Lemma 53.25.3. Let $k$ be an algebraically closed field. Let $X$ be an at-worst-nodal, proper, connected $1$-dimensional scheme over $k$. Assume the genus of $X$ is at least $2$ and that $X$ has no rational tails or bridges. Then $\text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X) = 0$.

Proof. Let $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$. Let $X^\nu$ be the normalization of $X$. Let $D' \in \text{Der}_ k(\mathcal{O}_{X^\nu }, \mathcal{O}_{X^\nu })$ be the element corresponding to $D$ via Lemma 53.25.2. Let $C \subset X^\nu$ be an irreducible component. If the genus of $C$ is $> 1$, then $D'|_{\mathcal{O}_ C} = 0$ by Lemma 53.25.1 part (1). If the genus of $C$ is $1$, then there is at least one closed point $c$ of $C$ which maps to a node on $X$ (since otherwise $X \cong C$ would have genus $1$). By the correspondence this means that $D'|_{\mathcal{O}_ C}$ fixes $c$ hence is zero by Lemma 53.25.1 part (2). Finally, if the genus of $C$ is zero, then there are at least $3$ pairwise distinct closed points $c_1, c_2, c_3 \in C$ mapping to nodes in $X$, since otherwise either $X$ is $C$ with two points glued (two points of $C$ mapping to the same node), or $C$ is a rational bridge (two points mapping to different nodes of $X$), or $C$ is a rational tail (one point mapping to a node of $X$). These three possibilities are not permitted since $C$ has genus $\geq 2$ and has no rational bridges, or rational tails. Whence $D'|_{\mathcal{O}_ C}$ fixes $c_1, c_2, c_3$ hence is zero by Lemma 53.25.1 part (3). $\square$

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