For a commutative ring $A$ with multiplicative subset $0\notin S$, if $P$ is a maximal element in the set of ideals not intersect $S$, it is prime.

Question

For a commutative ring $A$ with multiplicative subset $0\notin S$, if $P$ is a maximal element in the set of ideals not intersect $S$, it is prime.

I'm not sure what kind of approach I should take to set off.

Shoud I play with $S^{-1}A$?

Answer 1

Hint. Consider a maximal ideal $I$ with respect to not intersecting $S$. Let $a,b\in A$ satisfy $ab\in I$, and suppose neither of $a,b$ belong to $I$. Then, by maximality of $I$, the ideals $I_a$ and $I_b$ generated by $I$ and $a$ and $I$ and $b$ respectively must both intersect $S$.

Thus there are $i,j\in I$ and $x,y\in A$ such that $i+xa$ and $j+yb$ belong to $S$. $S$ being multiplicatively stable by hypothesis, we have $s=(i+xa)(j+yb)\in S$ i.e. $$s=\underbrace{ij+iyb+jxa}_{\in I}+\underbrace{xy\overbrace{ab}^{\in I}}_{\in I}\in I\cap S=\emptyset$$ which is absurd