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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.

user26857
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Alexy Vincenzo
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    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. – rschwieb Apr 24 '14 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 – GraduateStudent May 21 '20 at 07:34

1 Answers1

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$M$ 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...

Stefano
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    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)

     – Alexy Vincenzo Apr 21 '14 at 18:46
  • Exactly. You're welcome :) – Stefano Apr 21 '14 at 18:47
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    @AlexyVincenzo you meant $\phi(R)$ – Daniel Apr 23 '17 at 02:13
  • Can we say something about R in this case?. I think R should not be noetherian, but what counterexample would work here (if I am right)? – Stiven G Sep 06 '18 at 19:58
  • @StivenG You can have a look at the answer by Manny Reyes to this question: https://math.stackexchange.com/questions/258131/noetherian-module-implies-noetherian-ring – Stefano Sep 08 '18 at 03:35
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    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 – GraduateStudent May 21 '20 at 07:32
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    @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. – Stefano Aug 20 '20 at 16:09
  • What os the meaning of "factor through"? – Victor Apr 13 '23 at 22:01