# 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.

• For commutative rings this is true (Stefano's proof being a good solution), but there exists a noncommutative ring which isn't Noetherian and which has a faithful simple module. That module is of course finitely generated (being cyclic) and its annihilator is zero, so the quotient by the annihilator is clearly not a Noetherian ring or module. Apr 24, 2014 at 13:20
• Regarding @stefano's answer ,after checking the kernel to be $\frac{A}{\operatorname{Ann}M}$, I find that $\frac{A}{\operatorname{Ann}M}$ is a Noetherian A-module. However the question asks us to show that $\frac{A}{\operatorname{Ann}M}$ is Noetherian Ring, i.e $\frac{A}{\operatorname{Ann}M}$ is Noetherian $\frac{A}{\operatorname{Ann}M}$-module May 21, 2020 at 7:34

$M$ 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 $1 \mapsto \left( m_{1} , \ldots , m_{k} \right)$
• Thank you. Define $\phi:R \to M^k$ by $\phi(r)=(rm_1,...rm_k)$ Then, $\text{ker}(\phi)= \text{Ann}(M)$ and $R/\text{Ann}(M) \cong \phi(M).$ Since $M$ is Noetherian, $M^k$ is also Noetherian and hence $\phi(M)$ is Noetherian. (Qed) Apr 21, 2014 at 18:46
• @AlexyVincenzo you meant $\phi(R)$ Apr 23, 2017 at 2:13
• So after checking the kernel to be $\frac{A}{\operatorname{Ann}M}$, I find that $\frac{A}{\operatorname{Ann}M}$ is a Noetherian A-module. However the question asks us to show that $\frac{A}{\operatorname{Ann}M}$ is Noetherian Ring, i.e $\frac{A}{\operatorname{Ann}M}$ is Noetherian $\frac{A}{\operatorname{Ann}M}$-module May 21, 2020 at 7:32
• @SunShine The action of $A$ on $A/\mathrm{Ann}(M)$ factors through $A/\mathrm{Ann}(M)$, so the fact that $A/\mathrm{Ann}(M)$ is a noetherian $A$-module immediately implies the same fact as an $A/\mathrm{Ann}(M)$-module. Aug 20, 2020 at 16:09