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Let $R$ be a commutative ring with identity and $S$ be a multiplicative subset of $R$.

I am not sure that if $S^{-1}J(R)\subseteq J(S^{-1}R)$ is true.

Since $J(R)=\bigcap_{N\, is\, maximal\, ideal\, of\, R} N$, we have $S^{-1}J(R)\subseteq \bigcap S^{-1}N$.

For every maximal ideal $K$ of $S^{-1}R$, there is an ideal $L=\{r\in R | \frac{r}{s}$ for some $s\in S\}$ such that $S^{-1}L = K$. Consequently, $J(S^{-1}R)\subseteq \bigcap_{N\, is\, maximal\, ideal\, of\, R} S^{-1}N$.

I have tried some examples, such as $R=\mathbb{Z}_n, R=\mathbb{Z}_n\times \mathbb{Z}_m, R=\mathbb{Z}$. But all of them seem to show that $S^{-1}J(R)\subseteq J(S^{-1}R)$ is right.

Here is a related question Jacobson Radical in Localization..

Mod.esty
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The answer is no. For instance, suppose $R$ is a local domain that is not a field; examples include the formal power series ring $k[[x]]$ over your favorite field $k$, and the ring $\mathbb{Z}_{(p)}=\{a/b\in\mathbb{Q}:p\nmid b\}$ for a prime $p\in\mathbb{Z}$. Then take $S=R\setminus\{0\}$. The ring $S^{-1}R$ is the field of fractions of $R$, and hence has Jacobson radical $(0)$. But $J(R)\neq(0)$ since $R$ is local and not a field (for instance, $J(k[[x]])=\langle x\rangle$ and $J(\mathbb{Z}_{(p)})=\langle p\rangle$), so $S^{-1}J(R)$ is all of $S^{-1}R$.

Atticus Stonestrom
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