Lemma 53.23.5. Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume
the singularities of $X$ are at-worst-nodal,
$X$ does not have a rational tail (Example 53.22.1),
$X$ does not have a rational bridge (Example 53.23.1),
the genus $g$ of $X$ is $\geq 2$.
Then $\omega _ X$ is ample.
Proof. It suffices to show that $\deg (\omega _ X|_ C) > 0$ for every irreducible component $C$ of $X$, see Varieties, Lemma 33.44.15. If $X = C$ is irreducible, this follows from $g \geq 2$ and Lemma 53.8.3. Otherwise, set $k' = H^0(C, \mathcal{O}_ C)$. This is a field and a finite extension of $k$ and $[k' : k]$ divides all numerical invariants below associated to $C$ and coherent sheaves on $C$, see Varieties, Lemma 33.44.10. Let $X' \subset X$ be the closure of $X \setminus C$ as in Lemma 53.4.6. We will use the results of this lemma and of Lemmas 53.19.16 and 53.19.17 without further mention. Then we get a short exact sequence
See Lemma 53.4.6. We conclude that
Hence, if the lemma is false, then
Since $C \cap X'$ is nonempty and by the divisiblity mentioned above, this can happen only if either
$C \cap X'$ is a single $k'$-rational point of $C$ and $H^1(C, \mathcal{O}_ C) = 0$, and
$C \cap X'$ has degree $2$ over $k'$ and $H^1(C, \mathcal{O}_ C) = 0$.
The first possibility means $C$ is a rational tail and the second that $C$ is a rational bridge. Since both are excluded the proof is complete. $\square$
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