Lemma 52.11.8. In Situation 52.9.3 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 52.11.8. In Situation 52.9.3 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 20.21.7) 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 52.9.9. 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 52.9.9 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 52.11.5). Thus $\dim _ k H^1(C, \mathcal{L}) > 0$ which finishes the proof. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2949 by Maciek Zdanowicz on

Comment #3076 by Johan on