Proposition

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?

Solution

Since $M$ is Noetherian, it is finitely generated by Proposition 6.2 [Atiyah]. Let $M = \ev{x_1, \cdots, x_n}$. $A / \Ann(x_i)$ is isomorphic to $Ax_i$ as $A$-modules through the module isomorphism $a + \Ann(x_i) \mapsto ax_i$. Since $Ax_i$ is a submodule of $M$, we have the exact sequence $0 \rightarrow Ax_i \rightarrow M \rightarrow M / Ax_i \rightarrow 0$. By Proposition 6.3(i), $Ax_i$ must be Noetherians, so $A / \Ann(x_i)$ is Noetherian.

This implies that $A / \Ann(x_i)$ is Noetherian for every $i$. By Problem 6-3, $A / (\Ann(x_1) \cap \cdots \cap \Ann(x_n))$ is Noetherian. Since $\Ann(M) = \Ann(x_1) \cap \cdots \cap \Ann(x_n)$, $A / \Ann(M) = A / a$ is Noetherian.

Let $G$ be the subgroup of $\mathbb{Q}/\mathbb{Z}$ consisting of all elements whose order is a power of 2. As mentioned on P.74[Atiyah], $G$ satisfies the descending chain condition, so $G$ is an Artinian $Z$-module.

Let $k$ be a nonzero integer. Choose a large $m \in \mathbb{N}$ such that $2^m \nmid k$. Then $k \cdot \frac{1}{2^m} \ne 0$ in $G$, so $k$ is not an annihilator. Therefore, the annihilator $a$ of $G$ is $(0)$.

Then $A / a = \mathbb{Z} / (0) = \mathbb{Z}$ and $\mathbb{Z}$ is not Artinian because $(2) \supsetneq (2^2) \supsetneq (2^3) \supsetneq \cdots$.