and R are the only ideals of R;
(ii) an ideal I CR is maximal iff R/I is a field.
Proof. (i) Assume R is a field and let I CR be a non-zero ideal. Choose r ∈ I\{0}, then r hasan inverse r−1 ∈ R. Hence r−1r = 1 ∈ I, so I = R.
Conversely suppose {0} and R are the only ideals of R, and choose r ∈ R\{0}. Then 〈r〉 = Rand so there exists some s ∈ R such that sr = 1, i.e. r has an inverse r−1 = s. Therefore Ris a field.
(ii) If I is maximal then by Proposition 2.7, R/I has no ideals other than {I} and R/I. ThereforeR/I is a field by (i).
If now R/I is a field then again by Proposition 2.7 and (i), any ideal of R which contains Imust either be I or R, so I is maximal.
Remark. Let ϕ : R→ S be a ring homomorphism. Unlike the situation with prime ideals, m CSmaximal does not imply that ϕ−1(m) is maximal. For instance, let ϕ : Z → Q be the inclusionmap. Then {0Q} C Q is maximal as Q is a field, but ϕ−1({0Q}) = {0Z} ( 2Z �C Z, so ϕ−1({0Q})is not maximal.
However we do have the following result which is analogous to Remark 3.5:
Proposition 4.5. The bijection between ideals of R/I and ideals of R containing I restricts to abijection between maximal ideals of R/I and maximal ideals of R containing I.
Proof. Exercise.
We will soon show that every proper ideal is contained in some maximal ideal. In order toprove this however, we must take a brief diversion into set theory.
A partially ordered set or poset (Σ,6) is a set Σ and a binary relation 6 ⊆ Σ× Σ which is:
(i) reflexive, i.e. x 6 x ∀x ∈ Σ;
(ii) transitive, i.e. x 6 y and y 6 z =⇒ x 6 z ∀x, y, z ∈ Σ;
(iii) antisymmetric, i.e. x 6 y and y 6 x =⇒ x = y ∀x, y ∈ Σ.
A subset S ⊆ Σ is totally ordered if for all s, t ∈ S we have either s 6 t or t 6 s (or both).Given a subset S ⊆ Σ, an element u ∈ Σ is an upper bound for S if s 6 u for all s ∈ S.A maximal element of Σ is an element m ∈ Σ such that there is no s ∈ S with m 6 s and m 6= s.
Example. A poset without a maximal element is the set (Z,6).
Theorem (Zorn’s Lemma). Suppose that (Σ,6) is a non-empty poset and that any totally orderedsubset S ⊆ Σ has an upper bound in Σ. Then Σ has a maximal element.
This is equivalent to the Axiom of Choice, and we take it as an axiom in ZFC (where we generallydo maths).
We can now prove the following:
Proposition 4.6. Let R be a non-zero ring. Then every proper ideal I is contained in a maximalideal.
Proof. Let Σ be the set of ideals J �C R containing I, ordered by inclusion ⊆. Then (Σ,⊆) is a non-empty poset, since I ∈ Σ. If {Jλ : λ ∈ Λ} is a totally ordered subset of Σ then clearly J∗ = ∪λ∈Λ isa proper ideal of R containing I, and moreover J∗ is an upper bound for {Jλ : λ ∈ Λ}. By Zorn’sLemma, Σ then has a maximal element. But a maximal element of Σ is an ideal m 6= R containingI with no proper ideals J containing it, so is a maximal ideal containing I.
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This proposition shows that we usually have lots of maximal ideals, even if they can be hardto find.
Example 4.7. Let K be a field, R = K[x1, . . . , xn] and a1, . . . , an ∈ K. Then m = 〈x1 −a1, . . . , xn − an〉 is a maximal ideal. If it wasn’t, then there would exist a polynomial f ∈ R suchthat f 6= m and 〈f〉+ m ( R. Applying the division algorithm n times gives
f = f1(x1 − a1) + · · ·+ fn(xn − an) + b,
where fi ∈ K[xi, xi+1, . . . , xn] ⊆ R for each 1 6 i 6 n and b ∈ K. Since f /∈ m, we must have b 6= 0and so b has an inverse b−1. Therefore 1 = b−1 (f − fi(x1 − a1)− · · · − fn(xn − an)) ∈ 〈f〉 + mand so 〈f〉+ m = R, a contradiction.
Are these the only maximal ideals of K[x1, . . . , xn]? The answer is yes when K is algebraicallyclosed, but we need a bit more theory in order to prove this.
In some cases, there are far fewer maximal ideals.
Definition 4.8. A ring R is called a local ring if it has precisely one maximal ideal m. We usuallydenote this ring by the pair (R,m).
5 Localisation
We can construct Q from Z by inverting all non-zero elements. Formally this is done by viewingQ as a set of equivalence classes in Z× (Z\{0}) via the relation
(r, a) ∼ (s, b) ⇐⇒ as = br.
We then write ra for the equivalence class of (r, a). Addition and multiplication of equivalence
classes is defined byr
a+s
b=as+ br
aband
r
a
s
b=rs
ab. (∗)
We also have 0Q = 01 and 1Q = 1
1 . It is easy to check that provided r 6= 0, ar is a multiplicative
inverse for ra .
We wish to repeat the above for a general ring R. Notice from (∗) that if we invert a and bthen we have also inverted ab. This motivates the following.
Definition 5.1. Let R be a ring and A ⊆ R be a subset. We say A is multiplicatively closed if:
(i) 1R ∈ A;
(ii) a, b ∈ A =⇒ ab ∈ A.
Definition 5.2. Let R be a ring and A ⊆ R be multiplicatively closed. The localisation of R at A,denoted A−1R or R[A−1], is the set of equivalence classes of R×A under the equivalence relation
(r, a) ∼ (s, b) ⇐⇒ asc = brc for some c ∈ A.
We will again usually write the equivalence class of (r, a) as ra , with addition and multiplication
defined as in (∗).
Lemma 5.3. Let R be a ring and A ⊆ R a multiplicatively closed subset. Then the localisationA−1 of R at A is also a ring via the sum and product (∗), and 0A−1R = 0R
1Rand 1A−1R = 1R
1R.
Moreover there is a ring homomorphism
i : R→ A−1R
r 7→ r
1,
with kernel Ker i = {r ∈ R : ra = 0 for some a ∈ A}.
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In some cases, such as the construction of Q above, we wish to invert as many things as possible.
Definition 5.4. Let R be an integral domain. The quotient field or field of fractions of R, denotedFrac(R), is the localisation
Frac(R) = (R\{0})−1R.
Example 5.5. In each of the following, A is a multiplicatively closed subset of a ring R.
(i) RA is the zero ring if and only if 0 ∈ A.
(ii) Let a ∈ A. We write Ra for the localisation of R at the set {an : n > 0}.
(iii) Let P be a prime ideal of R. Then A = R\P is multiplicatively closed and we write RP forA−1R.
(iv) Let p ∈ Z be prime. Then
Zp ={ab∈ Q : b is a power of p
},
Z〈p〉 ={ab∈ Q : p - b
},
Frac(Z) = Q.
Since A−1R is a ring, we can talk about its ideals and how they relate to the ideals of R.
Definition 5.6. Given an ideal I of R, we define the localisation of the ideal I to be the set
A−1I ={xa
: x ∈ I, a ∈ A}.
Proposition 5.7. Let R be a ring, A ⊆ R a multiplicatively closed subset, and I CR an ideal.
(i) A−1I is an ideal of A−1R. Moreover, if I is generated by a set X, then A−1I is generatedby{x1 : x ∈ X
}.
(ii) We have xa ∈ A
−1I if and only if there is some b ∈ A with xb ∈ I.
(iii) A−1I = A−1R if and only if I ∩A 6= ∅.
(iv) The map I 7→ A−1I commutes with forming finite sums, products and intersections, andquotients.
Proof. See Exercise Sheet 1.
This leads to a correspondence theorem for between ideals of R and ideals of A−1R.
Theorem 5.8. There is a bijection
{ideals J CA−1R} ↔ {ideals I CR such that no element of A is a zero divisor in R/I},
sending J 7→ i−1(J) and I 7→ A−1I, where i−1 is the preimage of the homomorphism from Lemma5.3.
Moreover, this restricts to a bijection
{prime ideals QCA−1R} ↔ {prime ideals P CR with P ∩A = ∅}.
Proof. Suppose J C A−1R is an ideal. Then i−1(J) is an ideal, being the preimage of an idealunder a ring homomorphism. By definition we have
i−1(J) ={x ∈ R :
x
1∈ J
},
13
and therefore A−1(i−1(J)) ⊆ J (see Definition 5.6). Conversely if xa ∈ J then x
1 = a1xa ∈ J , so
x ∈ i−1(J). Thus xa ∈ A
−1(i−1(J)) hence J ⊆ A−1(i−1(J)), and therefore J = A−1(i−1(J)).We have shown that the maps are inverses to one another, so we must determine the image
of J 7→ i−1(J). We claim that I is in the image if and only if I = i−1(A−1I). Indeed, such anideal is certainly in the image of i−1, whereas if I = i−1(J) then A−1I = A−1(i−1(J)) = J , and soi−1(A−1I) = i−1(J) = I.
Now we always have I ⊆ i−1(A−1I), so I 6= i−1(A−1I) if and only if there is some x /∈ I suchthat x
1 ∈ A−1I. By Proposition 5.7(ii), this is equivalent to there being some x /∈ I and b ∈ A with
xb ∈ I. That is, there exists b ∈ A and x + I 6= I = 0R/I in R/I with (b + I)(x + I) = I = 0R/I ,i.e. some element of A is a zero divisor in R/I.
For the second part, observe first that if P C R is prime then R/P is an integral domain(Theorem 3.8), so A contains a zero divisor in R/P if and only if A ∩ P 6= ∅. It is thereforeenough to show that prime ideals always map to prime ideals. Recall from Proposition 3.3 that ifQCA−1R is prime, then i−1(Q)CR is prime. On the other hand if P CR is prime and P ∩A = ∅,then R/P is an integral domain and A ⊆ R/P does not contain 0R/P , so by Proposition 5.7(iv)we have
A−1R/A−1P ∼= A−1
(R/P ) ⊆ Frac(R/P ).
Since Frac(R/P ) is a field, it contains no non-zero zero divisors. Therefore as a subring neitherdoes A−1R/A−1P , i.e. it is an integral domain, and so A−1P CA−1R is a prime ideal.
The following corollary then gives an insight into the name “localisation”.
Corollary 5.9. Let P C R be a prime ideal. Then the prime ideals of RP are in bijection withthe prime ideals of R contained in P . In particular RP has a unique maximal ideal PP , and hence(RP , PP ) is a local ring.
Proof. By Theorem 5.8, the prime ideals of RP are in bijection with the prime ideals P ′ of R thatdo not intersect R\P . But this is precisely the condition that P ′ ⊆ P .
The maximality and uniqueness of PP follows from the fact that the bijection is inclusionpreserving. In particular if Q1 ⊆ Q2 are ideals of RP then i−1(Q1) ⊆ i−1(Q2), and if P1 ⊆ P2 areideals of R then (P1)P ⊆ (P2)P . The largest prime ideal of R contained in P is P itself, and thisis the unique ideal with this property, therefore PP is the unique maximal ideal of RP .
6 The radical, nilradical and Jacobson radical
Definition 6.1. Let R be a ring. An element r ∈ R is nilpotent if there exists an integer n > 1such that rn = 0.
Example 6.2. (i) In the ring M2(R) of 2× 2 real matrices,(0 10 0
)2
=(
0 00 0
).
(ii) Let K be a field, ∅ 6= X be a set and FX = KX = {functions f : X → K}. Then FX has nonon-zero nilpotent elements. Indeed, suppose some f ∈ F is nilpotent. If f 6= 0, then there issome x ∈ X such that f(x) 6= 0. Now let m > 1 be the smallest integer such that f(x)m = 0.Then 0 = f(x)f(x)m−1 and f(x), f(x)m−1 6= 0. Hence K has non-trivial zero divisors, whichis a contradiction to K being a field.
Definition 6.3. The nilradical of a ring R, denoted nil(R), is the set of all nilpotent elements ofR.
Theorem 6.4. Let R be a ring. Then nil(R) is an ideal of R, and moreover is the intersection ofall prime ideals of R.
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Proof. If r, s ∈ nil(R) then there exist n,m ∈ N such that rn = sm = 0. By the binomial theoremwe have
(r + s)n+m =n+m∑i=0
(n+m
i
)risn+m−i,
and for all 0 6 i 6 n+m we have either i > n or n+m− i > m, so either ri = 0 or sn+m−i = 0.Hence (r + s)n+m = 0 and r + s ∈ nil(R). Now for t ∈ R, (tr)n = tnrn = 0. Finally 0 ∈ nil(R) sonil(R) 6= ∅, and nil(R) is an ideal of R.
We now show that nil(R) ⊆ P for all prime ideals P , therefore giving containment one way.Indeed, let P be a prime ideal. Then for any r ∈ nil(R) there exists some n ∈ N such thatrn = 0 ∈ P , but since P is prime we must then have r ∈ P .
Finally, we show that the intersection of all prime ideals is contained in the nilradical. In fact,we will prove the contrapositive. Suppose r is not nilpotent. Then 0 /∈ {ri : i > 1} and the set
S = {I CR : I is an ideal and ri /∈ I for all i > 1}
is non-empty as {0} ∈ S. We turn S into a poset by inclusion, and then any totally ordered subsetof S has an upper bound, namely the union of all its elements (cf. proof of Proposition 4.6). ByZorn’s Lemma, there is a maximal element J ∈ S. That J is an ideal is immediate, so we now provethat it is prime. Suppose ab ∈ J but a /∈ J and b /∈ J . Then 〈a〉+J and 〈b〉+J are strictly greaterthan J , so rm ∈ 〈a〉+ J and rn ∈ 〈b〉+ J for some m,n ∈ N. Thus rn+m ∈ (〈a〉+ J)(〈b〉+ J) ⊆ J ,contradicting the choice of J . Therefore J is a prime ideal and moreover r /∈ J (set i = 1 in theabove), so r /∈
⋂P prime
P .
Definition 6.5. Let I be an ideal of a ring R. Define the radical of I, denoted rad(I), by
rad(I) = {r ∈ R : rn ∈ I for some n > 1}.
Theorem 6.6. Let I be an ideal of a ring R. Then rad(I) is an ideal of R, and moreover is theintersection of all prime ideals in R which contain I.
Proof. Consider the quotient homomorphism ϕ : R → R/I. Then r ∈ rad(I) iff ϕ(r) ∈ nil(R/I),thus rad(I) = ϕ−1(nil(R/I)) and hence is an ideal.
For the second statement we see that
rad(I) = ϕ−1(nil(R/I))
= ϕ−1
⋂PCR/I prime
P
=
⋂PCR/I prime
ϕ−1(P )
=⋂
PCR primeI⊆P
P,
where we have again used Proposition 2.7 in the last step.
Example 6.7. (i) Working in Z, we have rad(4Z) = 2Z and rad(3Z) = 3Z.
(ii) Again in Z,rad(12Z) =
⋂Pprime12Z⊆P
P.
The prime ideals in Z are pZ, and those containing 12Z are 2Z and 3Z. Hence rad(12Z) =2Z ∩ 3Z = 6Z.
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(iii) Let I = 〈x + y, y2〉 C R[x, y]. Then y ∈ rad(I), and x2 = y2 + (x − y)(x + y) ∈ I so alsox ∈ rad(I). Then rad(I) = 〈x, y〉.
Definition 6.8. Let R be a ring. The Jacobson radical, denoted J(R), is defined to be the set
J(R) =⋂
mCR maximal
m.
Remark. Note that in a local ring (R,m) (see Definition 4.8), the Jacobson radical is equal to themaximal ideal, i.e. J(R) = m.
Lemma 6.9. Let R be a ring and x ∈ R. Then x ∈ J(R) if and only if 1 + rx is invertible for allr ∈ R.
Proof. See Exercise Sheet 1.
7 Modules
Definition 7.1. Let R be a ring. An abelian group M = (M,+) (with identity 0) is an R-module(or just a module if it is clear from context) if there exists a multiplication map · : R ×M → M ,(r,m) 7→ rm such that for all r, s ∈ R and m,n ∈M :
(i) r(sm) = (rs)m;
(ii) r(m+ n) = rm+ rn;
(iii) (r + s)m = rm+ sm;
(iv) 1Rm = m.
Example 7.2. (i) If R is a field then an R-module is simply a vector space. The axioms for amodule are the same as a vector space except R is not necessarily a field.
(ii) Ideals in a ring R are also R-modules.
(iii) For a ring R, the set Rn of n-tuples of elements of R is an R-module.
(iv) R[X] is an R-module.
(v) R is a module over itself.
(vi) Any abelian group is a Z-module (and vice versa!).
(vii) If S ⊆ R is a subring then R is an S-module.
Modules therefore generalise the idea of vector spaces to rings.
Definition 7.3. A map ϕ : M → N between R-modules M and N is an R-module homomorphism(or R-homomorphism) if ϕ is an R-linear map, i.e. ϕ(rm + sn) = rϕ(m) + sϕ(n) for all r, s ∈ Rand m,n ∈ M . An R-module isomorphism is a bijective R-homomorphism. The set of all R-homomorphisms from M to N is denoted HomR(M,N).
Proposition 7.4. The set HomR(M,N) is an R-module, via the action (rϕ)(m) = rϕ(m) for allr ∈ R, ϕ ∈ HomR(M,N) and m ∈M .
Proof. Exercise.
If R is a field, then R-module homomorphisms are simple linear maps between vector spaces.
Definition 7.5. A submodule U of an R-module M is a subgroup (U,+) of (M,+), closed underthe restricted action of the multiplication, i.e. ru ∈ U for all r ∈ R and u ∈ U .
16
Note that the inclusion map U ↪→M is an R-module homomorphism.
Example 7.6. (i) Let I CR be an ideal and M an R-module. Then
IM =
{n∑i=1
aimi : n > 1, ai ∈ I, mi ∈M
}is a submodule of M .
(ii) If U, V ⊆M are submodules, then U ∩ V is a submodule of U , V and M .
The factor group M/U is also an R-module, via the action r(m+U) = (rm)+U . The quotientmap ϕ : M →M/U is an R-homomorphism, and this allows us to talk about I/J for ideals I andJ of a ring R.
Example 7.7. The quotient group Z/6Z is a Z-module. Note that 2(3 + 6Z) = 6 + 6Z = 0 inZ/6Z, hence multiplication of non-zero elements of a module by non-zero scalars may result inzero. This is in contrast to the situation in vector spaces.
For a general R-homomorphism ϕ : M → N , we can define Kerϕ and Imϕ in the usual way,and these are submodules of M and N respectively.
Definition 7.8. The cokernel of an R-homomorphism ϕ : M → N is the set
Cokerϕ = N/Imϕ.
Let U, V be submodules of an R-module M . Then the set
U + V = {u+ v : u ∈ U, v ∈ V }is also a submodule of M . This is used in the following theorem.
Theorem 7.9 (Isomorphism theorems). Let R be a ring and M,N be R-modules. We have thefollowing:
(i) if ϕ : M → N is an R-module homomorphism then
M/Kerϕ ∼= Imϕ;
(ii) if L ⊆M ⊆ N are submodules then
(N/L)/(M/L) ∼= N/M,
via the map (m+ L) +M/L 7→ m+M ;
(iii) if N is a module and L,M are submodules then
M/(M ∩ L) ∼= (M + L)/L,
via the map m+M ∩ L 7→ m+ L.
These isomorphisms are canonical (i.e. require no choices in their definition).
Proof. Exercise Sheet 2.
Definition 7.10. Let R be a ring and M an R-module. Let Γ be a subset of M . The submoduleof M generated by Γ, denoted 〈Γ〉 or
∑g∈ΓRg, is the set
〈Γ〉 =
{n∑i=1
rigi : n > 1, ri ∈ R, gi ∈ Γ
}.
The module M is finitely generated if there exists a finite set Γ ⊆M such that 〈Γ〉 = M .
Example 7.11. (i) Let R be a ring and I C R an ideal, then the R-module R/I is finitelygenerated. In fact it is cyclic, i.e. generated by one element, namely 1 + I.
(ii) If R is an integral domain and 0 6= f ∈ R, then
R[ 1f ] = R+R 1
f +R 1f2 + . . .
is usually not finitely generated as an R-module.
17
8 Nakayama’s Lemma
Definition 8.1. A minimal generating set for an R-module M is a subset Γ ⊆ M such that Γgenerates M but no proper subset of Γ generates M .
Example 8.2. Consider Z6 = Z/6Z, then {1+6Z} and {2+6Z, 3+6Z} are both minimal generatingsets. Contrast this with vector spaces, where the number of elements in any two minimal generatingsets of a given vector space are equal.
Theorem 8.3 (Nakayama’s Lemma). Let M be a finitely generated R-module, and I ⊆ J(R) anideal of R. If M = IM , then M = 0.
Proof. Suppose M 6= 0. Since M is finitely generated there exists a finite minimal generating setΓ = {g1, . . . , gn} say. Now M = IM =⇒ g1 ∈ IM , so there exists a1, . . . , an ∈ I such that
g1 =n∑i=1
aigi
and so
(1− a1)g1 =n∑i=2
aigi.
But a1 ∈ I ⊆ J(R), so by Lemma 6.9, 1− a1 is a unit of R. Thus
g1 = (1− a1)−1n∑i=2
aigi
and {g2, . . . , gn} is a generating set for M strictly smaller than Γ, a contradiction.
Corollary 8.4. Let M be a finitely generated R-module and N ⊆ M a submodule. Let alsoI C J(R) be an ideal of R. Then M = N + IM =⇒ M = N .
Proof. Take the equality M = N + IM and quotient both sides by the submodule N to obtainM/N = (N + IM)/N . By Theorem 7.9, we have (N + IM)/N ∼= IM/(N ∩ IM). Now the map
IM → I(M/N)n∑i=1
aimi 7→n∑i=1
ai(mi +N)
is a surjective R-module homomorphism, and its kernel is (IM) ∩N . Therefore
I(M/N) ∼= IM/(IM ∩N) ∼= (N + IM)/N.
Therefore we have M/N = I(M/N). Since M is finitely generated so too is M/N , and hence byNakayama’s Lemma we have M/N = 0, i.e. M = N .
Example 8.5. Consider K[x, y] for some field K and let m = 〈x, y〉. Let R = K[x, y]m, thelocalisation at the ideal m. Then R is a local ring, with maximal ideal mm. We will show that theideal
I = 〈x+ x2y + 3y2 + x4, y + 2y3 + y4 + 4x7〉m CR
is equal to mm. Note first that since R is local it has a unique maximal ideal, hence J(R) = mm.Now
I + mmmm = 〈x+ x2y + 3y2 + x4, y + 2y3 + y4 + 4x7, x2, xy, y2〉m= 〈x, y, x2, xy, y2〉m= 〈x, y〉m= mm.
So by Nakayama’s Lemma, I = mm.
18
Recall from earlier that we had an issue with minimal generating sets for modules, in that thenumber of elements in such a set is not well defined. Nakayama’s Lemma allows us to fix this incertain cases.
Theorem 8.6. Let (R,m) be a local ring and M a finitely generated R-module. If Γ ⊆M is a setof elements whose images in M/mM form a basis of M/mM as an R/m-vector space, then Γ is aminimal generating set of M as an R-module.
Proof. As M/mM is generated by the images of the elements of Γ, we have M = 〈Γ〉+mM . So byCorollary 8.4 to Nakayama’s Lemma, we have M = 〈Γ〉. If Γ′ ( Γ, then 〈Γ′〉+mM 6= 〈Γ〉+mM =M , and so Γ′ is not a generating set.
9 Exact sequences
Definition 9.1. A sequence of R-modules and R-module homomorphisms
· · · −→M0f1−→M1
f2−→M2 −→ · · ·fn−→Mn −→ · · ·
is called exact at Mi if Ker fi+1 = Im fi. A sequence which is exact at Mi for all i is called anexact sequence.
Example 9.2. (i) The sequence 0 −→ Lf−→M is exact if and only if f is injective.
(ii) The sequence Mg−→ N −→ 0 is exact if and only if g is surjective.
(iii) The sequence 0 −→Mg−→ N −→ 0 is exact if and only if g is an isomorphism.
Definition 9.3. A short exact sequence is an exact sequence of the form
0 −→ Lf−→M
g−→ N −→ 0.
Remark. This is equivalent to insisting that f is injective, g is surjective and Ker g = Im f .
Short exact sequences appear in many different sub-branches of algebra, and are very powerfulobjects.
Example 9.4. (i) Let R be a ring, M an R-module and N ⊆M a submodule. Then
0 −→ Ni−→M
π−→M/N −→ 0,
where i is the natural inclusion map and π is the canonical quotient map, is a short exactsequence.
(ii) Let K be a field and
0 −→ Lf−→M
g−→ N −→ 0
be a short exact sequence of K-modules. Then each module is a K-vector space, and usingfacts from linear algebra we have
dimKM = dimK Ker g + dimK Im g
= dimK Im f + dimK N
= dimK L+ dimK N.
More generally, if0 −→M0
f1−→M1f2−→M2 −→ · · ·
fn−→Mn −→ 0
is an exact sequence of K-vector spaces, then∑ni=0(−1)i dimKMi = 0.
19
Theorem 9.5 (Snake Lemma). Suppose the following commutative diagram of R-modules andR-module homomorphisms
L M N 0
0 L′ M ′ N ′
f
α
g
β γ
f ′ g′
has exact rows. Then there exists a homomorphism δ : Ker γ → Cokerα such that
Kerα −→ Kerβ −→ Ker γ δ−→ Cokerα −→ Cokerβ −→ Coker γ
is exact.Furthermore, if f is injective then so too is Kerα → Kerβ, and if g′ is surjective then so too
is Cokerβ → Coker γ.
The name of this theorem comes from the following diagram:
Kerα Kerβ Ker γ
L M N 0
0 L′ M ′ N ′
Cokerα Cokerβ Coker γ
δ
f
α
g
β γ
f ′ g′
Proof. We will first define all of the necessary maps, then prove exactness at each site.The map f |Kerα : Kerα→ Kerβ is given by the restriction of f to Kerα. Note that if ` ∈ Kerα
then β(f(`)) = f ′(α(`)) = 0 by the commutativity of the diagram. Therefore f(Kerα) ⊆ Kerβ.That this is a R-homomorphism follows from the fact that f itself is. Similarly the map g|Ker β :Kerβ → Ker γ is given by the restriction of g to Kerβ.
The map f : Cokerα → Cokerβ is induced from f ′, by setting f(`′ + Imα) = f ′(`′) + Imβ.This is well defined, as if `′1 + Imα = `′2 + Imα then `′1 − `′2 ∈ Imα, so `′1 − `′2 = α(`) for some` ∈ L. Then
f ′(`′1)− f ′(`′2) = f ′(`′1 − `′2)= f ′(α(`))= β(f(`))∈ Imβ,
so f ′(`′1) + Imβ = f ′(`′2) + Imβ. That f is a homomorphism follows from the fact that f ′ is. Wesimilarly define g : Cokerβ → Coker γ.
We now construct the connecting homomorphism δ : Ker γ → Cokerα by a process known as“diagram chasing”. Take n ∈ Ker γ ⊆ N . Since g is surjective, there exists some m ∈M such that
20
n = g(m). Then
0 = γ(n)= γ(g(m))= g′(β(m))
by the commutativity of the diagram, so β(m) ∈ Ker g′. By the exactness of rows, Ker g′ = Im f ′,so β(m) = f ′(`′) for some `′ ∈ L′. We then define
δ(n) = `′ + Imα ∈ Cokerα.
We must show that this is well defined. Since f ′ is injective, the only ambiguity in our processlies in our choice of m. Suppose then that g(m1) = g(m2) = n, and `′1, `
′2 ∈ L′ are the unique
elements such that β(m1) = f ′(`′1) and β(m2) = f ′(`′2). We must show that `′1 − `′2 ∈ Imα. Notethen that m1−m2 ∈ Ker g, and so by exactness of rows is equal to f(`) for some ` ∈ L. Thereforeβ(m1 −m2) = β(f(`)) = f ′(α(`)). By the injectivity of f ′, we then see that α(`) = `′1 − `′2. Thatδ is a homomorphism is left as an easy exercise.
We now prove exactness at each site.The composition g|Ker β ◦ f |Kerα = 0 follows from the fact that Im f = Ker g, therefore
Im f |Kerα ⊆ Ker g|Ker β . Suppose now that m ∈ Kerβ with g|Ker β(m) = 0. Then g(m) = 0so m ∈ Ker g = Im f , say m = f(`), and it remains to show that ` ∈ Kerα. But
f ′(α(`)) = β(f(`))= β(m)= 0
as m ∈ Kerβ, and since f ′ is injective we must have α(`) = 0.For exactness at Ker γ, we first calculate δ(g|Ker β(m)) for m ∈ Kerβ. Following our construc-
tion of δ above, we have gKer β(m) = g(m), and so `′ is chosen so that β(m) = f ′(`′). But β(m) = 0,so by the injectivity of f ′ we also have δ(g|Ker β(m)) = 0 and hence Im g|Ker β ⊆ Ker δ. Converselyif n ∈ Ker γ is such that δ(n) = 0, then the corresponding `′ is in Imα, say `′ = α(`). Therefore ifm is such that n = g(m), we have β(m) = f ′(α(`′)) = β(f(`)), and hence m− f(`) ∈ Kerβ. Theng|Ker β(m− f(`)) = g(m)− g(f(`)) = n.
For exactness at Cokerα, note that f(δ(n)) = f ′(`′) + Imβ = β(m) + Imβ = 0 in Cokerβ.Therefore Im δ ⊆ Ker f . Conversely if l′ + Imα ∈ Cokerα is such that f(l′ + Imα) = 0, thenf ′(`′) ∈ Imβ, say f ′(`′) = β(m). But then δ(g(m)) = `′ + Imα.
Finally, for exactness at Cokerβ we see first that g(f(`′+Imα)) = g(f ′(`′)+Imβ) = g′(f ′(`′))+Im γ = 0 since g′ ◦ f ′ = 0. Therefore Im f ⊆ Ker g. Conversely, if m′ + Imβ ∈ Cokerβ is suchthat g(m′ + Imβ) = 0, then g′(m′) ∈ Im γ, say g′(m′) = γ(n). Since g is surjective, there issome m ∈ M such that g(m) = n, so g′(m′) = γ(g(m)). Commutativity of the diagram thengives g′(m′) = g′(β(m)), so m′ − β(m) ∈ Ker g′ = Im f ′, say m′ − β(m) = f ′(`′). But nowf(`′ + Imα) = f ′(`′) + Imβ = m′ − β(m) + Imβ = m′ + Imβ.
We leave the last statement as an exercise.
Example 9.6. We reprove part (ii) of Theorem 7.9. Let L ⊆M ⊆ N be a sequence of submodulesand consider the following diagram:
0 M N N/M 0
0 M/L N/L (N/L)/(M/L) 0
f
α
g
β γ
f ′ g′
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The maps f, g and f ′, g′ are pairs of inclusion and quotient maps, so the rows are short exactsequences. We have α : M → M/L and β : N → N/L also quotient homomorphisms, and for allm ∈M
β(f(m)) = β(m)= m+ L
= f ′(m+ L) since m ∈M= f ′(α(m)),
so the first square commutes. Now define γ : N/M → (N/L)/(M/L) by γ(n+M) = (n+L)+M/L.This is well defined since if n+M = n′ +M then n− n′ ∈M so
γ(n)− γ(n′) = ((n+ L) +M/L)− ((n′ + L) +M/L)= (n− n′ + L) +M/L
= M/L = 0(N/L)/(M/L) since n− n′ ∈M.
It is also a homomorphism (easy check since it is the composition of two quotient maps). Finallywe check that the diagram commutes: for all n ∈ N we have
γ(g(n)) = γ(n+M)= (n+ L) +M/L, and
g′(β(n)) = g′(n+ L)= (n+ L) +M/L.
By the Snake Lemma, we therefore have an exact sequence
0→ Kerα→ Kerβ → Ker γ → Cokerα→ Cokerβ → Coker γ → 0.
Clearly Kerα = Kerβ = L and Cokerα = Cokerβ = 0. Therefore our exact sequence is equal to
0→ L→ L→ Ker γ → 0→ 0→ Coker γ → 0.
By exactness we immediately see that Ker γ = Coker γ = 0. Thus γ is both injective and surjective,so is an isomorphism between N/M and (N/L)/(M/L).
10 Free modules
Let R be a ring, Λ a set and Mλ an R-module for each λ ∈ Λ.
Definition 10.1. The direct product of {Mλ}λ∈Λ, denoted∏λ∈Λ
Mλ, consists of all sequences
(mλ)λ∈Λ with mλ ∈Mλ for each λ ∈ Λ. This is a module, with addition
(mλ)λ∈Λ + (nλ)λ∈Λ = (mλ + nλ)λ∈Λ
and for any r ∈ R,r(mλ)λ∈Λ = (rmλ)λ∈Λ.
The direct sum of {Mλ}λ∈Λ, denoted⊕λ∈Λ
Mλ, consists of all sequences (mλ)λ∈Λ with mλ ∈Mλ for
each λ ∈ Λ, and all but finitely many of the mλ are zero. This is again a module, with additionand scalar multiplication as before.
Note that if Λ is finite then∏λ∈ΛMλ =
⊕λ∈ΛMλ. For instance, R⊕ R ∼= R2.
Proposition 10.2. If U, V are submodules of M , then M = U ⊕ V ⇐⇒ M = U + V andU ∩ V = {0}.
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Proof. Exercise.
Remark. Care needs to be taken when dealing with direct products. For instance, for rings Rand S their direct product R × S has identity (1, 1). Then the natural map ϕ : R→ R × S givenby ϕ(r) = (r, 0) is not a ring homomorphism, since ϕ(1) = (1, 0) 6= (1, 1).
Definition 10.3. An R-module is called free if it is isomorphic to⊕
λ∈ΛR for some set Λ. Weadopt the convention the the zero module is free, with index set Λ = ∅.
Example 10.4. (i) Rn = R⊕R⊕ · · · ⊕R is clearly free.
(ii) The ring of m× n matrices over a ring R is free and isomorphic to Rmn.
(iii) The polynomial ring R[X] is free, as R[X] ∼= R⊕RX ⊕RX2 ⊕ . . . .
Recall that in contrast to vector spaces, not every module has a basis. However free modulesdo.
Proposition 10.5. An R-module is free iff there exists a set of generators {mλ}λ∈Λ of M suchthat whenever r1mλ1 + . . . rnmλn = 0 with ri ∈ R and λi ∈ Λ for all i, we have r1 = · · · = rn = 0.
Proof. The “only if” direction is clear.Conversely, assume we have a set of generators as above and define a map
ϕ :⊕λ∈Λ
→M
(rλ)λ∈Λ 7→∑λ∈Λ
rλmλ.
It is then straightforward to check that this is an isomorphism of R-modules.
Definition 10.6. A set of generators as in Proposition 10.5 is called a free basis, or just a basis.The rank of a free module is the cardinality of Λ, equivalently the number of basis elements.
Example 10.7. (i) 1, X,X2, . . . is a basis of R[X].
(ii) The rank of Rn is n.
(iii) A K-vector space has a basis and so is a free K-module.
(iv) Consider the maximal ideal m = 〈x, y〉 of R = K[x, y]. This is generated by two elementsbut is not free, for instance as −yx+ xy = 0 is a non-trivial dependence relation.
(v) Z2 is not free as a Z-module, since it is generated by 1 + 2Z but 2(1 + 2Z) = 2 + 2Z = 0Z2 ,so this is a non-trivial dependence relation.
Proposition 10.8. Let R be a ring and M an R-module. Then there exists a free module F anda surjective homomorphism of R modules ϕ : F →M . Furthermore if M is finitely generated thenF can be chosen to have finite rank.
Proof. Any R-module can be written as 〈Γ〉 for some Γ ⊆M , for instance by setting Γ = M . Thenlet F be the free module with basis Γ. Now define
ϕ : F →M
(rg)g∈Γ 7→∑g∈Γ
rgg.
Note that this sum is finite since F is a direct sum of copies of R. It is an easy exercise to see thatthis is a surjective R-module homomorphism.
If M is finitely generated, say by {mg}g∈Γ then we similarly define F to be the free modulewith finite basis Γ, and ϕ : F → M by ϕ((rg)g∈Γ) =
∑g∈Γ rgmg. It is again easy to check that
this is a surjective homomorphism.
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Example 10.9. Let M1, . . . ,Mn be R-modules. Then the sequence
0 −→M1 −→M1 ⊕ · · · ⊕Mn −→M2 ⊕M3 ⊕ · · · ⊕Mn −→ 0
is exact.
11 Noetherian rings and modules
Being finitely generated is obviously a good property for a module to have. But if M is a finitelygenerated R-module then there is no guarantee that its submodules will be.
Example 11.1. Let R = K[x1, x2, x3, . . . ]. Then R is an R-module and is finitely generated by{1}. However the submodule 〈x1, x2, x3 . . .〉 is not.
This motivates the following:
Definition 11.2. A module M is called a Noetherian module if every submodule of M is finitelygenerated. A ring R is called a Noetherian ring if it is a Noetherian module over itself (i.e. allideals are finitely generated).
Examples are hard to give without a bit of extra theory, so we present this first.
Theorem 11.3. Let M be an R-module. Then the following are equivalent:
(i) all submodules of M are finitely generated;
(ii) M satisfies the ascending chain condition (ACC), i.e. every chain of submodules
M1 ⊆M2 ⊆M3 ⊆ . . .
of M is stationary, that is there exists some N with Mn = MN for all n > N ;
(iii) every non-empty set of submodules of M has a maximal element.
Proof. (i) =⇒ (ii) : The union⋃iMi is a submodule of M , so is finitely generated by assumption.
Each of these generators must lie in some Mj , and taking N to be the maximum of these j wehave
⋃iMi = MN . Hence Mn = MN for all n > N .
(ii) =⇒ (iii) : Let S be a non-empty set of submodules of M and suppose S has no maximalelement. Since S is non-empty we can take some M1 ∈ S. Since M1 is not maximal we canfind some M2 ∈ S with M1 ( M2. Repeating this argument we can construct inductively anon-stationary ascending chain of submodules of M , contradicting (ii).
(iii) =⇒ (i) : Let U be a submodule of M and S the set of finitely generated submodules ofU . This is non-empty as it contains the zero module, so has a maximal element U ′ = 〈u1, . . . , un〉.Now take any v ∈ U , then U ′ + 〈v〉 = 〈u1, . . . , un, v〉 is a finitely generated submodule of U , so bymaximality must equal U ′. Hence U = U ′ is finitely generated.
We can now give some examples of Noetherian rings and modules.
Example 11.4. (i) Let R be a field, then the only ideals of R are R and {0} which are finitelygenerated. Therefore R is a Noetherian ring.
(ii) Modules and rings with a finite number of elements are Noetherian.
(iii) Any principal ideal domain is a Noetherian ring. Therefore Z, Z[i] and K[x] (K a field) areNoetherian rings (as they are Euclidean domains).
(iv) Finite dimensional K-vector spaces are Noetherian K-modules, since any subspace (submod-ule) has a finite basis.
Theorem 11.5. Let 0 −→ L −→M −→ N −→ 0 be an exact sequence of R-modules. Then M isNoetherian iff both L and N are Noetherian.
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Proof. Note that the property of being Noetherian is preserved by isomorphisms, thus it is sufficientto prove the theorem in the case L ⊆M and N = M/L.
Suppose first that M is Noetherian and let L′ be a submodule of L. Then L′ is a submoduleof M so is finitely generated, and hence L is Noetherian. Next, any submodule N ′ of M/L is ofthe form M ′/L for some submodule M ′ of M . Therefore M ′ is finitely generated, and reductionof these generators modulo L shows that N ′ is also finitely generated.
Conversely suppose that both L and N are Noetherian and consider a submodule M ′ ⊆ M .Then the submodules M ′∩L ⊆ L and M ′/L ⊆ N are both finitely generated, say by x1, . . . , xn andy1+L, . . . , ym+L respectively. Now for anym ∈M ′ we havem+L = (b1y1+· · ·+bmym)+L for somebi ∈ R, thus m−(b1y1+· · ·+bmym) ∈ L. But also m, y1, . . . , ym ∈M ′, so m−(b1y1+· · ·+bmym) =a1x1 + · · ·+ anxn for some ai ∈ R. Hence m = a1x1 + · · ·+ anxn + b1y1 + · · ·+ bmym, and so M ′
is finitely generated. Therefore M is Noetherian.
Proposition 11.6. Let R be a Noetherian ring and M an R-module. Then M is Noetherian iffM is finitely generated.
Proof. The “only if” direction is by definition.Suppose M is finitely generated, then there is a surjection ϕ : Rn → M for some n > 0. The
sequence 0 −→ Kerϕ −→ Rn −→ M −→ 0 is then exact, and since Rn is Noetherian then so toois M by Theorem 11.5.
Proposition 11.7. Let R be a Noetherian ring.
(i) Let I CR be an ideal. Then R/I is a Noetherian ring.
(ii) Let A ⊆ R be a multiplicatively closed subset. Then A−1R is a Noetherian ring.
Proof. (i) Let J be an ideal or R/I. Its preimage under the canonical quotient map is finitelygenerated, therefore so too is J .
(ii) Similarly for an ideal J of A−1R, its preimage under the natural map R→ A−1R is finitelygenerated. Therefore so too is J .
12 Primary decomposition
Definition 12.1. We call an ideal I C R irreducible if it cannot be written as I1 ∩ I2, where I1and I2 are proper ideals of R which strictly contain I.
Example 12.2. (i) 〈x2 + 1〉C R[x] is irreducible.
(ii) 〈(y − x2)(y2 − x3)〉 = 〈y − x2〉 ∩ 〈y2 − x3〉CR[x, y] is reducible.
Proposition 12.3. Every proper ideal of a Noetherian ring R is the intersection of finitely manyirreducible ideals.
Proof. Let S be the set of all ideals which are not the intersection of finitely many irreducible ideals.If S 6= ∅ then by Theorem 11.3(iii) it has a maximal element, J say. Now J is not irreducible, soJ = J1 ∩ J2 for some ideals J1, J2 ) J . By the maximality of J , it must be possible to write J1
and J2 as the intersection of finitely many irreducible ideals, and therefore we can also write J assuch. This is a contradiction, so S = ∅ and the result follows.
Definition 12.4. A proper ideal I CR is called primary if xy ∈ I =⇒ either x ∈ I or yn ∈ I forsome n > 1.
Remark 12.5. (i) A prime ideal is a generalisation of a prime number. In turn, a primary idealis a generalisation of the power of a prime number. This will allow us to talk about “uniquefactorisation” of ideals in much the same way we do for integers or polynomials say.
25
(ii) I is primary if and only if R/I 6= {0} and every zero divisor in R/I is nilpotent.
Example 12.6. If I is prime, then I is primary.
Proposition 12.7. Let I CR be a primary ideal, then rad(I) is a prime ideal.
Proof. Suppose xy ∈ rad(I), then xnyn ∈ I for some n > 1. Since I is primary we have eitherxn ∈ I or (yn)m ∈ I for some m > 1, and hence either x ∈ rad(I) or y ∈ rad(I).
Example 12.8. (i) {0} and 〈pn〉 for p a prime, n > 1 are the primary ideals in Z. These arethe only ideals with prime radical, and it is then clear that they are primary.
(ii) Let R = K[x, y] for some field K and I = 〈x, y2〉. Then R/I ∼= K[y]/〈y2〉 and all zero divisorsin this ring are nilpotent, hence by Remark 12.5(ii) I is primary. Note that 〈x, y〉2 ( I (〈x, y〉 = rad(I), so a primary ideal is not necessarily a power of a prime ideal.
(iii) If I is prime then In is not necessarily primary, even when I = rad(I). For instance, letR = K[x, y, z]/J where K is a field and J = 〈xy − z2〉. Then I = 〈x + J, z + J〉 is primeas R/I ∼= K[y] which is an integral domain. However (x + J)(y + J) = z2 + J ∈ I, butx+ J /∈ I2 and y + J /∈ rad(I2) = I, so I2 is not primary.
Proposition 12.9. If rad(I) = m is maximal, then I is primary.
Proof. Exercise.
For the next proposition we need to recall the quotient ideal
(I : J) = {r ∈ R : rJ ⊆ I}
for ideals I, J C R from Proposition 2.4. It is an easy exercise to show that (I : J1 + J2) = (I :J1) ∩ (I : J2) and (I1 ∩ I2 : J) = (I1 : J) ∩ (I2 : J), which allows us to prove:
Proposition 12.10. Irreducible ideals in Noetherian rings are primary.
Proof. Let R be Noetherian. We first show that if the zero ideal is irreducible then it is primary.Let xy = 0 with y 6= 0 and consider the chain
(0 : 〈x〉) ⊆ (0 : 〈x〉) ⊆ (0 : 〈x〉) ⊆ . . . .
By ACC this is stationary, i.e. (0 : 〈xn〉) = (0 : 〈xn+1〉) = . . . for some n > 1. It follows that〈xn〉 ∩ 〈y〉 = {0}, for if a ∈ 〈y〉 then ax = 0 so if also a ∈ 〈xn〉 then a = bxn and ax = bxn+1 = 0.Hence b ∈ (0 : 〈xn+1〉) = (0 : 〈xn〉), so bxn = a = 0. Since {0} is irreducible and 〈y〉 6= 0 we musttherefore have xn = 0, i.e. {0} is primary.
Now let I C R be irreducible. Then R/I is Noetherian by Theorem 11.5 and the zero ideal{0 + I}CR/I is irreducible by Proposition 2.7. Therefore {0 + I} is primary, so for any x, y ∈ Rwe have xy ∈ I implies that (x+ I)(y + I) ∈ {0 + I}, thus either x+ I = 0 + I or yn + I = 0 + Ifor some n. But this is equivalent to having either x ∈ I or yn ∈ I, hence I is primary.
Corollary 12.11. Every proper ideal of a Noetherian ring can be written as an intersection offinitely many primary ideals.
Proof. Exercise, use Propositions 12.3 and 12.10.
Example 12.12. Let R = K[x, y], I = 〈xy, y2〉 for some field K (we will see later that R isNoetherian). Then
I = 〈x, y2〉 ∩ 〈y〉, the first component is primary by Example 12.8(ii),
= 〈x, y〉2 ∩ 〈y〉, and powers of maximal ideals are primary by Proposition 12.9.
Since 〈y〉 is prime, it is also primary. Hence we have written I as the intersection of finitely manyprime ideals.
26
Definition 12.13. A primary decomposition of an ideal I in a ring R is an expression of I as afinite intersection of primary ideals
I =n⋂i=1
Qi.
The decomposition is minimal if:
(i) rad(Qi) are distinct for all i;
(ii)⋂
16j6nj 6=i
Qj 6⊆ Qi for all 1 6 i 6 n.
Definition 12.14. Let P CR be a prime ideal. We say an ideal I CR is P -primary if rad(I) = Pand I is primary.
Theorem 12.15. Let Q1, . . . , Qn be P -primary ideals in R. Then Q1 ∩ · · · ∩Qn is P -primary.
Proof. As rad(Q1∩· · ·∩Qn) = rad(Q1)∩· · ·∩ rad(Qn) = P , we need only check that Q1∩· · ·∩Qnis primary. Assume x, y ∈ R are such that xy ∈ Q1 ∩ · · · ∩Qn. If x /∈ Q1 ∩ · · · ∩Qn then x /∈ Qjfor some 1 6 j 6 n. Now xy ∈ Qj and since Qj is primary we have ym ∈ Qj for some m > 1, i.e.y ∈ rad(Qj) = P = rad(Q1 ∩ · · · ∩Qn), and the result follows.
Corollary 12.11 tells us that primary decompositions always exist, and now Theorem 12.15allows us to reduce this to a minimal decomposition. We will soon prove a uniqueness statementfor these minimal decompositions, but first we need the following lemma.
Lemma 12.16. Let Q be a primary ideal in R. Then for any x ∈ R
rad((Q : 〈x〉)) =
{R if x ∈ Q,rad(Q) if x /∈ Q.
Proof. Exercise.
Theorem 12.17. Let R be a Noetherian ring, I CR an ideal and
I = Q1 ∩ · · · ∩Qn
a minimal primary decomposition of I. Then the set {rad(Q1), . . . , rad(Qn)} is equal to the set ofprime ideals of R of the form rad((I : 〈x〉)) for some x ∈ R. In particular for any two minimalprimary decompositions
I = Q1 ∩ · · · ∩Qn = Q′1 ∩ · · · ∩Q′mwe have n = m and (possibly after reordering) rad(Qi) = rad(Q′i) for all 1 6 i 6 n.
Proof. Suppose first that rad((I : 〈x〉)) is prime for some x ∈ R. Then we have
rad((I : 〈x〉)) = rad((Q1 ∩ · · · ∩Qn : 〈x〉))= rad((Q1 : 〈x〉)) ∩ · · · ∩ rad((Qn : 〈x〉)).
Recall from Theorem 3.9 that I1 ∩ · · · ∩ In ⊆ P ⇐⇒ Ij ⊆ P for some j, where Ii are ideals and Pis prime. It is an easy exercise to show that in the “only if” direction, the subsets can be replacedby equalities, and hence rad((I : 〈x〉)) = rad((Qj : 〈x〉)) for some j. Since rad((I : 〈x〉)) 6= R wemust have rad((I : 〈x〉)) = rad((Qj : 〈x〉)) = rad(Qj) by Lemma 12.16. Therefore the set of primeideals of the form rad((I : 〈x〉)) is a subset of {rad(Q1), . . . , rad(Qn)}.
Now consider rad(Qi). By minimality of the primary decomposition we can choose x ∈ Qj forall j 6= i but x /∈ Qi. But then we have
rad((I : 〈x〉)) = rad((Q1 ∩ · · · ∩Qn : 〈x〉))= rad((Q1 : 〈x〉)) ∩ · · · ∩ rad((Qn : 〈x〉))= rad(Qi).
27
Thus {rad(Q1), . . . , rad(Qn)} is a subset of the set of prime ideals of the form rad((I : 〈x〉)), andthe equality is established. The final statement follows immediately, since the set of primes of theform rad((I : 〈x〉)) is independent of any choice of primary decomposition.
Definition 12.18. For any ideal I of a Noetherian ring R, the associated primes of I is the set
Ass(I) = {rad(Qi) : 1 6 i 6 n, I = Q1 ∩ · · · ∩Qn is a minimal primary decomposition}.
Proposition 12.19. For any ideal I of a Noetherian ring R, the set
{x+ I : x ∈ P for some P ∈ Ass(I)}
is precisely the set of zero divisors of R/I.
Proof. Exercise.
Remark. An ideal is primary if and only if it has one associated prime.
Example 12.20. (i) R = Z, I = 〈12〉 = 〈3〉∩〈4〉. Then Q1 = 〈4〉, Q2 = 〈3〉 which have radicals〈2〉 and 〈3〉 respectively. Therefore Ass(〈12〉) = {〈2〉, 〈3〉}.
(ii) Suppose I = 〈f〉 C K[x1, . . . , xn], and f = fn11 . . . fnr
r is the factorisation into irreduciblesover K. Then I = 〈fn1
1 〉 ∩ · · · ∩ 〈fnrr 〉 is a minimal primary decomposition, with associated
primes {〈f1〉, . . . , 〈fr〉}.
(iii) Consider I = 〈x, y2〉∩ 〈y〉CK[x, y]. Then Q1 = 〈x, y2〉, Q2 = 〈y〉 have radicals 〈x, y〉 and 〈y〉respectively, so Ass(I) = {〈x, y〉, 〈y〉}. Note that 〈y〉 ⊆ 〈x, y〉, which motivates the followingdefinition.
Definition 12.21. Let I be an ideal of a Noetherian ring R. An associated prime P of I is calledminimal if there is no associated prime P ′ such that P ′ ( P , otherwise P is called embedded.
13 Hilbert’s Basis Theorem
Theorem 13.1. If R is Noetherian, then the polynomial ring R[x] is Noetherian.
Proof. Suppose there exists an ideal I C R[x] which is not finitely generated. Choose a sequencef1, f2, f3, . . . of polynomials in R[x] such that
f1 ∈ I,f2 ∈ I\〈f1〉,f3 ∈ I\〈f1, f2〉, . . .
of minimal possible degree. If di = deg(fi), say fi = aixdi+ lower terms, then d1 6 d2 6 d3 6 . . .
and〈a1〉 ⊆ 〈a1, a2〉 ⊆ 〈a1, a2, a3〉 ⊆ . . .
is an ascending chain of ideals in R. Since R is Noetherian this chain is stationary, i.e. thereis some N such that 〈a1, . . . , aN 〉 = 〈a1, . . . , aN+1〉. Hence aN+1 =
∑Ni=1 biai for some suitable
bi ∈ R. Now consider
g = fN+1 −N∑i=1
bixdN+1−difi
= aN+1xdN+1 −
(N∑i=1
biai
)xdN+1 + lower terms.
Since fN+1 ∈ I\〈f1, . . . , fN 〉, it follows that g ∈ I\〈f1, . . . , fN 〉 is a polynomial of degree smallerthan dN+1, a contradiction to the choice of fN+1.
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Corollary 13.2. If R is Noetherian, then R[x1, . . . , xn] is Noetherian. In particular, if K is afield then K[x1, . . . , xn] is Noetherian.
Proof. Exercise (easy induction).
Example 13.3. Note that R needs to be a ring in the above theorem, that is if I is an idealthen I[x1, . . . , xn] may not be Noetherian. For instance 2ZC Z is an ideal of Z (but not a ring as1 /∈ 2Z). Then 2Z has ACC since Z does, and so is a Noetherian Z-module. Now if 2Z[x] wereNoetherian then it would be finitely generated, say by f1, . . . , fn. Then let N be the maximumof the degrees of the fi. If g ∈ 〈f1, . . . , fn〉, then either deg(g) 6 N or every coefficient of g isdivisible by 4. Therefore 2xN+1 /∈ 〈f1, . . . , fn〉 but 2xN+1 ∈ 2Z[x], a contradiction.
14 Noether normalisation
Let R be a ring.
Definition 14.1. An R-algebra is a ring S with an R-homomorphism ϕ : R → S. We say S is afinite R-algebra if it is finitely generated as an R-module, i.e. there exist x1, . . . , xn ∈ S such that
S = Rx1 + · · ·+Rxn.
If also R is a field then we say S is a finite dimensional R-algebra.We say S is a finitely generated R-algebra if there exist x1, . . . , xn ∈ S such that S =
R[x1, . . . , xn].
Example 14.2. R[x] is an R-algebra via the natural inclusion map. It is not finite, but it isfinitely generated.
The homomorphism ϕ turns S into an R-module, where multiplication is defined by r·s = ϕ(r)sfor all r ∈ R, s ∈ S.
When R ⊆ S, we call S an extension ring of R. If in addition R and S are fields, then we callS an extension field of R.
Definition 14.3. Let S be an R-algebra. An element s ∈ S is integral over R if there exists amonic polynomial
f(x) = xn + an−1xn−1 + an−2x
n−2 + · · ·+ a0 ∈ R[x]
such that f(s) = 0.We say S is integral over R if every s ∈ S is integral over R. If also R ⊆ S, then we call S an
integral extension.
Example 14.4. (i) The integral elements of Q over Z are the integers.
(ii) K[x2] ⊆ K[x] is an integral extension.
(iii) Z ⊆ Z[α] is integral for α = 12 (1+
√5) as α2−α−1 = 0. It is not integral for α = 1
2 (1+√
3).
Proposition 14.5. (i) Let R ⊆ S ⊆ T be rings. If S is a finite R-algebra and T is a finiteS-algebra, then T is a finite R-algebra.
(ii) If R ⊆ S is a finite R-algebra and t ∈ S, then t satisfies a monic polynomial over R.
(iii) If S is an R-algebra and t ∈ S is integral over R, then R[t] is a finite R-algebra.
Proof. (i) Exercise.
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(ii) Suppose S =∑ni=1Rsi. Then for each i, tsi ∈ S so there exist rij ∈ R such that
tsi =n∑j=1
rijsj =⇒n∑j=1
(tδij − rij)sj = 0,
where δij is the Kronecker Delta, taking value 1 if i = j and 0 otherwise. Now let A be thematrix with Aij = tδij − rij , and set ∆ = detA and s = (s1, . . . , sn)ᵀ. Then As = 0, hence0 = (Aadj)As = ∆s where Aadj is the adjoint matrix. Therefore ∆si = 0 for all i. But 1 ∈ Sis a linear combination of the si, so in particular we have ∆ = ∆ · 1 =. Therefore the monicpolynomial det(xδij − rij) over R is satisfied by t.
(iii) Exercise.
Corollary 14.6. Let S be a field and R a subring of S such that S is a finite R-algebra. Then Ris a field.
Proof. For any 0 6= r ∈ R, the inverse r−1 exists in S, so we must show r−1 ∈ R. Now byProposition 14.5(ii), r−1 satisfies a monic polynomial over R, say
r−n + an−1r−n+1 + · · ·+ a1r
−1 + a0 = 0
for some ai ∈ R. Then multiply by rn−1 to get
r−1 = −(an−1 + an−2r + · · ·+ a0rn−1) ∈ R,
so R is a field.
Lemma 14.7. Let K be an infinite field and f ∈ K[x1, . . . , xn] be a non-zero polynomial. Thenthere exist α1, . . . , αn ∈ K such that f(α1, . . . , αn) 6= 0.
Proof. We prove this by induction on n, with the case n = 0 being trivial. If now n = 1 then anynon-zero f ∈ K[x1] has at most deg(f) roots. Since K is infinite, we can choose α1 not equal toany of these roots and thus f(α1) 6= 0.
Assume now that n > 1 and the result holds for n − 1. Let f ∈ K[x1, . . . , xn] be non-zero. Iff ∈ K[x1, . . . , xn−1] then we are done, so assume this is not the case. Then we can write
f = grxrn + · · ·+ g1xn + g0
for some gi ∈ K[x1, . . . , xn−1] with gr 6= 0. Now by induction, there exist α1, . . . , αn−1 ∈ K suchthat gr(α1, . . . , αn−1) 6= 0. Therefore f(α1, . . . , αn−1, xn) ∈ K[xn] is a non-zero polynomial, so bythe n = 1 case above we see that there exists αn ∈ K with f(α1, . . . , αn) 6= 0.
Theorem 14.8 (Noether Normalisation). Let K be an infinite field and S a finitely generatedK-algebra. Then there exist z1, . . . , zm ∈ S such that:
(i) z1, . . . , zm are algebraically independent over K, i.e. there is no non-zero polynomial f ∈K[x1, . . . , xm] such that f(z1, . . . , zm) = 0;
(ii) S is finite over R = K[z1, . . . , zm].
Proof. Suppose S = K[y1, . . . , yn] and f ∈ K[x1, . . . , xn] is such that f(y1, . . . , yn) = 0, i.e.y1, . . . , yn are algebraically dependent over K. Then choose α1, . . . , αn−1 ∈ K and set zi = yi−αiynfor 1 6 i 6 n− 1. Now let g ∈ K[x1, . . . , xn] be such that
g(z1, . . . , zn−1, yn) = f(z1 + α1yn, . . . , zn−1 + αn−1yn, yn) = 0.
30
If f has degree d then let fd be the sum of all monomials of f of degree d (the homogeneous pieceof f of degree d). Then
fd(z1 + α1yn, . . . , zn−1 + αn−1yn, yn) = fd(α1yn, . . . , αn−1yn, yn) + lower order terms in yn
= fd(α1, . . . , αn−1, 1)ydn + lower order terms in yn.
Therefore considering g as a polynomial in yn over K[z1, . . . , zn−1] we have
g(z1, . . . , zn−1, yn) = fd(α1, . . . , αn−1, 1)ydn + lower order terms in yn,
Since fd 6= 0 (as deg(f) = d), we have by Lemma 14.7 that there exist α1, . . . , αn−1 such thatfd(α1, . . . , αn−1, 1) 6= 0. For this choice we have
fd(α1, . . . , αn−1, 1)−1g(z1, . . . , zn−1, yn) = 0,
a monic polynomial overK[z1, . . . , zn−1] satisfied by yn. Therefore yn is integral overK[z1, . . . , zn−1].The proof of the theorem is now by induction on the number n of generators of S. Suppose
S = k[y1, . . . , yn] is such that y1, . . . , yn are algebraically independent, then we are done. Otherwisethere exists some f ∈ K[x1, . . . , xn] such that f(y1, . . . , yn) = 0. Then by the above we can choosez1, . . . , zn−1 ∈ S such that yn is integral over S∗ = K[z1, . . . , zn−1] and S = S∗[yn]. By theinduction hypothesis applied to S∗ there exist elements w1, . . . , wm ∈ S∗ that are algebraicallyindependent over K with S∗ finite dimensional over R = K[w1, . . . , wm]. Now since yn is integralover S∗ it follows by Proposition 14.5(iii) that S∗[yn] is a finite S∗-algebra. Since both extensionsR ⊆ S∗ and S∗ ⊆ S are finite, it follows by Proposition 14.5(i) that the extension R ⊆ S is finiteas required.
Remark. (i) In fact Theorem 14.8 does hold for finite fields, but an alternative proof is needed(for instance, see Reid’s Undergraduate Commutative Algebra).
(ii) Theorem 14.8 shows that any finitely generated extension K ⊆ S can be written as a com-posite
K ⊆ K[z1, . . . , zm] ⊆ S,
where the first extension is polynomial and the second is finite.
Example 14.9. Let R = K[x, y]/〈xy−1〉. Using the notation above, we take f = xy−1, so d = 2and f2 = xy. Then f2(1, 1) = 1 6= 0, so we set z = x− y. Then g(z, y) = f(z+ y, y) = y2 + zy− 1.Hence R = K[y, z]/〈y2 + zy − 1〉 is integral over S = K[z].
Theorem 14.10 (Weak Nullstellensatz). Let K be a field and S a finitely generated K-algebra.If S is also a field, then S is finitely generated as a K-module.
In particular, if K is algebraically closed then every maximal ideal of K[x1, . . . , xn] is of theform 〈x1 − a1, . . . , xn − an〉 for some a1, . . . , an ∈ K.
Proof. Using Theorem 14.8 (Noether Normalisation) there exists a polynomial subalgebra R =K[x1, . . . , xr] of S, over which S is a finite algebra. If S is a field then so is R by Corollary 14.6.If r > 1 then 〈x1〉 is a proper ideal in R, a contradiction. Therefore S is finitely generated as anR-module.
For the second part, suppose R = K[x1, . . . , xn] and m C R is a maximal ideal. Then by thefirst part of the theorem we have that R/m is a finite dimensional K-algebra. So given α ∈ R/mwe have m(α) = 0 for some monic polynomial m ∈ K[t] of degree r by Proposition 14.5(ii). SinceK is algebraically closed, we can write m = (t− αr) . . . (t− αr) for some α1, . . . , αr ∈ K. As R/mis a field and m(α) = 0 we have α = αi for some i. Therefore α ∈ K and so R/m = K. Thusxi + m = ai + m for some ai ∈ K, and so 〈x1− a1, . . . , xn− an〉 ⊆ m. Since both sides are maximalideals, this is an equality.
31
Part II
Algebraic Geometry
15 Algebraic sets and varieties
Let K be a field (we will usually assume it to be algebraically closed) and consider the polynomialring K[x1, . . . , xn]. Normally we deal with polynomials simply as elements in the ring, but we willnow consider them as maps from Kn to K by substituting the variables x1, . . . , xn with elementsof K.
Definition 15.1. Let S be a subset of K[x1, . . . , xn]. The vanishing locus of S is the set
V(S) = {(a1, . . . , an) ∈ Kn : f(a1, . . . , an) = 0 for all f ∈ S}.
A set X ⊆ Kn is called an algebraic set if X = V(S) for some such S. The set Kn is often denotedAnK and is called affine n-space (this is done to avoid giving 0 special status).
Example 15.2. (i) V(x2 + y2 − 1) ⊆ A2R is a circle.
(ii) V(xyz) ⊆ A3R is the union of the three planes {x = 0}, {y = 0} and {z = 0}.
(iii) V((x2 + y2)3 − 4x2y2) is the four leaf clover:
-1.0 -0.5 0.5 1.0x
-1.0
-0.5
0.5
1.0
y
(iv) V(z − x3, y − x2) is a curve in A3R and is a twisted cubic t 7→ (t, t2, t3):
32
-1.0-0.5
0.00.5
1.0x
0.00.5
1.0y
-1.0
-0.5
0.0
0.5
1.0
z
(v) V(y2−x3−ax− b) ⊆ A2C gives an elliptic curve. These are very important in many branches
of mathematics.
(vi) V(xz − y2, x3 − yz, z2 − x2y) ⊆ A3R gives a singular twisted cubic t 7→ (t3, t4, t5). Note that
this has codimension 2, but has 3 generators. In fact this this set cannot be 2-generated.
(vii) If a1, . . . , an ∈ K, then V(x1 − a1, . . . , xn − an) ⊆ AnK is the point (a1, . . . , an).
(viii) Spirals r = cos θ are not algebraic sets, as they give a polynomial with an infinite number ofzeros.
Note that for a set S of polynomials, V(S) coincides with V(〈S〉), so every algebraic set X inAnK is of the form V(I) for some ideal I C K[x1, . . . , xn]. By Hilbert’s Basis Theorem, X is thealgebraic set generated by V(〈f1, . . . , fr〉) for some polynomials fi. Note also that
V(f1, . . . , fr) =r⋂i=1
V(fi),
so every algebraic set in the intersection of a finite number of hypersurfaces, algebraic sets generatedby a single non-zero polynomial. In particular, the algebraic subsets of A1
K are just the finite subsetsplus all of K (as V({0}) = K).
Let us then consider V as a map
V : {Ideals in K[x1, . . . , xn]} → {Algebraic sets in AnK}.
Proposition 15.3. Let R = K[x1, . . . , xn]. Then:
(i) V({0}) = AnK and V(R) = ∅;
(ii) I ⊆ J =⇒ V(I) ⊇ V(J) for ideals I, J of R;
(iii) V(IJ) = V(I ∩ J) = V(I) ∪ V(J) for ideals I, J of R;
(iv) for any set {Iλ}λ∈Λ of ideals of R,
V
(∑λ∈Λ
Iλ
)=⋂λ∈Λ
V(Iλ).
Proof. (i) Exercise.
33
(ii) Exercise.
(iii) Since I and J both contain I ∩ J , which in turn contains IJ , we see from (ii) that
V(I) ∪ V(J) ⊆ V(I ∩ J) ⊆ V(IJ).
Now if x /∈ V(I) ∪ V(J) then there exists f ∈ I and g ∈ J such that f(x) 6= 0 and g(x) 6= 0.Hence (fg)(x) 6= 0 so x /∈ V(IJ). Thus V(IJ) ⊆ V(I) ∪ V(J).
(iv) Since Iµ ⊆∑λ∈Λ Iλ for all µ ∈ Λ, we have from (ii) that
V
(∑λ∈Λ
Iλ
)⊆⋂λ∈Λ
V(Iλ).
Now if x ∈⋂λ∈Λ V(Iλ) and f ∈
∑λ∈Λ then f =
∑mi=1 fλi
for some m ∈ N, λi ∈ Λ andfλi∈ Iλi
. Then we have f(x) =∑mi=1 fλi
(x) = 0.
Example 15.4. Consider A3R. Then
V(xz, yz) = V(〈z〉 ∩ 〈x, y〉) = V(z) ∪ V(x, y)
is the union of the (x, y)-plane and the z-axis.
Remark. Proposition 15.3 can be used to show that the sets V(S) for S ⊆ K[x1, . . . , xn] definethe closed sets for a topology on AnK . We call this topology the Zariski topology. Facts fromcommutative algebra, e.g. Hilbert’s Basis Theorem, properties of Noetherian rings etc., can beused to prove statements about the Zariski topology, for instance any closed subset of AnK iscompact.
We now introduce an “inverse” to the map V:
I : {Subsets of AnK} → {Ideals in K[x1, . . . , xn]}.
Definition 15.5. For a subset X ⊆ AnK let the vanishing ideal of X be the set
I(X) = {f ∈ K[x1, . . . , xn] : f(x) = 0 for all x ∈ X}.
That this is an ideal is clear.
Proposition 15.6. (i) I(∅) = K[x1, . . . , xn]. If K is infinite then I(AnK) = {0}.
(ii) X ⊆ Y ⊆ AnK =⇒ I(X) ⊇ I(Y ).
(iii) X,Y ⊆ AnK =⇒ I(X ∪ Y ) = I(X) ∩ I(Y ).
Proof. (i) The first part is clear. The second follows from Lemma 14.7.
(ii) Straightforward: if X ⊆ Y then we need more functions to define it.
(iii) We have
f ∈ I(X ∪ Y ) ⇐⇒ f(x) = 0 for all x ∈ X ∪ Y⇐⇒ f(x) = 0 for all x ∈ X and for all x ∈ Y⇐⇒ f ∈ I(X) ∩ I(Y ).
Remark. Note that in (i) the assumption that K is infinite is necessary. For instance if p ∈ Z isprime, K = Zp and f(x) = xp − x ∈ K[x], then f ∈ I(A1
K).
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Proposition 15.7. Let I be an ideals of K[x1, . . . , xn] and X a subset of AnK . Then:
(i) X ⊆ V(I(X)), with equality if and only if X is an algebraic set;
(ii) I ⊆ I(V(I)).
Proof. The two inclusions are mostly tautological, for instance if I(X) is defined to be the set offunctions vanishing on X then for any point x ∈ X all functions in I(X) vanish on it.
If X = V(I(X)) then X is algebraic as it is of the form X = V(J) for some ideal J . Converselyif X is algebraic then X = V(J) for some ideal J . But J ⊆ I(X) so V(I(X)) ⊆ V(J) = X.
We would like a condition to ensure equality in Proposition 15.7(ii). This is not so easy, as twotypes of problems can occur:
(i) 〈xn〉 ( I(V(xn)) = 〈x〉 for all n > 2, so non-reduced elements present a challenge, and
(ii) in R[x], 〈x2 + 1〉 ⊆6= I(V(x2 + 1)) = I(∅) = R[x], so the ideal may product no zeroes.
We can attempt to solve (i) by using the radical rad(I) from Definition 6.5, but even this is notenough. In fact, (ii) gives a strict inclusion here too as rad(〈x2 + 1〉) = 〈x2 + 1〉. The correct wayto fix this is using Hilbert’s Nullstellensatz.
Recall that a field K is called algebraically closed if every non-constant polynomial in K[x] hasa root in K.
Theorem 15.8 (Nullstellensatz). Let K be an algebraically closed field and I CK[x1, . . . , xn] anideal. Then:
(Weak) I 6= K[x1, . . . , xn] =⇒ V(I) 6= ∅;
(Strong) I(V(I)) = rad(I).
Proof. (Weak): If I is a proper ideal of K[x1, . . . , xn] then I is contained in some maximal idealm by Proposition 4.6. But we know from the weak algebraic form of the Nullstellensatz (Theorem14.10) that m = 〈x1 − a1, . . . , xn − an〉 for some a1, . . . , an ∈ K. Now I ⊆ m =⇒ V(m) ⊆ V(I)and V(m) = {(a1, . . . , an)} 6= ∅.
(Strong): Note first that if f ∈ rad(I) then fm ∈ I for some m > 1. But since K[x1, . . . , xn]is an integral domain, the set of zeros of fm is the same as the set of zeros of f (counted withoutmultiplicity). Thus f ∈ I(V(I)).
We now show that for all f ∈ I(V(I)) we have f ∈ rad(I). This is obvious for f = 0 so assumef 6= 0. Let f1, . . . , fm generate I and set
J = 〈f1, . . . , fm, yf − 1〉 ⊆ K[x1, . . . , xn, y]
for a new variable y. Then
V(J) = V(〈f1, . . . , fm, yf − 1〉)= V(〈f1, . . . , fm〉) ∩ V(〈yf − 1〉)= V(I) ∩ V(〈yf − 1〉)
by Proposition 15.3. But since f ∈ I(V(I)), any point in V(I) will not be in V(〈yf−1〉). ThereforeV(J) = ∅ and by the weak Nullstellensatz we have 1 ∈ J . Hence
1 =m∑i=1
gi(x1, . . . , xn, y)fi + h(x1, . . . , xn, y)(yf − 1)
35
for some gi, h ∈ K[x1, . . . , xn, y]. Let z = 1y and chooseN > max{deg(g1), . . . ,deg(gm),deg(h)+1}.
Then
zN =m∑i=1
zNgi(x1, . . . , xn, y)fi + zN−1h(x1, . . . , xn, y)z(yf − 1)
=m∑i=1
gi(x1, . . . , xn, z)fi + h(x1, . . . , xn, z)(f − z)
in K[x1, . . . , xn, z]. Substituting f for z then gives
fN =m∑i=1
gi(x1, . . . , xn, f)fi ∈ I,
so f ∈ rad(I).
This theorem says that we have correspondences
{Ideals I CK[x1, . . . , xn]} {Subsets X ⊆ AnK}
{Radical ideals} {Algebraic subsets}
{Prime ideals} ?
{Maximal ideals} {Points p ∈ AnK}
⋃ ⋃⋃ ⋃⋃ ⋃
Lemma 15.9. Every non-empty set of algebraic subsets of AnK has a minimal element.
Proof. Suppose that Σ is a non-empty set of algebraic subsets of AnK with no minimal element.Then we can find a strictly descending chain X1 ) X2 ) X3 ) . . . . Recall that X1 ⊇ X2 =⇒I(X1) ⊆ I(X2), and note that if the left subset is strict then so too is the right. Therefore we havea strictly ascending chain of ideals
I(X1) ( I(X2) ( I(X3) ( . . . ,
in K[x1, . . . , xn]. But K[x1, . . . , xn] is Noetherian, so this is a contradiction.
Definition 15.10. An algebraic set X ⊆ AnK is called irreducible if for all decompositions X =X1 ∪X2 with X1, X2 ⊆ X algebraic sets, we have either X = X1 or X = X2. Irreducible algebraicsets are sometimes called algebraic varieties.
Example 15.11. V(xy) ⊆ A2R is the two coordinate axes which can be written as the union
V(x) ∪ V(y), so is reducible.
Proposition 15.12. (i) Let X ⊆ AnK be an algebraic set and I(X) the vanishing ideal of X.Then X is irreducible if and only if I(X) is prime.
(ii) Any algebraic set has an expression
X = X1 ∪ · · · ∪Xr,
unique up to reordering of the Xi, with Xi irreducible and Xi 6⊆ Xj for i 6= j. The Xi arecalled the irreducible components of X.
36
Proof. (i) We prove that X is reducible if and only if I(X) is not prime. Indeed, supposeX = X1 ∪ X2 is a non-trivial decomposition of X into algebraic sets. Then X1, X2 ( Xmeans that there is some f1 ∈ I(X1)\I(X) and some f2 ∈ I(X2)\I(X). The product f1f2
vanishes at all points of X, so f1f2 ∈ I(X). Therefore I(X) is not prime.
Conversely, suppose that I(X) is not prime. Then there exists f1, f2 /∈ I(X) such thatf1f2 ∈ I(X). Let X1 = V(I(X) + 〈f1〉) and X2 = V(I(X) + 〈f2〉). Then by Proposition 15.3
X1 = V(I(X)) ∩ V(f1)= X ∩ V(f1) since X is an algebraic set( X since f1 /∈ I(X),
similarly X2 ( X, and both are algebraic sets. So X1 ∪X2 ⊆ X, and moreover
(I(X) + 〈f1〉)(I(X) + 〈f2〉) = I(X)2 + 〈f1〉I(X) + 〈f2〉I(X) + 〈f1f2〉⊆ I(X),
so by Propositions 15.3 and 15.7 we have X = V(I(X)) ⊆ V((I(X) + 〈f1〉)(I(X) + 〈f2〉)) =X1 ∪X2. Thus X = X1 ∪X2, but neither component is equal to X, so X is reducible.
(ii) Let Σ be the set of algebraic subsets of AnK which do not have such a decomposition. IfΣ = ∅ then we are done, otherwise by Lemma 15.9 there is a minimal element X ∈ Σ. IfX is irreducible, then X /∈ Σ, a contradiction. Otherwise X has a non-trivial decompositionX = X1 ∪X2, and the minimality of X shows that X1, X2 /∈ Σ and so have a decompositioninto irreducibles. But then putting these decompositions together gives a decomposition ofX, so X /∈ Σ, another contradiction. Therefore Σ = ∅ and the existence is proved.
Uniqueness is left as an exercise.
Remark. The decomposition of X into irreducibles Xi corresponds to a minimal primary decom-position of I(X). The associated primes in the latter case are the prime ideals I(Xi).
Example 15.13. We will decompose the algebraic set X = V(x2 − yz, xz − x) ⊆ A3K into its
irreducible components, assuming that the field K is infinite. We begin by considering (p1, p2, p3) ∈X, and note that if p1 = 0 then we must also have p2p3 = 0 (so either p2 = 0 or p3 = 0). Thispart corresponds to the algebraic set V(x, yz) = V(x, y) ∪ V(x, z).
If now p1 6= 0 then p1p3 − p1 = 0 =⇒ p3 = 1, and thus p21 = p2. Therefore this part
corresponds to the algebraic set V(x2 − y, z − 1), and we can decompose
X = V(x, y) ∪ V(x, z) ∪ V(x2 − y, z − 1).
We will prove using Proposition 15.12 that each of the three components is irreducible. By thestrong Nullstellensatz we have I(V(x, y)) = rad(〈x, y〉), but 〈x, y〉 is prime as K[x, y, z]/〈x, y〉 ∼=K[z] so rad(〈x, y〉) = 〈x, y〉. Thus by Proposition 15.12(i) we see that V(x, y) is irreducible.Simililarly V(x, z) is irreducible. Finally, K[x, y, z]/〈x2 − y, z − 1〉 ∼= K[x] so 〈x2 − y, z − 1〉 is alsoprime so this component is also irreducible.
16 Intersection multiplicity
Polynomials in one variable have solutions over C. If we count these with multiplicity, then thenumber of solutions is equal to the degree of the polynomial. Moving to polynomials with twovariables, with two equations we expect a finite number of solutions as long as there are no commonfactors between the two. Ideally, given two polynomials f, g ∈ K[x, y] without common factors wewant the number of solutions to f = g = 0 to be (deg f)(deg g). However we need to deal withmultiplicities of solutions.
37
Definition 16.1. Given a point p = (p1, p2) ∈ A2K we define the local ring of A2
K at p to be
Op(A2K) = K[x, y]mp ,
the localisation of K[x, y] at the maximal ideal mp = 〈x− p1, y − p2〉.
Elements of Op(A2K) are generated by elements of the form f
g , where f, g ∈ K[x, y], g(p) 6= 0.
Definition 16.2. Let f, g ∈ K[x, y]. The intersection multiplicity of f and g is at the point p,denoted Ip(f, g) is
Ip(f, g) = dimK(Op(A2K)/〈f, g〉mp).
Example 16.3. (i) Let f(x, y) = y−x3, g(x, y) = y, p = (0, 0) and m = 〈x, y〉. Then Op(A2K) =
K[x, y]m and 〈f, g〉 = 〈y − x3, y〉 = 〈x3, y〉. Therefore
Op(A2K)/〈f, g〉m = K[x, y]m/〈x3, y〉m
∼= K〈1, x, x2〉 as a K-vector space,
so Ip(f, g) = 3.
(ii) Let f(x, y) = y − x3, g(x, y) = x, p = (0, 0) and m = 〈x, y〉. Then Op(A2K) = K[x, y]m and
〈f, g〉 = 〈y − x3, x〉 = 〈y, x〉 = m. Therefore Op(A2K)/〈f, g〉m ∼= K, so Ip(f, g) = 1.
(iii) Let f(x, y) = y2 − x2(x+ 1), g(x, y) = y2 + x3.
Intersection points are (0, 0), (− 12 ,
14
√2) and (− 1
2 ,−14
√2). Let p = (0, 0), then m = 〈x, y〉
38
and Op(A2K) = K[x, y]m. Now
〈f, g〉m = 〈y2 − x3 − x2, y2 + x3〉m= 〈y2 − x3 − x2, y2 + x3, (y2 − x3 − x2)− (y2 + x3)〉m= 〈y2 − x3 − x2, y2 + x3,−x2(2x+ 1)〉m
= 〈y2 − x3 − x2, y2 + x3, x2〉m as − 12x+ 1
∈ Op(A2K)
= 〈y2, x2〉m.
Thus Op(A2K)/〈f, g〉m ∼= K〈1, x, y, xy〉 as a K-vector space, so Ip(f, g) = 4.
We now prove some properties of Ip(f, g).
Proposition 16.4. (i) Ip(f, g) = 0 ⇐⇒ p /∈ V(f) ∩ V(g).
(ii) Ip(f, g) depends only on the components of V(f) and V(g) passing through p.
(iii) Ip(f, g) = Ip(g, f).
(iv) Ip(f, g) = Ip(f, g + αf) for all α ∈ K[x, y].
Proof. These are clear, mainly due to the fact that Ip(f, g) depends on the ideal 〈f, g〉 as opposedto f and g themselves.
Remark. Note that property (iv) shows that the intersection number is not quite the same as ourintuitive notion.
Our aim now is to bound the value of Ip(f, g) by some function of the degrees of f and g, butfirst we need a lemma.
Lemma 16.5. Let f, g ∈ K[x, y] with deg(f) = m and deg(g) = n. If f and g do not share acommon factor then dimK K[x, y]/〈f, g〉 6 mn.
Proof. Let Vd be the K-vector space of polynomials in x and y of degree at most d. Let r(d) =dimK Vd = 1
2 (d+ 1)(d+ 2). Now for d > max{m,n}, consider
Vd−m × Vd−nϕ−→ Vd
π−→ Vd/〈f, g〉 −→ 0,
where ϕ(u, v) = uf+vg and π is the natural quotient map. Then as π◦ϕ = 0 we have Imϕ ⊆ Kerπ.From the short exact sequence
0 −→ Kerπ −→ Vdπ−→ Vd/〈f, g〉 −→ 0
we get dimK Vd = dimK Kerπ + dimK(Vd/〈f, g〉) (see Example 9.4(ii)), so
dimK Vd − dimK Vd/〈f, g〉 = dimK Kerπ> dimK Imϕ
= dimK(Vd−m × Vd−n)− dimK Kerϕ.
Thereforer(d)− dimK Vd/〈f, g〉 > r(d−m) + r(d− n)− dimK Kerϕ.
But for d > m+ n, Kerϕ = {(αg,−αf) : α ∈ Vd−m−n}, so dimK Kerϕ = r(d−m− n). Thus weobtain dimK Vd/〈f, g〉 6 mn.
Now if dimK K[x, y]/〈f, g〉 > mn then there are more than mn linearly independent elementsin K[x, y]/〈f, g〉, but for sufficiently large d these must lie in Vd/〈f, g〉, contradicting the above.
Theorem 16.6. Let f, g ∈ K[x, y] with deg(f) = m and deg(g) = n. If f and g do not share acommon factor then Ip(f, g) 6 mn.
39
Proof. Suppose we have at least mn + 1 linearly independent elements in Op(A2K)/〈f, g〉mp , then
we can take a common denominator and hence haveu1
v, . . . ,
umn+1
v∈ Op(A2
K)
whose images in Op(A2K)/〈f, g〉mp
are linearly independent. Then u1, . . . , umn+1 are linearly inde-pendent in K[x, y]/〈f, g〉 as 1
v ∈ Op(A2K). This contradicts Lemma 16.5.
Remark. This shows that if f and g have no common factor, then Ip(f, g) is a well-definednon-negative integer. In fact, if there is a common factor then Ip(f, g) is infinite.
Definition 16.7. An affine change of coordinates is a map T : A2K → A2
K of the form(x
y
)7→M
(X
Y
)+(α1
α2
),
where M is a non-singular 2× 2 matrix with entries in K.
Theorem 16.8. Suppose T : A2K → A2
K is an affine change of coordinates, then
IT−1(p)(f ◦ T, g ◦ T ) = Ip(f, g).
Proof. An exercise in isomorphisms of local rings, for instance see Fulton [1, pp. 22].
A nice consequence of this theorem is that we can make calculations easier by changing coor-dinates so that p = (0, 0).
We continue to prove properties about intersection multiplicity.
Definition 16.9. Let f ∈ K[x1, . . . , xn]. A point p ∈ V(f) is called singular if
∂f
∂x1(p) = · · · = ∂f
∂xn(p) = 0,
and non-singular otherwise.
Example 16.10. (i) Let f(x, y) = y2 + x3, then fx = 3x2 and fy = 2y. So fx(p) = fy(p) = 0if and only if p = (0, 0), and (0, 0) ∈ V(f) so this is a singular point.
(ii) Let f(x1, . . . , xn) = xd+11 + x2
2 + · · ·+ x2n, then fx1 = (d+ 1)xd1 and fxi
= 2xi for 2 6 i 6 n.We see that p is singular if and only if p = (0, . . . , 0).
Recall from the algebraic version of the Nullstellensatz (Theorem 14.10) that maximal idealsof K[x1, . . . , xn] are of the form 〈x1 − p1, . . . , xn − pn〉 for some p = (p1, . . . , pn) ∈ AnK . We willhenceforth refer to such an ideal by mp (as we have been doing in the n = 2 case).
Proposition 16.11. Let f ∈ K[x1, . . . , xn] and p ∈ AnK . Then f is non-singular at p if and onlyif f ∈ mp\m2
p.
Proof. Let p = (p1, . . . , pn), and suppose f is non-singular at p. Then p ∈ V(f) so we can write
f(x1, . . . xn) =n∑i=1
αi(xi − pi) + f(x1, . . . , xn)
for some f ∈ m2p and αi ∈ K for all i. Thus fxi
= αi + higher order terms, so if fxi(p) 6= 0 for
some i then αi 6= 0, i.e. f ∈ mp\m2p.
Conversely if f ∈ mp\m2p then αi 6= 0 for some i, hence fxi(p) 6= 0 and p is non-singular.
40
Definition 16.12. If f ∈ K[x1, . . . , xn] is non-singular at a point p = (p1, . . . , pn) ∈ AnK then thetangent space of f at p is the set
Tf (p) =
{(a1, . . . , an) ∈ AnK :
n∑i=1
fxi(p)(ai − pi) = 0
}.
For n = 2, the tangent space is called the tangent line.
Example 16.13. Let f(x, y) = y − x2, then fx = −2x and fy = 1. The tangent line at p =(a, b) = (a, a2) is given by those pairs satisfying
fx(a, b)(x− a) + fy(a, b)(y − b) = 0
=⇒ − 2a(x− a) + (y − a2) = 0
=⇒ y = 2ax− a2
as expected.
Remark. Note that some take the tangent space to be the vector space given by first translatingp to the origin.
Theorem 16.14. Suppose that f, g ∈ K[x, y] are non-singular at p with distinct tangents. ThenIp(f, g) = 1.
Proof. By an affine change of coordinates we can assume that p = (0, 0), and that the tangentsare the x and y axes (see Theorem 16.8). Thus,
f(x, y) = x+ f(x, y)g(x, y) = y + g(x, y)
for some f , g ∈ m2p = 〈x2, xy, y2〉. Then working in the local ring at p and in the context of
Nakayama’s Lemma (Theorem 8.3), letting M = mpmp, N = 〈f, g〉mp
⊆ mpmpand I = mpmp
wehave
N + IM = 〈f, g〉mp+ m2
p
= 〈x+ f , y + g, x2, xy, y2〉mp
= 〈x+ f , y + g, x2, xy, y2,−f ,−g〉mpas f , g ∈ m2
p
= 〈x, y〉mp = M
Thus by Nakayama’s Lemma we have N = M , so
Op(A2K)/〈f, g〉mp = Op(A2
K)/mpmp
∼= K
and Ip(f, g) = 1.
Example 16.15. Recall Example 16.3(iii), where f(x, y) = y2 − x2(x+ 1) and g(x, y) = y2 + x3.The other two intersection points were p± = (− 1
2 ,±14
√2). Now
fx = −3x2 − 2x gx = 3x2
fy = 2y gy = 2y
so both p± are non-singular, and as fx(p±) 6= gx(p±) the tangent lines are distinct. HenceIp±(f, g) = 1.
Theorem 16.16. Suppose f1, f2, g ∈ K[x1, . . . , xn] are such that f1, f2 and g have no commonfactors. Then for p ∈ AnK ,
Ip(f1f2, g) = Ip(f1, g) + Ip(f2, g).
41
Proof. Consider the inclusion 〈f1f2〉 ⊆ 〈f1〉. This induces a surjective map µ : Op(A2K)/〈f1f2〉mp →
Op(A2K)/〈f1〉mp
. Then we have a commutative diagram
0 Kerπ1 Kerπ2
0 Op(A2K) Op(A2
K) 0
0 Kerµ Op(A2K)/〈f1f2〉mp Op(A2
K)/〈f1〉mp 0
Cokerλ 0 0.
ϕ
λ
∼
π1 π2
µ
The columns and middle rows are clearly exact. As λ is the zero map, we have Kerµ ∼= Cokerλ,and so by the Snake Lemma (Theorem 9.5) we have an exact sequence
0 −→ 〈f1f2〉mp−→ 〈f1〉mp
−→ Cokerλ −→ 0 −→ 0,
thus Cokerλ ∼= 〈f1〉mp/〈f1f2〉mp . Since f1 and f2 share no factors we have 〈f1f2〉 = 〈f1〉 ∩ 〈f2〉,hence
〈f1〉mp/〈f1f2〉mp = 〈f1〉mp/(〈f1〉mp ∩ 〈f2〉mp) ∼= (〈f1, f2〉mp)/〈f2〉mp∼= Op(A2
K)/〈f2〉mp ,
where the last isomorphism is due again to the fact that f1 and f2 share no factors (define asurjective homomorphism Op(A2
K)→ 〈f1, f2〉mp/〈f2〉mp
by h 7→ hf1 + 〈f2〉mp). We therefore have
a short exact sequence
0 −→ Op(A2K)/〈f2〉mp
−→ Op(A2K)/〈f1f2〉mp
−→ Op(A2K)/〈f1〉mp
−→ 0.
We can multiply by g to get a commutative diagram
0 Op(A2K)/〈f2〉mp
Op(A2K)/〈f1f2〉mp
Op(A2K)/〈f1〉mp
0
0 Op(A2K)/〈f2〉mp
Op(A2K)/〈f1f2〉mp
Op(A2K)/〈f1〉mp
0.
·g ·g ·g
The kernels of the vertical maps are trivial as g shares no factors with f1, f2, and so the cokernelsform a short exact sequence
0 −→ Op(A2K)/〈f2, g〉mp
−→ Op(A2K)/〈f1f2, g〉mp
−→ Op(A2K)/〈f1, g〉mp
−→ 0.
As K-vector spaces we thus have
dimK Op(A2K)/〈f1f2, g〉mp
= dimK Op(A2K)/〈f1, g〉mp
+ dimK Op(A2K)/〈f2, g〉mp
,
from which the result follows.
Definition 16.17. The multiplicity of f ∈ K[x1, . . . , xn] at p ∈ AnK , denoted mp(f), is the integer` such that f ∈ m`
p\m`+1p , where m0
p = K[x1, . . . , xn] and mp = 〈x1 − p1, . . . , xn − pn〉 as before.
42
Consider p = (0, 0) ∈ A2K and let f = f` + f`+1 + · · · + f`+r, where fi is a homogeneous
polynomial of degree i and f` 6= 0 (we call such sums forms). Then mp(f) = `. Moreover if K isalgebraically closed then we can factorise f` into a product of linear factors f` =
∏Lmii . Then the
tangent lines to f at zero are the lines V(Li), each with multiplicity mi.
Example 16.18. (i) f is non-singular at p if and only if mp(f) = 1.
(ii) Let f(x, y) = y2−x2(x+1) = y2−x3−x2. Then m0(f) = 2 and f2 = y2−x2 = (x−y)(x+y),so the tangent lines to f at zero are V(x+ y) and V(x− y).
(iii) Let g(x, y) = y2 + x3, then m0(g) = 2 and g2 = y2. Thus the tangent lines to g at zero areV(y) with multiplicity 2.
For non-origin points p 6= 0, we can translate to the origin, find multiplicities and tangent lines,then translate back. However this needs the following proposition.
Proposition 16.19. Multiplicity is invariant under an affine change of coordinates, i.e. forT : A2
n → A2n we have
mT−1(p)(f ◦ T ) = mp(f).
Proof. Exercise.
Theorem 16.20. Let mp(f) = µ and mp(g) = ν. Then Ip(f, g) > µν, with equality if and only iff and g have no common tangents at p.
Proof. By Theorem 16.8 and Proposition 16.19 we may assume that p = 0, so m = 〈x, y〉. Definethe homomorphism ϕ : Op(A2
K)/mµ+νm → Op(A2
K)/(〈f, g〉m+mµ+νm ) by ϕ(α+mµ+ν
m ) = α+〈f, g〉m+mµ+ν
m . We then have an exact sequence
0 −→ Kerϕ −→ Op(A2K)/mµ+ν
mϕ−→ Op(A2
K)/(〈f, g〉m + mµ+νm ) −→ 0. (∗)
Now if α + mµ+νm ∈ Kerϕ, then α ∈ 〈f, g〉m + mµ+ν
m . Hence we can write α = βf + γg + m forsome β, γ ∈ Op(A2
K), m ∈ mµ+νm , and thus α + mµ+ν
m ∈ 〈f, g〉m/mµ+νm (by this we mean the ideal
generated by f and g in the ring Op(A2K)/mµ+ν – this is also what we mean wherever the object in
the denominator is not a subobject of the numerator). Therefore dimK Kerϕ 6 dimK〈f, g〉m/mµ+νm ,
and moreover we have a natural surjection 〈f〉m/mµ+νm ⊕ 〈g〉m/mµ+ν
m → 〈f, g〉m/mµ+νm . We deduce
thatdimK
(〈f〉m/mµ+ν
m
)+ dimK
(〈g〉m)/mµ+ν
m
)> dimK Kerϕ (∗∗)
Now the map ψ : Op(A2K)/mν
m → 〈f〉m/mµ+νm defined by ψ(α+mν
m) = αf+mµ+νm is an isomorphism
(linearity and surjectivity are clear):
• (Well defined) if α+ mνm = α′ + mν
m, then f(α− α) ∈ mµ+νm ;
• (Injective) if fα ∈ mµ+νm then since f ∈ mµ\mµ+1 we must have α ∈ mν
m.
Similarly the map ψ′ : Op(A2K)/mµ
m → 〈g〉m/mµ+νm defined by ψ′(α + mµ
m) = αg + mµ+νm is an
isomorphism. Therefore
dimK Op(A2K)/〈f, g〉m > dimK Op(A2
K)/(〈f, g〉m + mµ+νm ) since 〈f, g〉m ⊆ 〈f, g〉m + mµ+ν
m (∗ ∗ ∗)= dimK Op(A2
K)/mµ+νm − dimK Kerϕ by (∗)
> dimK Op(A2K)/mµ+ν
m −(dimK
(〈f〉m/mµ+ν
m
)+ dimK
(〈g〉m/mµ+ν
m
))= dimK Op(A2
K)/mµ+νm −
(dimK
(Op(A2
K)/mνm
)+ dimK
(Op(A2
K)/mµm
))=(µ+ ν + 1
2
)−(ν + 1
2
)−(µ+ 1
2
)= µν.
This will be an equality iff (∗∗) and (∗ ∗ ∗) are equalities.
43
Suppose now that f and g have no common tangents at p. Then let fµ and gν be the terms ofdegree µ and ν in f and g respectively. Then f and g have no common tangents at p if and onlyif fµ and gν have no common components. Now consider the set
{Af +Bg : A homogeneous of degree ν − 1, B homogeneous of degree µ− 1}.
Working modulo mµ+ν , this set is Afµ + Bgν , and is a vector space of forms of degree µ+ ν − 1.It is not hard to check that the dimension of all expressions Afµ + Bgν mod mµ+ν is µ + ν,which implies that mµ+ν−1 ⊆ 〈f, g〉+ mµ+ν . Then by Nakayama’s Lemma (Theorem 8.3) we havemµ+ν−1 ⊆ 〈f, g〉. Therefore mµ+ν
m ⊆ 〈f, g〉m and (∗ ∗ ∗) is an equality.We now show that the surjection from (∗∗) is injective iff fµ and gν have no common components
(iff f and g have a common tangent). Indeed, since localisation takes exact sequences to exactsequences (see Exercise Sheet 2) it suffices to prove the the map
π : 〈f〉/mµ+ν ⊕ 〈g〉/mµ+ν → 〈f, g〉/mµ+ν
is injective if and only if f and g have no common tangents. Therefore, suppose that 0 = αf +βg + mµ+ν , that is all terms in αf + βg are of degree µ+ ν or greater. Writing
α = αr + αr+1 + . . . , β = βs + βs+1 + . . .
as a sum of homogeneous pieces with αr 6= 0 if α 6∈ mν and βs 6= 0 if β 6∈ mµ, we have
αf + βg = αrfµ + βsgν + higher order terms.
Since we are assuming αf + βg ∈ mµ+ν , we must have either:
(i) r + µ = s+ ν < µ+ ν and αrfµ + βsgν = 0, or
(ii) αrfµ, βsgν ∈ mµ+ν .
In the former case, since fµ and gν do not share a common component we see that fµ divides βsand gν divides αr. But due to the inequality this can only happen if αr = βs = 0, and henceα ∈ mν and β ∈ mµ. In the latter case we have either r > ν or αr = 0, and either s > µ or βs = 0,and thus once more we have α ∈ mν and β ∈ mµ. Therefore π is injective.
If fµ and gν share a common component then let h be a multiple of both fµ and gν of degreeµ+ ν− 1. Then hf/fµ+mµ+ν = h+mµ+ν and −hg/gν +mµ+ν = −h+mmu+ν are both non-zero,but the pair is mapped to zero under π, so π is not injective.
Hence (∗∗) and (∗ ∗ ∗) are equalities if f and g do not share a tangent, and (∗∗) is a strictinequality if they do.
Example 16.21. (i) Let f = y2 − x2(x + 1) and g = y2 + x3 as in Example 16.3(iii). Thetangents at 0 are distinct, and m0(f) = m0(g) = 2. Hence I0(f, g) = 4, verifying theprevious calculation.
(ii) Let f = (x2 + y2)3 − 4x2y2 be the four-leaf clover, then m0(f) = 4. Now we find theintersection multiplicity of f with the line V(x):
〈(x2 + y2)3 − 4x2y2, x〉 = 〈y6, x〉,
so Op(A2K)/〈y6, x〉m ∼= K〈1, y, . . . , y5〉 and I0(f, x) = 6. To find the intersection multiplicity
with V(x− y)} then we use I0(f, x− y) = m0(f)m0(x− y) = 4, since the curves do not sharea tangent at 0.
Theorem 16.22. Let K be algebraically closed, I an ideal in K[x1, . . . , xn] and V(I) = {p1, . . . , pm}.Then
K[x1, . . . , xn]/I ∼=m⊕i=1
Opi(AnK)/Im.
44
Proof. See Fulton [1, Section 2.9].
This theorem allows us to prove the following, which gives a “global–local” property.
Corollary 16.23. Suppose K is algebraically closed and f, g define curves in A2K with no common
components. Then dimK K[x, y]/〈f, g〉 =∑p∈A2
KIp(f, g).
Proof. Since f and g have no common component, V(〈f, g〉) is a finite set {p1, . . . , pm}, so we get
dimK K[x, y]/〈f, g〉 =m∑i=1
dimK Opi(A2K)/〈f, g〉mpi
=∑p∈A2
K
Ip(f, g).
17 Projective space
Theorem 16.6 gives us a bound on the intersection multiplicity of two functions in terms of theirdegree. Our ideal result would be for that bound to be an equality, and indeed we can rectify thisby considering curves in projective, rather than affine, space.
Definition 17.1. The projective n-space over the field K, denoted PnK , is the set
PnK = {lines through the origin in An+1K }.
We call one of these lines a point in PnK .We can also define PnK in the following way. Let ∼ be the equivalence relation on An+1
K \{0}defined by
(x0, . . . , xn) ∼ (y0, . . . , yn) ⇐⇒ (x0, . . . , xn) = λ(y0, . . . , yn) for some 0 6= λ ∈ K.
Then PnK = (An+1K \{0})/ ∼. Any equivalence class p of PnK is represented by a line L through the
origin, and a representative is given by any point (a0, . . . , an) ∈ L\{0}. We write p = [a0 : a1 :· · · : an] and call the (n+ 1)-tuple (a0, . . . , an) homogeneous coordinates for p.
Example 17.2. Let p = [1 : 2 : 3] ∈ P2K , then also p = [2 : 4 : 6].
Note that [0 : 0 : · · · : 0] is not a point in PnK as (0, 0, . . . , 0) does not determine a line throughthe origin.
We can identify AnK with [a0 : · · · : an−1 : 1].
••
So for P2K we have a copy of the affine plane inside, given by {[a0 : a1 : 1]}. We call P2
K theprojective K-plane. The line at infinity is the set {[a0 : a1 : 0]}, note that this is one dimensionalover K since all points will be equivalent under ∼ to either [a : 1 : 0] or [1 : a : 0], which has onedegree of freedom.
45
Definition 17.3. A polynomial F ∈ K[x0, . . . , xn] is homogeneous if for all 0 6= λ ∈ K,
F (λx0, . . . , λxn) = λdF (x0, . . . , xn)
for some d, called the degree of F .
Example 17.4. F (x, y, z) = x2 + y2− z2 is homogeneous of degree 2, but F (x, y, z) = x2 + y2− zis not homogeneous.
Homogeneous polynomials are sums of monomials of the same degree.
Definition 17.5. Suppose f ∈ K[x1, . . . , xn], then the homogenisation of f is the homogeneouspolynomial F ∈ K[x0, x1, . . . , xn] given by the following. Let d = deg f and write f =
∑di=0 fi
where fi is homogeneous of degree i. Then
F (x0, . . . , xn) =d∑i=0
fi(x1, . . . , xn)xd−i0 .
Alternatively we can define F (x0, . . . , xn) = xd0f(x1x0, . . . , xn
x0). Note that we can recover f from F
via f(x1, . . . , xn) = F (1, x1, . . . , xn).
Example 17.6. (i) Let f(x, y) = x2 + y2 − 1, then F (x, y, z) = x2 + y2 − z2.
(ii) Let f(x, y) = x2y + y2 + x+ 3y, then F (x, y, z) = x2y + y2z + xz2 + 3yz2.
If f ∈ K[x, y, z], then f is not necessarily a function on P2K . However if F ∈ K[x, y, z]
is homogeneous, then F (λx, λy, λz) = λdF (x, y, z). Therefore we can talk about homogeneouspolynomials defining curves in P2
K , as the zero sets are well defined. If F is irreducible then we saythat the curve is irreducible.
Note that f(x, y) = F (x, y, 1) defines a curve in A2K , similarly for F (x, 1, z) and F (1, y, z).
Definition 17.7. A projective change of coordinates has the formxyz
= M
XYZ
,
where M is a non-singular 3× 3 matrix with entries in K.
Example 17.8. The function f = y−x2 defines a parabola in A2K and homogenises to F = yz−x2.
The transformation xyz
=
1 0 00 −1 10 1 1
XYZ
takes yz − x2 to
(Y + Z)(−Y + Z)−X2 = −Y 2 + Z2 −X2,
which is the homogenisation of the circle x2 + y2 − 1. In other words, the parabola and the circleare the same in projective geometry.
Definition 17.9. We say curves C and D are projectively equivalent if there exists a projectivechange of coordinates that takes C to D.
We saw earlier that not every f ∈ K[x, y, z] defines a function on P2K , but by considering
only homogeneous polynomials we could define an action. We will extend this to functions on P2K
defined by ratios F/G where F and G are homogeneous of some degree d and d′ respectively:
F
G(λx, λy, λz) =
λdF
λd′G(x, y, z) = λd−d
′ F
G(x, y, z).
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The local ring Op(P2K) is the set of functions F/H where H(p) 6= 0 and F,H are homogeneous
polynomials of the same degree.Consider the local equation of a curve at p ∈ P2
K , say fp = F/H with F (p) = 0, H(p) 6= 0.This is not well defined as we can take non-zero scalar multiples of H, so suppose F/H is suchthat H(p) 6= 0. Then H/H ∈ Op(P2
K) as H(p) 6= 0 and
H
H
F
H=F
H.
Hence the ideal generated by 〈fp〉 in Op(P2K) is well defined.
Definition 17.10. The intersection multiplicity at p of two projective curves defined by F and Gis
Ip(F,G) = dimKOp(P2
K)〈fp, gp〉Op(P2
K),
where fp and gp are local defining functions for the curves at p.
In practice, we can just restrict to affine pieces of P2K , as in the following example.
Example 17.11. Consider f = y − x3, g = x. Then we have seen that I0(f, g) = 1, but do thecurves intersect at infinity? We homogenise to obtain F (x, y, z) = yz2− x3 and G(x, y, z) = x. Tofind the intersection points we solve yz2 − x3 = x = 0, i.e. x = 0 and yz2 = 0, so either y = 0 orz = 0. Now the solution x = y = z = 0 does not define a point in P2
K , but if z 6= 0 then we havea solution [0 : 0 : 1] ∈ A2
K , which corresponds to the existing solution, and if y 6= 0 then we have[0 : 1 : 0], which is on the line at infinity.
If now we fix y = 1 we get a copy of A2K which intersects the line at infinity at [0 : 1 : 0]. This
gives new equations f ′(x, z) = F (x, 1, z) = z2 − x3 and g′(x, z) = G(x, 1, z) = x. Then
I[0:1:0](F,G) = I0(f ′, g′)
= dimKO0(A2
K)〈f ′, g′〉
= dimKO0(A2
K)〈z2 − x3, x〉
= dimK K〈1, z〉= 2.
Note then that∑
p∈V(F )∩V(G)
Ip(F,G) = 1 + 2 = 3 = (degF )(degG).
Theorem 17.12 (Bezout’s Theorem). Suppose K be algebraically closed, and let F and G definecurves in P2
K with no common components. Then∑p∈P2
K
Ip(F,G) = (degF )(degG).
Remark. This generalises the Fundamental Theorem of Algebra, which states that the numberof solutions (with multiplicity) of a single variable polynomial over an algebraically closed field isequal to the degree of the polynomial.
Let mp(F ) = mp(fp) for some local definition of F at p. Then we have:
Corollary 17.13. If F and G have no common component then∑p∈V(F )∩V(G)
mp(F )mp(G) 6 (degF )(degG).
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Corollary 17.14. If F and G meet in (degF )(degG) distinct points, then the intersection mul-tiplicity of all these points is 1.
Corollary 17.15. If two curves of degree m and n meet in more than mn points then they havea common component.
Definition 17.16. A point p of a projective curve C in P2K defined by F is called singular if
∂F
∂x(p) =
∂F
∂y(p) =
∂F
∂z(p) = 0.
The curve is called non-singular if it has no singular points.
There is an interesting relationship between the derivatives of F and F itself.
Theorem 17.17 (Euler’s Relation). If F is homogeneous of degree d, then
x∂F
∂x(x, y, z) + y
∂F
∂y(x, y, z) + z
∂F
∂z(x, y, z) = dF (x, y, z).
Proof. Differentiate the identity F (λx, λy, λz) = λdF (x, y, z) with respect to λ and set λ = 1.
This allows us to prove:
Theorem 17.18. Let p = [x0 : y0 : z0] be a point of the curve C ⊆ P2K , defined by F (x, y, z) = 0.
If z0 6= 0 then p is non-singular if and only if(x0z0, y0z0
)is a non-singular point of the affine curve
C defined by F (x, y, 1) = 0.Furthermore, the projective line
x∂F
∂x(p) + y
∂F
∂y(p) + z
∂F
∂z(p) = 0
can be identified with the tangent line at(x0z0, y0z0
).
Proof. Exercise, use Theorem 17.17.
We can use Bezout’s Theorem to obtain bounds on the number of singular points of a curve.
Theorem 17.19. Let K be algebraically closed and C be a curve of degree d, defined by F , withno repeated factors. Moreover, let p1, . . . , pr be the singular points of C, and µi = mp(F ). Then
r∑i=1
(µi2
)6
(d
2
).
Proof. As C has no multiple components we have that the singular set is finite and given by
∂F
∂x(p) =
∂F
∂y(p) =
∂F
∂z(p) = 0.
Consider the forms of degree d− 1
G = a∂F
∂x+ b
∂F
∂y+ c
∂F
∂z, a, b, c ∈ K.
For “most” values of a, b, c this has no common component with C. Then mpi(F ) > m =⇒
mpi(G) > m− 1, so
r∑i=1
µi(µ1 − 1) 6r∑i=1
mpi(F )mpi(G) 6 (degF )(degG) = d(d− 1).
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Bibliography
[1] W. Fulton. Algebraic curves. Advanced Book Classics. Addison-Wesley, 1989.
[2] M. Reid. Undergraduate algebraic geometry, volume 12 of LMS Student Texts. CUP, 1988.
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