Lemma 55.9.6. In Situation 55.9.3 given $\mathcal{L}$ an invertible $\mathcal{O}_ X$-module and $a = (a_1, \ldots , a_ n) \in \mathbf{Z}^{\oplus n}$ we define

$\langle a, \mathcal{L} \rangle = \sum a_ i\deg (\mathcal{L}|_{C_ i})$

Then $\langle , \rangle$ is bilinear and for $b = (b_1, \ldots , b_ n) \in \mathbf{Z}^{\oplus n}$ we have

$\left\langle a, \mathcal{O}_ X(\sum b_ i C_ i) \right\rangle = \left\langle b, \mathcal{O}_ X(\sum a_ i C_ i) \right\rangle$

Proof. Bilinearity is immediate from the definition and Varieties, Lemma 33.44.7. To prove symmetry it suffices to assume $a$ and $b$ are standard basis vectors in $\mathbf{Z}^{\oplus n}$. Hence it suffices to prove that

$\deg (\mathcal{O}_ X(C_ j)|_{C_ i}) = \deg (\mathcal{O}_ X(C_ i)|_{C_ j})$

for all $1 \leq i, j \leq n$. If $i = j$ there is nothing to prove. If $i \not= j$, then the canonical section $1$ of $\mathcal{O}_ X(C_ j)$ restricts to a nonzero (hence regular) section of $\mathcal{O}_ X(C_ j)|_{C_ i}$ whose zero scheme is exactly $C_ i \cap C_ j$ (scheme theoretic intersection). In other words, $C_ i \cap C_ j$ is an effective Cartier divisor on $C_ i$ and

$\deg (\mathcal{O}_ X(C_ j)|_{C_ i}) = \deg (C_ i \cap C_ j)$

by Varieties, Lemma 33.44.9. By symmetry we obtain the same (!) formula for the other side and the proof is complete. $\square$

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