Let $R$ be a commutative ring and $\mathfrak a\triangleleft R$ a proper ideal of $R$. I wish to prove that $T^{-1}\mathfrak a$ is included in the Jacobson ideal of $R$, where $T=1+\mathfrak a$.
In order to complete my proof, I am missing one crucial step. Namely, I'd like to show that if an ideal $T^{-1}\mathfrak p\triangleleft T^{-1}R$ is maximal, then necessarily $\mathfrak p\triangleleft R$ is maximal. That this is the case has been mentioned here, for instance.
Using the characterisation of prime ideals of $T^{-1}R$, one easily sees that the prime ideal $\mathfrak p$ is maximal among all ideals which are disjoint from $T$.
The way I thought to complete this proof is by contradiction. Assume that $\mathfrak p$ isn't maximal. Then we find some maximal ideal $\mathfrak m\triangleleft R$ properly containing $\mathfrak p$. By the maximality property of $\mathfrak p$, the ideal $\mathfrak m$ must contain $1+a$ for some $a\in\mathfrak a$.
Of course, it would suffice to prove that $\mathfrak a\subseteq\mathfrak p$ (because then, $1=(1+a)-a\in\mathfrak m$, contradiction). But it isn't clear to me if this is the way to do it.
