finitely generated projective

An MO user had a question about the “if” part of the following claim:

- A module $P$ over a commutative ring $k$ is finitely generated projective iff the canonical map $\psi: P^\vee \otimes_k P \to \hom(P, P)$ is an isomorphism.

The question got shut down rapidly as “not research level”. But I thought it deserved an answer. Anyone who has a simpler proof should feel free to contact me (my email should be readily available through my MO user page).

The idea of proof is to establish an adjunction $P \otimes_k - \dashv P^\vee \otimes_k -$; see the proposition below. Since we also have $P \otimes_k - \dashv \hom(P, -)$ and right adjoints are unique up to canonical isomorphism, we get an isomorphism

$P^\vee \otimes_k - \cong \hom(P, -).$

But $P^\vee \otimes_k -$, being itself left adjoint to $\hom(P^\vee, -)$, preserves colimits, so $\hom(P, -)$ preserves colimits. The fact that it preserves coequalizers means $P$ is projective (i.e., $\hom(P, -)$ is right exact). To see $P$ is finitely generated, write $P$ as the (filtered) colimit of the system of its finitely generated submodules $F$ and inclusions between them. Then the comparison map

$colim_F \hom(P, F) \to \hom(P, colim_F F) \cong \hom(P, P)$

is an isomorphism. In particular, some element on the left, represented by an element of some $\hom(P, F)$, gets mapped to $1_P \in \hom(P, P)$. That is to say, there is some $\phi: P \to F$ whose composition with the inclusion $F \hookrightarrow P$ is $1_P$. This forces $\phi$ to be a monomorphism of $P$ into the f.g. submodule $F \hookrightarrow P$, i.e., $F \hookrightarrow P$ and $1_P: P \hookrightarrow P$ are mutually submodules of one another, with $\phi$ the inverse of the inclusion $F \hookrightarrow P$, forcing $P \cong F$ to be finitely generated.

Given that the canonical map $P^\vee \otimes P \to \hom(P, P)$ is an isomorphism, there is an adjunction $P \otimes - \dashv P^\vee \otimes -$.

For this we need a unit and a counit. The counit is given by the evaluation map $c: P \otimes_k P^\vee \to k$. The unit $u: k \to P^\vee \otimes_k P$ is given by the composite

$k \stackrel{[1_P]}{\to} \hom(P, P) \stackrel{\psi^{-1}}{\to} P^\vee \otimes_k P$

and the proof is completed by verifying the triangular equations for an adjunction, viz. that the following composites are identity maps:

$P \cong P \otimes_k k \stackrel{P \otimes_k u}{\to} P \otimes_k P^\vee \otimes_k P \stackrel{c \otimes P}{\to} k \otimes P \cong P,$

$P^\vee \cong k \otimes_k P^\vee \stackrel{u \otimes_k P^\vee}{\to} P^\vee \otimes_k P \otimes_k P^\vee \stackrel{P^\vee \otimes_k c}{\to} P^\vee \otimes_k k \cong P^\vee.$

We leave this verification to the reader.

Created on October 14, 2014 at 02:56:19
by
Todd Trimble