Math 572 – Lecture Notes Chapter 15 – Commutative Algebra and Algebraic Geometry

Lecture 32

§15.1 Noetherian Rings and Aﬃne Algebraic Sets

In this chapter, R will usually be a commutative ring with 1 6= 0.

Definition A commutative ring R is Noetherian or satisfies the ascending chain condition if there are no infinite increasing chains of ideals in R. That is, whenever

I1 ⊆ I2 ⊆ I3 ⊆ · · ·

is an increasing chain of ideals, there exists some m such that Ik = Im for all k ≥ m.

Proposition 1 If I is an ideal in a Noetherian ring R, then R/I is Noetherian. Any homomorphic image of a Noetherian ring is Noetherian.

Proof. Any inﬁnite ascending chain of ideals in R/I would correspond to an inﬁnite ascending chain of ideals in R. The homomorphic image is isomorphic to R/I where I is the kernel of the homomorphism.

1 Theorem 2 The following are equivalent:

(1) R is Noetherian.

(2) Every nonempty set of ideals of R contains a maximal element under in- clusion.

(3) Every ideal of R is ﬁnitely generated.

Proof. (1)⇒(2). Assume R is Noetherian. Let Σ by any nonempty collection of ideals, but assume Σ has no maximal element. Choose I1 ∈ Σ. Since I1 is not maximal, there exists I2 ∈ Σ such that

I1 ( I2.

Since I2 is not maximal, there exists I3 ∈ Σ such that

I1 ( I2 ( I3. Continuing in this fashion produces an inﬁnite ascending chain in Σ:

I1 ( I2 ( I3 ( ··· ( Ik ( Ik+1 ( ··· . This contradicts the fact that R is Noetherian. Hence Σ must have a maximal element.

(2)⇒(3). Assume (2) holds. Let I be any ideal of R. If I is not ﬁnitely generated, there exist elements x1, x2, x3,... ∈ I such that

I1 = (x1) ( I2 = (x1, x2) ( I3 = (x1, x2, x3) ( ··· Let Σ be the collection of ideals in the sequence above. By (2), Σ contains a 0 0 maximal element I . But then I = In for some n, which means that I was actually ﬁnitely generated, a contradiction. So, every ideal of R is ﬁnitely generated.

(3)⇒(1). Assume every ideal of R is ﬁnitely generated. Let

I1 ⊆ I2 ⊆ I3 ⊆ · · ·

2 be a chain of ideals, and let ∞ [ I = Ii. i=1 Then I is an ideal, which by (3), is ﬁnitely generated

I = (x1, x2, . . . , xk)

Then xi ∈ Imi for some mi. Let m = max(m1, . . . , mk). Then x1, . . . , xk ∈ In for all n ≥ m. Thus, In = I for all n ≥ m and the ascending chain of ideals is ﬁnite. Hence, R is Noetherian.

Note In the previous proof, we did not every use the assumption of commutativity or that the ring contained 1. Hence, we could have deﬁned a Noetherian ring as a ring (not necessarily commutative and not necessarily containing 1) satisfying the ascending chain condition. Then the previous theorem still holds.

Example Every ideal in a PID is ﬁnitely generated (by a single element). Hence, any PID is Noetherian.

Z, Z[i], F [x] for F a ﬁeld are Noetherian.

Z[x1, x2, x3,...] is not Noetherian since the ideal (x1, x2, x3,...) is not ﬁnitely generated.

Z[x1, . . . , xn] is Noetherian (by Hilbert’s basis theorem).

Example Noetherian ring may having inﬁnitely long descending chains. For example, in Z 2 3 (2) ) (2 ) ) (2 ) ) ···

Theorem 3 (Hilbert’s Basis Theorem) If R is a Noetherian ring, then so is R[x].

3 Proof. Let I be any ideal in R[x]. We’ll show that I is ﬁnitely generated, and so R[x] will be Noetherian.

Let L ⊆ R be the set of all leading coeﬃcients of polynomials in I. Thus,

axm + lower order terms ∈ I ⇒ a ∈ L.

We’ll show that L is an ideal of R. Since the zero polynomial is in I, 0 ∈ L. Let

f = axd + ··· and g = bxe + ··· be polynomials in I. For any r ∈ R, either ar − b = 0 or ar − b is the leading coeﬃcient of rxef − xdg. This shows that L is an ideal in R.

Since R is Noetherian, L is ﬁnitely generated:

L = (a1, . . . , an) ai ∈ R.

For each ai choose fi ∈ I such that

ei fi = aix + ···

Set N = max(e1, . . . , en).

For each d ∈ {0, 1,...,N − 1}, let

Ld = “set of leading coeﬃcients of elements of degree d in I along with 0”. Let f = axd + ··· and g = bxd + ··· be polynomials of degree d in I. For any r ∈ R, either ra − b = 0 or ra − b is the leading coeﬃcient of rf − g

Hence, Ld is also an ideal of R, and so is ﬁnitely generated.

Ld = (bd1, bd2, . . . , bdnd ).

For each bdi choose fdi ∈ I such that

d fdi = bdix + ··· .

4 Let I0 be the ideal in R[x] deﬁned by

0 I = {f1, . . . , fn} ∪ {fdi : 0 ≤ d < N, 1 ≤ i ≤ nd} . By construction I0 ⊆ I. We’ll show I0 = I and so I is ﬁnitely generated.

If I 6= I0, there exists some nonzero polynomial f ∈ I \ I0 of minimal degree m. Write f = axm + ··· where deg(f) = m and a is the leading coeﬃcient.

Case 1. Suppose m ≥ N. Since a ∈ L

a = r1a1 + ··· + rnan for some ri ∈ R. Then

m−e1 m−en 0 g = r1x f1 + ··· rnx fn ∈ I has degree m and leading coeﬃcient a. Then f − g ∈ I \ I0 has degree < m. By the minimality of m, f − g = 0 and so f = g ∈ I0, a contradiction.

Case 2. Suppose m = d < N. In this case, a ∈ Ld. Write

a = r1bd1 + ··· + rnd bdnd , ri ∈ R. Then 0 g = r1fd1 + ··· + rnd fdnd ∈ I had degree m = d and has the same leading coeﬃcient as f. As before, we have a contradiction.

Hence, I = I0.

This proves that R[x] is Noetherian.

Corollary

If R is Noetherian, R[x1, . . . , xn] is Noetherian. If k is a ﬁeld k[x1, . . . , xn] is Noetherian.

5 Lecture 33

Lecture 34

Deﬁnition Let k be a ﬁeld. A ring R is called a k-algebra if k is contained in the center of R and the identity of k is the identity of R.

Deﬁnition (1) The ring R is a ﬁnitely generated k-algebra if R is generated as a ring

by k together with some ﬁnite set r1, r2, . . . , rn of elements of R.

(2) Let R and S be k-algebras. A map ψ : R → S is a k-algebra homomorphism if ψ is a ring homomorphism that is the identity on k.

Corollary 5 The ring R is a ﬁnitely generated k-algebra if and only if there is some surjective k-algebra homomorphism

φ: k[x1, x2, . . . , xn] → R.

Hence, any ﬁnitely generated k-algebra is Noetherian.

Proof. If R is generated as a k-algebra by r1, . . . , rn, we deﬁne the map φ: k[x1, . . . , xn] → R by φ(xi) = ri and φ(a) = a for all a ∈ k. Then φ extends uniquely to a surjective homomorphism.

Conversely, given a surjective homomorphism, the image of x1, . . . , xn generates R as a k-algebra, proving that R is ﬁnitely generated. Then R is isomorphic to the quotients of the Noetherian ring k[x1, . . . , xn] modulo the kernel, which is again Noetherian.

Aﬃne Algebraic Sets

6 An is the set of n-tuples of the field k is called affine n-space. (Apply the forgetful functor to the vector space kn to get affine space.)

n Polynomials f ∈ k[x1, . . . , xn] may be regarded as functions f : A → k by evalua- tion:

f :(a1, . . . , an) 7→ f(a1, . . . , an) ∈ k. This gives a ring of functions, denoted k[An], called the coordinate ring of An.

Deﬁnition If S ⊆ k[An], then

n Z(S) := {(a1, . . . , an) ∈ A : f(a1, . . . , an) = 0 for all f ∈ S}

is called an aﬃne algebraic set. V = Z(S) is called the locus of S in An.

Examples 1. If n = 1 and f(x) ∈ k[x], then Z(f) is the set of roots of f.

n 2. One point subsets {(a1, . . . , an)} of A are Z(x1 − a1, . . . , xn − an).

3. If f ∈ k[x1, . . . , xn] is a nonconstant polynomial, Z(f) is called a hyper- surface.

In R2, Z(y−x2) is the parabola y = x2, Z(xy−1) is the hyperbola y = 1/x.

In R3, Z(x2 + y2/4 + z2/9 − 1) is an ellipsoid.

7 Basic Properties of Z Let S,T ⊆ k[An]. (1) If S ⊆ T , then Z(T ) ⊆ Z(S).

(2) Z(I) = Z(S) where I = (S) is the ideal generated by S.

(3) Z(S) ∩ Z(T ) = Z(S ∪ T ).

(4) For ideals I,J ⊆ k[An], Z(I) ∪ Z(J) = Z(IJ).

(5) Z(0) = An and Z(1) = ∅.

By (2), we may consider the map

n n Z : {ideals of k[A ]} → {aﬃne algebraic sets in A }.

Since k[x1, . . . , xn] is Noetherian, an ideal I = (f1, . . . , fq) is ﬁnitely generated, and from (3)

Z(I) = Z(f1) ∩ Z(f2) ∩ · · · ∩ Z(fq). So, each aﬃne algebraic set is the intersection of ﬁnitely many hypersurfaces.

Deﬁnition For any A ⊆ An,

I(A) := {f ∈ k[x1, . . . , xn]: f(a1, . . . , an) = 0 for all (a1, . . . , an) ∈ A}.

I(A) is an ideal and is the largest ideal that vanishes on A.

8 Basic Properties of I Let A, B ⊆ An (6) If A ⊆ B, then I(B) ⊆ I(A).

(7) If I(A ∪ B) = I(A) ∩ I(B).

n (8) I(∅) = k[x1, . . . , xn] and, if k is inﬁnite, I(A ) = 0. (9) A ⊆ Z(I(A)). If I is any ideal, I ⊆ I(Z(I)).

(10) If V = Z(I) is an aﬃne algebraic set, Z(I(V )) = V .

If I = I(A) is the ideal of a set, then I(Z(I)) = I.

Deﬁnition If V ⊆ An is an aﬃne algebraic set, k[An]/I(V ) is the coordinate ring of V , and is denoted k[V ].

Note that two polynomials f, g ∈ k[An] deﬁne the same function on V if and only if f − g ∈ I(V ). Thus all elements of the coset f¯ = f + I(V ) deﬁne the same function on V .

Example ???????

9 Lecture 35 — Continuation of §15.1

Deﬁnition Let V and W be algebraic sets. A map φ: V → W is a morphism (or polyno-

mial map or regular map) if there are polynomials φ1, . . . , φm ∈ k[x1, . . . , xn] such that

φ((a1, . . . , an)) = (φ1(a1, . . . , an), . . . , φm(a1, . . . , an)) ∈ W

for all (a1, . . . , an) ∈ V .

φ: V → W is an isomorphism if there is a morphism ψ : W → V with

φ ◦ ψ = 1W and ψ ◦ φ = 1V .

Note

In general φ1, . . . , φm are not uniquely deﬁned. For example, f = x and g = x + (xy − 1) both deﬁne the same morphism from V = Z(xy − 1) to W = A1.

10 Theorem 6 Let V ⊆ An and W ⊆ Am be aﬃne algebraic sets. There is a bijective correspondence morphisms V → W k-algebra homomorphisms ←→ k[W ] → k[V ] In particular,

(1) Every morphism φ: V → W induces a k-algebra homomorphism φ˜: k[W ] → k[V ] deﬁned by φ˜(f) = f ◦ φ.

(2) Every k-algebra homomorphism Φ: k[W ] → k[V ] is induced by a unique morphism φ: V → W , i.e., Φ = φ˜.

(3) If φ: V → W and ψ : W → U are morphisms, then

ψ]◦ φ = φ˜ ◦ ψ˜: k[U] → k[V ].

(4) φ: V → W is an isomorphism if and only if φ˜: k[W ] → k[V ] is a k-algebra isomorphism.

Proof of (1). Let F ∈ k[x1, . . . , xm] and φ1, . . . , φm ∈ k[x1, . . . , xn]. Then

F ◦ φ = F (φ1, . . . , φm) ∈ k[x1, . . . , xn].

If g ∈ I(W ) and φ: V → W is a morphism, then we have the diagram φ V W g g◦φ 0 Since

(g ◦ φ)(a1, . . . , an) = 0 for all (a1, . . . , an) ∈ V, it follows that g ◦ φ ∈ I(V ).

Thus, φ induces a well-deﬁned mapping from k[x1, . . . , xm]/I(W ) to k[x1, . . . , xn]/I(V ) given by φ˜: k[W ] → k[V ]

11 f 7→ f ◦ φ.

Here, f is a coset of I(W ). So if f = f + I(W ) where f ∈ k[x1, . . . , xm], then f ◦ φ = f ◦ φ + I(V ). To be a little bit more careful, suppose f = f 0 + g where 0 0 f, f ∈ k[x1, . . . , xm] and g ∈ I(W ). Then f ◦φ = f ◦φ+g◦φ. For (a1, . . . , an) ∈ V , we have

0 (f ◦ φ)(a1, . . . , an) = (f ◦ φ)(a1, . . . , an) + (g ◦ φ)(a1, . . . , an) . | {z } =0 0 Thus, f ◦ φ = f ◦ φ ∈ k[x1, . . . , xn]/I(V ) = k[V ].

This completes the proof of part (a).

Proof of (2). Conversely, let Φ: k[W ] → k[V ] be a k-algebra homomorphism. Let Fi ∈ k[x1, . . . , xn] be a coset representative of Φ(¯xi) as follows: Φ(xi mod I(W )) = Fi mod I(V ). Then

φ = (F1,...,Fm) deﬁnes a polynomial map An → Am. We’ll show that φ: V → W is a morphism. We need to show that φ(a1, . . . , an) ∈ W for all (a1, . . . , an) ∈ V .

Let g ∈ I(W ) ⊂ k[x1, . . . , xm]. In k[W ],

g(x1 + I(W ), . . . , xm + I(W )) = g(x1, . . . , xm) + I(W ) = I(W ) = 0 ∈ k[W ].

2 [Could insert example here like g(x1, x2) = x1x2.] Since Φ maps I(W ) into I(V ), we have Φ g x1 + I(W ), . . . , xm + I(W ) = 0 ∈ k[V ]. 2 2 [Example: Φ(g(x1, x2) = Φ(x1x2)) = Φ(x1)Φ(x2) = g(Φ(x1, x2)).] Since Φ is a k-algebra homomorphism, we may rewrite the previous expression as g Φ x1 + I(W ) ,..., Φ xm + I(W ) = 0 ∈ k[V ].

By deﬁnition, Φ(xi + I(W )) = Fi + I(V ), and so g F1 + I(V ),...,Fm + I(V ) = 0 ∈ k[V ],

12 and hence

g(F1,...,Fm) ∈ I(V ).

It follows that, for every (a1, . . . , an) ∈ V , g F1(a1, . . . , an),...,Fm(a1, . . . , an) = 0.

This shows that if (a1, . . . , an) ∈ V , then every polynomial in I(W ) vanishes on φ(a1, . . . , an).

By property (10) of the maps Z and I, this means that

φ(a1, . . . , an) ∈ Z(I(W )) = W, which proves that φ maps points of V to points of W . Therefore, φ = (F1,...,Fm) is a morphism V → W .

We need to show that φ is unique. Since each Fi is well-deﬁned modulo I(V ), this morphism does not depend on the choice of the Fi. The morphism φ: V → W ˜ induces Φ, i.e., φ = Φ, since both homomorphisms take the values Fi + I(V ) on xi + I(W ). This proves part (b).

Proof of (c). Let φ: V → W and ψ : W → U be morphism. Let f ∈ k[U]. Then

(^ψ ◦ φ)(f) = f ◦ (ψ ◦ φ) = (f ◦ ψ) ◦ φ = φ˜(f ◦ ψ) = φ˜(ψ˜(f)) = (φ˜ ◦ ψ˜)(f). Hence, ψ]◦ φ = φ˜ ◦ ψ˜.

Proof of (d). Assume φ: V → W is an isomorphism. Then there exists a morphism

ψ : W → V such that φ ◦ ψ = 1W and ψ ◦ φ = 1V . By part (c), we have ˜ ˜ ˜ ˜ ψ ◦ φ = 1k[W ] and φ ◦ ψ = 1k[V ]. Hence φ˜: k[W ] → k[V ] is an isomorphism.

Assume φ˜: k[W ] → k[V ] is an isomorphism. Then for some morphism ψ : W → V we have ˜ ˜ ˜ ˜ φ ◦ ψ = ψ]◦ φ = 1k[V ] = 1fV and ψ ◦ φ = φ]◦ ψ = 1k[W ] = 1fW and so

ψ ◦ φ = 1V and φ ◦ ψ = 1W . Hence, φ: V → W is an isomorphism.

13 Example. For an inﬁnite ﬁeld k, let V = A1 and W = Z(x3 − y2) = {(a2, a3): a ∈ k}. The map

φ: V → W a 7→ (a2, a3) is a morphism V → W . Note that φ is a bijection.

The coordinate rings are k[V ] = k[x] and k[W ] = k[x, y]/(x3 − y2). Then

φ˜: k[W ] → k[V ] x 7→ x2 y 7→ x3.

The image of φ˜ is the subalgebra k[x2, x3] = k + x2k[x], so in particular φ˜ is not surjective. Hence φ˜ is not an isomorphism of coordinate rings, and so φ is not an isomorphism V → W , even though φ is bijective. The inverse map is give by ψ(0, 0) = 0 and ψ(a, b) = b/a for b 6= 0, and this cannot be achieved by a polynomial map.

14 Lecture 36

§15.2 Radicals and Aﬃne Varieties

Deﬁnition Let I be an ideal of a commutative ring R.

(1) The radical of I, denoted rad I, is

rad I = {a ∈ R: ak ∈ I for some k ≥ 1}.

(2) The radical of the zero ideal is the nilradical of R.

(3) An ideal I is a radical ideal if I = rad I.

Proposition 11 Let I be an ideal in a commutative ring R. Then rad I is an ideal containing I, and (rad I)/I is the nilradical of R/I.

In particular, R/I has no nilpotent elements if and only if I = rad I is a radical ideal.

Proof. Clearly I ⊆ rad I. The nilradical consist of elements in the quotient some power of which is 0, which is just rad I/I. (Under the lattice isomorphism theorem, rad I corresponds to rad I/I which is the nilradical of R/I.) Hence to show that rad I is an ideal, it suﬃces to show that nilradical of any commutative ring is an ideal.

Let N be the nilradical of R. 0 ∈ N and so N 6= ∅. If a ∈ N and r ∈ R, then an = 0 for some n ≥ 1. Then (ra)n = rnan = 0. So ra ∈ N. We need to show that if a, b ∈ N, then a + b ∈ N. Suppose an = 0 and bm = 0. Then n+m n+m X n+m k n+m−k (a + b) = k a b . k=0 For each k, either k ≥ n (implying ak = 0) or n+m−k ≥ m (implying bn+m−k = 0). Hence (a + b)n+m = 0 so that a + b ∈ N.

15 Proposition 12 The radical of any proper ideal I is the intersection of all prime ideals containing I. In particular, the nilradical is the intersection of all prime ideals of R.

Proof. Passing to the quotient R/I, the previous proposition shows that it suﬃces to prove the result for I = 0. Thus we need to show that the nilradical N equals the intersection N 0 of all prime ideals of R.

Let a ∈ N. Since ak = 0 ∈ P for some k ≥ 1, there is some smallest power n such that an ∈ P . Then an−1 ∈ P or a ∈ P . This implies a ∈ P (otherwise we would contradict the minimality of n). Thus, N ⊆ N 0.

We’ll show N 0 ⊆ N by showing that if a∈ / N, then a∈ / N 0. If a ∈ R\N, let S be the collection of all proper ideals not containing a power of a. Since 0 ∈ S,

S= 6 ∅. If I1 ⊆ I2 ⊆ · · · is a chain in S, its union is an upper bound not containing ak for any k ≥ 1. By Zorn’s lemma, S contains a maximal element, say P . We’ll show P is prime. Suppose, to the contrary, that x, y∈ / P but xy ∈ P . By the maximality of P , an ∈ (x) + P and am ∈ (y) + P for some m, n ≥ 1. Then am+n ∈ (xy) + P = P , contradicting the choice of P as an element of S. So, P is a prime ideal not containing a, and so a∈ / N 0. This completes the proof.

Corollary 13 Prime (and hence maximal) ideals are radical.

Proof. If P is a prime ideal, it is the intersection of all prime ideals containing P , so P = rad P by the preceding proposition.

Example

a1 a2 ar In the ring of integers Z, the radical of the ideal (a) = (p1 p2 ··· pr ) is rad(a) = (p1p2 ··· pr).

16 Proposition 14 If R is Noetherian and I is an ideal, some positive power of rad I is contained in I. In particular, the nilradical N of a Noetherian ring is nilpotent: N k = 0 for some k ≥ 1.

Proof. If I is an ideal of the Noetherian ring R, then rad I = (a1, . . . , am) for some ki generators a1, . . . , am. Then ai ∈ I for some ki ≥ 1. Let k = max(k1, . . . , km). The km d1 d2 dm ideal (rad I) is generated by the elements a1 a2 ··· am where d1 +···+dm = km. di di Each of these products has a factor ai with di ≥ k. Then ai ∈ I, and so each generator of (rad I)km belongs to I, and so (rad I)km ⊆ I.

In the preceding section, if we let V = Z(I) and I = I(V ), then on such sets I and Z are inverses of each other:

Z(I(V )) = V and I(Z(I)) = I.

The elements of k[An]/I(V ) give k-valued functions on V , and since k has no nilpotent elements, powers of nonzero functions are nonzero. Thus, k[An]/I(V ) has no nilpotent elements, which means that I(V ) is a radical ideal.

Example For arbitrary ﬁelds, not every radical ideal is the ideal of an algebraic set V . (x2 + 1) ⊂ R[x] is maximal, hence radical, but is not the ideal of an algebraic set. (Degree 1 polynomials over nonalgebraically closed ﬁelds turn out to be the obstacle.)

Hilbert’s Nullstellensatz Let E be an algebraically closed ﬁeld. Then I(Z(I)) = rad I for every ideal I

of E[x1, . . . , xn]. Moreover, the maps Z and I in the correspondence

−→I {aﬃne algebraic sets} {radical ideals} ←− Z are bijections that are inverses of each other.

17 Proof. Proved in the next section §15.2.

18 Lecture 37

Example The maps I and Z are defined over any field, but are not bijections when k is not algebraically closed. However, for any field k, Z is always surjective and I is always injective. (Exercise 15.2.9)

Note By Hilbert’s nullstellensatz, for any proper ideal Z(I) 6= ∅ since rad I 6= k[An]. Hence, for any proper ideal there always exists at least one common zero of all polynomials in the ideal (over algebraically closed ﬁelds).

Aﬃne Varieties

Deﬁnition A nonempty aﬃne algebraic set V is irreducible if it cannot be written as

V = V1 ∪ V2 where V1 and V2 are proper algebraic sets in V . An irreducible aﬃne algebraic set is called an aﬃne variety.

Proposition 17 (1) The aﬃne algebraic set V is irreducible if and only if I(V ) is a prime ideal.

(2) Every nonempty aﬃne algebraic set V may be written uniquely in the form

V = V1 ∪ V2 ∪ · · · ∪ Vq

where each Vi is irreducible and Vi * Vj for all j 6= i.

Proof of (1). Let I = I(V ) and suppose V = V1 ∪ V2 is reducible, where V1 and V2 are proper subsets and are aﬃne algebraic sets. Since V1 6= V , there is some f1 ∈ I(V1)\I. Similarly, there is some f2 ∈ I(V2)\I. Then f1f2 vanishes on V1 ∪ V2 = V . So, f1f2 ∈ I which shows that I is not a prime ideal.

n Conversely, if I is not a prime ideal, there exist f1, f2 ∈ k[A ] such that f1f2 ∈ I but f1 ∈/ I and f2 ∈/ I. Let V1 = Z(f1) ∩ V and V2 = Z(f2) ∩ V . Then V1 and

19 V2 are algebraic sets. Since neither f1 nor f2 vanishes on V , V1 ( V and V2 ( V . Because f1f2 ∈ I, V ⊆ Z(f1f2) = Z(f1) ∪ Z(f2), and so V is reducible.

Proof of (2) – Existence. Let S be the collection of nonempty algebraic sets that cannot be written as a finite union of irreducible algebraic sets. Suppose, BWOC, that S= 6 ∅. Let I0 be a maximal element in the corresponding set of ideals n {I(V ): V ∈ S} (which exists since k[A ] is Noetherian). Then V0 = Z(I0) is a minimal element of S. Since V0 ∈ S it is not irreducible. If V0 = V1 ∪ V2 for some proper algebraic subsets V1,V2 of V0, then by the minimality of V0, both V1 and V2 are finite unions of irreducible algebraic sets. Then V0 is the union of finitely many irreducible algebraic sets, a contradiction. Hence, S = ∅. Hence, every affine algebraic set has a decomposition into affine varieties.

Proof of (2) – Uniqueness. Suppose V has two decompositions into aﬃne varieties:

V = V1 ∪ V2 ∪ · · · ∪ Vr = U1 ∪ U2 ∪ · · · ∪ Us, where there are no redundant entries. V1 is contained in ∪iUi. This gives a decomposition of V1 into algebraic sets:

V1 = (V1 ∩ U1) ∪ (V1 ∩ U2) ∪ · · · ∪ (V1 ∩ Us).

Since V1 is irreducible, V1 = V1 ∩ Ui for some j, and so V1 ⊆ Uj. By a symmetrical 0 0 argument Uj ⊆ Vj0 for some j . Then V1 ⊆ Vj0 and so j = 1 and V1 = Uj. Applying a similar argument for each Vi gives r = s and {V1,...,Vr} = {U1,...,Us}.

Corollary 18 An aﬃne algebraic set V is a variety if and only if its coordinate ring k[V ] is an integral domain.

Proof. This is because I(V ) is a prime ideal if and only if the quotient k[V ] = k[An]/I(V ) is an integral domain.

20 Definition If V is a variety, the field of fractions of the integral domain k[V ] is the field of rational functions on V and is denoted k(V ). Then dimension of a variety V , denoted dim V , is the transcendence degree of k(V ) over k.

Recall: A transcendence base for E/F is a maximal subset (with respect to inclu- sion) of E which is algebraically independent over F . Any two transcendence bases have the same cardinality, and this cardinality is called the transcendence degree of E/F .

Examples n (1) Single points (a1, . . . , an) ∈ A are aﬃne varieties.

V = {(a1, . . . , an)} = Z(x1 − a1, . . . , xn − an). ∼ The ideal I(V ) = (x1−a1, . . . , xn−an) is maximal and k[x1, . . . , xn]/I(V ) = k. Any ﬁnite set is the union of its one point subsets, and this is its unique decomposition into aﬃne varieties.

2 ∼ (2) The x-axis in R is irreducible since its coordinate ring is R[x, y]/(y) = R[x] which is an integral domain. Similarly, any line in R2 is irreducible.

(3) The union of the x and y axes in R2, namely Z(xy), is not a variety. Z(xy) = Z(x) ∪ Z(y) is its unique decomposition into subvarieties. The corresponding coordinate ring R[x, y]/(xy) has zero divisors.

(4) The hyperbola xy = 1 in R2 is a variety. Earlier, we showed that its coordinate ring is the integral domain R[x, 1/x]. The two disjoint branches (deﬁned by x > 0 and x < 0) are not subvarieties.

(5) Let V = Z(l1, l2, . . . , lm) be the zero set of linear polynomials l1, . . . , lm ∈ k[x1, . . . , xn] and V 6= ∅. Then V is an aﬃne variety (called a linear variety). Determining whether V 6= ∅ is a linear algebra problem.

21 Lecture 38

Deﬁnition A proper ideal Q in the commutative ring R is primary if whenever ab ∈ Q and a∈ / Q, then bn ∈ Q for some n ≥ 1. Equivalently, if ab ∈ Q and a∈ / Q, then b ∈ rad Q.

Proposition 19 Let R be a commutative ring with 1.

(1) Prime ideals are primary.

(2) The ideal Q is primary if and only if every zero divisor in R/Q is nilpotent.

(3) If Q is primary, then rad Q is a prime ideal, and is the unique smallest prime ideal containing Q.

(4) If Q is an ideal whose radical is a maximal ideal, then Q is a primary ideal.

(5) Suppose M is a maximal ideal and Q is an ideal with M n ⊆ Q ⊆ M for some n ≥ 1. Then Q is a primary ideal with rad Q = M.

Proof of (1). Immediate from the deﬁnition of primary.

Proof of (2). Let Q be primary. Assume a∈ / Q and b∈ / Q but ab ∈ Q. Then for some m, n ≥ 1, am ∈ Q and bn ∈ Q. Thus the zero divisors a + Q and b + Q in R/Q are nilpotent.

Conversely, let a + Q and b + Q be zero divisors in R/Q. If these are nilpotent, then am ∈ Q and bn ∈ Q for some m, n ≥ 1. Hence, Q is primary.

Proof of (3). Suppose ab ∈ rad Q. For some m ≥ 1, (ab)m = ambm ∈ Q. Since Q is primary, either am ∈ Q, in which case a ∈ rad Q, or (bm)n ∈ Q, in which case b ∈ rad Q. Hence, rad Q is prime. It follows that rad Q is the smallest prime containing Q. (This is by Proposition 12, which says, the radical of a proper ideal I is the intersection of all prime ideals containing I.)

22 Proof of (4). Let Q be an ideal such that rad Q is maximal. We pass to the quotient R/Q. By (2) it suﬃces to show that every zero divisor in R/Q is nilpotent. We are reduced to the case where Q = (0) and M = rad Q = rad(0), which is the nilradical, is a maximal ideal. If d were a zero divisor, then (d) would be a proper ideal, hence contained in a maximal ideal. Hence d ∈ M, and so every zero divisor is in M and is therefore nilpotent.

Proof of (5). Let M be a maximal ideal and suppose M n ⊆ Q ⊆ M for some n ≥ 1. Then Q ⊆ M implies rad Q ⊆ rad M = M. Conversely, M n ⊆ Q shows that M ⊆ rad Q. So, rad Q = M is a maximal ideal, and Q is primary by part (4).

Deﬁnition If Q is a primary ideal, then the prime ideal P = rad Q is the associated prime to Q, and Q is said to belong to P (or to be P -primary).

Deﬁnition (1) An ideal I in R has a primary decomposition if it may be written as a ﬁnite intersection of primary ideals:

m \ I = Qi Qi a primary ideal. i=1

(2) The primary decomposition above is minimal and the Qi are called the primary components of I if

(a) no primary ideal contains the intersection of the remaining primary ideals, i.e., Qi + ∩j6=iQj for all i, and

(b) the associated prime ideals are all distinct: rad Qi 6= rad Qj for i 6= j.

Deﬁnition A proper ideal I in a commutative ring R is irreducible if I = J ∩ K implies I = J or I = K.

23 Proposition 20 Let R be a Noetherian ring. Then

(1) every irreducible ideal is primary, and

(2) every proper ideal in R is a ﬁnite intersection of irreducible ideals.

Proof of (1). Let Q be an irreducible ideal and suppose ab ∈ Q and b∈ / Q. We must show that an ∈ Q for some n ≥ 1. For each k deﬁne

k Ak = {x ∈ R: a x ∈ Q}.

Ak is clearly an ideal. Also,

A1 ⊆ A2 ⊆ A3 ⊆ · · · .

Since R is Noetherian, this chain stabilizes An = An+1 = ··· for some n. Consider the ideals I = (an) + Q and J = (b) + Q. Note that Q ⊆ I and Q ( J and so Q ⊆ I ∩ J. If y ∈ I ∩ J, we may write y = anz + q for some z ∈ R and q ∈ Q. Since ab ∈ Q, aJ ⊆ Q, and in particular ay ∈ Q. Then

n+1 a z = ay − aq ∈ Q ⇒ z ∈ An+1 = An. ⇒ anz = y − q ∈ Q ⇒ y ∈ Q ⇒ I ∩ J ⊆ Q ⇒ I ∩ J = Q. Since Q is irreducible and (b) + Q 6= Q (since b∈ / Q), we have (an) + Q = Q and so an ∈ Q. Hence, Q is primary.

Proof of (2). Let S be the collection of ideals in R that cannot be written as an intersection of ﬁnitely many irreducible ideals. If S 6= ∅, since R is Noetherian, S contains a maximal element, say I. Since I ∈ S, I is not irreducible, and so I = J ∩ K for some ideals J and K distinct from I. Then I ( J and I ( K. By the maximality of I, J/∈ S and K/∈ S. Hence, J and K can be written and the ﬁnite intersection of irreducible ideals and therefore so can I, a contradiction. So, S = ∅, which proves the proposition.

24 Theorem 21 (Primary Decomposition Theorem) Let R be a Noetherian ring. Then every proper ideal I in R has a minimal primary decomposition. If

m n \ \ 0 I = Qi = Qj i=1 j=1 are two minimal primary decompositions for I, then the sets of associated primes are the same:

0 0 {rad Q1,..., rad Qm} = {rad Q1,..., rad Qn}.

Moreover, the primary components Qi belonging to the minimal elements of this set of associated primes are uniquely determined by I.

Proof. By the previous proposition, any proper ideal I has a primary decomposition

n \ I = Qi i=1 where the Qi are primary. If for some i, Qi ⊇ ∩j6=iQj, we may eliminate that term from the intersection. Hence, we may assume the decomposition satisfies condition (a) in the definition of a minimal decomposition. Since a finite intersection of P - primary ideals is again P -primary (exercise 15.2.31), replacing the primary ideals in the decomposition with the intersections of all those primary ideals belonging to the same prime, we may assume the decomposition satisfies (b) in the definition of a minimal decomposition.

The proof of uniqueness is given in the exercises and in section 15.4.