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Let $R$ be a commutative ring with unity, $S \subset R$ a multiplicative closed subset and $I \subseteq R$ be an ideal. Show that$$S^{-1} \sqrt{I}=\sqrt{S^{-1}I}$$ and $$S^{-1}J(R)=J(S^{-1}R),$$ where $ J(R)$ is the Jacobson radical of $R.$

For the first part I used the fact that $$\sqrt{I}=\{x\in R\mid x^{n}\in I\text{ for some }n\ge1\},$$ and I think I managed to do it, but I'm stuck on the second part. I'm trying to use the fact$$J(R)=\{x\in R \mid 1+rx \hspace 2mm\text{is a unit}\hspace 2mm \forall r\in R\},$$ but I don't think I'm there yet. Can you help?

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Note that Jacobson ideal of $R$ is the intersection of all maximal ideals of $R.$ – Ehsan M. Kermani Jan 30 '13 at 7:19
@ehsanmo Yes I know, but I think using units will be easier I don't know.. – i.a.m Jan 30 '13 at 7:33
You should be aware of the correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ with empty intersection with $S.$ – Ehsan M. Kermani Jan 30 '13 at 7:47
@ehsanmo I'm aware of that but the question does not say that $S$ and $I$ are disjoint or not.. – i.a.m Jan 30 '13 at 8:10
If we take $R=\mathbb Z$, $I=p \mathbb Z$ and $S=\mathbb Z \setminus p\mathbb Z$ then we have $S^{-1}Jac(R)=(0)$ but $Jac(S^{-1}R)=Jac(\mathbb Z_{(p)})=p\mathbb Z_{(p)}$. Maybe if you assume that $S$ has empty intersection with every maximal ideal... – JSchlather Jan 30 '13 at 8:32
up vote 5 down vote accepted

The question is wrong as currently stated and there doesn't seem to be a good way to fix it. For an easy counterexample we can consider the localization of $\mathbb Z$ at the prime ideal $p\mathbb Z$. That is we take $S=\mathbb Z \setminus p \mathbb Z$ and $R=\mathbb Z$. Then it's easy to see that $Jac(R)=0$ since no number is divisible by an infinite number of primes for instance. Thereby $S^{-1}Jac(R)=0$. But we know that $S^{-1}R=(\mathbb Z \setminus p\mathbb Z)\mathbb Z=\mathbb Z_{(p)}$ is a local ring with its maximal ideal given by $p\mathbb Z_{(p)}$ so it has non-trivial Jacobson radical.

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