Lemma 55.9.9. In Situation 55.9.3 let $d = \gcd (m_1, \ldots , m_ n)$. Let $D = \sum (m_ i/d) C_ i$ as an effective Cartier divisor. Then there exists a sequence of effective Cartier divisors

\[ (X_ k)_{red} = Z_0 \subset Z_1 \subset \ldots \subset Z_ m = D \]

such that $Z_ j = Z_{j - 1} + C_{i_ j}$ for some $i_ j \in \{ 1, \ldots , n\} $ for $j = 1, \ldots , m$ and such that $H^0(Z_ j, \mathcal{O}_{Z_ j})$ is a field finite over $k$ for $j = 0, \ldots m$.

**Proof.**
The reduction $D_{red} = (X_ k)_{red} = \sum C_ i$ is connected (Lemma 55.9.4) and proper over $k$. Hence $H^0(D_{red}, \mathcal{O})$ is a field and a finite extension of $k$ by Varieties, Lemma 33.9.3. Thus the result for $Z_0 = D_{red} = (X_ k)_{red}$ is true. Suppose that we have already constructed

\[ (X_ k)_{red} = Z_0 \subset Z_1 \subset \ldots \subset Z_ t \subset D \]

with $Z_ j = Z_{j - 1} + C_{i_ j}$ for some $i_ j \in \{ 1, \ldots , n\} $ for $j = 1, \ldots , t$ and such that $H^0(Z_ j, \mathcal{O}_{Z_ j})$ is a field finite over $k$ for $j = 0, \ldots , t$. Write $Z_ t = \sum a_ i C_ i$ with $1 \leq a_ i \leq m_ i/d$. If $a_ i = m_ i/d$ for all $i$, then $Z_ t = D$ and the lemma is proved. If not, then $a_ i < m_ i/d$ for some $i$ and it follows that $(Z_ t \cdot Z_ t) < 0$ by Lemma 55.9.7. This means that $(D - Z_ t \cdot Z_ t) > 0$ because $(D \cdot Z_ t) = 0$ by the lemma. Thus we can find an $i$ with $a_ i < m_ i/d$ such that $(C_ i \cdot Z_ t) > 0$. Set $Z_{t + 1} = Z_ t + C_ i$ and $i_{t + 1} = i$. Consider the short exact sequence

\[ 0 \to \mathcal{O}_ X(-Z_ t)|_{C_ i} \to \mathcal{O}_{Z_{t + 1}} \to \mathcal{O}_{Z_ t} \to 0 \]

of Divisors, Lemma 31.14.3. By our choice of $i$ we see that $\mathcal{O}_ X(-Z_ t)|_{C_ i}$ is an invertible sheaf of negative degree on the proper curve $C_ i$, hence it has no nonzero global sections (Varieties, Lemma 33.44.12). We conclude that $H^0(\mathcal{O}_{Z_{t + 1}}) \subset H^0(\mathcal{O}_{Z_ t})$ is a field (this is clear but also follows from Algebra, Lemma 10.36.18) and a finite extension of $k$. Thus we have extended the sequence. Since the process must stop, for example because $t \leq \sum (m_ i/d - 1)$, this finishes the proof.
$\square$

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