If $M$ is a Noetherian $R$-module, then $R/\text{Ann}(M)$ is a Noetherian ring [closed]
Let $M$ be an $R$-module and $\text{Ann}(M)=\{r \in R: rm =0 , \forall m \in M\}.$ Suppose $M$ is Noetherian. Could anyone advise me on how to prove $R/\text{Ann}(M)$ is also Noetherian?
Hints will suffice. Thank you.
$M$ is finitely generated because it is noetherian, say by $\lbrace m_{1} , \ldots , m_{k} \rbrace$. Consider $M^{k}$, which is noetherian, and define a map
$R \rightarrow M^{k}$ which sends $1 \mapsto \left( m_{1} , \ldots , m_{k} \right)$
Check the kernel...