|May6-09, 04:10 AM||#1|
1. The problem statement, all variables and given/known data
Show that in a PID, every ideal is contained in a maximal ideal.
Hint: Use the Ascending Chain Condition for Ideals
2. Relevant equations
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.
3. 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.
|May6-09, 09:28 AM||#2|
Twice you used the fact that a nonzero nonunit has an irreducible factor. This fact is often proved using ACC.
You might want to double check that your textbook's proof of "nonzero nonunit has an irreducible factor" didn't use as one of its steps "every ideal is contained in a maximal ideal" and leave the proof as an exercise. :)
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