# Unique maximal ideal implies set of non-units is an ideal

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.

• 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. – tylerc0816 Oct 29 '13 at 21:43
• Correct. ${}{}$ – anon Oct 29 '13 at 21:50