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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