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Let $A$ be a commutative ring with $1$ and let $M,N$ be $A$-modules.

Since there is a map $f: A \rightarrow S^{-1}A$, defined by $a \mapsto \frac{a}{1}$ then given any $S^{-1}A$-module we can view it as a $A$ module via restriction of scalars right?

Now $S^{-1}M$ and $S^{-1}N$ are $S^{-1}A$-modules.

My question is if the following isomorphism holds?

$S^{-1}M \otimes_{A} S^{-1}N \cong S^{-1}M \otimes_{S^{-1}A} S^{-1}N$

Is the above valid because any $S^{-1}A$ module is an $A$-module or why? (or perhaps it is false), can you please help?

Following Daniel's hint:

$S^{-1}(M \otimes_{A} N) \cong S^{-1}A \otimes_{A} (M \otimes_{A} N) \cong (S^{-1}A \otimes_{S^{-1}A}) (S^{-1}A \otimes_{A} (M \otimes _{A} N) )$

After this I end up with $(S^{-1}A \otimes _{S^{-1}A} N) \otimes_{A} S^{-1}M$ which is isomorphic to $S^{-1}N \otimes_{A} S^{-1}M$. Where's the error?

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What you've written doesn't make sense--you write $S^{-1}A\otimes_{S^{-1}A}N$, but $N$ is not an $S^{-1}A$-module, so this is meaningless. You've skipped some steps, which is where your error is--try writing things out line-by line. –  Daniel Litt Apr 10 '11 at 6:25
    
(In any case, I didn't realize this was homework, so I've deleted my answer.) –  Daniel Litt Apr 10 '11 at 6:26
    
@Daniel Litt: It isn't, but thanks, your hint was useful. –  user6495 Apr 11 '11 at 5:18

1 Answer 1

up vote 5 down vote accepted

Congratulations on having noticed this subtle point, rarely discussed in textbooks.
As is often the case, a more general statement is clearer; for your question take $P=S^{-1}M, Q=S^{-1}N$ in the following
General statement Suppose $P,Q$ are $S^{-1}A$-modules. Then there is a canonical $S^{-1}A$- isomorphism $P \otimes _A Q\to P \otimes_ {S^{-1}A} Q$

Preliminary remark An $A$-module $E$ can have at most $one$ $S^{-1}A$-module structure compatible with its $A$-module structure.
Proof of Preliminary remark: we must have $\frac{a}{s} \ast e = (s\bullet)^{-1} (ae)$ (The existence of an $S^{-1}A$-module structure on $E$ forces multiplication by $s$ to be an $A$-linear automorphism $(s\bullet)$ of the $A$-module $E$)

Proof of General statement The preliminary remark shows that the $S^{-1}A$-module structures on $P \otimes_A Q$ coming from $P$ or from $Q$ coincide. Hence there are canonical ${S^{-1}A}$- morphisms
$P \otimes _A Q\to P \otimes_ {S^{-1}A} Q: p\otimes q\mapsto p\otimes q$ and
$P \otimes_ {S^{-1}A} Q \to P \otimes _A Q\ : p\otimes q\mapsto p\otimes q$ which are mutually inverse $S^{-1}A$- isomorphisms ; this proves the General statement.

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There is a cautionary tale in this context. Let f:A --> B be a ring homomorphism. It lets any B-module be thought of as an A-module through restriction of scalars. If we have two B-modules P and Q, both can be viewed as A-modules and we can form the A-module tensor product $P \otimes_A Q$. This can be made into a B-module in two ways, using either the B-module structure on P or on Q. When B is a localization of A then the two ways of viewing $P \otimes_A Q$ as a B-module are exactly the same, but in general these two B-module structures on $P \otimes_A Q$ need not be isomorphic. –  KCd Apr 10 '11 at 16:46
    
An example of the above situation is in Example 6.19 on page 35 in www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf –  KCd Apr 10 '11 at 16:47
    
Yes, exactly. All these subtle caveats around the tensor product should be incorpotated into a textbook and it seems that you have already written the kernel of one such... –  Georges Elencwajg Apr 10 '11 at 17:47
    
thank you very much. –  user6495 Apr 11 '11 at 5:18
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Note that the Preliminary remark is in fact equivalent to the statement that "$A \to S^{1}A$ is an epimorphism in the category of (commutative) rings." [Unless I am mistaken, the statement remains the same with or without the word "commutative."] Thus, the General statement applies with $S^{-1}A$ replaced by any $A$-algebra $B$ such that $A \to B$ is an epimorphism of rings. In particular, this holds if $B = A/I$. –  Charles Staats Jul 10 '11 at 21:06

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