I am trying to solve this exercise from Sharp's book Steps in Commutative Algebra:
Let $M$ be a finitely generated Artinian module over the commutative ring $R$. Show that $R/\mathrm{Ann}(M)$ is an Artinian ring.
It is easy to show that if $M$ is a Noetherian $R$-module then $R/\mathrm{Ann}(M)$ is a Noetherian ring, but I can't prove the result for Artinian modules.
Let $m_1,\dots,m_k$ be a set of generators for $M$; then $\operatorname{Ann} M=\bigcap_{i=1}^k\operatorname{Ann} m_i$. We prove the desired result by induction on $k$. If $k=1$, then $R\big/\operatorname{Ann}M\cong M$ as $R$-modules via the map $\bar{r}\mapsto rm_1$, so the result is clear. Thus assume $k>1$. Since submodules of Artinian modules are Artinian, the submodule $N:=\sum_{i=1}^{k-1}Rm_i$ of $M$ is Artinian, and hence by the inductive hypothesis $R\big/\operatorname{Ann}N$ is Artinian. Also, by the case $k=1$, $R\big/\operatorname{Ann}m_k$ is Artinian. Hence by Lemma 2 the quotient of $R$ by $\operatorname{Ann}N\cap \operatorname{Ann}m_k$ is Artinian. But this latter intersection is simply $\operatorname{Ann}M$, so the result follows.
Lemma 1: If $N\leqslant M$ are $R$-modules and both $N$ and $M\big/N$ are Artinian, then $M$ is Artinian.
Proof: Suppose $M$ is not Artinian, and let $$M=M_0>M_1>M_2>\dots$$ be any strictly descending chain of submodules of $M$. This chain naturally induces descending chains \begin{align} N&=N_0\geqslant N_1\geqslant N_2\geqslant\dots \\ M\big/N&=\bar{M}_0\geqslant\bar{M}_1\geqslant\bar{M}_2\geqslant\dots, \end{align} where $N_i=M_i\cap N$ and $\bar{M}_i=(M_i+N)\big/N$. Since $N$ and $M\big/N$ are Artinian, there must exist some $k$ where the chains above stabilize, ie such that $N_l=N_k$ and $\bar{M}_l=\bar{M}_k$ for each $l\geqslant k$. In particular, $M_{k+1}\cap N=M_k\cap N$ and $M_{k+1}+N=M_k+N$. But now \begin{align} M_{k+1}&=M_{k+1}\cap (M_{k+1}+N) \\ &=M_{k+1}\cap (M_{k}+N) \\ &= M_k+(M_{k+1}\cap N) \\ &= M_k+(M_{k}\cap N)=M_k, \end{align} where the third equality follows from the "modular law". This is a contradiction, so we are done. $\blacksquare$
Lemma 2: If $I,J$ are ideals of $R$ such that $R\big/I$ and $R\big/J$ are Artinian, then $R\big/(I\cap J)$ is Artinian.
Proof: Let $M=R\big/(I\cap J)$ and let $N$ be the submodule $I\big/(I\cap J)$ of $M$. Note that $M\big/N\cong R\big/I$, which is Artinian, and that $N\cong (I+J)\big/J$, which is a submodule of the Artinian module $R\big/J$ and hence Artinian. So by Lemma 1 $M$ is Artinian, as desired. $\blacksquare$