localization of a ring

**student2011** · April 12th 2011, 01:46 AM

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

**NonCommAlg** · April 13th 2011, 01:20 AM

Originally Posted by student2011

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

if $a \in \sqrt{I}$ , then $a^n \in I$ for some integer $n$ . so if $s \in S$ , then $(s^{-1}a)^n=s^{-n}a^n \in S^{-1}I$ . thus $s^{-1}a \in \sqrt{S^{-1}I}$ .
this proves that $S^{-1} \sqrt{I} \subseteq \sqrt{S^{-1}I}$ .
for the converse, let $s \in S$ and $a \in R$ be such that $s^{-1}a \in \sqrt{S^{-1}I}$ . then $s^{-n}a^n \in S^{-1}I$ , for some integer $n$ . so $s^{-n}a^n = t^{-1}b$ , for some $t \in S$ and $b \in I$ . thus $uta^n = us^n b \in I$ , for some $u \in S$ . thus $(uta)^n \in I$ and hence $uta \in \sqrt{I}$ .
therefore $s^{-1}a=(sut)^{-1}uta \in S^{-1} \sqrt{I}.$

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