Lemma 55.11.8. In Situation 55.9.3 assume $X$ is a minimal model, $\gcd (m_1, \ldots , m_ n) = 1$, and $H^0((X_ k)_{red}, \mathcal{O}) = k$. Then the map

is surjective and has a nontrivial kernel as soon as $(X_ k)_{red} \not= X_ k$.

Lemma 55.11.8. In Situation 55.9.3 assume $X$ is a minimal model, $\gcd (m_1, \ldots , m_ n) = 1$, and $H^0((X_ k)_{red}, \mathcal{O}) = k$. Then the map

\[ H^1(X_ k, \mathcal{O}_{X_ k}) \to H^1((X_ k)_{red}, \mathcal{O}_{(X_ k)_{red}}) \]

is surjective and has a nontrivial kernel as soon as $(X_ k)_{red} \not= X_ k$.

**Proof.**
By vanishing of cohomology in degrees $\geq 2$ over $X_ k$ (Cohomology, Proposition 20.20.7) any surjection of abelian sheaves on $X_ k$ induces a surjection on $H^1$. Consider the sequence

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

of Lemma 55.9.9. Since the field maps $H^0(Z_ j, \mathcal{O}_{Z_ j}) \to H^0((X_ k)_{red}, \mathcal{O}_{(X_ k)_{red}}) = k$ are injective we conclude that $H^0(Z_ j, \mathcal{O}_{Z_ j}) = k$ for $j = 0, \ldots , m$. It follows that $H^0(X_ k, \mathcal{O}_{X_ k}) \to H^0(Z_{m - 1}, \mathcal{O}_{Z_{m - 1}})$ is surjective. Let $C = C_{i_ m}$. Then $X_ k = Z_{m - 1} + C$. Let $\mathcal{L} = \mathcal{O}_ X(-Z_{m - 1})|_ C$. Then $\mathcal{L}$ is an invertible $\mathcal{O}_ C$-module. As in the proof of Lemma 55.9.9 there is an exact sequence

\[ 0 \to \mathcal{L} \to \mathcal{O}_{X_ k} \to \mathcal{O}_{Z_{m - 1}} \to 0 \]

of coherent sheaves on $X_ k$. We conclude that we get a short exact sequence

\[ 0 \to H^1(C, \mathcal{L}) \to H^1(X_ k, \mathcal{O}_{X_ k}) \to H^1(Z_{m - 1}, \mathcal{O}_{Z_{m - 1}}) \to 0 \]

The degree of $\mathcal{L}$ on $C$ over $k$ is

\[ (C \cdot -Z_{m - 1}) = (C \cdot C - X_ k) = (C \cdot C) \]

Set $\kappa = H^0(C, \mathcal{O}_ C)$ and $w = [\kappa : k]$. By definition of the degree of an invertible sheaf we see that

\[ \chi (C, \mathcal{L}) = \chi (C, \mathcal{O}_ C) + (C \cdot C) = w(1 - g_ C) + (C \cdot C) \]

where $g_ C$ is the genus of $C$. This expression is $< 0$ as $X$ is minimal and hence $C$ is not an exceptional curve of the first kind (see proof of Lemma 55.11.5). Thus $\dim _ k H^1(C, \mathcal{L}) > 0$ which finishes the proof. $\square$

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## Comments (2)

Comment #2949 by Maciek Zdanowicz on

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