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.

  • $\begingroup$ Three comments: (1) You don't insist on one thing that's common for trace maps: that $Tr(a)=a$ for $a\in A$. Do you want to? This imposes annoying restrictions on characteristic; essentially, you have to assume that the rank of $B$ is invertible in $A$. (2) I would be very surprised if your definition of $Tr$ was nonzero when $B$ has finite proj. dim. Do you have examples of this? (3) Since you're assuming $B$ is module-finite over $A$, Tor-dimension is the same as projective dimension. You mention this, but as a matter of communication, you might consider the more common formulation. $\endgroup$ Dec 3, 2009 at 13:04
  • 1
    $\begingroup$ Thanks for the comments! The trace map I defined as "the usual thing" does not satisfy $\mathrm{Tr}(a)=a$ for $a\in A$. Just think about the case that $B$ is free of rank $m$ over $A$; then $\mathrm{Tr}(a)=ma$. Maybe I'd better impose the conditions that $A$ and $B$ are reduced, and that $\mathrm{Frac}(B)$ (total ring of fractions) is separable over $\mathrm{Frac}(A)$. I don't understand your second comment: the case of finite projective dimension includes that of finite flat, so there are lots of nonzero examples (i.e. definition 2) recovers definition 1) in the flat case). $\endgroup$
    – B. Cais
    Dec 3, 2009 at 13:20
  • $\begingroup$ Yes, I should have said "finite nonzero proj. dim." Essentially, having finite positive proj. dim. should force $B$ to have rank zero as an $A$-module, and (my intuition is) those alternating sums should cancel out. For example, if $B = A/xA$ for a nzd $x \in A$, I don't think your definition 2) does what you want. Or does it, and I'm confused? $\endgroup$ Dec 3, 2009 at 13:25
  • $\begingroup$ To expand on that last comment: if $B=A/xA$, then there are no nonzero $A$-module homomorphisms from $B$ to $A$, so the trace is the zero map. Is that what you expect? $\endgroup$ Dec 3, 2009 at 13:32
  • $\begingroup$ Yes, this is as t should be: the composite of $\mathrm{Tr}:B\rightarrow A$ with $A\rightarrow B$ should be multiplication by the "rank" of $B$ as an $A$-module. Isn't it the case that every finltely generated module over a noetherian regular local ring has finite projective dimension? So $B$ is by no means forced to have $A$ "rank" zero. $\endgroup$
    – B. Cais
    Dec 3, 2009 at 13:53

4 Answers 4


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:

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 $f:A\rightarrow B$ 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):

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?


With regards to 4), I think the condition you described involving the conductor ($f'(C_A) \subseteq (C_B)$) is not sufficient (although I do agree that it is necessary).

For the trace map at the level of the total field of fractions, I just applied trace to each field extension separately and added the results. If this is not the right thing to do here, let me know (the extension of total fields of fractions need not turn an extended ring into a free modules, so I don't see what else to do).

Consider the following example:

I'll define the normal rings first.

$$A' = k[x^2] \oplus k[y], B' = k[x] \oplus k[y].$$

I'll use the obvious inclusion $A' \subseteq B'$. Set $A$ to be the subring of $A'$ made up of all pairs $(s,t)$ such that the constant term of $s$ and the constant term of $t$ agree (for example, $(3 + x^2 + x^4, 3 + y)$ would be in $A$). Set $B$ to be the subring of $B'$ made up of all pairs $(u,v)$ such that constant terms of $u$ and $v$ agree. Clearly we have an inclusion $A \subseteq B$.

In other words, $\text{Spec} A$ is the node obtained by gluing the two components of $\text{Spec} A'$ together at their respective origins and $\text{Spec} B$ is the same (but with gluing the components of $\text{Spec} B'$ together at the origin).

The conductor of $A$ in $A'$ is the ideal made up of all pairs $(s, t)$ with zero constant term (it corresponds to the origin on each ring). Likewise the conductor of $B$ in $B'$ is the ideal made up of all pairs $(u, v)$ with no constant term.

Clearly the trace map sends the conductor to the conductor but $Tr(B) \not\subset A$ because $Tr(1,1) = (2,1) \notin A$.

  • $\begingroup$ (Sorry if this is too elementary, but I want to see this example!) $A'=k[x^2]\oplus k[y]$ is direct sum of two rings $k[x^2]$ and $ k[y]$, right? (I was initially thinking it as $k[x^2,y]$; then I confused a little and so asking this doubt.) $\endgroup$
    – Soluble
    May 20, 2018 at 7:40
  • $\begingroup$ @Soluble I probably should have said direct product. But yes, it is the direct sum/product of the two rings. It is not a domain, there are orthogonal idempotents, etc. $\endgroup$ May 23, 2018 at 1:13

I had pretty much forgotten, but Craig Huneke and I needed to know something about traces a few years ago, and couldn't find a good reference, so we put some facts as an appendix in this paper. Our context forced us to define the trace of a module, rather than an algebra, but of course you can just ignore the algebra structure on $B$.

The setup for us was this: Let $R$ be a Noetherian ring and $M$ a finitely generated torsion-free $R$-module. The trace of $M$ is an $R$-homomorphism $\mathrm{End}_R(M) \to R$, which matches the one you're describing if $M$ is free. It's obtained as follows: let $\alpha : Hom(M,R)\otimes_R M \rightarrow End(M)$ be the natural homomorphism defined by $\alpha(f\otimes x)(y) = f(y) \cdot x$. If the dual of $\alpha$, that is $\alpha^*$, is an isomorphism from $Hom(End_R(M),R)$ to $End_R(Hom(M,R))$, then the trace of $M$ is $(\alpha*)^{-1}$ of the identity on $Hom(M,R)$. We say then that $M$ "has a trace."

[Exercise: if $M$ has $R$-torsion, then the trace of $M$ is the same as the trace of $M/\mathrm{torsion}$, so defining the trace in the torsion-free situation is enough.]

Now, here's the sufficient condition we needed (Prop A.2): if $R$ satisfies $(S_2)$ and $M_p$ is free over $R_p$ for all height-one primes $p$, then $M$ has a trace.

This is a jazzed-up version of your (3), and maybe helps with (4).

(Parenthetically, we also needed to know when the trace was surjective, which is where the characteristic comes in, and I conflated it with the trace existing in one of my comments earlier.)

  • $\begingroup$ Thanks for the wonderful reference! This is almost strong enough for my purposes, but the requirement that the localizations at ht-1 primes be flat (=free) is too strong for me... I have a specific situation in which I have a morphism of generically smooth and CM curves over a field $k$ which is not flat at the non-smooth points. These points have codimension 1, so I won't get the desired trace map from your result, unfortunately. Nor does it always exist in this kind of a situation (see my earlier example). However, when the conductor ideals work out correctly, my hunch is that it does... $\endgroup$
    – B. Cais
    Dec 3, 2009 at 18:50
  • $\begingroup$ In order to have a map $B to A$, one needs the algebra structure. We are using the fact that $B$ maps to $Hom_A(B,B)$. $\endgroup$ Dec 3, 2009 at 19:17
  • $\begingroup$ Yes, Long's right. $\endgroup$ Dec 3, 2009 at 19:51

The trace ideal of $f$ is defined under very general condition: see Appendix of Auslander-Goldman paper: "Maximal order",TAMS (97), No1, 1960 (p1-24)

(See page 20) It is basically what Graham described above. I don't think you need any condition to define trace in this sense (even work over non-commutative rings). However, for the ideal to be interesting, you probably need some sort of non-degenerate conditions like in Graham-Craig paper.

EDIT: Of course, you need a map $B \to A$, so things are more subtle then the references. Here is a summary of what I can think of: Let $(-)^* =Hom_A(-,A)$. Of course, one can always embed $B \to Hom_A(B,B)$.

Now there is a natural map $Hom_A(B,B) \to Hom_A(B,B)^{**}$.

Under mild conditions, the map $(B\otimes B^*)^{**} \to Hom_A(B,B)^{**}$ (induced by $B\otimes B^* \to Hom(B,B)$ is an isomorphism. (1)

But the map $B\otimes B^* \to A$ by evaluation induces a map: $(B\otimes B^*)^{**} \to A^{**}=A$.

So the critical issue now is to check (1), i.e when $(B\otimes B^*)^{**} \to Hom_A(B,B)^{**}$ is isomorphic.

Since both modules are reflexive, we only need to check that the map is an isomorphism in codimension 1. Of course, if $B$ is free in codimension 1, then it is true, but I think a much weaker condition, like $B$ having constant rank on the minimal primes of A would suffice.

EDIT: I just thought about it a little more. So far, we have shown that if $A$ is normal and $B$ is $A$-free locally in codimension $1$, then one can construct a map. Unfortunately, in dimension $1$ which seems to be your main concern, this implies flatness. Actullay, here the situation seems to be quite subtle. The map in (1) is quite often injective, but I think I can prove:

Proposition: If $A$ is dimension $1$, Gorenstein and $B$ is torsion-free as $A$-module, then the map (1) is an isomorphism implies that $A$ is locally a direct summand of $B$.

So I am now convinced that the situation is interesting in dimension 1! May be some other people can help?

  • $\begingroup$ Can you explain more why $B\otimes B^*\rightarrow A$ gives a map $(B\otimes B^*)^{**}\rightarrow A$? Isn't the natural map $A\rightarrow A^{**}$ and not the other way around? You'd have to invert $A\rightarrow A^{**}$... remind me what conditions on $A$ ensure that this is possible? $\endgroup$
    – B. Cais
    Dec 3, 2009 at 16:25
  • $\begingroup$ I think $A$ is isomorphic to $A^{**}$. $\endgroup$ Dec 3, 2009 at 17:00
  • $\begingroup$ Ah, right... I was being silly. The covariant functor $\mathrm{Hom}_A(A,\cdot)$ from the category of $A$-modules to itself is naturally isomorphic to the identity. So in fact $A^*$ is isomorphic to $A$ as an $A$-module via the canonical map. $\endgroup$
    – B. Cais
    Dec 3, 2009 at 18:44
  • $\begingroup$ That's right (: $\endgroup$ Dec 3, 2009 at 19:35
  • $\begingroup$ Also, I think one has to be careful: I think it is a standard fact that the natural map $$M^*\otimes_A M\rightarrow \mathrm{Hom}_A(M,M)$$ is an isomorphism if and only if $M$ is a projective (=free when $A$ is local) module. I'm sure that after one unwinds all the double duals, one recovers my 1) in the original posting. I'd sort of doubt that double dualizing the map above will produce an isomorphism if the this map is not already an isom to begin with. $\endgroup$
    – B. Cais
    Dec 3, 2009 at 19:36

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