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

where $\omega _ X$ is as above.

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. $\square$

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