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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?

user6495
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  • I think the equation you have is not technically correct - shouldn't it be something like $$S^{-1}(T^{-1}M) \cong S^{-1}(T^{-1}A) \otimes_{T^{-1}A} T^{-1}M

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

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

     – Zev Chonoles Jul 24 '11 at 17:01
  • @Zev Chonoles: thanks, but what to do after the next step you wrote, also where do we use the assumption $P \subset Q$? – user6495 Jul 24 '11 at 17:07
  • 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. – Zev Chonoles Jul 24 '11 at 17:12

3 Answers3

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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 $(A_Q)_P$.

John M
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    Shouldn't you start with $(M_Q)_P = A_P \otimes_A M_Q$ ? – Bruno Joyal Jul 24 '11 at 18:13
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    @Bruno Joyal: No I don't think so, because here $M_Q$ is an $A_Q$-module. – John M Jul 24 '11 at 18:33
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    It is, but the definition is still $(M_Q)P = A_P \otimes_A M_Q$... I think what you wrote is correct, but perhaps missing a couple of steps to show that $A_P \otimes_A M_Q \simeq A_P \otimes{A_Q} M_Q$. I guess it depends on how picky one wants to be... :) – Bruno Joyal Jul 24 '11 at 18:40
  • @ Bruno Joyal: Hi thanks Bruno. I've made an edit to make it more clear. – John M Jul 24 '11 at 18:46
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    @John M: The crux of the matter seems to me to be that $(A_Q)_P\cong A_P$, so I think it would be helpful to the OP if you added an explanation of why this is true. – Zev Chonoles Jul 24 '11 at 19:45
  • @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... – Pierre-Yves Gaillard Jul 25 '11 at 08:37
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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.

Bruno Joyal
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  • I don't understand how you went from the second-to-last line to the last line, could you add a bit of explanation? I can see that $((T^{-1}A\otimes_A T^{-1}A)\otimes_{T^{-1}A}T^{-1}A\cong T^{-1}A\otimes_A T^{-1}A$, but why should that be isomorphic to $T^{-1}A$? – Zev Chonoles Jul 24 '11 at 19:43
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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. – Bruno Joyal Jul 24 '11 at 19:57
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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 $a \mapsto a \otimes 1 = 1 \otimes a$.

Glorfindel
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Dylan Moreland
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