If $M$ is a Noetherian $A$-module then show that $\frac{A}{\operatorname{Ann}M} $ is a Noetherian Ring.
This one shouldn't be be a Duplicate since I am unable to understand the other solutions. What I cannot understand has been given below
I had read on MSE , which shows that $\frac{A}{\operatorname{Ann}M}$ is isomorphic to $M^k$ where $M$ is generated by $\{m_1,m_2 \dots m_k\}.$ Since $M$ is Noetherian, then so will be $M^k$, thus we conclude that $\frac{A}{\operatorname{Ann}M}$ is Noetherian.
Since $M$ is an $A$ module, I find that $\frac{A}{\operatorname{Ann}M}$ is a Noetherian A-module, but 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.
How should I solve?
Let $I=\operatorname{Ann}(M)$.
The $A$-submodules of $A/I$ correspond to the ideals of $A$ which contain $I$.
Thus, since $A/I$ is Noetherian as an $A$-module, the ideals of $A$ which contain $I$ satisfy the ascending chain condition, so $A/I$ is a Noetherian ring.