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Let $\phi: A \rightarrow B$ be a ring homomorphism. Let $M$ be an $A$-module. We can think $B$ as $A$-module via the map $\phi$ defined by $\phi:A\times B \rightarrow B$, $(a,b)\mapsto\phi(a)\cdot b$.

So we can construct the $A$-module $M \otimes_A B$. Furthermore, $M \otimes_AB$ can be thought of as a $B$-module via $B\times (B \otimes _AM)\rightarrow (B\otimes _A M)$ defined by $(b',(b\otimes m))\mapsto b'b\otimes m$.

Question: Why is $\operatorname{Hom}_A(M,N)=\operatorname{Hom}_B(B\otimes_A M,N)$ where $N$ is an $B$-module and in the left hand side $N$ is considered as a $A$-module as described above?

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I suppose that $N$ is a $B$-module, that can be considered also an $A$-module via $(a,n)\mapsto\phi(a)n$. –  egreg Jan 15 at 15:38
    
I agree with egreg. Actually $N$ is a $B$-module, and if $N|_A$ denotes the underlying $A$-module, then the claim is $\hom_A(M,N|_A) \cong \hom_B(B \otimes_A M,N)$, i.e. that $B \otimes_A - $ is left adjoint to $(-)|_A$. –  Martin Brandenburg Jan 15 at 15:39
    
Thanx. have edited the question. –  Susobhan Jan 15 at 17:12

1 Answer 1

Hint: Given an $A$-linear map $f : M \to N$, define $B \times M \to N$ by $(b,m) \mapsto b \, f(m)$. Check that this is $A$-bilinear, hence lifts to an $A$-linear map $h : B \otimes_A M \to N$ characterized by $h(b \otimes m)= b \, f(m)$. Check that it is actually $B$-linear. Conversely, given a $B$-linear map $h : B \otimes_A M \to N$, check that $f : M \to N$, $m \mapsto h(1 \otimes m)$ is $A$-linear. Finally, prove that these constructions are inverse to each other.

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I tried exactly the same thing. I am stuck somewhere in checking of the A-linearity of the second map. I will try it again. –  Susobhan Jan 15 at 17:16
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This is clear since $m \mapsto 1 \otimes m$ is $A$-linear by the very definition of the tensor product (and bilinear maps) and $h$ is $A$-linear (since it is $B$-linear). –  Martin Brandenburg Jan 15 at 18:41

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