$S^{-1}J(A)\subseteq J(S^{-1}A)$

Question

Suppose that $A$ is a (commutative unital) ring, and suppose that $I$ is an ideal of $A$. If $J(R)$ denotes the Jacobson radical of a ring $R$, then for $S= 1+ I$ we have $S^{-1}J(A)\subseteq J(S^{-1}A)$. The (usual?) proof of this relies on $AI$ being contained in $I$.

Is there an alternative proof not relying on this fact, so that we can relax the statement and only assume that $I$ is a subring of $A$?

Answer 1

No, the statement isn't true if all you know is that $I$ is a subrng. For instance, let $A=\mathbb{Z}_{(2)}$ and let $I=3\mathbb{Z}$. Since $-2\in 1+I$, $S^{-1}A=\mathbb{Q}$, so $J(S^{-1}(A))=0$. But $J(A)=2A$, and $S^{-1}J(A)=\mathbb{Q}$.