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Let $R$ be a commutative ring with $1$. We call $R$ a local ring when $R$ has exactly one maximal ideal $m$. Let $I \subset R$ be a proper ideal of $R$, i.e. $I \neq R$.

Question: Is $I$ contained in $m$?

Approach: I think this is true. If $I$ is a proper ideal of $R$, then $I$ does not contain any units of $R$ (otherwise we would have $I=R$, contradiction!). But this is the only thing I can think of.

Could you please help me solving my question? Thank you!

Diglett
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    OK, I know the linked duplicate isn't exactly what was asked. But it is a trivial abstraction, so that is my rationale for linking. Based on the comment to the solution, this is exactly what the user needed. – rschwieb Jun 28 '19 at 10:27

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A maximal ideal is maximal by set-inclusion. In a local ring, there is just one maximal ideal. So any other ideal must be contained in it.

Wuestenfux
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  • I do not understand, why every ideal must be contained in the maximal ideal. Isn't it possible that some ideal is not contained in any maximal ideal at all (something like an "infinite chain argument")? – Diglett Jun 28 '19 at 10:24
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    @Diglett Not in a ring with identity, as you apparently are assuming (when you mentioned units.) Apparently your missing bit of information is exactly what I thought, and you'll find it in the duplicate. – rschwieb Jun 28 '19 at 10:28