This is not for homework, but I would just like a hint please. The question asks

If a commutative ring $R$ (with $1$) has a unique maximal ideal, then the set of non-units in $R$ is an ideal.

This is actually an 'if and only if', but I have shown one direction. I'm not sure how to go about proving this direction, though. I don't think contradiction or contrapositive are useful, because assuming the set of non-units is not an ideal doesn't seem to give anything useful. However, showing this directly seems difficult too, because we don't know anything about the set of non-units in an arbitrary ring (it may not even be closed under addition). I also have the following fact at my disposal:

When $R$ is a nonzero ring (with $1$), then every ideal of $R$ except $R$ itself is contained in a maximal ideal.

I would really appreciate a hint about how to go about approaching this.


Every nonunit is contained in a maximal ideal.

  • 6
    $\begingroup$ So, as a sketch, I could take some non-unit $n$ and by assumption $(n) \subseteq M$, where $M$ is the unique maximal ideal in $R$. In this way, I can say that the set of non-units is contained in $M$. Now $M$ cannot contain any units because otherwise $M = R$. Hence, the non-units exhaust $M$, and so form an ideal in their own right. $\endgroup$ – tylerc0816 Oct 29 '13 at 21:43
  • 1
    $\begingroup$ Correct. ${}{}$ $\endgroup$ – anon Oct 29 '13 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.