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This is the first time I'm asking a question, hope it's not a silly question.
I'm studying through Ravi Vakil's notes, and I came up to this 1.5E exercise that reads like this:

Suppose $A \to B$ is a morphism of rings. If $M$ is a $B$-module, you can create an $A$-module $M_A$ by considering it as an $A$-module. This gives a functor $\newcommand\Mod{\mathop{\textrm{Mod}}\nolimits}(\,\cdot_A) : \Mod(B) \to \Mod(A)$. Show that this functor is right-adjoint to $( \cdot \otimes_A B)$. In other words, describe a bijection

$\newcommand\Hom{\mathop{\textrm{Hom}}\nolimits} \Hom_B(N \otimes_A B,M) \cong \Hom_A(N,M_A)$ functorial in both arguments.

I can clearly see why $N \otimes_A B$ is a $B$-module (it's enough to define multiplication $b(n\otimes c)=(n \otimes_b c))$. I've thought about it but I just can't find a bijection.

Thanks to all in advance.

egreg
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Tyche
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1 Answers1

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If $M_B$ is a $B$-module, then $M_A\cong\operatorname{Hom}_B(B,M)$, where $B$ is considered as an $A$-$B$-bimodule, so this is the standard adjunction between tensor and Hom.

A direct way of seeing the bijection is to consider $f\colon N\otimes_AB\to M$ and sending it to $\tilde{f}\colon N\to M$ defined by $$ \tilde{f}(n)=f(n\otimes 1) $$ which is a morphism of $A$-modules: $$ \tilde{f}(na)=f(na\otimes 1)=f(n\otimes a)=f((n\otimes 1)a)=\tilde{f}(n)a $$ Similarly, one can define, given $g\colon N_A\to M_A$, $$ \hat{g}\colon N\times B\to M $$ by $$ g'(n,b)=g(n)b $$ which is $A$-balanced: $$ g'(na,b)=g(na)b=g(n)ab=g'(n,ab) $$ and so $g'$ defines a unique morphism $\hat{g}\colon N\otimes_A B\to M$ (of $B$-modules).

It's a simple verification that we get two maps inverse of each other.

egreg
  • 238,574
  • Oh, I see you point!! This way I can use the fact that HomB(N⊗AB,M)=HomA(N,HomA(B,M)). A little difference from your statament would be that MA=HomA(B,M) and not HomB(B,M) – Tyche Dec 06 '13 at 11:26
  • @Tyche $\operatorname{Hom}_B(N\otimes_AP,M)\cong\operatorname{Hom}_A(N,\operatorname{Hom}_B(P,M))$, where $_AP_B$ is a bimodule. – egreg Dec 06 '13 at 11:29
  • I'm sorry, I haven't touched math for about 2 years and I feel very rusty :S
    I was trying to write a bijection between Ma=HomB(B,M). I thought to define
    f:MA ->HomB(B,M) by
    m |--> f(m) where f(m):B->M and sends each b|-->bm
    is this a good idea?     – Tyche Dec 06 '13 at 11:53