# Faithful $S$-Noetherian module

I found a problem in this paper on Example 2.

Let $$R$$ be a commutative ring, $$S$$ a multiplicative subset of $$R$$, and $$M$$ a nonzero $$S$$-Noetherian $$R$$-module. For $$R'=R/\mathrm{Ann}_R(M)$$, we have that $$M$$ is a faithful $$S$$-Noetherian $$R'$$-module. How to prove that $$R'$$ is $$S$$-Noetherian?

• I think this could inspire you. Commented Dec 28, 2019 at 17:19
• thank you very much ^^ Commented Dec 28, 2019 at 17:29

We prove that if $$M$$ is faithful $$R$$-module and $$M$$ is $$S$$-Noetherian, then $$R$$ is $$S$$-Noetherian.

Fix $$s_0\in S$$ such that $$Ms_0\subseteq N$$ for some finitely generated $$R$$-submodule of $$M$$. Write $$K = \{x\in R\,|\,s_0x=0\}$$ Note that $$\mathrm{Ann}_R(N) \subseteq K$$. Indeed, if $$Nx=0$$, then $$0 = Ms_0x$$ and hence (by the fact that $$M$$ is faithful) $$s_0x = 0$$. Thus $$x\in K$$. Pick generators $$n_1,...,n_k$$ of $$N$$ and consider a morphism $$\phi:R \rightarrow \bigoplus_{i=1}^kn_iR = L$$ given by $$R\ni x\mapsto (n_ix)_{1\leq i\leq k}\in \bigoplus_{i=1}^kn_iR$$. Clearly $$L$$ is $$S$$-Noetherian as a finite direct sum of $$S$$-Noetherian modules $$n_iR\subseteq M$$ for $$1\leq i\leq k$$. Now the kernel of $$\phi$$ is $$\mathrm{Ann}_R(N)$$. This implies that $$R/\mathrm{Ann}_R(N) = R/\mathrm{ker}(\phi)$$ is $$S$$-Noetherian as an $$R$$-submodule of $$S$$-Noetherian module $$L$$.

Now pick any ideal $$I\subseteq R$$. By bold statement above there exist $$s\in S$$ and a finitely generated ideal $$J$$ of $$R$$ such that $$Is\subseteq J+\mathrm{Ann}_R(N)\subseteq J+ K$$ Then $$Iss_0 \subseteq Js_0$$ Now $$ss_0\in S$$ and $$Js_0$$ is a finitely generated ideal of $$R$$. Thus $$R$$ is $$S$$-Noetherian.

Remark.

$$K = \mathrm{Ann}_R(s_0)$$ where $$s_0$$ is treated as an element of $$R$$-module $$R$$.

• Is $s$ in $Ann_R(N)$ the same as $s \in S$ (there exist $s \in S$....)? Also, you said that kernel $\phi$ is $Ann_R(N)$, but why in the implication, kernel $\phi$ is $Ann_R(s)$? Commented Dec 17, 2019 at 12:31
• @R Yes it is the same. Because $Ann_R(N) = \{x\in R\,|\,sx=0\} = Ann_R(s)$. The last equality is precisely the definition.
– Slup
Commented Dec 17, 2019 at 13:08
• Sorry, but I don't understand why $Ann_R(N) = \{ x \in R | sx = 0\}$? Commented Dec 17, 2019 at 15:07
• @RANGGAJAYACIPTAWAN We have $N = Ms$. Then $x\in \mathrm{Ann}_r(N)$ iff $0= Nx=Msx$ iff $sx=0$ (because $M$ is faithful).
– Slup
Commented Dec 18, 2019 at 6:40
• Why $N = Ms$ ? we just know the fact that $Ms \subseteq N$. Commented Dec 19, 2019 at 6:51