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I read about for any separable morphism of non-singular varieties $f:X'\to X$, one can define a homomorphism $\text{Tr}:f_*(\Omega_{X'}^q) \to \Omega_{X}^q$,so that the map $\Omega_{X}^q \to f_*(\Omega_{X'}^q) $splits. But I didn't find a reference about how is it done?

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  • $\begingroup$ I believe that Serre's "Algebraic Groups and Class Fields" contains this. $\endgroup$ May 20, 2014 at 15:20
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    $\begingroup$ @JesseSilliman: I believe Serre does this when $X$ and $X'$ are curves; in particular, $q$ is at most $1$. In fact, this suffices to compute the trace maps at codimension $1$ points of $X$, i.e., away from a closed subset of codimension $2$. If the sheaves involved are $S2$, as they are in this question, that suffices. This is essentially what de Jong and I say in our erratum. Unfortunately, in the original article, the sheaves are not always $S2$. $\endgroup$ May 20, 2014 at 19:03
  • $\begingroup$ Just to make sure I completely understand this... Doesn't one need to require that the degree of $f$ is not divisible by the characteristic of the field one is working on? Otherwise the natural map $\mathcal{O}_X\longrightarrow f_\ast\mathcal{O}_{X'}$ can't be split by the trace. $\endgroup$
    – pozio
    Dec 30, 2020 at 9:49

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If you do not mind a reference that contains a serious mistake, you can use the following.

MR1716049 (2000h:13016) Reviewed
Zannier, Umberto
A note on traces of differential forms. (English summary)
J. Pure Appl. Algebra 142 (1999), no. 1, 91–97.
13N05 (14F10)
article

Part of the mistake is corrected in the following erratum.

MR2507247 (2010h:14007) Reviewed
de Jong, A. J.(1-CLMB); Starr, Jason(1-SUNYS)
Erratum: Cubic fourfolds and spaces of rational curves [MR2085418].
Illinois J. Math. 52 (2008), no. 1, 345–346.
14C05 (14E08)

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  • $\begingroup$ To spell out the counterexample in the paper a bit better, let $\mathrm{char(k)} \neq 0$, let $X' = \mathrm{Spec} k[u,v]$ and let $X = \mathrm{Spec} k[u^2, uv, v^2]$, with the obvious map. Then $Tr(u dv)$ should be $2 u dv$. We have $2 u dv = (u/v) d(v^2) = 2 d(uv) - (v/u) d(u^2)$, so $2 u dv$ is regular on the smooth part of $X$, but it isn't regular at the singular point. $\endgroup$ Nov 15, 2014 at 16:42
  • $\begingroup$ Do I understand correctly that the proof in Zannier would be correct if Zannier imposed that $X$ was non-singular, not just normal? (And, therefore, Zannier's paper is a correct answer to the stated question.) $\endgroup$ Nov 15, 2014 at 20:04
  • $\begingroup$ @DavidSpeyer: Yes, I think that is correct. However, for smooth schemes, I think one can just use Serre's "Algebraic Groups and Class Fields", as Jesse Silliman notes above. $\endgroup$ Nov 17, 2014 at 18:25

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