Lemma 53.25.2. Let $k$ be an algebraically closed field. Let $X$ be an at-worst-nodal, proper, connected $1$-dimensional scheme over $k$. Let $\nu : X^\nu \to X$ be the normalization. Let $S \subset X^\nu $ be the set of points where $\nu $ is not an isomorphism. Then

**Proof.**
Let $x \in X$ be a node. Let $x', x'' \in X^\nu $ be the inverse images of $x$. (Every node is a split node since $k$ is algebriacally closed, see Definition 53.19.10 and Lemma 53.19.11.) Let $u \in \mathcal{O}_{X^\nu , x'}$ and $v \in \mathcal{O}_{X^\nu , x''}$ be uniformizers. Observe that we have an exact sequence

This follows from Lemma 53.16.3. Thus we can view $u$ and $v$ as elements of $\mathcal{O}_{X, x}$ with $uv = 0$.

Let $D \in \text{Der}_ k(\mathcal{O}_ X, \mathcal{O}_ X)$. Then $0 = D(uv) = vD(u) + uD(v)$. Since $(u)$ is annihilator of $v$ in $\mathcal{O}_{X, x}$ and vice versa, we see that $D(u) \in (u)$ and $D(v) \in (v)$. As $\mathcal{O}_{X^\nu , x'} = k + (u)$ we conclude that we can extend $D$ to $\mathcal{O}_{X^\nu , x'}$ and moreover the extension fixes $x'$. This produces a $D'$ in the right hand side of the equality. Conversely, given a $D'$ fixing $x'$ and $x''$ we find that $D'$ preserves the subring $\mathcal{O}_{X, x} \subset \mathcal{O}_{X^\nu , x'} \times \mathcal{O}_{X^\nu , x''}$ and this is how we go from right to left in the equality. $\square$

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