Maximal element of the set of ideals

[fireisland27]

Homework Statement

Let R be a commutative ring with unity and S a subset of R which is closed under multiplication. If P is a maximal element of the set of ideals which do not intersect S nontrivially, then I want to show that P is a prime ideal of R.

Homework Equations

The Attempt at a Solution

I was trying to use a similar proof to the one that shows that every maximal ideal in R is prime. That is, given a maximal ideal M and an element xy in M consider the ideal M+(x) and show that since this ideal is either M or the whole ring, y must be in M. I'm having trouble adjusting it to work here though, if indeed this is even the correct strategy. Any sugestions?

 

[mighty2000]

Your approach is on the right track. To show that P is a prime ideal, we can use the same idea as the proof for maximal ideals. Consider the ideal P+(x) for some x in R. Since P is maximal among ideals that do not intersect S nontrivially, P+(x) must either be P or the entire ring R. If P+(x) is P, then x must be in P since P is an ideal. If P+(x) is R, then 1 must be in P+(x), which means that there exists some y in P and some s in S such that xy+s=1. This implies that xy=1-s, which means that x is a unit in R. But since S is closed under multiplication, xy is in S, so x must also be in S. This contradicts the fact that P is maximal among ideals that do not intersect S nontrivially. Therefore, P must be a prime ideal of R.