Lemma 49.15.2. Let f : Y \to X be a finite flat morphism of Noetherian schemes. The map
\text{Trace}_ f : f_*\mathcal{O}_ Y \longrightarrow \mathcal{O}_ X
of Section 49.3 corresponds to a map \mathcal{O}_ Y \to f^!\mathcal{O}_ X (see proof). Denote \tau _{Y/X} \in H^0(Y, f^!\mathcal{O}_ X) the image of 1. Via the isomorphism H^0(f^!\mathcal{O}_ X) = \omega _{X/Y} of Lemma 49.15.1 this agrees with the construction in Remark 49.4.7.
Proof.
The functor f^! is defined in Duality for Schemes, Section 48.16. Since f is finite (and hence proper), we see that f^! is given by the right adjoint to pushforward for f. In Duality for Schemes, Section 48.11 we have made this adjoint explicit. In particular, the object f^!\mathcal{O}_ X consists of a single cohomology sheaf placed in degree 0 and for this sheaf we have
f_*f^!\mathcal{O}_ X = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X)
To see this we use also that f_*\mathcal{O}_ Y is finite locally free as f is a finite flat morphism of Noetherian schemes and hence all higher Ext sheaves are zero. Some details omitted. Thus finally
\text{Trace}_ f \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X) = \Gamma (X, f_*f^!\mathcal{O}_ X) = \Gamma (Y, f^!\mathcal{O}_ X)
On the other hand, we have f^!\mathcal{O}_ X = \omega _{Y/X} by the identification of Lemma 49.15.1. Thus we now have two elements, namely \text{Trace}_ f and \tau _{Y/X} from Remark 49.4.7 in
\Gamma (Y, f^!\mathcal{O}_ X) = \Gamma (Y, \omega _{Y/X})
and the lemma says these elements are the same.
Let U = \mathop{\mathrm{Spec}}(A) \subset X be an affine open with inverse image V = \mathop{\mathrm{Spec}}(B) \subset Y. Since f is finite, we see that A \to B is finite and hence the \omega _{Y/X}(V) = \mathop{\mathrm{Hom}}\nolimits _ A(B,A) by construction and this isomorphism agrees with the identification of f_*f^!\mathcal{O}_ Y with \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X) discussed above. Hence the agreement of \text{Trace}_ f and \tau _{Y/X} follows from the fact that \tau _{B/A} = \text{Trace}_{B/A} by Lemma 49.4.3.
\square
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