Let $f:A\rightarrow B$ be a homomorphism of noetherian rings which
makes $B$ into a finite $A$-module. Under what conditions on $f$, $A$,
$B$ can one associate to this map a canonical "trace map"
$$\mathrm{Tr}_f:B\rightarrow A,$$
i.e. a homomorphism of $A$-modules $B\rightarrow A$ which is compatible with base change (perhaps with restrictions on the kind of allowable base changes), localization, and
which recovers the "usual" thing when $B$ is free over $A$ (i.e.
$\mathrm{Tr}_f(b) =$ the trace of the $A$-linear endomorphism given by
multiplication by $b$ on the finite $A$-module $B$)?
Here are my thoughts so far:
1) If $f$ is flat, one always has $\mathrm{Tr}_f$. Just work locally,
using that finite flat over a noetherian local ring is free.
2) More generally, if $f$ is of finite Tor-dimension, I can construct
$\mathrm{Tr}_f$ by taking a finite projective resolution
$$0\rightarrow P_n\rightarrow \cdots\rightarrow P_0\rightarrow B\rightarrow 0$$
of $B$ as an $A$-module: lift multiplication by $b$ on $B$ to a map of complexes
$b:P_{\bullet}\rightarrow P_{\bullet}$ and define
$$\mathrm{Tr}_f(b) := \sum_i (-1)^i \mathrm{Tr}_i(b)$$
where $\mathrm{Tr}_i(b)$ is the trace of the (lift of the)
endomorphism $b$ of $B$ to $P_i$.
One chekcs this is independent of the choice of projective resolution.
It commutes with "tor-independent" base change (sometimes called
"cohomologically transverse" base change).
3) If $A$ and $B$ are normal, I can construct $\mathrm{Tr}_f$ as follows: the localization of $f$ at any height-1 prime ideal is automatically flat by Matsumura 23.1 as the corresponding localizations of $A$ and $B$ are regular and the dimensions work out (the fiber ring is 0-dimensional as $f$ is finite). Thus, one has a canonical trace map on each localization, and since $A$ and $B$ are normal, they are the intersections of these localizations so we win.
4) If $A$ and $B$ are only assumed reduced, one can look at the injections $A\rightarrow A'$
and $B\rightarrow B'$ with $A'$ and $B'$ the normalizations of $A$ and $B$ in their total rings of fractions. Let $f':A'\rightarrow B'$ be the corresponding map. By 3), we get a trace map
for $f'$ and the whole problem of constructing
the trace map for $f$
comes down to showing that $\mathrm{Tr}_{f'}$ carries $B$ into $A$. Letting $C_A:=\mathrm{ann}_{A'}(A'/A)$
be the conductor ideal (with $C_B$ defined similarly), I think that a necessary condition for
$\mathrm{Tr}_{f'}(B)$ to be contained in $A$ is
$$f'(C_A) \supseteq C_B.$$
Is this condition sufficient? As an example of how things can go wrong if this condition is
violated, consider the finite map between reduced $k$-algebras ($k$ a field)
$$k[x,y]/(xy) \rightarrow k[x]$$
given by sending $y$ to 0. The normalization of $k[x,y]/(xy)$ is the product $k[x]\times k[y]$
and the trace map attached to $f':k[x]\times k[y]\rightarrow k[x]$ sends $b\in k[x]$ to
$(b,0)$. But $(b,0)\in k[x]\times k[y]$ lies in the image of $k[x,y]/(xy)\rightarrow k[x]\times k[y]$
if and only if $b(0)=0$. It follows that the trace map on normalizations doesn't restrict to a trace map on the original rings.
I'd be happy to assume that $A$ and $B$ are flat $R$-algebras, for a regular local ring $R$, and that $f:A\rightarrow B$ is an $R$-algebra homomorphism. I'd also be happy to assume that $A$, $B$, and $f$ are local, with $A$ and $B$ reduced complete intersections over $R$. I'm wondering if there is a framework for trace maps in this context that is general enough to handle the different constructions given in 1) — 4) above.
Best Answer
This is really a response to Karl's beautiful example; I'm posting it as an "answer" only because there isn't enough room to leave it as a comment.
The condition on conductor ideals is one that I had come across by thinking about the dual picture. Namely, let $f:Y\rightarrow X$ be a finite map of 1-dimensional proper and reduced schemes over an algebraically closed field $k$. Then $Y$ and $X$ are Cohen-Macaulay by Serre's criterion, so the machinery of Grothendieck duality applies. In particular, the sheaves $f_*O_Y$ and $f_*\omega_Y$ are dual via the duality functor $\mathcal{H}om(\cdot,\omega_X)$, as are $O_X$ and $\omega_X$. Here, $\omega_X$ and $\omega_Y$ are the ralative dualizing sheaves of $X$ and $Y$, respectively. Thus, the existence of a trace morphism $f_*O_Y\rightarrow O_X$ is equivalent by duality to the existence of a pullback map on dualizing sheaves $\omega_X\rightarrow f_*\omega_Y$.
In the reduced case which we are in, one has Rosenlicht's explicit description of the dualizing sheaf: for any open $V$ in $X$, the $O_X(V)$-module $\omega_X(V)$ is exactly the set of meromorphic differentials $\eta$ on the normalization $\pi:X'\rightarrow X$ with the property that $$\sum_{x'\in \pi^{-1}(x)} res_{x'}(s\eta)=0$$ for all $x\in V(k)$ and all $s\in O_{X,x}$.
It is not difficult to prove that if $C$ is the conductor ideal of $X'\rightarrow X$ (which is a coherent ideal sheaf on $X'$ supported at preimages of non-smooth points in $X$), then one has inclusions $$\pi_*\Omega^1_{X'} \subseteq \omega_X \subseteq \pi_*\Omega^1_{X'}(C).$$ Since $X'$ and $Y'$ are smooth, so one has a pullback map on $\Omega^1$'s, our question about a pullback map on dualizing sheaves boils down the following concrete question:
By looking at the above inclusions, I was led to conjecture the necessity of conductor ideal containment as in my original post. As Karl's example shows, this containment is not sufficient.
Here is Karl's example re-worked on the dual side:
Set $B:=k[x,y]/(xy)$ and $A:=k[u,v]/(uv)$ and let $f:A\rightarrow B$ be the $k$-algebra map taking $u$ to $x^2$ and $v$ to $y$. Writing $B'$ and $A'$ for the normalizations, we have $B'$ and $A'$ as in Karl's example, and the conductor ideals are $(x,y)$ and $(u,v)$. Now the pullback map on meromorphic differentials on $A'$ is just $$(f(u)du,g(v)dv)\mapsto (2xf(x^2)dx,g(y)dy).$$ The condition of being a section of $\omega_A$ is exactly $$res_0(f(u)du)+res_0(g(v)dv)=0,$$ and similarly for being a section of $\omega_B$. But now we notice that $$res_0(2xf(x^2)dx)+res_0(g(y)dy) = 2 res_0(f(u)du) + res_0(g(v)dv) = res_0(f(u)du)$$ if $(f(u)du,g(v)dv)$ is a section of $\omega_A$. Thus, as soon as $f$ is not holomorphic (i.e. has nonzero residue) the pullback of the section $(f(u)du,g(v)dv)$, as a meromorphic differential on $B'$, will NOT lie in the subsheaf $\omega_B$.
Clearly what goes wrong is that the ramification indices of the map $f:A'\rightarrow B'$ over the two preimages of the nonsmooth point are NOT equal. With this in mind, I propose the following addendum to my original number 4):
Certainly this rules out Karl's example. I think another way of stating the condition is that the map $f':Spec(B')\rightarrow Spec(A')$ should be "equi-ramified" over the nonsmooth locus of $Spec(A)$, i.e. that the ramification indices of $f'$ over all $x'\in Spec(A')$ which map to the same nonsmooth point in $Spec(A)$ are all equal.
Is this the right condition?