Let $D$ be a principal ideal domain. Show that every proper ideal of $D$ is contained in a maximal ideal of $D$.

Question

I know that a PID must satisfy the Ascending chain condition. So Im guessing its going to involve that in the argument some way but Im not sure how to prove it.

Answer 1

Recall that in a Noetherian ring, any non-empty set of ideals has a maximal element (with respect to inclusion). And a PID is certainly Noetherian, since every ideal is generated by one element.

So given a PID $D$ and a proper ideal $I$ of $D$, let $S$ be the set of proper ideals containing $I$. $S$ is non-empty since it contains $I$, hence has a maximal element $M$.

If $M$ is not a maximal ideal, then there is a proper ideal $N$ with $M\subset N$, but then $M$ is not a maximal element of $S$. Hence $M$ is in fact a maximal ideal of $D$.

Answer 2

If you know that the ascending chain property holds for PIDs then you may prove your claim by contradiction.

Let $R$ be a PID. Assume there exists a proper ideal $J \subsetneq R$ which is not contained in a maximal ideal. Set $I_0=J$. For every finite sequence

$$ J=I_0 \subsetneq I_1 \subsetneq \dots \subsetneq I_n $$

exists an ideal $I_{n+1}$ such that $I_n\subsetneq I_{n+1}$, otherwise $I_n$ was maximal, which is not possible, as $J$ is not contained in a maximal ideal. Hence, we constructed a ascending chain of ideal, which never gets stationary. This contradicts the ascending chain property of our ring.

The general case when $R$ is just assumed to be unital can be proved using Zorn's lemma.