Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When I did some reading on localization, I stumbled upon a thought which I'm not able to prove. This might be right or wrong in a trivial fashion, but I'm stuck nonetheless.

Let A be an integral domain and suppose $S \subset A$ is multiplicatively closed, but $S^{-1} A \neq Q(A)$. Thus there exists $P \in Spec(A)$ so that $S^{-1}P$ is a prime ideal in $S^{-1}A$, and consequently, we may look at the localization $S^{-1}A_{S^{-1}P}$. Does $S^{-1}A_{S^{-1}P} \cong A_P$ hold, and if yes, how can I prove it?

share|improve this question
add comment

3 Answers 3

up vote 5 down vote accepted

Yes, $S^{-1}A_{S^{-1}P} \cong A_P$ holds. Here is why.

Put $T=A\setminus P$. Your hypothesis that $P $ survives in $S^{-1}A$ amounts to $P\cap S=\emptyset$, or equivalently that $S\subset T$.
But then inverting first elements of $S$, obtaining $S^{-1} A$ and then inverting the image of $T$ obtaining $(S^{-1}A)_{S^{-1}P} $ amounts to the same as directly inverting all of $T$ in one fell swoop, obtaining $A_P$.

You can find the details in Bourbaki, Commutative algebra, chap.II , §2, Prop.7.

share|improve this answer
add comment

A localization of a localization is itself a localization. Essentially for the same reason that a fraction of fractions is just a fraction.

Proposition. Let $A$ be a commutative ring with $1$; let $S$ be a multiplicative subset of $A$, and let $R=S^{-1}A$ be the localization of $A$ at $S$. Let $T$ be a nonempty multiplicative subset of $R$, and let $U\subseteq A$ be the set $$U = \left\{ u\in A\ \left|\ \exists s\in S\text{ such that }\frac{u}{s}\in T\right\}\right..$$ Then $U$ is a multiplicative subset of $A$, and $$\frac{(r/s')}{(u/s)} \longmapsto \frac{rs}{us'}$$ is an isomorphism $T^{-1}R\to U^{-1}A$.

I'll give two proofs: a "pedestrian" one and a fancy one.

Proof the Pedestrian. First, we prove that $U$ is a multiplicative subset of $A$: if $u,u'\in A$, then there exist $s,s'\in S$ such that $\frac{u}{s},\frac{u'}{s'}\in T$. Since $T$ is multiplicative, $\frac{uu'}{ss'}\in T$, hence $uu'\in U$. So $U$ is multiplicative.

Second, if $u\in U$ and $s'\in S$, then $us'\in U$; that is, $U$ absorbs multiplication by elements of $S$: for there exists $s\in S$ such that $\frac{u}{s}\in T$. Then, since $\frac{us'}{ss'} = \frac{u}{s}$, and $ss'\in S$, then $us'\in T$, as claimed.

Therefore, $\frac{rs}{us'}$ makes sense as an element of $U^{-1}A$.

Next, we show that the map is well-defined. If $$\frac{r/s'}{u/s} = \frac{\rho/\sigma'}{\vartheta/\sigma}$$ in $T^{-1}R$, then there exists $\frac{x}{y}\in T$ such that $$\frac{x}{y}\left(\frac{r\vartheta}{s'\sigma} - \frac{u\rho}{s\sigma'} \right)= 0_R,$$ which is equivalent to $$\frac{x(r\vartheta s\sigma' - u\rho s'\sigma)}{yss'\sigma\sigma'} = 0_R,$$ which in turn is equivalent to saying that there exists $z\in S$ such that $$zx(r\vartheta s\sigma' - u\rho s'\sigma) = 0_A.$$ But since $\frac{x}{y}\in T$, then $x\in U$; and since $z\in S$, then $zx\in U$. Therefore, the equation written above implies that in $U^{-1}A$ we have $$\frac{rs}{us'} = \frac{\rho \sigma}{\vartheta\sigma'},$$ which is exactly what wee need for the map to be well defined.

Showing that the map is a homomorphism is now easy, just as doing fraction arithmetic. We have: $$\begin{align*} \frac{r/s'}{u/s} + \frac{x/y}{z/w} &= \frac{\frac{rz}{s'w} + \frac{xu}{ys}}{\frac{uz}{sw}}\\ &= \frac{\quad\frac{rzys + xus'w}{s'wys}}{\frac{uz}{sw}}\\ &\longmapsto \frac{(rzys + xus'w)(sw)}{uzs'wys} = \frac{rzys + xus'w}{uzs'y}.\\ \frac{rs}{s'u} + \frac{xw}{yz} &= \frac{rsyz + xws'u}{s'uyz}.\\ \frac{r/s'}{u/s} \times \frac{x/y}{z/w} &= \frac{\quad \frac{rx}{s'y}\quad}{\frac{uz}{sw}}\\ &\longmapsto \frac{rxsw}{s'yuz}.\\ \frac{rs}{s'u}\times \frac{xw}{yz} &= \frac{rsxw}{s'uyz}. \end{align*}$$

Now assume that $\frac{r/s'}{u/s}$ maps to $0$. That means that $\frac{rs}{us'} = 0_{U^{-1}A}$, so there exists $\vartheta\in U$ such that $\vartheta(rs)=0$. But then there exists $s''\in S$ such that $\frac{\vartheta}{s''}\in T$, and $$\frac{\vartheta}{s''}\cdot\frac{r}{s'} = \frac{\vartheta r}{s''s} = 0_{R}$$ because $s\in S$ and $s(\vartheta r) = 0$. So the map is one-to-one.

Finally, given $\frac{a}{u}\in U^{-1}A$, there exists $s\in S$ such that $\frac{u}{s}\in T$; then $\frac{a/s}{u/s}\mapsto \frac{as}{us} = \frac{a}{u}$, so the map is onto. Thus, we have an isomorphism. $\Box$

The proposition can also be proven by invoking the Universal Property of the Localization: if $S$ is a multiplicative subset of $A$ and $\psi\colon A\to S^{-1}A$ is the canonical map, then for every commutative ring $R$ and every ring homomorphism $\varphi\colon A \to R$ such that $\varphi(s)$ is a unit in $R$ for every $s\in S$, there exists a unique $\Phi\colon S^{-1}A\to R$ such that $\varphi=\Phi\circ\psi$.

The map $\Phi$ is necessarily given by $\Phi(\frac{a}{s}) = \varphi(a)\varphi(s)^{-1}$.

Proof the Fancy. Let $A$, $S$, $R$, $T$, and $U$ be as in the statement of the proposition, let $\varphi\colon A\to R$ be the canonical map, $\varphi(a) = \frac{as_0}{s_0}$ (for some fixed $s_0\in S$, say); and $\psi\colon A\to U^{-1}A$ be the canonical map $\psi(a)=\frac{au_0}{u_0}$. Let $\theta\colon R\to T^{-1}R$ be the canonical map, $\theta(r)=\frac{rt_0}{t_0}$. Write $t_0 = \frac{\rho_0}{\sigma_0}$.

First, we would prove, as above, that $U$ is a multiplicative subset and absorbs multiplication by elements of $S$.

Now consider the composition $A\to S^{-1}A\to T^{-1}R$. If $u\in U$, then there exists $s\in S$ such that $\frac{u}{s}\in T$; then $\frac{u}{s}$ is mapped to a unit in $T^{-1}R$; but $s$ is mapped to a unit in $S^{-1}A$, so $\frac{ss}{s}\in S^{-1}A$ is mapped to a unit in $T^{-1}R$. Thus, the product $\frac{uss}{ss}$ is mapped to a unit, but this is the image of $u$ in $T^{-1}R$. Hence the compositum $A\to S^{-1}A\to T^{-1}R$ maps every $u\in U$ to a unit, so the map factors through $U^{-1}A$; this gives the map $\Psi\colon U^{-1}A\to T^{-1}R$.

Now map $S^{-1}A$ to $U^{-1}A$ by selecting $u\in U$ and sending $\frac{a}{s}$ to $\frac{au}{su}\in U^{-1}A$. If $\frac{t}{s'}\in T$, then $\frac{tu}{s'u}$ is a unit in $U^{-1}A$, since $tu\in U$. Thus, the map factors through $T^{-1}(S^{-1}A) = T^{-1}R$, giving a map $T^{-1}R\to U^{-1}A$. Now just verify that the compositions of the maps $T^{-1}R\to U^{-1}A\to T^{-1}R$ and $U^{-1}A\to T^{-1}R\to U^{-1}A$ are the identity. $\Box$


Now, looking at your special case, note that $U$ consists precisely of all the "numerators" in $S^{-1}P$, and it is easy to see that they are precisely the elements of $P$.

share|improve this answer
    
Wow, I'm baffled by how helpful this community is. Thanks a bunch! –  tommy Dec 11 '11 at 19:49
add comment

Well, there is a one-to-one correspondence between prime ideals of $S^{-1}A$ and prime ideals of $A$ that have an empty intersection with $S$, via the mapping $P\mapsto S^{-1}P$.

So, let $S^{-1}P\in Spec(S^{-1}A)$ for some $P\in Spec(A)$ with $P\cap S=\emptyset$. Consider the mapping $f:S^{-1}A_{S^{-1}P}\rightarrow A_P$ given by:

$f\left(\frac{a_1}{s_1}/(\frac{a_2}{s_2})\right)=\frac{a_1 s_2}{s_1 a_2}$.

Note that in $\frac{a_1}{s_1}/(\frac{a_2}{s_2})\in S^{-1}A_{S^{-1}P}$, we have $s_i\in S$, $a_1\in A$, and $a_2\in A\setminus P$. Thus, $s_1a_2\notin P$ (since neither of $s_1$ or $a_2\in P$) and, obviously, $a_1 s_2\in A$. Thus, $f$ really does take elements to $A_P$. I'll let you show that $f$ is, in fact a homomorphism.

Given $\frac{a}{s}\in A_P$, $f\left(\frac{a}{s}/(\frac{1}{1})\right)=\frac{a}{s}$, so $f$ is onto. Finally, if $f\left(\frac{a_1}{s_1}/(\frac{a_2}{s_2})\right)=0$, then $\frac{a_1s_2}{a_2s_1}=\frac{0}{1}$, implying that $a_1s_2=0$, and hence $a_1=0$. From that, it follows that, in $S^{-1}A_{S^{-1}P}$, $\frac{a_1}{s_1}/(\frac{a_2}{s_2})=0$, and thus $f$ is injective.

Therefore, $S^{-1}A_{S^{-1}P}\cong A_P$.

share|improve this answer
    
Thanks, I think it's great when isomorphy can be described explicitly and nice like this. (sadly I cannot upvote with a guest account) –  tommy Dec 9 '11 at 17:34
    
@tommy - You're most welcome. –  user5137 Dec 9 '11 at 19:14
    
Could you tell me why $a_2 \notin P$ is true? It's the only thing that keeps me from understanding this proof. –  tommy Dec 11 '11 at 20:05
    
It's because of the bassackwards notation for localizing at a prime ideal. If $R$ is commutative with identity and $P$ is a prime ideal of $R$, then $R_P=\{a/b\,\vert\,a\in R, b\in R\setminus P\}$. So, since we're localizing $S^{-1}A$ at the prime ideal $S^{-1}P$, denominators of $S^{-1}A_{S^{-1}P}$ are elements of $S^{-1}A$ that are not in $S^{-1}P$. –  user5137 Dec 12 '11 at 1:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.