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So I have a commutative ring with unity, and I also have an ideal I of the ring such that every element not in the ideal is a unit of R. I have already proven that I is maximal by showing that R/I is a field, but I'm really not sure how to sure that it is unique. Let's say I have any maximal ideal of R (possibly I itself). Now I want to show that it is equal to I. How would I do that? I'm quite stuck, so any hints would be appreciated.

Thanks in advance.

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Let $\mathfrak{m} \subseteq R$ be a maximal ideal. Can $\mathfrak{m}$ have any units?

Next step: Is $\mathfrak{m}$ always contained in some ideal that you already know? How do we now use maximality?

Rings with a unique maximal ideal are called local, by the way.

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    $\begingroup$ $R^\times$ is the union of the maximal ideals, if there are two maximal ideals then they are comaximal so there are two non-units such that $a+b=1$, thus local ring (only one maximal ideal) means $\mathfrak{m}=R-R^\times$ is an ideal and $(R-R^\times)R$ is not $R$. Then the difficulty is to classify the valued local rings, those with many prime ideals, those with non-principal maximal ideal. $\endgroup$ – reuns Oct 3 at 2:16
  • $\begingroup$ Perfect. Thanks $\endgroup$ – Tim Oct 3 at 18:34
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I'm really not sure how to sure that it is unique.

Any other ideal which isn't I would contain a unit by your hypothesis and would therefore equal R.

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