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Let $A$ be a commutative ring with $1$. Suppose that $P \subseteq Q$ are prime ideals in $A$ and that $M$ is an $A$-module. Prove that the localization of the $A$-module $M_{Q}$ at $P$ is the localization $M_{P}$, i.e $(M_{Q})_{P} = M_{P}$.

Hint from the book: Use the fact that $S^{-1}A \otimes _{A} M \cong S^{-1}M$ as $S^{-1}A$ modules.

Here's what I have:

First set $S=A \setminus P$ and $T=A \setminus Q$, then by assumption $T \subset S$.

So using the hint:

$S^{-1}(T^{-1}M) \cong S^{-1}A \otimes_{A} T^{-1}M \cong S^{-1}A \otimes_{A} (T^{-1}A \otimes_{A} M) \cong (S^{-1}A \otimes_{A} T^{-1}A) \otimes_{A} M$

From here I'm stuck. Can you please help?

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    It should suffice to prove that $S^{-1}(T^{-1}A)\cong S^{-1}A$ as $A$-modules, hence $S^{-1}(T^{-1}M)\cong S^{-1}(T^{-1}A)\otimes_A M\cong S^{-1}A\otimes_A M\cong S^{-1}M.$ I imagine the universal property of localization, plus the fact that $P\subseteq Q$, should do the trick, but I'm not thinking clearly today - I'm sure someone else who will be able to help you will come along shortly.2011-07-24

3 Answers 3

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How about: Noting that $(A_Q)_P=A_P$, $(M_Q)_P=(A_Q)_P \otimes_{A_Q} M_Q = A_P \otimes_{A_Q} (A_Q \otimes_A M)=(A_P \otimes_{A_Q} A_Q) \otimes_A M =A_P \otimes_A M = M_P$

Added later: The prove $(A_Q)_P=A_P$, map $A_P \rightarrow (A_Q)_P$ by $a/s \mapsto (a/1)/(s/1)$ This is injective, and if we choose (a/t)/(s/t') \in (A_Q)_P, then this is hit by at'/st \in A_P, since (at'/1)/(st/1)=(a/t)/(s/t') in $(A_Q)_P$.

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    @John M - If I were you, I’d define a morphism $(A_Q)_P\to A_P$ by the obvious formula, and check that the morphisms are inverse. Of course it’s a matter of personal taste...2011-07-25
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This is a little messy, but it works:

Note that $S^{-1}A$ and $T^{-1}A$ are naturally $(A,T^{-1}A)$-bimodules, so that, as $(A, T^{-1}A)$-bimodules,

$S^{-1}A \otimes_A T^{-1}A \simeq (T^{-1}A \otimes_{T^{-1}A}S^{-1}A) \otimes_A(T^{-1}A \otimes_{T^{-1}A}T^{-1}A)$

$\simeq (S^{-1}A \otimes_{T^{-1}A}T^{-1}A) \otimes_A(T^{-1}A \otimes_{T^{-1}A}T^{-1}A)$

$\simeq S^{-1}A \otimes_{T^{-1}A}((T^{-1}A \otimes_AT^{-1}A) \otimes_{T^{-1}A}T^{-1}A)$

$\simeq S^{-1}A \otimes_{T^{-1}A}T^{-1}A \simeq S^{-1}A$

Now just substitute this in your sequence of isomorphisms and use the definition.

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    You can prove that directly: the map $T^{-1}A \times T^{-1}A \to T^{-1}A$ given by $(u,v) \mapsto uv$ is $A$-bilinear and surjective, and it factors through $T^{-1}A \otimes_A T^{-1}A$ by definition. On the other hand, the map $T^{-1}A \to T^{-1}A \otimes_A T^{-1}A$ given by $u \mapsto 1 \otimes u$ is an an $A$-module homomorphism which is an explicit inverse for this map.2011-07-24
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It should be easy to show, using the universal property of tensor products of algebras and that of localization, that for each $A$-algebra $f\colon A \to B$ with $f(S) \subset B^*$ there is a unique homomorphism of $A$-algebras $S^{-1}A \otimes_A T^{-1}A \to B$. Here the $A$-algebra structure $A \to S^{-1}A \otimes_A T^{-1}A$ is given by $a \mapsto a \otimes 1 = 1 \otimes a$.