The Stacks project

Lemma 115.25.1. In Semistable Reduction, Situation 55.9.3 the dualizing module of $C_ i$ over $k$ is

\[ \omega _{C_ i} = \omega _ X(C_ i)|_{C_ i} \]

where $\omega _ X$ is as above.

Proof. Let $t : C_ i \to X$ be the closed immersion. Since $t$ is the inclusion of an effective Cartier divisor we conclude from Duality for Schemes, Lemmas 48.9.7 and 48.14.2 that we have $t^!(\mathcal{L}) = \mathcal{L}(C_ i)|_{C_ i}$ for every invertible $\mathcal{O}_ X$-module $\mathcal{L}$. Consider the commutative diagram

\[ \xymatrix{ C_ i \ar[r]_ t \ar[d]_ g & X \ar[d]^ f \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ s & \mathop{\mathrm{Spec}}(R) } \]

Observe that $C_ i$ is a Gorenstein curve (Semistable Reduction, Lemma 55.9.2) with invertible dualizing module $\omega _{C_ i}$ characterized by the property $\omega _{C_ i}[0] = g^!\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}$. See Algebraic Curves, Lemma 53.4.1, its proof, and Algebraic Curves, Lemmas 53.4.2 and 53.5.2. On the other hand, $s^!(R[1]) = k$ and hence

\[ \omega _{C_ i}[0] = g^! s^!(R[1]) = t^!f^!(R[1]) = t^!\omega _ X \]

Combining the above we obtain the statement of the lemma. $\square$

Comments (0)

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Your email address will not be published.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 0C6E. Beware of the difference between the letter 'O' and the digit '0'.