Let $A$ be a Noetherian commutative ring and Let $A\rightarrow B$ be a finite flat homomorphism of rings. We can thus form the so called "trace" $\mathrm{Tr_{B/A}}:B\rightarrow A$, which is a homomorphism of $A$ – modules defined as follows:
Every $b\in B$ acts on $B$ (when viewed as an $A$ – module) by multiplication. Since $B$ is finite flat over $A$ and $A$ is Noetherian, $B$ is a locally free $A$ – module and hence multiplication by $b$ is given locally (on a principal open subset $\mathrm{Spec}(A_s)\subseteq \mathrm{Spec}(A),s\in A$ and under some isomorphism $B_s\cong A_s^n$) by multiplication by a matrix. We define $\mathrm{Tr_{B/A}}(b)$ to be the trace of this matrix. Since the trace of a matrix is independent of the choice of basis this homomorphism of $A$ – modules glues nicely and is well defined.
In the case $A\rightarrow B$ is finite etale one can even show that this morphism is nondegenerate, i.e.: Induces an isomorphism $B\overset{\sim}\rightarrow
\mathrm{Hom}_A(B,A)$ by adjunction (this is a well known claim of Galois theory in the case where $A\rightarrow B$ is a finite seperable field extension).
My (rather ill-formulated) questions are the following:
1) Are there any other algebraic/geometric constructions I should think of as similar to this one?
2) Is there a deeper reason for the existence of such trace morphisms (for example some categorical phenomena that this is a special case of)? what's so special about finite flat homomorphisms that makes this happen? It seems to me pretty mysterious that such a homomorphism even exists, and I do not seem to completely grasp it's geometric meaning.
3) What is the geometric intuition behind the cannonical isomorphism of $A$ – modules $B\overset{\sim}\rightarrow \mathrm{Hom}_A(B,A)$ in case $A\rightarrow B$ is finite etale?
[edit] Let me try to be abit more specific about what bothers me:
Given a ring homomorphism $\phi :A\rightarrow B$ there's an obvious adjunction: $\mathrm{Forget}:\mathsf Mod_B \substack{\longrightarrow\\\perp \\\longleftarrow \\}\mathsf Mod_A:\mathrm{Hom}_A(B,-)$ which, by evaluating the counit at $A$, gives a map $\mathrm{Tr}_\phi:\mathrm{Hom}_A(B,A)\rightarrow A$ in $\mathsf{Mod}_A$.
Also, given a proper map $f:X\rightarrow Y$ between reasonable schemes (for example essentially finite type schemes over a field) we have the adjunction given by Grothendieck duality $Rf_*:\mathsf D^b_c(X) \substack{\longrightarrow\\\perp \\\longleftarrow \\}\mathsf D^b_c(Y):Rf^!$ which induces (again by evaluating the counit at $\mathcal{O}_Y$) a morphism $\mathrm{Tr}_f:Rf_*Rf^!\mathcal{O}_Y\rightarrow\mathcal{O}_Y$ in $\mathsf{D}_c^b(Y)$
What bothers me is that my original trace, unlike the two trace maps I just mentioned, does not seem to come as naturally from some adjunction or anything like that. Where is it coming from? What is it?
Best Answer
Regarding 2) in the question. If $\varphi \colon A \to B$ is a finite flat homomorphism of rings, then the corresponding map of schemes $f \colon X \to Y$ is a finite flat map of schemes, therefore proper. Here $X:= Spec(B)$ and $Y:= Spec(A)$. From Grothendieck duality one has an adjunction $f_* \dashv f^!$, where $f_* \colon Qco(X) \to Qco(Y)$ and $f^!$ is its right adjoint. We don't need to take derived categories because we are in relative dimension 0. it turns out that one can describe this adjoint as: $$ f^!(\mathcal{G}) = \mathcal{H}om_{\mathcal{O}_Y}(f_*\mathcal{O}_X,\mathcal{G})^\sim $$ with $\mathcal{G} \in Qco(Y)$ and where $(-)^\sim$ denotes the equivalence between $\mathcal{O}_X$-modules and $f_*\mathcal{O}_X$-modules over $Y$, being $f$ an affine morphism. Now the counit of the adjunction $$ \int_f \colon f_*f^! \mathcal{G} \longrightarrow \mathcal{G} $$ applied to $\mathcal{O}_Y$ yields the map $$ \mathrm{Tr}_{\varphi} \colon B \longrightarrow A $$ that is mentioned in the question.
One of the most fascinating aspects of Grothendieck duality is this interrelation between very abstract concepts (the adjunction) together with very concrete descriptions (the matrix trace).
In higher dimensions one need to add derived functors (and most conveniently, derived categories) in the abstract part and higher dimensional residues in the concrete part.
Regarding 3) If $f$ is étale then it has a "trivial relative dualizing sheaf", in other words
$$ f^! \mathcal{O}_Y \cong f^* \mathcal{O}_Y \cong \mathcal{O}_X $$
which illustrates the isomorphism $B\overset{\sim}\rightarrow Hom_A(B,A)$ in the question.
Regarding the last question: your original trace is an explicit computation of both descriptions of duality for a finite flat map. The uniqueness of adjoints forces the trace to agree with the counit of the adjunction. In a philosophical way, the counit is a way of integrating, and the trace of a matrix is another, and in this case, they both agree as they should.