Let be a commutative integral domain with identity 1. Let be the set of all prime ideals of , and let be the set of all prime ideals of containing the subset of .

My task is to show that .

Here's my thinking on this, and what I've done:

"To show that , it suffices to show that an ideal of is proper if and only if it doesn't contain 1, since by definition all prime ideals are proper ideals.

Proof by contrapositive. Suppose where is an ideal of . Since an ideal is a subgroup under addition, and further since 1 is an additive generator for , it follows that thus is not a proper ideal.

Suppose that . Since by definition, it immediately follows that and thus is proper.

Thus there is no prime ideal that contains 1, since 1 is not contained in any proper ideal."