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The Stacks project

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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