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

\[ g_ C = 1 + \sum \nolimits _{i = 1, \ldots , n} m_ i\left([\kappa _ i : k] (g_ i - 1) - \frac{1}{2}(C_ i \cdot C_ i)\right) \]

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

\[ 1 - g_ C = \chi (C, \mathcal{O}_ C) = \chi (X_ k, \mathcal{O}_{X_ k}) \]

Choose a sequence of effective Cartier divisors

\[ X_ k = D_ m \supset D_{m - 1} \supset \ldots \supset D_1 \supset D_0 = \emptyset \]

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

\begin{align*} 1 - g_ C & = \chi (X_ k, \mathcal{O}_{X_ k}) \\ & = \sum \nolimits _ j \left(-(D_ j \cdot C_{i_ j}) + \chi (C_{i_ j}, \mathcal{O}_{C_{i_ j}})\right) \\ & = - \sum \nolimits _ j (C_{i_1} + C_{i_2} + \ldots + C_{i_{j - 1}} \cdot C_{i_ j}) + \sum \nolimits _ j \chi (C_{i_ j}, \mathcal{O}_{C_{i_ j}}) \\ & = -\frac{1}{2}\sum \nolimits _{j \not= j'} (C_{i_{j'}} \cdot C_{i_ j}) + \sum m_ i \chi (C_ i, \mathcal{O}_{C_ i}) \\ & = \frac{1}{2} \sum m_ i(C_ i \cdot C_ i) + \sum m_ i \chi (C_ i, \mathcal{O}_{C_ i}) \end{align*}

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

\[ 0 = \sum \nolimits _{j \not= j'} (C_{i_{j'}} \cdot C_{i_ j}) + \sum m_ i(C_ i \cdot C_ i) \]

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

\[ g_ C = - \frac{1}{2}\sum m_ i(C_ i \cdot C_ i) + \sum m_ i[\kappa _ i : k](g_ i - 1) + 1 \]

which is what the lemma says too. $\square$

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