# The localization of an ideal is equal to the localization of the ring

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 $$st \notin m$$.
• (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 $ab\notin m$). – user419669 Feb 12 at 2:53
• 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...) – André 3000 Feb 12 at 3:02
• Ah, that should be $ti$ in my last comment, not $si$. But it looks like you understood me nonetheless. :) – André 3000 Feb 12 at 3:11