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In Gallian's Contemporary Abstract Algebra, one of the exercises is to show that if $D$ is a principal ideal domain, then show that every proper ideal of $D$ is contained in a maximal ideal of $D$.

My thinking is to just assume for contradiction that there is some proper ideal $I$ of $D$ such that for any maximal ideal $M$, $I \not\subset M$. But then $I$ is a maximal ideal and is contained in itself.

My question is why is being a PID necessary? Is my proof wrong, or does this just hold for all integral domains?

EDIT: So I figured out a proof of the statement with the PID condition:

Consider a proper ideal $I$ of $D$. Then $I = \langle p \rangle$ and we can assume $p$ is reducible, otherwise $I$ is maximal. So there exist $q,r \in D$ such that $q$ is irreducible, $r$ is not a unit, and $p=qr$. Then $I \subset \langle q \rangle \subset D$.

So is Zorn's Lemma necessary to prove the more general statement or is there a way around it?

Ebearr
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  • In any ring every ideal is contained in some maximal ideal – Bhaskar Vashishth May 05 '15 at 00:49
  • @Bhaskar: Only if you follow the convention that all rings are unital; not everyone does. – Brian M. Scott May 05 '15 at 01:33
  • Sometimes the most general setting where a theorem works may be a bit beyond the scope of the course or textbook. So only a special case is treated, perhaps proved by taking advantage of properties for that special case. For instance, many theorems about PIDs also work in Dedekind domains, but using harder proofs. Just because there is a proof that every proper ideal is in a maximal ideal for a PID does not mean the result breaks down if you remove that condition; the proof for PIDs, however, might no longer work and you'd need to find a new proof covering the broader setting. – KCd May 05 '15 at 01:56
  • ohh ya gallian does not consider unity in Rings – Bhaskar Vashishth May 05 '15 at 02:02

1 Answers1

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The fact that $I$ is not contained by a maximal ideal does not necessarily imply $I$ is maximal. If you want to prove $I$ is maximal you have to prove $I$ is not properly contained by any ideal, not just the maximal ones.

I know of a proof by Zorn's lemma.Let $R$ be a ring Let $\{I_\alpha | \alpha \in J\}$ be a chain of ideals of $R$ that contain $I$, where $I$ is an arbitrary ideal of $R$. Then $\bigcup\limits_{\alpha \in J} I_\alpha$ is a proper ideal of $R$ that contains each $I_\alpha$ (Prove this, you need to use the fact that the ideals form a chain). We have proved that every chain of ideals containing $I$ is bounded from above. By Zorn's lemma the set of ideals containing $I$ has maximal ideals.

Of course if an ideal is maximal inside of the set of ideals containing $I$ it is also maximal in the set of ideals of $R$ (because any ideal in $R$ which could contain it would have to also contain $I$).

Asinomás
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  • This is generally correct, but be careful. To apply Zorn's lemma, it is very important that the union of any increasing chain of proper ideals is a proper ideal, not just chains indexed by $\mathbb{N}$ (i.e., countable chains). It is true for all chains, though, so we're good. –  May 05 '15 at 00:44
  • Thank you very much! Would you say it looks good now? – Asinomás May 05 '15 at 00:48
  • Almost. Rather than show that the union of a chain of ideals is an ideal, you actually need to show that the union of a chain of proper ideals is a proper ideal. That works, though, since, if the union weren't proper, it would contain a unit. That unit would have to appeared in one of the terms of the chain (by the definition of union). It couldn't have, however, since the chain was a chain of proper ideals. –  May 05 '15 at 00:51
  • Oh right, I remember my teacher did this a couple of weeks ago, but I wasn't paying much attention. Thank you very much, this process has been very profitable for myself also. – Asinomás May 05 '15 at 00:53
  • Sure thing; I'm glad I could help. –  May 05 '15 at 00:54
  • @Gamamal: Thanks! So is Zorn's Lemma necessary to prove this for non PID's? – Ebearr May 05 '15 at 01:02
  • I don't know, this is the proof I know, but I don't think you can prove it straight-forwardly. – Asinomás May 05 '15 at 01:08