Lemma 55.11.2. In Situation 55.9.3 we have
where $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$, $g_ i$ is the genus of $C_ i$, and $g_ C$ is the genus of $C$.
Lemma 55.11.2. In Situation 55.9.3 we have
where $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$, $g_ i$ is the genus of $C_ i$, and $g_ C$ is the genus of $C$.
Proof. Our basic tool will be Derived Categories of Schemes, Lemma 36.32.2 which shows that
Choose a sequence of effective Cartier divisors
such that $D_{j + 1} = D_ j + C_{i_ j}$ for each $j$. (It is clear that we can choose such a sequence by decreasing one nonzero multiplicity of $D_{j + 1}$ one step at a time.) Applying Lemma 55.11.1 starting with $\chi (\mathcal{O}_{D_0}) = 0$ we get
Perhaps the last equality deserves some explanation. Namely, since $\sum _ j C_{i_ j} = \sum m_ i C_ i$ we have $(\sum _ j C_{i_ j} \cdot \sum _ j C_{i_ j}) = 0$ by Lemma 55.9.7. Thus we see that
by splitting this product into “nondiagonal” and “diagonal” terms. Note that $\kappa _ i$ is a field finite over $k$ by Varieties, Lemma 33.26.2. Hence the genus of $C_ i$ is defined and we have $\chi (C_ i, \mathcal{O}_{C_ i}) = [\kappa _ i : k](1 - g_ i)$. Putting everything together and rearranging terms we get
which is what the lemma says too. $\square$
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