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Just a quick note to say it’s almost entirely obvious that if we have $f:A \rightarrow B$ a ring map, an $A$-module $M$ and a $B$-module $N$, then giving a map $b:M \otimes_A B \rightarrow N$ is exactly the same as giving a map $a:M \rightarrow N_A$ (where this denotes $N$ considered as an $A$-module via the map $f$). In other words, extension of scalars is a left adjoint to restriction of scalars.

To see this, note that given $b$ we can define $a$ by $m \mapsto b(m\otimes 1)$, and given $a$ we can define $b$ by $m \otimes x \mapsto x.a(m)$, and these operations are obviously mutually inverse.

Not sure why I felt like blogging about such a triviality. Maybe I think it’s a nice example of something that confused me slightly while solving a problem today and now, with the language of adjoint functors, I should never have to waste time thinking slightly about it again. I guess if you care this is also the basic case of the pullback-pushforward adjunction on quasicoherent sheaves.