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 $$b \ne 0$$ such that $$S[b^{-1}]$$ is a field, then $$S$$ is a field

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.