Algebra II, Take Home Exam to Be Submitted on Canvas by 10:00 P.M

Algebra II, Take Home Exam

To be submitted on Canvas by 10:00 P.M. on Wednesday, May 6. You may use material from Jacobson and from the class notes in your solutions to these problems. If you use a result from these sources you should cite it (e.g. “The proposition on slide 6 of Lecture 5”). You may not consult with other people or use other references. 1) Let A be a commutative ring in which every element x satisﬁes xn = x for some n > 1 (depending on x). Show that every prime ideal in A is maximal.

Solution. Suppose that P is a prime ideal of R. For x ∈ R, let x be the class of x in R/P . Suppose x 6= 0. xn = x for some n > 1 implies xn = x which implies x(xn−1 − 1) = 0. Thus xn−1 = 1 since R/P is an integral domain. Now n > 1 implies x is a unit in R/P , with inverse xn−2. Thus R/P is a ﬁeld so P is a maximal ideal.

2) Let A be a commutative ring with nilradical Nil(A). Show that the following are equivalent. i) A has exactly one prime ideal. ii) Every element of A is either a unit or a nilpotent. iii) A/Nil(A) is a ﬁeld.

Solution. i) ⇒ ii). Suppose A has exactly one prime ideal P . By Proposition 8, on slide 10, lecture 12, the nilradical Nil(A) is the intersection of the prime ideals of A so Nil(A) = P . Thus every element of P is nilpotent. If x ∈ A \ P , then x cannot be contained in a maximal ideal of A since P is the unique maximal ideal of A. Thus x is a unit in A. ii) ⇒ iii). Suppose x ∈ A. Let x be the class of x in A/Nil(A). Now if x is nilpotent then x = 0 in A/Nil(A) and if x is not nilpotent then x is a unit in A which implies x is a unit in A/Nil(A). Thus A/Nil(A) is a ﬁeld. iii) ⇒ i). The assumption of iii) implies m = Nil(A) is a maximal ideal of A. By Proposition 8, slide 10 of Lecture 12, Nil(A) = ∩Q where the intersection is over all prime ideals of A. Thus m ⊂ Q for all prime ideals Q of A. Thus Q = m is the only prime ideal of A since m is maximal. Thus m is the unique prime ideal of A.

3) Let A be a (commutative) local ring, M and N be ﬁnitely generated A-modules. Prove that if M ⊗A N = 0 then M = 0 or N = 0. Hint: Apply the Corollary of Slide 6 of Lecture 10 when I = m is the maximal ideal of A. You may also use the following Lemma without proof, but you must explain when you use it.

Lemma 0.1. Suppose that ϕ : A → B is a homomorphism of commutative rings and E, E0 are two A-modules. Then there is an isomorphism of B-modules

0 ∼ 0 (E ⊗A B) ⊗B (E ⊗A B) = (E ⊗A E ) ⊗A B.

A proof of this lemma can be found in Proposition 3, page AII.8.3 of Algebra, by N. Bourbaki. 1 Solution (to Problem 3). Let F = A/m which is a ﬁeld. Suppose M ⊗A N = 0. Then ∼ 0 = (M ⊗A N) ⊗A A/m = (M ⊗A A/m) ⊗A/m (N ⊗A A/m)

by the Lemma. But since F is a field, M ⊗A A/m and N ⊗A A/m are vector spaces over F and thus are free F -modules. M is a finitely generated F -module since the images of a finite set of generators of M as an A-module in M ⊗A A/m are generators of M ⊗A A/m as an A/m-module. Similarly, N ⊗A A/m is a finitely generated A/m-module. Thus both M ⊗A A/m and N ⊗A A/m have finite ranks. Thus

0 = rank((M ⊗A A/m) ⊗A/m (N ⊗A A/m)) = rank(M ⊗A A/m)rank(N ⊗A A/m)

by the corollary on slide 20 of Lecture 9. Thus one of M ⊗A A/m or N ⊗A A/m has rank 0 and is thus zero. Without loss of generality, M ⊗A A/m = 0. Now ∼ M ⊗A A/m = M/mM by the corollary of slide 6 if Lecture 10. M/mM = 0 implies mM = M. Since A is a local ring with maximal ideal m (so that m is the Jacobson radical of A) we have that M = 0 by Nakayama’s Lemma (Proposition 16, slide 5 of Lecture 13).

4) Let A be a commutative ring, M be a ﬁnitely generated A-module and ϕ : M → An be a surjective homomorphism. Show that Ker ϕ is a ﬁnitely generated A-module.

Solution. We have a short exact sequence of A-modules ϕ (1) 0 → Ker ϕ →ι M → An → 0. An is a projective module over A since it is a free module (by the theorem on slide 17 of Lecture 5). Thus (1) is split exact by the theorem on slide 19 of Lecture 5. Thus there is a homomorphism ψ : M → Ker ϕ such that ψι = idKer ϕ by the theorem on slide 3 of Lecture 5. Thus ψ is surjective. Since M is a ﬁnitely generated A-module its image Ker ϕ is thus a ﬁnitely generated A-module. Explicitely, if y1, . . . , yk generate M as an A-module, then ψ(y1), . . . , ψ(yk) generate Ker ϕ as an A-module.

5) Let A be a commutative ring, M be an A-module and I an ideal of A. Suppose that Mm = 0 for all maximal ideals m which contain I. Prove that M = IM.

∼ Solution. Suppose m is a maximal ideal of A. Then (M/IM)m = Mm/(IM)m by Proposi- tion 20, slide 2 of Lecture 14. Suppose I 6⊂ m. Then there exists y ∈ I ∩ (A \ m). Suppose z z 1 yz x ∈ Mm. Then x = s for some z ∈ M and s ∈ A \ m and so x = s = y s ∈ (IM)m. Thus (IM)m = Mm and so (M/IM)m = 0. Now suppose I ⊂ M. Then Mm = 0 by assumption and so (M/IM)m = 0. Thus (M/IM)m = 0 for all maximal ideals m of A and so M/IM = 0 by Proposition 22, slide 5 of Lecture 14. Thus M = IM.