Lemma 55.12.1. In Situation 55.9.3 assume that $C_ n$ is an exceptional curve of the first kind. Let $f : X \to X'$ be the contraction of $C_ n$. Let $C'_ i = f(C_ i)$. Write $X'_ k = \sum m'_ i C'_ i$. Then $X'$, $C'_ i$, $i = 1, \ldots , n' = n - 1$, and $m'_ i = m_ i$ is as in Situation 55.9.3 and we have
for $i, j < n$ we have $(C'_ i \cdot C'_ j) = (C_ i \cdot C_ j) - (C_ i \cdot C_ n) (C_ j \cdot C_ n) /(C_ n \cdot C_ n)$,
for $i < n$ if $C_ i \cap C_ n \not= \emptyset $, then there are maps $\kappa _ i \leftarrow \kappa '_ i \rightarrow \kappa _ n$.
Here $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ and $\kappa '_ i = H^0(C'_ i, \mathcal{O}_{C'_ i})$.
Proof.
By Resolution of Surfaces, Lemma 54.16.8 we can contract $C_ n$ by a morphism $f : X \to X'$ such that $X'$ is regular and is projective over $R$. Thus we see that $X'$ is as in Situation 55.9.3. Let $x \in X'$ be the image of $C_ n$. Since $f$ defines an isomorphism $X \setminus C_ n \to X' \setminus \{ x\} $ it is clear that $m'_ i = m_ i$ for $i < n$.
Part (2) of the lemma is immediately clear from the existence of the morphisms $C_ i \to C'_ i$ and $C_ n \to x \to C'_ i$.
By Divisors, Lemma 31.32.11 the pullback $f^{-1}C'_ i$ is defined. By Divisors, Lemma 31.15.11 we see that $f^{-1}C'_ i = C_ i + e_ i C_ n$ for some $e_ i \geq 0$. Since $\mathcal{O}_ X(C_ i + e_ i C_ n) = \mathcal{O}_ X(f^{-1}C'_ i) = f^*\mathcal{O}_{X'}(C'_ i)$ (Divisors, Lemma 31.14.5) and since the pullback of an invertible sheaf restricts to the trivial invertible sheaf on $C_ n$ we see that
\[ 0 = \deg _{C_ n}(\mathcal{O}_ X(C_ i + e_ i C_ n)) = (C_ i + e_ i C_ n \cdot C_ n) = (C_ i \cdot C_ n) + e_ i(C_ n \cdot C_ n) \]
As $f_ j = f|_{C_ j} : C_ j \to C_ j$ is a proper birational morphism of proper curves over $k$, we see that $\deg _{C'_ j}(\mathcal{O}_{X'}(C'_ i)|_{C'_ j})$ is the same as $\deg _{C_ j}(f_ j^*\mathcal{O}_{X'}(C'_ i)|_{C'_ j})$ (Varieties, Lemma 33.44.4). Looking at the commutative diagram
\[ \xymatrix{ C_ j \ar[r] \ar[d]_{f_ j} & X \ar[d]^ f \\ C'_ j \ar[r] & X' } \]
and using Divisors, Lemma 31.14.5 we see that
\[ (C'_ i \cdot C'_ j) = \deg _{C'_ j}(\mathcal{O}_{X'}(C'_ i)|_{C'_ j}) = \deg _{C_ j}(\mathcal{O}_ X(C_ i + e_ i C_ n)) = (C_ i + e_ i C_ n \cdot C_ j) \]
Plugging in the formula for $e_ i$ found above we see that (1) holds.
$\square$
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