Localization of a module ($S^{-1}M$)

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

Letbe the semigroup algebra inside $k[t^{\mathbb{Q}}]$, and let $I$ be generated by the $t^q$ with $q>0$. This integral domain is not Noetherian, sinceMoreover $I^2=I$, because $t^q=(t^{q/2}{)}^2$. Thus ${\bigcap }_{j\ge 1}{I}^j=I\ne 0$, whereas the map $R\to (1+I{)}^{-1}R$ is injective because $R$ is a domain.