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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 $I\subset m$...

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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 $st \notin m$.

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    $\begingroup$ (1) Just to make it more explicit: since $I\not\subset m$, take $i\in I, i\not\in m$. Then $i/1\in I_m\subset R_m$. But since $I_m$ is an ideal of $R_m$ and $1/i\in R_m$ (here we use that $i\not\in m$), we also have $1=i/1\cdot 1/i\in I_m$, so $I_m=R_m$. (2) Well, I want to use primeness of $m$ but I don't know how. We know $ti\notin m,t\notin m$, but it doesn't follow from primeness that $i\not\in m$ (primeness says that if $a\notin m$ and $b\notin m$, then $ab\notin m$). $\endgroup$ – user419669 Feb 12 at 2:53
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    $\begingroup$ The first part of your comment is right on. For the second part, the contradiction is simpler than that. We've assumed that $I \subseteq m$. But $st \notin m$ and at the same time $st = si \in I \subseteq m$. (I guess technically you need primeness of $m$ to say that $S$ is multiplicatively closed, but that's necessary to define the localization in the first place...) $\endgroup$ – André 3000 Feb 12 at 3:02
  • $\begingroup$ Ah, that should be $ti$ in my last comment, not $si$. But it looks like you understood me nonetheless. :) $\endgroup$ – André 3000 Feb 12 at 3:11

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