Todd Trimble
finitely generated projective

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

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 kP kP \otimes_k - \dashv P^\vee \otimes_k -; see the proposition below. Since we also have P khom(P,)P \otimes_k - \dashv \hom(P, -) and right adjoints are unique up to canonical isomorphism, we get an isomorphism

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

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

colim Fhom(P,F)hom(P,colim FF)hom(P,P)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)\hom(P, F), gets mapped to 1 Phom(P,P)1_P \in \hom(P, P). That is to say, there is some ϕ:PF\phi: P \to F whose composition with the inclusion FPF \hookrightarrow P is 1 P1_P. This forces ϕ\phi to be a monomorphism of PP into the f.g. submodule FPF \hookrightarrow P, i.e., FPF \hookrightarrow P and 1 P:PP1_P: P \hookrightarrow P are mutually submodules of one another, with ϕ\phi the inverse of the inclusion FPF \hookrightarrow P, forcing PFP \cong F to be finitely generated.


Given that the canonical map P Phom(P,P)P^\vee \otimes P \to \hom(P, P) is an isomorphism, there is an adjunction PP 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 kP kc: P \otimes_k P^\vee \to k. The unit u:kP kPu: k \to P^\vee \otimes_k P is given by the composite

k[1 P]hom(P,P)ψ 1P kPk \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:

PP kkP kuP kP kPcPkPP,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 k kP u kP P kP kP P kcP kkP .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