More formally stated:

Prove that if $R$ is a commutative ring with $1$, then every element of $R$ that is not invertible is contained in a proper maximal ideal.

I know I have to assume Zorn's Lemma, but I don't see how non-invertible elements must lie in a proper maximal Ideal. Any hints?

Thank you for your time.

  • $\begingroup$ This follows from showing that noninvertible elements lie in some proper ideal and that all proper ideals are contained in maximal ideals. In this vein, what is an ideal containing noninvertible $x$? It is proper? Now, consider the set of all ideals containing this ideal. How might you express the supremum of a chain $I_0\subset I_1\subset\dots I_n\subset\dots$? $\endgroup$ – yearning4pi Jan 23 '13 at 8:50
  • $\begingroup$ I don't see why this should be tagged as [axiom-of-choice]. There is no question about the necessity of the use of Zorn's in the proof, there is a question about the proof itself. @BDub: ping. $\endgroup$ – Asaf Karagila Jan 23 '13 at 9:38
  • $\begingroup$ @AsafKaragila I agree, to me, ring-theory and ideals seem like the most pertinent tags. I didn't retag the question with AC. If the edit log shows that I did, I think it's because I submitted a formatting edit on a version of the question before your retags took effect? Perhaps the tags reappeared when it was approved. $\endgroup$ – BW. Jan 23 '13 at 9:44
  • $\begingroup$ @BDub: You need 300 reputation for retagging, so I suppose that you just retagged without loading my revision, which caused an override. $\endgroup$ – Asaf Karagila Jan 23 '13 at 9:47

It is well known that every proper ideal is contained in a maximal ideal. If $a$ is a noninvertible element, then the generated ideal $(a)$ is not the whole ring. If it were, then $1\in (a)$, implying $ab=1$ for some $b$, a contradiction. As a proper ideal, $(a)$, and hence $a$, must then be contained in a maximal ideal.

  • 1
    $\begingroup$ it should be noted that this result uses the axiom of choice. $\endgroup$ – Ittay Weiss Jan 23 '13 at 9:41

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