Lemma 55.9.11. In Situation 55.9.3 given a pair of indices $i, j$ such that $C_ i$ and $C_ j$ are exceptional curves of the first kind and $C_ i \cap C_ j \not= \emptyset $, then $n = 2$, $m_1 = m_2 = 1$, $C_1 \cong \mathbf{P}^1_ k$, $C_2 \cong \mathbf{P}^1_ k$, $C_1$ and $C_2$ meet in a $k$-rational point, and $C$ has genus $0$.
Proof. Choose isomorphisms $C_ i = \mathbf{P}^1_{\kappa _ i}$ and $C_ j = \mathbf{P}^1_{\kappa _ j}$. The scheme $C_ i \cap C_ j$ is a nonempty effective Cartier divisor in both $C_ i$ and $C_ j$. Hence
\[ (C_ i \cdot C_ j) = \deg (C_ i \cap C_ j) \geq \max ([\kappa _ i: k], [\kappa _ j : k]) \]
The first equality was shown in the proof of Lemma 55.9.6. On the other hand, the self intersection $(C_ i \cdot C_ i)$ is equal to the degree of $\mathcal{O}_ X(C_ i)$ on $C_ i$ which is $-[\kappa _ i : k]$ as $C_ i$ is an exceptional curve of the first kind. Similarly for $C_ j$. By Lemma 55.9.7
\[ 0 \geq (C_ i + C_ j)^2 = -[\kappa _ i : k] + 2(C_ i \cdot C_ j) - [\kappa _ j : k] \]
This implies that $[\kappa _ i : k] = \deg (C_ i \cap C_ j) = [\kappa _ j : k]$ and that we have $(C_ i + C_ j)^2 = 0$. Looking at the lemma again we conclude that $n = 2$, $\{ 1, 2\} = \{ i, j\} $, and $m_1 = m_2$. Moreover, the scheme theoretic intersection $C_ i \cap C_ j$ consists of a single point $p$ with residue field $\kappa $ and $\kappa _ i \to \kappa \leftarrow \kappa _ j$ are isomorphisms. Let $D = C_1 + C_2$ as effective Cartier divisor on $X$. Observe that $D$ is the scheme theoretic union of $C_1$ and $C_2$ (Divisors, Lemma 31.13.10) hence we have a short exact sequence
\[ 0 \to \mathcal{O}_ D \to \mathcal{O}_{C_1} \oplus \mathcal{O}_{C_2} \to \mathcal{O}_ p \to 0 \]
by Morphisms, Lemma 29.4.6. Since we know the cohomology of $C_ i \cong \mathbf{P}^1_\kappa $ (Cohomology of Schemes, Lemma 30.8.1) we conclude from the long exact cohomology sequence that $H^0(D, \mathcal{O}_ D) = \kappa $ and $H^1(D, \mathcal{O}_ D) = 0$. By Lemma 55.9.10 we conclude
\[ 1 - g_ C = d[\kappa : k](1 - 0) \]
where $d = m_1 = m_2$. It follows that $g_ C = 0$ and $d = m_1 = m_2 = 1$ and $\kappa = k$. $\square$
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