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!
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.