MATH 831 HOMEWORK SOLUTIONS – ASSIGNMENT 6

Exercise 3.14. Let M be an A-module and 𝔞 an ideal of A. Suppose that M_{𝔪} = 0 for all maximal ideals 𝔪 ⊇ 𝔞. Prove that M = 𝔞M.

Proof. Let \overline{A} = A/𝔞, \overline{M} = M/𝔞M and \overline{𝔪} = 𝔪/𝔞 for all maximal ideals 𝔪 ⊇ 𝔞. Then \overline{M} is naturally an \overline{A}-module and it suffices to show \overline{M} = 0 to conclude that M = 𝔞M. As the maximal ideals of \overline{A} are exactly the ideals of the form \overline{𝔪} = 𝔪/𝔞 where 𝔪 are maximal ideals of A such that 𝔪 ⊇ 𝔞, it is enough to prove that \overline{M}\overline{𝔪} = 0 for all maximal ideals 𝔪 ⊇ 𝔞 by Proposition 3.8. But

\overline{M}\overline{𝔪} ≅ \frac{M_{𝔪}}{𝔞M_{𝔪}} (check for yourself).

Therefore \overline{M}\overline{𝔪} = 0 as M_{𝔪} = 0. □

Alternative proof. Suppose, on the contrary, that M ⊉ 𝔞M. Choose x ∈ M \ 𝔞M. Then, clearly, we have 𝔞 ⊆ (𝔞M :_{A} x) ⊊ (1) where (𝔞M :_{A} x) := {a ∈ A | ax ∈ 𝔞M} (which is an ideal). Choose a maximal ideal 𝔪 of A such that 𝔪 ⊇ (𝔞M :_{A} x) ⊇ 𝔞. Then the assumption that M_{𝔪} = 0 implies that x/1 = 0 ∈ M_{𝔪}, which implies that there exists s ∈ A \ 𝔪 such that sx = 0. But this definitely contradicts our choice that 𝔪 ⊇ (𝔞M :_{A} x). Hence M = 𝔞M. □

Note that, in the above two proofs of Exercise 3.14, all we need is that M_{𝔪} = 𝔞M_{𝔪} for all maximal ideals 𝔪 of A such that 𝔪 ⊇ 𝔞. Exercise 3.15. Let A be a ring, and let F be the A-module A^{n}. Show that every set of n generators of F is a basis of F.

Deduce that every set of generators of F has at least n elements.

Proof. Let x_{1}, x_{2}, …, x_{n} ∈ F = A^{n} be a set of n generators of F. Say x_{i} = (a_{i1}, a_{i2}, …, a_{in}) for

1 ≤ i ≤ n. Let e_{i} = (0, …, 0, 1^{i}, 0, …, 0) be the i-th standard basis element of F = A^{n}. Then, as x_{1}, x_{2}, …, x_{n} generate F, there exist b_{i1}, b_{i2}, …, b_{in} such that e_{i} = b_{i1}x_{1} + b_{i2}x_{2} + ⋯ + b_{in}x_{n}.

Let U = (a_{ij})_{n×n} and V = (b_{ij})_{n×n} be the two resulted matrices. Then we have V U = E_{n} where E_{n} is the n × n identity matrix. Hence UV = E_{n}. (Indeed, V U = E_{n} implies that det(U) is a unit in A therefore U^{−1} exists and therefore V = V UU^{−1} = U^{−1} just as in linear algebra.)

Now it is clear that x_{1}, …, x_{n} are linearly independent over A. Indeed, for any c_{1}, c_{2}, …, c_{n} ∈ A such that c_{1}x_{1} + c_{2}x_{2} + ⋯ + c_{n}x_{n} = 0, we have (c_{1}, c_{2}, …, c_{n})U = 0. Thus (c_{1}, c_{2}, …, c_{n}) = (c_{1}, c_{2}, …, c_{n})UV = 0. Hence x_{1}, x_{2}, …, x_{n} form a basis of F.

Consequently, every set of generators of F has at least n elements. If, on the contrary, F is generated by y_{1}, …, y_{m} for some m < n, then clearly F is also generated by y_{1}, …, y_{m}, y_{m+1} = 0, …, y_{n} = 0. But then by the above result, y_{1}, …, y_{m}, y_{m+1} = 0, …, y_{n} = 0 form a basis of F, which is impossible. □

Exercise 6.3. Let M be an A-module and let N_{1}, N_{2} be submodules of M. If M/N_{1} and M/N_{2} are Noetherian, so is M/(N_{1} ∩ N_{2}). Similarly with Artinian in place of Noetherian.

Proof. Define f : M → \frac{M}{N_{1}} ⊕ \frac{M}{N_{2}} by f(x) = (x + N_{1}, x + N_{2}) for any x ∈ M. It is easy to check that f is an A-linear homomorphism. Therefore f induces an injective A-homomorphism \overline{f} : \frac{M}{Ker(f)} → \frac{M}{N_{1}} ⊕ \frac{M}{N_{2}} . As M/N_{1} and M/N_{2} are Noetherian (Artinian), so is \frac{M}{N_{1}} ⊕ \frac{M}{N_{2}} by Corollary 6.4, and so is \frac{M}{Ker(f)} by Proposition 6.3.

Hence it is enough to show that Ker(f) = N_{1} ∩ N_{2}. Indeed, for any x ∈ M, we have that x ∈ Ker(f) ⟺ f(x) = (x + N_{1}, x + N_{2}) = (0, 0) ∈ \frac{M}{N_{1}} ⊕ \frac{M}{N_{2}} ⟺ x ∈ N_{1} and x ∈ N_{2} ⟺ x ∈ N_{1} ∩ N_{2}. □

Exercise 6.4. Let M be a Noetherian A-module and let 𝔞 be the annihilator of M in A. Prove that A/𝔞 is a Noetherian ring.

If we replace “Noetherian” by “Artinian” in this result, is it still true?

Proof. As M is Noetherian, it is finitely generated. Say M is generated by x_{1}, x_{2}, …, x_{n}. Define

f : A → \overbrace{M ⊕ ⋯ ⊕ M}^{n copies} = M^{n} by f(a) = (ax_{1}, ax_{2}, …, ax_{n}) for any a ∈ A. It is easy to see that f is an A-linear map. As M is Noetherian, so is M^{n} by Corollary 6.4, and so is A/ Ker(f) by Proposition 6.3.

Hence it is enough to show that Ker(f) = 𝔞, the annihilator of M. Indeed, for any a ∈ A, we have that a ∈ Ker(f) ⟺ f(a) = (ax_{1}, ax_{2}, …, ax_{n}) = (0, 0, …, 0) ∈ M^{n} ⟺ ax_{1} = ax_{2} = ⋯ = ax_{n} = 0 ⟺ aM = 0, i.e. a ∈ {a ∈ A | aM = 0} =: 𝔞.

However, the result is not true if we replace “Noetherian” by “Artinian”. For example, let A = ℤ and M = G as in Example 3) on page 74. That is, G is the subgroup of ℚ/ℤ consisting of all elements whose order is a power of fixed prime number p. Then G is an Artinian ℤ-module as shown in the textbook. Moreover, the annihilator of G is 0: Indeed, for any 0 ≠ n ∈ ℤ, there is \frac{1}{p^{|n|}} + ℤ ∈ ℚ/ℤ, which is in G, such that n(\frac{1}{p^{|n|}} + ℤ) = \frac{n}{p^{|n|}} + ℤ ≠ 0 ∈ G (since \frac{n}{p^{|n|}} ∉ ℤ). But A/0 ≅ A = ℤ is not Artinian as shown in Example 2) on page 74. □

Exercise 7.1. Let A be a non-Noetherian ring and let Σ be the set of ideals in A which are not finitely generated. Show that Σ has maximal elements and that the maximal elements of Σ are prime ideals.

Hence a ring in which every prime ideal is finitely generated is Noetherian (I. S. Cohen).

Proof. The existence of maximal elements of Σ is a consequence of Zorn’s lemma. Indeed, let Ω ⊆ Σ be a totally ordered subset of Σ (under containment). Let 𝔟 = ∪_{𝔞∈Ω}𝔞. Then 𝔟 is an ideal of A since Ω is totally ordered. If 𝔟 is finitely generated, say 𝔟 = (b_{1}, b_{2}, …, b_{r}), then b_{i} ∈ 𝔞_{i} for some 𝔞_{i} ∈ Ω. We may assume that 𝔞_{1} ⊆ 𝔞_{2} ⊆ ⋯ ⊆ 𝔞_{r} since Ω is totally ordered. Therefore 𝔟 = (b_{1}, b_{2}, …, b_{r}) ⊆ ∪_{i=1}^{r}𝔞_{i} = 𝔞_{r} ⊆ ∪_{𝔞∈Ω}𝔞 = 𝔟, which forces 𝔞_{r} = (b_{1}, b_{2}, …, b_{r}), which is contrary to the choice that 𝔞_{r} ∈ Ω ⊆ Σ. Therefore 𝔟 is not finitely generated. Hence 𝔟 ∈ Σ is an upper bound for Ω. So Σ has at least one maximal element.

Suppose that there is a maximal element 𝔞 of Σ such that 𝔞 is not a prime ideal of A. Then there exist x, y ∈ A \ 𝔞 such that xy ∈ 𝔞.

As x ∉ 𝔞, i.e. 𝔞 ⊊ 𝔞 + (x), and 𝔞 is a maximal element of Σ, we have 𝔞 + (x) ∉ Σ, i.e. 𝔞 + (x) is finitely generated. Say 𝔞 + (x) = (c_{1}, c_{2}, …, c_{n}). Write c_{i} = a_{i} + b_{i}x where a_{i} ∈ 𝔞, b_{i} ∈ A for each i = 1, 2, …, n and let 𝔞_{0} = (a_{1}, a_{2}, …, a_{n}). Then clearly we have 𝔞_{0} ⊂ 𝔞 and therefore 𝔞 + (x) = (c_{1}, c_{2}, …, c_{n}) ⊆ 𝔞_{0} + (x) ⊆ 𝔞 + (x), which forces 𝔞_{0} + (x) = 𝔞 + (x).

Let (𝔞 : x) = {a ∈ A | ax ∈ 𝔞}. Then obviously 𝔞_{0} + x(𝔞 : x) ⊆ 𝔞. On the other hand, for any a ∈ 𝔞, we can write a = a_{0} + bx with a_{0} ∈ 𝔞_{0}, b ∈ A. Therefore bx = a − a_{0} ∈ 𝔞, i.e. b ∈ (𝔞 : x). Hence a = a_{0} + bx ∈ 𝔞_{0} + x(𝔞 : x). So we have 𝔞 = 𝔞_{0} + x(𝔞 : x).

By our choice of x and y, we have y ∈ (𝔞 : x) and y ∉ 𝔞. Together with the easy fact that 𝔞 ⊂ (𝔞 : x), we know that (𝔞 : x) is strictly larger than 𝔞 and hence is finitely generated, say by z_{1}, z_{2}, …, z_{m}.

But then 𝔞 = 𝔞_{0} +x(𝔞 : x) = (a_{1}, …, a_{n})+x(z_{1}, …, z_{m}) = (a_{1}, …, a_{n}, xz_{1}, …, xz_{m}) is finitely generated, which is a contradiction. Hence 𝔞 has to be a prime ideal of A.

Hence a ring in which every prime ideal is finitely generated is Noetherian. □

Note: The exercises are from ‘Introduction to Commutative Algebra’ by M. F. Atiyah and I. G. Macdonald. All the quoted results are from the textbook unless different sources are quoted explicitly. For the convenience of the readers, the number of the chapter is included when a particular exercise is numbered. For example, Exercise m.n means the Exercise n from Chapter m.