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Let $f: S^{-1} M \to S^{-1}A \otimes_A M$ defined by $$m/s \to 1/s \otimes m$$ $g: S^{-1}A \otimes_A M \to S^{-1} M$ defined by $$a/s \otimes m \to am/s $$

Prove that $f$ and $g$ are well defined ?

How can we prove $f$ is an $S^{-1}A$ module homomorphism?

Here $A$ is commutative ring with identity and $M$ is module and $S$ multiplicative subset of $A$.

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You mean $S$ is a multiplicative subset of $A$ ? – user18119 Nov 18 '12 at 19:57
yes sorry thats what i meant – user48931 Nov 18 '12 at 20:05
the only problem I have is how to prove that f is well defined ? all the rest is ok – user48931 Nov 18 '12 at 20:39

Hint: To get the map $S^{-1}A \otimes_A M \rightarrow M$, define a suitable bilinear map $S^{-1}A \times M \rightarrow S^{-1}M$ and then use the universal property of the tensor product.

To show that the map $S^{-1}M \rightarrow S^{-1}A\otimes_A M$ is well defined, note that $m/s = n/t \Rightarrow \exists s' \in S| s'(tm - sn) = 0 \Rightarrow (1/s'st) \otimes (s'(tm - sn)) = 0 \Rightarrow (1/st) \otimes (tm - sn) = 0 \Rightarrow (1/s) \otimes m - (1/t) \otimes n = 0$

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got it thanks :) – user48931 Nov 18 '12 at 20:59

Prove that $f$ is well defined

means the following:

An element of $S^{-1}M$ is an equivalence class of "fractions" $m/s$. Prove that $f(m/s)$ remains the same no matter what representative for the equivalence class of $m/s$ you choose.

(This is more or less always what it means when an exercise tells you to prove that a function is well-defined.) So, two fractions $m/s$ and $n/t$ represent the same equivalence class if there is an $u\in S$ such that $u(tm - sn) = 0$. So, let's see what happens to the two fractions if we use $f$ on them: $$ f(m/s) = 1/s \otimes m \\ f(n/t) = 1/t\otimes n $$ Now, the question is, do they represent the same element in $S^{-1}A \otimes_A M$? The usual way to test this is to take the difference between the two results and verify that it equals $0$. But as they stand it's not easy to calculate the difference, since the two tensor products have no terms in common. We need to fix that.

In $S^{-1}A$, we have that $1/s = ut/ust$, and likewise, $1/t = us/ust$, so we get (by bilinearity of tensor product): $$ 1/s \otimes m = ut/ust \otimes m = 1/ust\otimes utm \\ 1/t \otimes n = us/ust \otimes n = 1/ust\otimes usn $$ Now we can calculate the difference between these two as $$ 1/ust\otimes utm - 1/ust\otimes usn = 1/ust\otimes (utm - usn) = 1/ust\otimes 0 = 0 $$ and thus the value of $f$ at $m/s$ in independent of the choice of representative, and $f$ is therefore well defined. Can you make a similar reasoning for $g$? Can you prove that $f\circ g$ and $g\circ f$ are the identity functions on their respective $A$-modules? What does this tell us about the relationship between $S^{-1}A \otimes_A M$ and $S^{-1}M$?

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Thanks for the help , Yes I proved that f 0g and g 0 f are identity and that means that S−1A⊗AM and S−1M are isomorphic to each others – user48931 Nov 18 '12 at 20:48
how can we prove f is S-1A mod homom ? – user48931 Nov 18 '12 at 21:01
Well, by definition, a group homomorphism $h:N\to N'$ of modules over a ring $R$ is an $R$-module homomorphism if $$ h(rm) = rh(m) $$ for any $r \in R$ and $m \in N$ (the fact that $h$ is a group homomorphism covers the case $r = 1$, if such an element exists in $R$). Since in this case, $h = f$, $R = S^{-1}A$, $N = S^{-1}M$ and $N' = S^{-1}A\otimes M$, you have to rewrite $r$ and $m$ as fractions, calculate, and see what you get on each side of the equality sign. If they turn out to be equal, then you have a module homomorphism. – Arthur Nov 18 '12 at 22:04

To prove that the maps are well-defined, you must show that if $m/s=m^\prime/s^\prime$, then $f(m/s)=f(m^\prime/s^\prime)$. For $f$, this means that you must show that $1/s^\prime \otimes m^\prime = 1/s \otimes m$. This follows from the following fact: $(ms^\prime-m^\prime)r=0$ for some $r \in A$. Now, multiply both sides of $1/s^\prime \otimes m^\prime = 1/s \otimes m$ with $ss^\prime$, and use that (since we tensor over $A$), you can move elements from $A$ on the other side of $\otimes$). Multiply on both sides by $r$. The result follows.

To prove that $g\circ f$ is the identity, just plug in and see what happens. A general element in $S^{-1}M$ is written as $m/s$. Now: $g(f(m/s))=g(1/s \otimes m)=m/s$, so $g \circ f$ is the identity. The composition $f \circ g$ is similar.

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how can we prove f is S-1A mod homom ? – user48931 Nov 18 '12 at 21:00
@user48931: That just means that $f$ is a homomorphism of abelian groups plus it preserves the $S^{-1}A$-module structure: $f(am/s)=(a/s)f(m)$. That can be checked. – Fredrik Meyer Nov 18 '12 at 21:12
dont we have to prove that f(a'/s' . m/s ) = a'/s' f(m/s) ? – user48931 Nov 18 '12 at 21:16
@user48931: That amounts to the same thing, because $f((a^\prime/s^\prime) m/s)=f((a/ss^\prime)m)$. – Fredrik Meyer Nov 18 '12 at 21:23
Then we have to prove f(am/s) = a f (m/s) right ? – user48931 Nov 18 '12 at 21:31

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