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Let M be a Noetherian module over commutative ring R.

Show that R/Ann(M) is a Noetherian ring.

I know M is Noetherian module over R then M is also Noetherian over the commutative ring R/Ann(M) . From this result can I say R/Ann(M) is Noetherian ring?

user26857
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Pradip
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  • Done, as M is finitely generated say generated by k elements then $M^k$ is also Noetherian and I have define a map R to $M^k$ – Pradip Sep 14 '19 at 06:33

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Since $M$ is Noetherian, it is finitely generated. Let $M=(x_1,...x_n)$. Define the map $f:R\to M\oplus\cdots\oplus M$ (n times) by $f(r)=(rx_1,...,rx_n)$. Clearly, $Ker(f)=\{r\in R\mid rx_i=0 $ for all $i$$\}=\cap_iann(x_i)=ann(M)$. Thus, $\bar{f}:R/ann(m)\to M\oplus\cdots\oplus M$ (n times) is a monomorphism, that is, $R/ann(M)$ can be consider as a submodule of $M\oplus\cdots\oplus M$. Now since $M$ is Noetherian, $M\oplus\cdots\oplus M$ is Noetherian, and so $R/ann(m)$ is Noetherian.

E.R
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