Spring 2016, lecture notes by Maksym Fedorchuk 51

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1 Spring 2016, lecture notes by Maksym Fedorchuk Problem Set 2 Solution Problem. Prove the following statements. (1) The nilradical of a ring R is the intersection of all prime ideals of R. (2) The radical of an ideal I is the intersection of all prime ideals of R containing I. Proof. For the quotient homomorphism : R! R/I, we have 1 nilrad(r/i) = {f 2 R (f) n = 0 for some n 2 N} = {f 2 R f n 2 I for some n 2 N} = rad(i). Therefore, the second claim follows from the first one, which we proceed to prove. The inclusion nilrad(r) \ p p2spec R is obvious, for f n = 0 implies that f 2 p for every prime p. Suppose f is not nilpotent. Then S = {f n } 1 n=0 is a multiplicative set not containing 0. Hence by Problem 2.6, there exists a prime ideal p disjoint from S. (See below for details.) In particular, f/2 p. This finishes the proof. Problem. Let k be a field. Describe the contraction of the maximal ideal (x 1,y 1) under the ring homomorphism f : k[z 1,z 2,z 3 ]! k[x, y] given by z 1 7! x 2, z 2 7! xy, z 3 7! y 2. We will do this problem by proving a much more general statement: Proposition Let R = k[z 1,...,z m ]/I and S = k[x 1,...,x n ]/J be finitely generated algebras over a field k. Suppose f : R! S is a k-algebra homomorphism defined by f(z i )= i (x 1,...,x n ) for some polynomials i 2 k[x 1,...,x n ]. (Note that 1,..., m have to satisfy F ( 1,..., m) 2 J for every F 2 I.) Then for every (a 1,...,a n ) 2 k n such that J (x 1 a 1,...,x n a n ), the contraction of the induced maximal ideal (x 1 a 1,...,x n a n ) S is a maximal ideal z 1 1 (a 1,...,a n ),...,z n n (a 1,...,a n ) R. Lemma Every k-algebra homomorphism ev a : k[x 1,...,x n ]! k, is defined by ev a (x i )=a i for some a =(a 1,...,a n ) 2 k n and we have ker(ev a )= (x 1 a 1,...,x n a n ). In particular, (x 1 a 1,...,x n a n ) is a maximal ideal for every (a 1,...,a n ) 2 k n.

2 52 Algebra II Commutative and Homological Algebra Proof. This follows from two observations. The first is that ev a F (x 1,...,x n ) = F (a 1,...,a n ) for every F 2 k[x 1,...,x n ]. The second is that F (x 1,...,x n )=F (a 1,...,a n ) mod (x 1 a 1,...,x n a n ). N.B. The above follows from the existence of the Taylor expansion of F (x 1,...,x n ) around (a 1,...,a n ). N.B. A k-algebra homomorphism from a finitely generated k-algebras to k is called an evaluation homomorphism. Lemma completely describes such homomorphisms. Proof. By Lemma , we have that (x 1 a 1,...,x n a n ) S is the kernel of some evaluation homomorphism ev a : S! k defined by ev a (x i )=a i. Consider now the composition homomorphism R f! S ev! k. Then ev a f is the evaluation homomorphism from R to k defined by z i 7! i(a 1,...,a n ). In particular, z 1 1 (a 1,...,a n ),...,z n n (a 1,...,a n ) = ker(ev f), again by Lemma This finishes the proof. Problem. Suppose k is a field (not necessarily algebraically closed; the case of algebraically closed field was done in class). For a homomorphism of finitely generated k-algebras R! S, prove that the contraction of a maximal ideal in S is a maximal ideal in R. Proof. Let n S be a maximal ideal and K = S/n be the corresponding quotient field. Then k K is a finitely generated field extension, hence k K is a finite field extension by Weak Nullstellensatz (Proposition 1.4.1). Let m := n \ R be the contraction of n in R. Then we have a sequence of ring extensions: k R/m K. Since k K is a finite ring extension, we must have that R/m K is a finite ring extension. But both of these rings are domains and K is a field. It follows that R/m is a field by Proposition Problem. Finish the proof of Proposition from the lecture notes by showing that for an irreducible a ne variety X k n, the ideal of polynomials vanishing on X is prime. Proof. Suppose that X is an irreducible a ne variety but the ideal I := I(X) is not prime. Then there exist f,g 2 k[x 1,...,x n ] such that f,g /2 I(X), but fg 2 I(X). Let X 1 = Z(I,f) Z(I(X)) = X and X 2 = Z(I,g) Z(I(X)) = X be algebraic sets. But X 1 [ X 2 = Z (I,f)(I,g) Z(I) =X. It follows that X 1 [ X 2 = X. But because X is irreducible, we conclude that either X 1 = X and X 2 = X. Suppose X 1 = X. Then rad(i,f) = rad(i) =I. (Note that I is automatically radical.) But then f 2 I. A contradiction!

3 Spring 2016, lecture notes by Maksym Fedorchuk 53 Problem. Let k be a field. For integers 1 apple m<n, consider the matrix x 1 x 2 x m 1 x m (10.2.3) x 1 x 2 x 3 x m x m+1 M m,n := B A. x n m x n 1 x n Let I n,m be the ideal in k[x 1,...,x n ] generated by the 2 2 minors of M m,n.prove that Z(I n,m )={(t, t 2,t 3,...,t n ) t 2 k} k n. Proof. Note that for i apple m and j apple n m, the polynomial x i+j x i x j belong to I n,m. Suppose (a 1,...,a n ) 2 Z(I n,m ). Then we must have a i+j = a i a j for all i apple m and j apple n m. It follows that a r = a r 1 for every 1 apple r apple n. Problem. For any ideal I R and a multiplicative set S disjoint from I, prove that there exists a prime ideal p I disjoint from S. Proof. Let R := R/I and let S be the image of S in R. ThenS is a multiplicative set in R that does not contain zero. In particular, the ring S 1 R is not a zero ring. Let q be a prime ideal in S 1 R. Then the contraction of q to R is a prime ideal avoiding S and containing I. (Alternatively, one can apply Zorn s lemma to the set of the ideals containing I and disjoint from S.) Problem. Let p be a prime ideal of a ring R. Prove that the residue field of R p is the fraction field of R/p: R p /pr p = Frac(R/p). Proof. Let : R! R/p be the quotient homomorphism. There is an obvious ring homomorphism : R p! Frac(R/p) that sends a/s 2 R p to (a)/ (s). It is easy to see that is well-defined (do this explicitly if you are still shaky with localization.) Clearly, is surjective: Every element of R/p can be written as (a) for a 2 R and every non-zero element of R/p can be written as (s) for s/2 p. Finally, (a/s) = 0 if and only if (a) = 0 if and only if a 2 p if and only if a/s 2 pr p. Hence ker = pr p. We are done. Problem. For an R-module M and a multiplicative set S R, prove that S 1 R R M ' S 1 M. Proof. We have an obvious R-bilinear map S 1 R M! S 1 M given by r s,m 7! rm s. By the universal property of the tensor product, this gives rise to an R-linear homomorphism f : S 1 R R M ' S 1 M,

4 54 Algebra II Commutative and Homological Algebra r that satisfies f s m = rm s. In fact, it is easy to see that f is also S 1 R-linear. It is a tedious but straightforward exercise to check that m g : S 1! S 1 R R M, g = 1 s s m is a well-defined R-module homomorphism (which is also an S 1 R-module homomorphism). Clearly, f and g are inverses of each other and so define an S 1 R-module isomorphism S 1 R R M ' S 1 M. Problem. For a finitely generated R-module M and a multiplicative set S R, prove that S 1 M = 0 if and only if there exists s 2 S such that sm = 0. Proof. If sm = 0 for some s 2 S, then for every element x = m/t 2 S 1 M,we have x = m t = sm st = 0 st =02 S 1 M. Conversely, suppose S 1 M = 0. Then for every m 2 M, there exists an element s m 2 S such that s m m = 0. (This follows by unwinding what it means for m = m/1 tobezeroins 1 M.) By the assumption, we can choose finitely many generators m 1,...,m n of M. Then it is easy to see that s := s m1 s mn satisfies sm = 0. Problem. Let R be a ring. Suppose that for each prime p, the local ring R p has no non-zero nilpotent elements. Prove that R has no non-zero nilpotent elements. If each R p is an integral domain, is R necessarily an integral domain? Proof. Suppose r 2 R is a nilpotent element. Then r = r remains nilpotent in 1 R p.thuswemusthave r 1 =02 R p, for every prime p. Thus for each primes p, there exists an element in R \ p that kills r. In other words, the annihilator of r Ann(r) :={a 2 R ar =0} is not entirely contained in any of the prime ideals. Since Ann(r) is an ideal itself, we conclude that Ann(r) =R = (1) and so r = 0.

5 Spring 2016, lecture notes by Maksym Fedorchuk 55 Example: Consider a product of two domains, say R = S T. Then a prime ideal in S T is of the form either S q, whereq T is prime, or p T, where p S is prime. It is easy to see that (S T ) S q ' T q. (Indeed, the complement of S q contains 0 1, which allows to kill all non-zero elements of S.) It follows that the localization of R is a domain for all primes of R, butr is clearly not a domain. Definition A ring R with no non-zero nilpotent elements (that is, satisfying nilrad(r) = (0)) is called reduced. Problem. Let R be a ring and S R a multiplicative set. For a ring homomorphism : R! S 1 R, prove that ] (Spec S 1 R)= \ Spec R f. Is the above set necessarily open? Proof. By definition, ] (Spec S 1 R) is the set of prime ideals in R that are contractions of prime ideals in S 1 R. The distinguished open Spec R f Spec R is defined to be the complement of V(f), and so consists of those prime ideals in R that do not contain f. A prime ideal p R is a contraction of a prime ideal in S 1 R if and only if S \ p = ; if and only if f /2 p for every f 2 S if and only if p 2 T f2s Spec R f. This finishes the proof. If we take p to be a prime ideal and S = R\p, then T f2s Spec R f is the set of all prime ideals contained in p. Such a set is generally not open in Zariski topology. For example, if R = Z and p =(p) for a prime p, thenspecr p = {(0), (p)}, which is not open in Spec Z because its complement is not closed. Another example would be R = k[x] and p =(x), etc. f2s