Rabinowitch's trick

Statement

Let $R$ be a commutative unital ring. The following are equivalent:

1. $R$ is a Jacobson ring
2. If $P$ is a prime ideal of $R$ and if $S:=R/P$ contains an element $b\neq 0$ such that $S[b^{-1}]$ is a field, then $S$ is a field

Proof

1 implies 2

As in the hypotheses, let $P$ be a prime ideal in $R$. Then $S:=R/P$ is an integral domain, and hence 0 is a prime ideal in this. Further, $S$ is also a Jacobson ring since Jacobson is quotient-closed, and thus the zero ideal in $S$ is an intersection of maximal ideals.

Now, the primes of $S[b^{-1}]$ correspond to the primes of $S$ that do not contain $b$. Since $S[b{-1}]$ is a field, $b$ is contained in every nonzero prime ideal of $S$. Thus if $S$ were not a field, zero would not be a maximal ideal, and hence $b$ would be contained in every maximal ideal, contradicting the fact that the intersection of all maximal ideals is zero. Thus $S[b^{-1}]$ is a field.

2 implies 1

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