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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?

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  • $\begingroup$ I think this could inspire you. $\endgroup$
    – user26857
    Commented Dec 28, 2019 at 17:19
  • $\begingroup$ thank you very much ^^ $\endgroup$
    – RANGGAJAYA CIPTAWAN
    Commented Dec 28, 2019 at 17:29

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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 $R$.

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  • $\begingroup$ 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 $Ann_R(s)$? $\endgroup$
    – RANGGAJAYA CIPTAWAN
    Commented Dec 17, 2019 at 12:31
  • $\begingroup$ @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. $\endgroup$
    – Slup
    Commented Dec 17, 2019 at 13:08
  • $\begingroup$ Sorry, but I don't understand why $Ann_R(N) = \{ x \in R | sx = 0\}$? $\endgroup$
    – RANGGAJAYA CIPTAWAN
    Commented Dec 17, 2019 at 15:07
  • $\begingroup$ @RANGGAJAYACIPTAWAN We have $N = Ms$. Then $x\in \mathrm{Ann}_r(N)$ iff $0= Nx=Msx$ iff $sx=0$ (because $M$ is faithful). $\endgroup$
    – Slup
    Commented Dec 18, 2019 at 6:40
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    $\begingroup$ Why $N = Ms$ ? we just know the fact that $Ms \subseteq N$. $\endgroup$
    – RANGGAJAYA CIPTAWAN
    Commented Dec 19, 2019 at 6:51

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