# What are traces?

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?

 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?

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.

• I think you messed up the order of composition in the counit, it should probably be $f_{*} f^{!} \mathcal{G} \rightarrow \mathcal{G}$ – Anonymous Coward Aug 9 '17 at 17:07
• There is something I don't understand about your answer. The counit gives you a map $\mathrm{Hom}_{A}(B,A)\rightarrow A$ (which is by the way completely tautological, it just takes a homomorphism and inputs "1"). but $\mathrm{Hom}_{A}(B,A)$ is not the same thing as $B$. There is not even an obvious map $B\rightarrow \mathrm{Hom}_{A}(B,A)$, so I do not see how you even get the map $B\rightarrow A$ – Anonymous Coward Aug 9 '17 at 18:10
• @AnonymousCoward Thank you for spotting the typo. As for the étale case, consider the map that carries $1 \in B$ to $\mathrm{Tr}_{\varphi}$. It can be shown that this map is an isomorphism precisely when $f$ is étale. – Leo Alonso Aug 9 '17 at 21:20
• I don't understand the downvote.In the initial question there was no mention to Grothendieck duality. – Leo Alonso Aug 9 '17 at 22:42
• Perhaps it was abit too harsh and I'm sorry, but (this part of) your answer sort of missed the point of my question, as $f_*f^!\mathcal{O}_Y$ is not the same thing as $B$, and I don't see how the map you constructed is related to my trace. Relating them in a natural way will completely satisfy me and and give you back the upvote. – Anonymous Coward Aug 9 '17 at 22:55

It seems worth mentioning the geometric setting: Let $A$ and $B$ be the rings of functions on varieties $X$ and $Y$ over an algebraically closed field $k$, so we have a map $\phi: Y \to X$. Let $d$ be the degree of $\phi$ and let $x \in X$ be a closed point over which $\phi$ is etale, so $\phi^{-1}(x) = \{ y_1, \ldots, y_d \}$. Let $x$ correspond to the maximal ideal $\mathfrak{p} \subset A$ and let $y_i$ correspond to the maximal ideal $\mathfrak{q}_i \subset B$. By the Chinese Remainder Theorem, $$B \otimes_A A/\mathfrak{p} = \bigoplus B/\mathfrak{q}_i.\quad (\ast)$$

Let $g \in B$. Then $\mathrm{Tr}(g)$, evaluated at the point $x$, is the same as the trace of the map $\bar{g} :B \otimes_A A/\mathfrak{p} \to B \otimes_A A/\mathfrak{p}$ induced by multiplication by $g$.

Under the isomorphism $(\ast)$, $\bar{g}$ acts on $B/\mathfrak{q}_i$ by $g(y_i)$. In particular, this action is diagonal!

So, for $x$ as above, we have $$\mathrm{Tr}(g)(x) = \sum g(y_i).$$ The nifty fact is that this extends to a regular function on the locus where $\phi$ is branched.

• To restate this in even simpler terms and expand slightly: suppose $\phi$ is surjective. In general a function $f$ on $Y$ will not descend to a function on $X$, but we can try to get it to do so. One method would be to take $f$ and produce a function $F$ by summing the value of $f$ across fibers above the points of $X$. This certainly ought to be linear over the functions pulled back from $X$. Defining the trace makes this process rigorous. If we'd prefer playing well with multiplicative structure to linearity, we could multiply across fibers - in this case the rigorous analogue is the norm – forget this Jan 26 '20 at 6:24
• All of this is pretty metaphorical, at least at my level of understanding, particularly if you want to leave the context of varieties over an algebraically closed field. For example, if $k \hookrightarrow K$ is a finite field extension, the associated morphism of schemes doesn't have interesting fibers. But if you think of $k$ as morally a function field of some $X$ and $K$ as the function field of a (possibly ramified) finite covering space $Y \to X$ then it all ties together rather nicely – forget this Jan 26 '20 at 6:35
• @SarahGriffith Thanks, that is a very helpful rewriting! Regarding your comments on non-algebraically closed fields, this is a situation where I find it much easier to think in terms of the functor of points. Let $k \hookrightarrow K$ be a finite field extension of degree $d$ and let $\bar{k}$ be an algebraic closure of $k$. Then $(\mathrm{Spec}\ K)(\bar{k})$ is $d$ points. The field $K$ can be identified with certain $K$-valued functions on those points, and the trace $K \to k$ is summing up the values on the $d$ points. – David E Speyer Jan 26 '20 at 13:21

I think I can partly answer this question (but only in a restricted setting: when $A$ is a field and $B$ is a central simple $A$-algbera with involution $\sigma$. My suspicion is that if you want $A$ to be an arbitrary noetherian commutative ring, then you will be generalizing the below from algebraic groups to group schemes).

When $A$ is a field and $B$ a CSA, there's a simpler definition of the trace map: over some field extension $F/A$, the base change $B_F$ is a matrix algebra, and the trace map can be defined as taking the composition $B\rightarrow B_F\rightarrow F$ where the last arrow is the usual trace map; the image lands in $A$ because the characteristic polynomial of an element $b\in B$ has coefficients in $A$.

From now on I'll write $(B,\sigma)$ for the central simple algebra $B$ with involution. A lot of effort goes into proving the statements of the next few paragraphs (a reference for this is The Book of Involutions by Knus, Merkurjev, Rost, and Tignol).

Let $M$ be a finitely generated $B$-module, and $h$ an hermitian form on $M$. The algebra $E=\text{End}_B(M)$ is Brauer-equivalent to $B$, and there's an involution $\sigma_h$ on $E$ such that $\sigma_h|_A=\sigma|_A$ and is defined as the adjoint $$h(x,f(y))=h(\sigma_h(f)(x),y)$$ for any $x,y\in M$. Moreover, this association is a one-to-one correspondence between nonsingular hermitian forms (up to a scaling-factor of $A^\times$) and involutions on $B$.

From here one can define similitudes. If $D$ is a division algebra (finite dimensional over $A$) with involution, and $V$ a vector space over $D$ with hermitian form $h$ (with respect to the involution on $D$), $(V,h)$, Then $\text{Sim}(V,h)$ are those linear maps $g:V\rightarrow V$ which preserve $h$ up to a scalar multiple. One can similarly define the similitudes of the corresponding central simple algebra with involution $(\text{End}_D(V),\sigma_h)$ as the set $\text{Sim}(\text{End}_D(V),\sigma_h)$ whose elements are $g\in \text{End}_D(V)$ such that $\sigma_h(g)g\in A^\times$. These happen to be equal.

The usefulness of these constructions comes when you take the functor of points pov for algebraic groups (or group schemes). Then, one can show that there is an algebraic group $G$ over $A$ whose $F$-points (for a field extension $F/A$) are exactly groups defined using these similitudes. That's purposefully vague since it doesn't answer your question but only gives motivation (maybe). An example would be the split case of a matrix algebra $M_{n,n}(A)$ with involution $\tau$ the transposition of matrices. Then there is an algebraic group $G$ over $A$ with $G(F)=\text{Sim}(M_{n,n}(A_F),\tau_F)$ and it is just the orthogonal group. The benefit of going this way however is to consider twisted forms of such groups.

Now to the trace map. From the csa with involution $(B,\sigma)$ and the trace map $Trd:B\rightarrow A$ defined above (the Trd is because this is usually called the reduced trace map), one can define a bilinear form $T_{(B,\sigma)}:B\times B\rightarrow A$ by $T_{(B,\sigma)}(x,y)=Trd(\sigma(x)y)$.

If $\sigma$ is particularly nice (which here means it's of the first kind, I didn't really go over what this means but essentially it corresponds to either an orthogonal or symplectic group and not a unitary group), then there is the following theorem:

There is an isomorphism $\sigma_*:B\otimes {^iB}\xrightarrow{\sim} \text{End}_A(B)$ such that $\sigma_*(a\otimes b)(x)=ax\sigma(b)$. Here $^iB$ is the same set as $B$ but it has a different algebra structure (defined so that there is a corresponding involution on the other side $^i\sigma$) and so that the tensor product $B\otimes {^iB}$ has a canonical involution $\sigma\otimes {^i\sigma}$. Finally, by the correspondence above $\sigma\otimes {^i\sigma}$ is the involution corresponding to the trace $T_{(B,\sigma)}$.

So:

1) I'm not sure if there are constructions which you should think about when seeing this but, my first thought was the trace maps defined in the theory of algebras with involution (only a very small portion of which was mentioned above; the whole theory uses trace maps everywhere).

2) I don't know how to generalize this categorically. But, it seems a more general approach would be through azumaya algebras (which are generalizations of central simple algebras to the case where the base is a ring and not a field). Then your assumptions would read $A$ is a commutative ring, $B$ is an $A$-algebra finitely generated over $A$ and projective as an $A$-module, and the canonical morphism from $B\otimes B^{op}\rightarrow \text{End}_A(B)$ is an isomorphism. Then you can at least see where the trace map (on the right) goes on the left (which is called the Goldman element).

I would hope you could get more detail and more mileage by considering some kind of involution on an Azumaya algebra but I've never seen it considered.

3) An \'etale algebra is really similar to a central simple algebra. Both are twisted forms of something. Really an \'etale algebra is a twisted form of products of central simple algebras. This actually fits into the above theory and is discussed throughout the Book of Involutions. (When there is an involution of the second kind on an algebra $B$ -- corresponding to a unitary group -- then the center of $B$ is actually a quadratic \'etale extension of $A$).

If you tensor the last isomorphism you found $B\xrightarrow{\sim} \text{Hom}_A(B,A)$ by $B$ then it is canonically isomorphic with the map $B\otimes B\xrightarrow{\sim} \text{Hom}_A(B,B)=\text{End}_A(B)$.

Of course, all the standard constructions like norm too will play a role and can be defined in appropriate settings. Even more importantly, trace makes sense sometimes even when $B$ is not flat. For example, assume $A, B$ are domains, then you do have a trace from $B$ to the fraction field of $A$, using standard field theory arguments. This in fact will be a map to $A$ if $A$ is integrally closed, no flatness is used.