# Thread: localization of a ring

1. ## localization of a ring

Show that if $\displaystyle S$ is any multiplicative closed subset of $\displaystyle R$ where $\displaystyle R$ is commutative ring with unity, $\displaystyle (\sqrt{I})_s=\sqrt{I_s}$

2. Originally Posted by student2011
Show that if $\displaystyle S$ is any multiplicative closed subset of $\displaystyle R$ where $\displaystyle R$ is commutative ring with unity, $\displaystyle (\sqrt{I})_s=\sqrt{I_s}$
if $\displaystyle a \in \sqrt{I}$, then $\displaystyle a^n \in I$ for some integer $\displaystyle n$. so if $\displaystyle s \in S$, then $\displaystyle (s^{-1}a)^n=s^{-n}a^n \in S^{-1}I$. thus $\displaystyle s^{-1}a \in \sqrt{S^{-1}I}$.
this proves that $\displaystyle S^{-1} \sqrt{I} \subseteq \sqrt{S^{-1}I}$.
for the converse, let $\displaystyle s \in S$ and $\displaystyle a \in R$ be such that $\displaystyle s^{-1}a \in \sqrt{S^{-1}I}$. then $\displaystyle s^{-n}a^n \in S^{-1}I$, for some integer $\displaystyle n$. so $\displaystyle s^{-n}a^n = t^{-1}b$, for some $\displaystyle t \in S$ and $\displaystyle b \in I$. thus $\displaystyle uta^n = us^n b \in I$, for some $\displaystyle u \in S$. thus $\displaystyle (uta)^n \in I$ and hence $\displaystyle uta \in \sqrt{I}$.
therefore $\displaystyle s^{-1}a=(sut)^{-1}uta \in S^{-1} \sqrt{I}.$