# ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS

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1 ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS UZI VISHNE The 11 problem sets below were composed by Michael Schein, according to his course. Take into account that we are covering slightly different material. (1) Modules (a) Prove the Five Lemma: Let R be a ring and consider the following commutative diagram of R-modules: g 1 g A 1 2 g A 2 3 g A 3 4 A 4 A 5 f 1 f 2 B 1 h 1 B 2 h 2 B 3 h 3 B 4 h 4 B 5 f 2 Suppose that the two rows are exact and that f 1, f 2, f 4, and f 5 are isomorphisms. Prove that f 3 is also an isomorphism. (b) Let M be a finitely generated R-module. Prove that M has a maximal submodule. (c) Recall that l(m) is the composition length of a module M. If M = M 1 M 2, prove that l(m) = l(m 1 ) + l(m 2 ). (d) Let R be a commutative ring and let M be an R-module generated by m elements. Suppose that there is a surjective map φ : M R (n). Prove that m n. (e) Let R be a commutative ring. An element s R is called regular if sr 0 for all 0 r R. Let M be an R-module, and define tor(m) = {m M : sm = 0 for some regular s R}. Prove the following: (i) tor(m) is a submodule of M, and tor(r) = 0. (ii) tor(m 1 M 2 ) tor(m 1 ) tor(m 2 ). (iii) If A = M/tor(M) is a free R-module, then M A tor(m). (f) Let N 1, N 2, and K be submodules of M such that N 1 N 2 and K N 1 = 0. Then (K + N 1 )/(K + N 2 ) N 1 /N 2. f 4 f 5 Date: December 19,

2 2 UZI VISHNE (g) Let 0 K M N 0 be an exact sequence of R- modules. If K is free of rank k and N is free of rank n, prove that M is free of rank k + n. (2) Modules and composition series (a) Recall that l(m) is the length of a composition series, if it exists, of a module M. If M = M 1 M 2, prove that l(m) = l(m 1 ) + l(m 2 ). (b) Let R be an arbitrary ring, and let M be a simple R- module. Prove that there exists a maximal left ideal I R such that M R/I as R-modules. (c) Let F be a field and n 1, and let T n (F ) be the ring of all upper triangular n n matrices with entries in F. Consider F n as a T n (F )-module in the obvious way: if v F n and A T n (F ), write v as a column and define the scalar multiplication A v to be the product of matrices Av. Find a composition series for F n as a T n (F )-module. (3) Prime ideals, localization (a) Let R be a commutative ring. Prove that an R-module M is Noetherian if and only if for every sequence f 1, f 2,... of elements of M there exists N N such that for every n 1 it is possible to write f n in the form f n = N i=1 r inf i, where r 1n, r 2n,... r Nn R. (b) Let p be a prime number. Let S = {p k : k N} and let M = S 1 Z/Z. In other words, { m } M = n + Z : n = pk, k N. Show that the Z-module M is Artinian but not Noetherian. (c) Let I be an ideal of a commutative ring R. Prove that the following conditions are equivalent: (i) I is a prime ideal (for any elements x, y R, we have xy I if and only if x I or y I). (ii) The quotient ring R/I is an integral domain. (iii) If J 1, J 2 are ideals of R with J 1 J 2 I, then J 1 I or J 2 I. (iv) If J 1, J 2,..., J n are ideal of R such that J 1 J 2 J n I, then J i I for some 1 i n. (v) The complement R\I is closed under multiplication. (d) Let R be a commutative ring, and let I R be an ideal. Given x R, consider the ideal J x = {r R : rx I}. Suppose that J x and Rx + I are both finitely generated ideals of R. Prove that I is finitely generated.

3 ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS 3 (e) Let R be a commutative ring, and let I R be an ideal. Suppose that I is not finitely generated, but that all ideals of R that strictly contain I are finitely generated. Prove that I is a prime ideal. (f) Let R be a commutative ring. Prove that R is Noetherian if and only if all prime ideals are finitely generated. (g) Consider the Z-module Q/Z. Is it Noetherian? Is it Artinian? (h) Let R be a ring. An element x R is called nilpotent if there exists n 1 such that x n = 0. Let N be the set of all nilpotent elements of R. If R is commutative, prove that N is an ideal. If R is commutative and Noetherian, prove that there exists a natural number m such that N m = 0. (i) Let R be a commutative ring. An ideal I R is called a nilpotent ideal if there exists n 1 such that I n = 0. Give an example of a commutative ring such that the ideal of all nilpotent elements is not a nilpotent ideal. (4) Affine algebras (a) Let F be a field. The polynomial F -algebra F [λ 1, λ 2 ] is clearly affine. But prove that the subalgebra generated by 1, λ 1 λ 2, λ 1 λ 2 2, λ 1 λ 3 2,... is not affine. This shows that a subalgebra of an affine algebra is not necessarily affine. (b) Prove that any affine F -algebra has countable dimension as an F -vector space. (c) Let F be an algebraically closed field and let R be an affine F -algebra. Prove that there is only one non-zero simple R- module up to isomorphism. (d) Let R be a commutative ring. For n 1, consider the set M n (R) of n n matrices with entries in R. It is a ring under the usual addition and multiplication of matrices. Show that every two-sided ideal I M n (R) is of the form I = M n (J), where J is an ideal of R. Hint: Take J to be the set of all elements of R that appear as the entry in the top left corner of an element of I. (e) Let F be a field and consider the affine algebra R = F [x]. For all n 1, show that the subalgebra F [x 2n 1, x 2n + x, x 2n+1 ] R is equal to R. (f) Prove that S = F [x 3 +x, x 2 ] is a proper subalgebra of F [x] by showing that x S. Hint: Show that every element of S may be written as f + g(x 3 + x), where f and g are polynomials that involve only even powers of x.

4 4 UZI VISHNE (5) Dedekind domains (a) Let A be a commutative ring and let R be a commutative A-algebra. Let r 1, r 2,..., r n be integral over A. Prove that A[r 1, r 2,..., r n ] is finitely generated as an A-module. (b) Let A be a Dedekind domain. Prove that it is a UFD (unique factorization domain) if and only if it is a PID (principal ideal domain). (c) Consider the field K = Q( d), where d is a square-free integer. Prove that { Z[ d] : d 2, 3 mod 4 O K = [ ] Z : d 1 mod d 2 (d) Let R = Q[x, y], and let I R be the ideal I = (x 2 y 3 ). Show that A = R/I is an integral domain, but that A is not integrally closed. Hint: Find a subring of the polynomial ring in one variable Q[t] that is isomorphic to A. (e) Prove that every UFD is integrally closed. (6) Dedekind domains (a) Let K be a number field (i.e. K is a finite extension of Q) and let O K be the integral closure of Z in K. Prove that FracO K = K. Hint: Prove the following more precise statement. Let y K. Then c n y n + c n 1 y n c 0 = 0 for suitable n and suitable c 0,..., c n Z. Let b = lcm(c 0,..., c n ). Show that y = a/b for some a O K. (b) An integral basis of K is a collection of elements β 1,..., β n O K such that: (i) K = β 1 Q + β 2 Q + + β n Q. (ii) O K = β 1 Z + β 2 Z + + β n Z. Every number field K has an integral basis, and you may assume this. Prove that O K is a Dedekind domain. Hint: Let P O K be a non-zero prime ideal. Show that P Z = pz for some prime p. The integral basis may help you prove that O K /P is a field. (c) Suppose that B and B are transcendence bases of R. Prove that they have the same cardinality. (We did this in class for B and B finite). Hint: Each element of B is algebraically dependent on a finite number of elements of B, and the union of these finite subsets is all of B.

5 ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS 5 (d) Prove the Noether Normalization Lemma for arbitrary fields F. Hint: This can be done using the following variation of the proof we saw in class for infinite fields F. Recall that R = F [a 1,..., a n ], and if tr.deg F (R) < n, let f F [λ 1,..., λ n ] be a polynomial such that f(a 1,..., a n ) = 0. Write f = γ i1,...,i n λ i 1 1 λ in n. Now let u j be the highest degree of λ j that appears in any monomial of f, and define u = 1 + max{u 1,..., u n }. Now set ˆf = f(λ 1 + λ un 1 n, λ 2 + λ un 2 n,..., λ n 1 + λ u n, λ n ) and define c i = a i a un 1 n for 1 i n 1. Then ˆf(c 1,..., c n 1, a n ) = 0. Set R = F [c 1,..., c n 1 ] and define h R [λ n ] by h(λ n ) = ˆf(c 1,..., c n 1, λ n ). Now show that h has an invertible leading coefficient. (7) Algebraic geometry (a) Let F be a field. Recall that if Z F n is an algebraic set, we can associate to it the ideal I(Z) = {f F [x 1,..., x n ] : f(a 1,..., a n ) = 0 (a 1,..., a n ) Z}. Similarly, given an ideal I F [x 1,..., x n ], one can associate to it the algebraic set Z(I) = {(a 1,..., a n ) F n : f(a 1,..., a n ) = 0 f I}. For any algebraic set Z F n, prove that Z = Z(I(Z)). (b) Prove that an algebraic set Z F n is irreducible if and only if I(Z) is a prime ideal. (c) Let R be a Noetherian ring, and let φ : R R be a ring homomorphism. Prove that if φ is surjective, then it is an isomorphism. (d) Let F be a field. Does the ring extension F [x] F [x, y] satisfy INC? (e) Let A R be an extension of commutative rings. Let Q R be any ideal. Show that if R is integral over A, then R/Q is integral over A/(Q A). (8) Integral extensions (a) Let F be a field, and suppose the integral domain R is an F -algebra. If r R is algebraic over F, prove that r 1 Frac R is also algebraic over F. (b) Let A R be integral domains, and suppose that R is integral over A. Prove that R is a field if and only if A is a field.

6 6 UZI VISHNE (c) Let F K be an extension of fields such that K is algebraic over F. Let α K. Then I α = {f F [x] : f(α) = 0} is a non-zero ideal in F [x]. Since F [x] is a PID, I α is principal. Therefore there is a uniquely defined monic polynomial f α such that I α = (f α ). This f α is called the minimal polynomial of α. Suppose that A is an integrally closed domain such that Frac A = F. Prove that α K is integral over A if and only if f α A[x]. Hint: One direction is trivial. To prove the other, let K E be the splitting field of f α. Then in E[x] we have f α (x) = n i=1 (x β i), where β 1 = α. Show that every β i is integral over A and express the coefficients of f α in terms of the elements β i. (d) The aim of the remaining exercises is to prove the Going- Down Theorem: Let A R be an extension of integral domains, where A is an integrally closed domain and R is integral over A. Then A R has the property GD. Recall that this means that, given prime ideals P 0 P 1 A and a prime ideal Q 1 spec R lying over P 1, there exists Q 0 spec R such that Q 0 Q 1 and Q 0 lies over P 0. Let S 0 = A P 0, and define S = {ar : a S 0, r R Q 1 } R. Show that S is a monoid under multiplication. (e) Let P 0 be the ideal of R generated by the set P 0 (note that P 0 is an ideal of A, but not necessarily of R). Suppose that there exists an element s S P 0. Then we may write s = ar, where a S 0 and r R Q 1. Similarly, we can write s = m i=1 p ir i for a suitable m, where p i P 0 and r i R. Show that there exist h 0, h 1,..., h m 1 P 0 such that g(s) = 0, where g(x) = x m + h m 1 x m h 1 x + h 0. Hint: Let M = A[r 1,..., r m ]. Prove that M is finitely generated as an A-module. Consider the map φ : M M given by φ(x) = sx (why is this indeed a homomorphism of modules?) and use the proof of Nakayama s lemma. (f) We use the notation of the previous question. Let F = Frac A, and let f s (x) = x n +d n 1 x n 1 + +d 1 x+d 0 F [x]

7 ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS 7 be the minimal polynomial of s. Show that d i P 0 for all 0 i n 1. Hint: The ring (A/P 0 )[x] is a UFD. (g) Let f r (x) = x n + d n 1x n d 1x + d 0 F [x] be the minimal polynomial of r. Why does f r indeed have degree n? Prove that d i = a n i d i for all 0 i n 1. (h) Prove that d i P 0 for all 0 i n 1 and deduce from this that r Q 1. On the other hand, we know that r R Q 1. It follows from this contradiction that S P 0 =. (i) Consider the set S = {Q R : S Q =, P 0 Q} of ideals of R. By the result of the previous question, S is non-empty. Prove that it contains a maximal element and that any maximal element is a prime ideal. (j) Let Q 0 be a maximal element of S. Prove that Q 0 Q 1 and that Q 0 A = P 0. Therefore the extension A R does indeed satisfy the property GD. (9) Localization (a) Let R be a ring and S a sub-monoid such that S Z(R). In class we defined an equivalence relation on R S as follows: (r 1, s 1 ) (r 2, s 2 ) if there exists s S such that r 1 s 2 s = r 2 s 1 s. We defined S 1 R to be the set of equivalence classes and claimed that it can be given a ring structure as follows: (r 1, s 1 ) + (r 2, s 2 ) = (r 1 s 2 + r 2 s 1, s 1 s 2 ) (r 1, s 1 ) (r 2, s 2 ) = (r 1 r 2, s 1 s 2 ). Show that these operations are well-defined. (b) Let R be a ring and I R a left ideal. Let R = R/I, and let S = {s + I : s S} R. Prove that S 1 R S 1 R/S 1 I. (c) Construct an integral domain with exactly 5772 maximal ideals. (d) Let F be a field and let R = F [[x]] be the ring of power series with coefficients in F. Show that R is a local ring. (e) Let R be a commutative ring and S a sub-monoid (not containing zero). If M is an R-module, we define an equivalence relation on M S by (m 1, s 1 ) (m 2, s 2 ) if there exists s S such that s(s 2 m 1 s 1 m 2 ) = 0. Show that this is an equivalence relation, that the set S 1 M of equivalence classes can be made into an abelian group by defining (m 1, s 1 ) + (m 2, s 2 ) = (s 2 m 1 + s 1 m 2, s 1 s 2 ), and that it is a

8 8 UZI VISHNE well-defined S 1 R-module under the scalar multiplication (r, s) (m, s ) = (rm, ss ). (f) Let R be a commutative ring and M an R-module. If P R is a prime ideal and S = R P is its complement, we write R P for S 1 R and M P for S 1 M. Prove that the following three statements are equivalent: (i) M P = 0 for all prime ideals P R. (ii) M Q = 0 for all maximal ideals Q R. (iii) M = 0. (10) Krull dimension (a) Let R be a commutative ring. Let I R be an ideal, and suppose that I r i=1 P i, where P 1, P 2,..., P r R are prime ideals. Prove that there exists some 1 j r such that I P j. (b) Let R be a Noetherian UFD. Let P R be a prime ideal such that ht R (P ) = 1. Prove that P is principal. (c) Suppose that R is a Noetherian domain. Prove that every non-zero element can be factored as a product of irreducible elements. (Recall that x R is irreducible if for all pairs y, z R such that x = yz, either y or z is a unit.) (d) Let R be a Noetherian domain. Prove that R is a UFD if and only if every prime ideal P R such that ht R (P ) = 1 is principal. (e) Let R be a Noetherian ring. Let P R be a prime ideal such that ht R (P ) = r. Prove that there exist a 1,..., a r P such that P is minimal over the ideal I = Ra 1 + Ra r and I cannot be generated by fewer than r elements. (f) Here is an example, due to Nagata, of a Noetherian ring with infinite Krull dimension. Let F be a field and let R = F [λ 1, λ 2,... ] be the commutative polynomial ring in (countably) infinitely many variables. Consider the ideal P i = λ i(i 1)/2+1,..., λ i(i+1)/2 1, λ i(i+1)/2. Show that P i is a prime ideal with i generators. Let S = R\( i=1 P i). Show that the ring S 1 R is Noetherian and that if M S 1 R is a maximal ideal, then M = S 1 P i for some i. Show that ht S 1 R(S 1 P i ) = i. Conclude that Kdim(R) =. (11) Algebraic geometry and Krull dimension (a) Let F be a field. If V F n is an irreducible algebraic variety of dimension n 1, prove that it is defined by a

9 ADVANCED COMMUTATIVE ALGEBRA: PROBLEM SETS 9 single polynomial, i.e. that there exists f F [λ 1,..., λ n ] such that V = {(a 1,..., a n ) F n : f(a 1,..., a n ) = 0}. (b) Let F be an algebraically closed field, and let Z F n be any algebraic set. Prove that there exists a finite number of irreducible algebraic sets V 1,..., V r such that Z = V 1 V 2 V r. Recall that an algebraic set Z is called irreducible if for any two algebraic sets Z 1, Z 2 such that Z = Z 1 Z 2, one always has either Z 1 Z 2 or Z 2 Z 1. (c) Let F be a field which is not algebraically closed. Prove that the Nullstellensatz fails for F, in other words that there exists a radical ideal I F [λ 1,..., λ n ] for which there is so algebraic set Z F n such that I = I(Z). (d) Let R be a Noetherian domain. Prove that R is a principal ideal domain (PID) if and only if every maximal ideal is principal. (e) Let C C 3 be the curve defined parametrically as follows: C = { (t 3, t 4, t 5 ) : t C }. Prove that C is an algebraic set. Let P C[x, y, z] be the ideal I(C), and justify our choice of notation by proving that P is prime. (f) Let C C 3 and the ideal P C[x, y, z] be as in the previous question. Prove that ht C[x,y,z] (P ) = 2 but that P cannot be generated by any two elements. This shows that the inequality in the Hauptidealsatz is sometimes a strict inequality.

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