ac.commutative algebra - Trace map attached to a finite homomorphism of noetherian rings

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 $f:Yrightarrow X$ 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 $f_*O_Yrightarrow O_X$
is equivalent by duality to the existence of a pullback map on dualizing sheaves $omega_Xrightarrow f_*omega_Y$.

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 $$sum_{x'in pi^{-1}(x)} res_{x'}(seta)=0$$
for all $xin V(k)$ and all $sin O_{X,x}$.

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:

When does the pullback map on meromorphic differentials $Omega^1_{k(X')}rightarrow pi_*Omega^1_{k(Y')}$ carry the subsheaf $omega_X$ into $pi_*omega_Y$?

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 $f:Arightarrow B$ 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):

In the notation of 4) above and of Karl's post, assume that $f'(C_A)=C_B^e$
for some positive integer $e$. Then the trace map $B'rightarrow A'$ carries $B$
into $A$.

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?