# ALGEBRA HW 4. M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are.

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1 ALGEBRA HW 4 CLAY SHONKWILER (a): Show that if 0 M f M g M 0 is an exact sequence of R-modules, then M is Noetherian if and only if M and M are. Proof. ( ) Suppose M is Noetherian. Then M injects into M, so M can be viewed as a submodule of M; since submodules of Noetherian modules are Noetherian, M is Noetherian. Also, since the sequence is exact, M M/f(M ); since quotients of Noetherian modules are Noetherian, this implies that M is Noetherian. ( ) On the other hand, suppose M and M are Noetherian. Let M M 2 be a chain of submodules of M. Then, by problem set 3 #2, f (M ) f (M 2 ) is a chain of submodules of M and g(m ) g(m 2 ) is a chain of submodules of M. Since M is Noetherian, eventually f (M k ) = f (M k+ ) for all k n for some n N. Similarly, since M is Noetherian, eventually g(m k ) = g(m k+ ) for all k n 2 for some n 2 = N. Now, let n = max{n, n 2 }. Then, for all k n, f (M k ) = f (M k+ ) and g(m k ) = g(m k+ ); using the result proved in problem set 3 #2(c), we have the following commutative diagram with exact rows for all k n: 0 f (M k ) f M k g g(m k ) 0 0 f (M k+ ) f M k+ g g(m k+ ) 0 Thus, by the five lemma, the map M k M k+ is an isomorphism; since M k M k+ and this map is simply the inclusion map, this implies that M k = M k+ for all k n. Since our choice of a chain of submodules of M was arbitrary, we see that all chains of submodules of M eventually stabilize, which is precisely what it means for M to be Noetherian. (b): Deduce that R-modules N and N 2 are each Noetherian if and only if N N 2 is.

2 2 CLAY SHONKWILER Proof. For any R-modules N and N 2, we have the natural exact sequence 0 N N N 2 N 2 0; by our result in part (a), this in turn means that N and N 2 are each Noetherian if and only if N N 2 is Noetherian. (c): Let S be a multiplicative subset of a commutative ring R. Show that if R is Noetherian then so is the localization S R. Does the converse also hold? Proof. Let I = ( r s, r 2 s 2,... ) be an ideal in S R. Let a I. Then a = i b i r i s i = i b i s i r i where all but finitely many b i are zero. Then b i s i S R, so we see that a is a linear combination in S R of the r i ; since our choice of a was arbitrary, we see that, in fact, I = (r, r 2,...). Now, (r, r 2,...) can also be viewed as an ideal of R; since R is Noetherian, this ideal is finitely generated. That is, the contraction of I to R is equal to (a,..., a n ) for some a i R. Consider (a,..., a n ) as an ideal in S R. Certainly (a,..., a n ) I. To see the other containment, let a I. Then each r i = n i= c ia i, so a = i b i s i r i = i b i s i n j= c j a j = i,j b i c j s i a j. Since all but finitely many of the b i are zero, we see that a is a finite linear combination of the a j and so a (a,..., a n ). Since our choice of a was arbitrary, we see that I (a,..., a n ). Having shown both containments, we conclude that I = (a,..., a n ), so I is finitely generated. Since our choice of I was arbitrary, we conclude that all ideals of S R are finitely generated, and so S R is Noetherian. The converse, on the other hand, fails to hold. To see why, consider R = R[x, x 2,...] and let S = {x, x 2,...}. Then, as we ve seen before, R[x, x 2,...] is not Noetherian (specifically, (x, x 2,...) is not finitely generated). However, S R = R[x, x 2,...] (x,x 2,...) = R(x, x 2,...) is a field and so is trivially Noetherian. 2 (a): Do there exist infinite strictly increasing chains I I 2 of ideals in C[x]? Do there exist finite strictly increasing chains I I 2 I n of ideals in C[x] with arbitrarily large n?

4 4 CLAY SHONKWILER fields V and W over a field k, then V k W 0 (specifically, if e and f are basis elements of V and W, then e f is part of a basis for V k W ). Hence, since M/mM R/m N/mN = 0, this implies that either M/mM = 0 or N/mN = 0. Suppose that M/mM = 0. Then 0 = M/mM = M R R/m and so, by Nakayama s Lemma, M = 0. Similarly, if N/mN = 0, then N = 0 Therefore, we conclude that either M = 0 or N = 0. (b): What if R is not local? Answer: The above statement does not necessarily hold if R is not local. For example, consider R = Z and the Z-modules Z/2 and Z/3. Then Z/2 Z Z/3 = 0, even though neither Z/2 nor Z/3 is 0. 4 Let M be an R-module and let 0 N N N 0 be an exact sequence of R-modules. Under each of the following four conditions (considered separately), either show that the sequence 0 M N M N M N 0 must be exact or else give a counterexample. (a): M is flat. Answer: By definition, if M is flat, then 0 M N M N M N 0 is exact. (b): N is flat. Counterexample: Consider the following exact sequence: 0 Z 2 Z (mod 2) Z/2 0. Let M = Z/2. Then, since Z/2 Z Z = Z/2 and Z/2 Z Z/2 = Z/2, the application of M Z yields the following sequence: 0 Z/2 2 Z/2 id Z/2 0. Now, in this sequence, multiplying by 2 kills everything in Z/2, so ker 2 = Z/2 0 = im 0, so this sequence is not exact. (c): N is flat. Counterexample: In the example given in (b) above, the middle term, Z, is flat, so the example in (b) is also a counterexample for this situation.

5 ALGEBRA HW 4 5 (d): N is flat. Answer: By the definition of Tor (M, ) as the universal δ- functor for M, Tor (M, N ) M N M N M N 0 is a long exact sequence. Now, since N is flat and Tor (, N) is also the universal δ-functor for N (that is, Tor is a balanced functor), it must be the case that Tor (M, N ) = 0; hence, the tail end of the above sequence reduces to 0 M N M N M N 0 which must also be exact. 5 Let R be a commutative ring with Jacobson radical J. For x R, let P x be the set of r R such that r (mod x) (i.e. such that r xr). Let R denote the multiplicative group of units in R. (a): Show that J = {x R P x R }. Proof. Let S = {x R P x R }. Let x S. Let m be a maximal ideal of R and suppose x / m. Since m is maximal, this implies that (x, m) = (), so there exists a R and b m such that = ax + b. Re-arranging this equation yields: b = ax xr, so b P x R. However, this implies that b is a unit and so that m = (), contradicting the fact that m is maximal and, therefore, proper. From this contradiction, then, we see that x m. Since our choice of maximal ideal m was arbitrary, this in turn implies that x m for all maximal ideals m and, hence, x m = J. m maximal Since our choice of x S was arbitrary, this implies that S J. On the other hand, suppose x J. Let r P x. Then r xr J. If r is not invertible, then (r) is a proper ideal and so r m for some maximal ideal m. However, since r J, r n for all maximal n; specifically, r m. Hence, r, r m and so = r (r ) m, meaning m = (), contradicting the fact that m is maximal and, therefore, proper. Hence, we see that r R. Since our choice of r P x was arbitrary, we see that P x R and so x S. Since our choice of x was arbitrary, this in turn implies

6 6 CLAY SHONKWILER that J S. Thus, having shown containment in both directions, we conclude that J = S. (b): Let M be a finitely generated R-module and let a,..., a n M. Show that the R-module M is generated by a,..., a n if and only if the R/J-module M/JM is generated by the images of these elements. Proof. Certainly, if a,..., a n generate M, then their images must generate M/JM. To prove the converse, we will need the following generalization of Nakayama s Lemma: Lemma 5.. Let R be a ring with Jacobson radical J and M a finitely generated R-module. If M R/J = 0, then M = 0. Proof. Suppose M 0. Let b,..., b k be a minimal generating set of M. Then k. Since 0 = M R/J = M/JM, we have that M = JM. Then b M and k b = j i b i for j i J. Hence, i= ( j )b = k j i b i. i=2 Now, ( j ) = j J, so, by the result proved in (a), j R. Let h = ( j ). Then multiplying both sides above by h, we see that k b = hj i b i. i=2 Thus, b is a linear combination of b 2,..., b k, so b,..., b k = b 2,..., b k, meaning that {b,..., b k } was not a minimal generating set. From this contradiction, then, we conclude that, in fact, M = 0. Now, suppose the images a,..., a n generate M/JM. Let N be the submodule of M generated by a,..., a n. Then M/N R R/J = 0, so, by Lemma 5., M/N = 0. Therefore, N = M, so we see that a,..., a n generate M. 6 Let M be a finitely generated R-module. Prove that the following conditions on M are equivalent: (i): M is locally free over R (i.e. M m is free over R m for all maximal ideals m R).

7 ALGEBRA HW 4 7 (ii): For every maximal ideal m R there is an f / m such that M f is free over R f. (iii): f,..., f n R such that (f,..., f n ) = and M fi is free over R fi for i =,..., n. Proof. (i) (ii): Suppose m is a maximal ideal of R. Then, by hypothesis, M m is free over R m. Let m,..., m n be a basis for M m over R m and let {g,..., g n } be a generating set for M. Then for each g i, n r i,j g i = m j s i,j j= for r i,j R and s i,j / m. Now, let h f = i,j Note that f / m. Let f N for some a i R. Hence, h n f N = i= a ig i f N s i,j. M f. Then h M and so h = n i= a ig i = i a i r i,j f N m j. s i,j However, since = f/s i,j s i,j f and, by construction, f/s i,j R, we see that h f N = i,j a i r i,j (f/s i,j ) f N+ m j. Note that a ir i,j (f/s i,j ) M f N+ f. Thus the m j form a generating set for M f. Since the m j are linearly independent in M m M f, we see that {m,..., m n } forms a basis for M f over R f, so M f is free over R f. (ii) (iii): Let S = {f R M f is free over R f }. Let (S) be the ideal generated by the elements of S, and suppose (S) (). Then, since (S) is a proper ideal of R, there exists a maximal ideal m containing (S). Then, by hypothesis, there exists f / m such that M f is free over R f. However, this implies that f S and, hence, f (S) m. From this contradiction, then, we see that (S) = (). Since M is finitely generated, there is some finite generating set f,..., f k S such that (f,..., f k ) = (S) = (). Since each f i S, M fi is free over R fi for all i =,..., n. (iii) (i): Let m be a maximal ideal of R. By hypothesis, there exist f,..., f n R such that (f,..., f n ) = () and M fi is free over R fi for all i. Now, since m is proper, there exists some f k such that f k / m and M fk is free over R fk. Since free implies locally free and m fk is maximal in R fk, this in turn implies that (M fk ) mfk is j

8 8 CLAY SHONKWILER free over (R fk ) mfk. However, since f k R\m, (M fk ) mfk = M m and (R fk ) mfk = R m, so we see that M m is free over R m. 7 Let M be an R-module. Prove that the following conditions on M are equivalent: (i): M is faithfully flat. (ii): For every complex of R-modules N N N, the given complex is exact iff M N M N M N is exact. (iii): For every homomorphism of R-modules φ : N N 2, φ is surjective iff φ : M N M N 2 is surjective. Proof. (i) (ii): If N N N is a complex, then it is also a sequence and so, since M is faithfully flat, N N N is exact if and only if M N M N M N is exact. φ (ii) (iii): Let φ : N N 2 be surjective. Then N N2 0 φ is exact, and so, by (ii), M N M N 2 0 is exact and, therefore, φ : M N M N 2 is surjective. On the other φ hand, if φ : M N M N 2 is surjective, then M N φ M N 2 0 is exact; hence, by (ii), N N2 0 is exact, meaning that φ : N N 2 is surjective. (iii) (i): 8 Let A be a flat R-algebra. Prove that the following conditions on A are equivalent: (i): A is faithfully flat over R. (ii): Every maximal ideal of R is the contraction of a maximal ideal of A. (iii): Spec A Spec R is surjective. (iv): For every R-module N, if A R N = 0, then N = 0. Proof. (i) (ii) Let m be a maximal ideal in R. Then m R, the inclusion map, is not surjective. Therefore, by the contrapositive of part of problem 7 above, m R A R R A is not a surjection. Now, m R A = m e, the extension of m to A. On the other hand, R R A = A, so we see that the inclusion m e A is not a surjection, so m e is a proper ideal of A and, therefore, is contained in some maximal ideal n of A. Now, if we denote the contraction of an ideal a of A to R by a c, then m m ec n c.

9 ALGEBRA HW 4 9 However, since n is a proper ideal, / n and so / n c, meaning n c is a proper ideal of R. Therefore, since m is maximal in R, it must be the case that m = n c, so m is the contraction of a maximal ideal of A. Since our choice of maximal ideal m was arbitrary, we conclude that every maximal ideal of R is the contraction of a maximal ideal of A. (ii) (iv) Suppose N 0 is an R-module such that N R A = 0. Let x 0 be an element of N. Then the map f : R xr given by a xa is surjective. Hence, if k = ker f, xr R/k. Now, consider the module xr R A R/k R A A/k e, where k e denotes the extension of the ideal k to A. Now, since f() = x 0, k is a proper ideal of R and so is contained in some maximal ideal m. By hypothesis, m = n c for some maximal n in A. Therefore, A n k e, so A/k e 0 and, hence, xr R A 0. Now, since A is flat and 0 xr R given by the inclusion map is exact, 0 xr R A R R A is exact; specifically, the map xr A R A is injective. Now, as we ve just seen, xr A 0, so this implies that R A 0. From this contradiction, then, we conclude that for every R-module N, if N R A = 0, then N = 0. (iv) (i) Suppose φ : N N 2 such that φ : A N A N 2 is surjective. Then is exact. Since A is flat, 0 φ(n ) N 2 N 2 /φ(n ) 0 0 A φ(n ) A N 2 A (N 2 /φ(n )) 0 is exact. Now, we can distribute the tensor across quotients, so A (N 2 /φ(n )) (A N 2 )/(A φ(n ). Recall that φ : A N A N 2 is surjective. Thus, A φ(n ) = im φ = A N 2, so 0 = (A N 2 )/(A φ(n )) = A (N 2 /φ(n )). Therefore, by hypothesis, N 2 /φ(n ) = 0, and so N 2 = φ(n ), meaning that φ : N N 2 is surjective. (i) (iii) Since we ve shown that (i) (ii) (iv), we should feel free to use (ii) or (iv) if they should come in handy. Now, first of all, let p be a prime ideal of R and consider the localization A p as a module over R p. We suppose first that A p is faithfully flat over

10 0 CLAY SHONKWILER R p (which we ll show in a minute). Note that pr p is the unique maximal ideal in p; by (ii) there exists a maximal m in A p such that m c = pr p. Consider the following diagram: R φ A R p φ A p where φ is the map induced by the map φ : R A. Then pr p = m c = φ (m). Let q = A m. Then φ (q) = φ (m A) = φ (m) φ (A) = p. Since our choice of p was arbitrary, we see that Spec A Spec R is surjective. Now, to justify our assumption that A p is faithfully flat over R p, suppose p Spec R. Then R p is flat over R and A p = A R R p. Now, suppose M and N are R p modules such that f : M N induces a surjective map f : A p Rp M A p Rp N. Then, since A p = A R R p, the following sequence is exact: (A R R p ) Rp M f (A R R p ) Rp N 0. Now (A R R p ) Rp M = A R (R p Rp M) = A R M and similarly for N, so the above sequence reduces to: A R M A R N 0. Since A is faithfully flat over R, this in turn implies that M N 0 is exact, which is to say that f : M N is surjective. (iii) (ii) Suppose m is maximal in R. Then m is prime in R, so m = p c for some prime ideal p of A by hypothesis. Now, p is contained in some maximal ideal n; since n is proper, it does not contain, so n c does not contain. Furthermore, since m = p c n c, we see that, in fact, m = n c. Since our choice of m was arbitrary, we conclude that every maximal ideal of R is the contraction of a maximal ideal of A. 9 For each of the following R-algebras A, determine whether A is a finitely generated R-module and whether it is a finitely generated R-algebra. Also determine whether the R-module A is flat and whether it is faithfully flat. (a): R = Z, A = Z[x]/(3x). Answer: A is not finitely generated as an R-module. To see why, suppose {f,..., f n } were a finite generating set. Let N =

11 max{deg(f i )}. Then for any a i R, n a i f i ALGEBRA HW 4 i= has degree at most N, so, for example, x N+ cannot be written as such a linear combination. Since x N+ A, this implies that A is not finitely generated as an R-module. On the other hand, {, x} serves as a generating set for A as an R-algebra, so A is finitely-generated as an R-algebra. Now, since x A and 3x = 0 in A, we see that A has 3-torsion elements; since a flat module over a domain cannot have torsion, this implies that A is not a flat module. (b): R = Z, A = Z[/5]. Answer: A is not finitely generated as an R-module. To see why, suppose {a,..., a n } were a finite generating set. Then each a i = k i j= b i,j 5 l i,j. Let N = max{l i,j }. Then 5 N+ cannot be written as a linear combination of the a i, so we see that A is not finitely-generated as an R-module. On the other hand, {, 5} does serve as a generating set of A as an R-algebra, so we see that A is finitely generated as an R-algebra. Furthermore, since A is an integral domain and is a ring extension of R, A is torsion-free as an R-module. Therefore, since torsion-free modules over a PID are flat and Z is a PID, we see that A is a flat module. Now, consider the maximal ideal (5) in Z. 5 is a unit in A = Z[/5], so we see that 5 cannot be contained in any ideals of A. Hence, (5) is not the contraction of any maximal ideal in A and so, by the criterion for faithful flatness given in #8(ii), A is not faithfully flat. (c): R = Z, A = Z[i]. Answer: Note that any element of the Gaussian integers is of the form a + bi for a, b Z. Hence, we see that {, i} forms a generating set for A as an R-algebra, so A is finitely-generated as an R-module (and, consequently, also as an R-algebra). Furthermore, if a+bi = 0 for some a, b Z, then a = b = 0, so we see that, in fact, A is a free R-module and, therefore, flat. Since {, i} is a basis for A, A R 2 and so, if N is an R-module such that 0 = A R N, then 0 = A R N R 2 R N = (R R N) (R R N) = N 2,

12 2 CLAY SHONKWILER so N = 0. Hence, by criterion 8(iv), A is faithfully flat. (d): R = Z, A = Z[i, /5]. Answer: Since we saw in (b) that Z[/5] is not finitely generated as a Z-module and since A = Z[i, /5] Z[/5][i], this implies that A{ is not finitely generated as a Z-module. However, it s clear that, 5, i} form a generating set for A as an R-algebra, so A is finitely generated as an R-algebra. Furthermore, since A is an integral domain and a ring extension of R, A is torsion-free and, therefore, flat. However, (5) is a maximal ideal in Z, whereas 5 is a unit in Z[i, /5], so 5 is not contained in any ideals of A and, therefore, (5) is not the contraction of any maximal ideal of A. Therefore, by criterion 8(ii), A is not faithfully flat. (e): R = Z, A = Z[i, /(2 + i)]. Answer: A is not finitely-generated as an R-algebra. To see why, suppose {a,..., a n } were a finite generating set. Then each a j = l j k= b j,k i j,k (2 + i) m j,k. Let N = max{m j,k }. Then cannot be written as a linear (2+i) N+ combination of the a j, so we see that A is not finitely-generated as an R-module. { } On the other hand,, i, 2+i is a generating set for A an an R-algebra, so A is finitely-generated as an R-algebra. Furthermore, since A is an integral domain and a ring extension of R, A is torsionfree and, therefore, flat. Finally, note that the maximal ideals in R are of the form (p) for p prime. Then, as we saw on PS8#r(a) from last semester, either p is prime in Z[i] or p = β p β p for β p and β p prime in Z[i]. So long as p 5, β p, β p 2 + i, so β p, β p are prime in A. Furthermore, as we saw in PS9#4(c), the contraction of (β p ) to Z is just (p), so we see that (p) is the contraction of a maximal ideal in A to R. Of course, 5 = (2 + i)(2 i); since 2 + i is a unit in A, we have to be a little more careful. However, 2 i / A and so 2 i is not a unit in A. In addition, 2 i is prime in Z[i] and is not divisible by 2+i or any of its multiples, ) so 2 i is prime in A. Furthermore, the contraction of is, by the same reasoning, simply (5). Therefore, we see that ( 2 i all maximal ideals of Z are contractions of maximal ideals in A and so, by criterion 8(ii), A is faithfully flat. (f): R = R[x], A = R[[x]].

13 ALGEBRA HW 4 3 Answer: A is not finitely-generated as an R-module. To see why, suppose {f (x),..., f n (x)} were a finite generating set for A. Let n h(x) = g i (x)f i (x) i= be a linear combination of the f i. Then h is simply a polynomial with real coefficients. However, there are elements of A that are not polynomials, so we see that this cannot be a generating set for A and, hence, A is not finitely-generated as an R-module. Since R[x] is a PID and A is torsion-free, A is a flat R-module. However, A is local and so has exactly one maximal ideal, whereas R has infinitely many maximal ideals (for example, (x a) is maximal for all a R), so we see that not every maximal ideal in R can be the contraction of a maximal ideal in A, so, by criterion 8(ii), A is not faithfully flat. (g): R = R[x] (x), A = R[[x]]. Answer: DRL 3E3A, University of Pennsylvania address:

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