Suppose $m\subset R$ is a maximal ideal. Suppose $I\subset R$ is an ideal. I'm trying to understand these claims: If $m$ does not contain $I$, then $I_m=R_m$ as localizations of $R$-modules. If $m$ contains $I$, then $I_m\ne R_m$.
First statement: the inclusion $I_m\subset R_m$ is obvious: if $r/s\in I_m$ ($r\in I, s\in R-m)$ them $r/s\in R_m$ because $I\subset R$ so $r\in I\subset R$. But I don't understand why $R_m\subset I_m$ holds. Consider $r/s\in R_m$; here $r\in R$. To show that $r/s\in I_m$, I need to prove that $r\in I$, right? I don't see how it follows from $I\not\subset m$ or from $I\cap (R-m)\ne \emptyset$.
Second statement: here I guess I need to find $r/s\in R_m$ such that $r\not\in I$, knowing that $I\subset m$ (then it will follow that $r/s\not\in I_m$). Can I take any $r\not\in I$? But this doesn't use the assumption $I\subset m$...
For notational convenience, let $S = R \setminus m$.
(1) In the localization $R_m$, every element of $R$ that is not in $m$ becomes a unit. An ideal containing a unit is the whole ring.
(2) For contradiction, suppose that $I_m = R_m$. Then $1 \in I_m$ so $1 = i/s$ for some $i \in I$ and $s \in S$. Then there exists a $t \in S$ such that $ts = ti \in I$. Can you derive a contradiction from here? Hint: $st \in S$ so $st \notin m$.