Lemma 55.11.3. In Situation 55.9.3 with $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ and $g_ i$ the genus of $C_ i$ the data
is a numerical type of genus equal to the genus of $C$.
Lemma 55.11.3. In Situation 55.9.3 with $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ and $g_ i$ the genus of $C_ i$ the data
is a numerical type of genus equal to the genus of $C$.
Proof. (In the proof of Lemma 55.11.2 we have seen that the quantities used in the statement of the lemma are well defined.) We have to verify the conditions (1) – (5) of Definition 55.3.1.
Condition (1) is immediate.
Condition (2). Symmetry of the matrix $(C_ i \cdot C_ j)$ follows from Equation (55.9.6.1) and Lemma 55.9.6. Nonnegativity of $(C_ i \cdot C_ j)$ for $i \not= j$ is part (1) of Lemma 55.9.7.
Condition (3) is part (3) of Lemma 55.9.7.
Condition (4) is part (2) of Lemma 55.9.7.
Condition (5) follows from the fact that $(C_ i \cdot C_ j)$ is the degree of an invertible module on $C_ i$ which is divisible by $[\kappa _ i : k]$, see Varieties, Lemma 33.44.10.
The genus formula proved in Lemma 55.11.2 tells us that the numerical type has the genus as stated, see Definition 55.3.4. $\square$
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