1
$\begingroup$

Recall the standard equivalence between the opposite of the category of affine varieties and the category of reduced and finitely-generated algebras (over a given algebraically-closed field): $$\mathsf{Aff}^{\mathrm{op}}\cong\mathsf{RedAlg}$$

Question: Is this an adjoint equivalence? If so, what is the underlying adjunction?

$\endgroup$
4
$\begingroup$

This is a weird question to ask, since really every equivalence is an adjoint equivalence. The question of whether an equivalence "is" an adjoint equivalence only makes sense if you have chosen specific natural isomorphisms $1\cong FG$ and $GF\cong 1$ (and given a choice of $1\cong FG$, there is always a unique choice of $GF\cong 1$ such that they give an adjunction). And the "underlying adjunction" of an adjoint equivalence is just the equivalence itself together with the choice of natural isomorphisms (which are the unit and counit).

That said, the "standard" choice of the isomorphisms $1\cong FG$ and $GF\cong 1$ for the equivalence $\mathsf{Aff}^{\mathrm{op}}\cong\mathsf{RedAlg}$ does make it an adjoint equivalence. Explicitly, $F:\mathsf{Aff}^{\mathrm{op}}\to\mathsf{RedAlg}$ is the functor taking an affine variety to its algebra of regular functions and $G:\mathsf{RedAlg}\to \mathsf{Aff}^{\mathrm{op}}$ is the functor taking an algebra to its maximal spectrum. Given a variety $V$, each point of $V$ determines a maximal ideal in $F(V)$, giving a map $V\to GF(V)$ which turns out to be an isomorphism of varieties. And given an algebra $A$, each element of $A$ can be considered as a regular function on the spectrum $G(A)$, giving a map $A\to FG(A)$ which turns out the be an isomorphism of algebras. It is straightforward to verify that these natural transformations satisfy the required identities to be the unit and counit of an adjunction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.