# Ring Localization Exercise

This exercise is taken from Falko Lorenz's Algebra, 4.12

Let $R$ be a commutative ring with unity and $S$ a multiplicative subset of $R$. Form the localization $S^{-1}R$ of $R$ relative to S, with canonical map $i:R\rightarrow S^{-1}R$. If $\alpha$ is an ideal of $R$ denote by $S^{-1}$ the ideal of $S^{-1}R$ generated by $i(\alpha)$. It is easy to check that $S^{-1}\alpha$ consists of all elements of the form a/s with $a\in\alpha$ and $s\in S$; moreover $S^{-1}\alpha=(1)$ iff $\alpha\cap S\not=\emptyset$. Conversely, if $\beta$ is an ideal of $S^{-1}R$, denote the ideal $i^{-1}(\beta)$ of $R$ by $\beta\cap R$. Then $\alpha$ is of the form $\alpha=i^{-1}(\beta)$ iff no element of $S$ gives rise to a zero divisor of R/$\alpha$. Prove that the maps $\mathbb{B}\rightarrow\mathbb{B}\cap R$ and $\mathbb{A}\rightarrow S^{-1}\mathbb{A}$ establish a one-to-one correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ that are disjoint from $S$.

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The Proposition, and the following Corollary, at page 87 of Reid's Undergraduate Commutative Algebra prove your statement (and it is explained certainly better than the way I could do that). If you can't look at the book comment here and we will find a solution! –  Giovanni De Gaetano Oct 10 '12 at 10:52
I found books.google.com/…, which has a preview of that page for free, but I'm still stuck on proving that $i$ is one-to-one. –  Chris Oct 10 '12 at 11:03

## 1 Answer

Throughout the answer I assume the validity of the Proposition at page 87 of Reid's "Undergraduate Commutative Algebra". If I understood your problem, we want to prove that the map $i$ induces an injective map between prime ideals of $R$ which are disjoint from $S$ and prime ideals of $S^{-1}R$.

Let $\alpha, \beta$ be prime ideals of $R$ such that $\alpha \cap S= \beta \cap S = \emptyset$, and such that $i(\alpha)\left(S^{-1}R\right)=i(\beta)\left(S^{-1}R\right)$ (i.e. the ideals generated by the images coincide). Then I claim $\alpha=\beta$.

By point $b)$ of the Proposition mentioned above: $$\{r\in R \;|\; rs\in \alpha \:\forall\; s\in S\}= \{r\in R \;|\; rs\in \beta \:\forall\; s\in S\}.$$

Using the primality of $\alpha$ we deduce that $rs\in \alpha \iff r\in \alpha\: \text{or}\: s\in \alpha$, but $\alpha\cap S= \emptyset$. So: $$\{r\in R \;|\; rs\in \alpha \;\forall \;s\in S\}= \alpha.$$

Playing the same trick with $\beta$ the equality is proved.

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I'm confused about the proof from the book you cited, how does it follow that $a\in i^{-1}(i(I))$ implies (a,1) is related to (b,t)? –  Chris Oct 10 '12 at 12:17
Because $a\in i^{-1}(i(I))$ means that $a\in IS^{-1}R \cap R$. Now let $b\in I$ and $\frac{r}{t} \in S^{-1}R$ such that $a= b\cdot\frac{r}{t}$ (because of the relation I just stated). Then $(a,1)$ is in relation with $(b\cdot r, t)$, but $b\cdot r \in I$ because it is an ideal. –  Giovanni De Gaetano Oct 10 '12 at 12:29
Gah, and now im stuck on showing its onto : ( –  Chris Oct 10 '12 at 18:15
Nevermind got it! Just showed that any prime ideal has an inverse such that every element in the target prime ideal is hit by the map of the inverse, and thus the inverse maps onto the target. –  Chris Oct 10 '12 at 18:31