The Stacks project

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. $\square$


Comments (2)

Comment #6316 by Bogdan on

It seems unclear why Lemma 0BT8 implies that corresponds to a map .

Comment #6422 by on

Tried to improve the explanation, but may only have made it worse. See here.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BVI. Beware of the difference between the letter 'O' and the digit '0'.