Show that in a PID, every ideal is contained in a maximal ideal.
Hint: Use the Ascending Chain Condition for Ideals
Every ideal in a PID is a principal ideal domain.
If p is an irreducible element of a PID, then <p> is a maximal ideal.
The Attempt at a Solution
Let D be a PID and let N = <a> be an ideal.
I assumed that N does not equal D, since no maximal ideal of D can contain D.
It follows that a is not a unit in D.
If a=0 and D contains no nonzero nonunit element, then D is a field and <a>=<0> is a maximal ideal. If a=0 and D contains a nonzero nonunit b, then b has an irreducible factor p. So N=<a>=<0> is contained in <p>, which is maximal.
If a is a nonzero nonunit which is itself irreducible, then <a> is a maximal ideal of D.
If a is a nonzero nonunit which is reducible, then a=cq where c is a nonunit and q is irreducible. It follows that N=<a> is contained in <q>, which is maximal.
This seems right to me. I'm writing this up because I never used the Ascending Chain Condition.