Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring.

Question

Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring.

I was trying to show that $R/I$ is a submodule of $M^n$ if $M=\langle f_1,...,f_n\rangle$. Now $M$ is an $R/I$-module too. Now how do we try?

Answer 1

Define $h:R\rightarrow M^n$ by $h(r)= (rf_1,...,rf_n)$. Its kernel is $I$ so $h$ factors by an injective morphism $g:R/I\rightarrow M^n$.