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
Comments (0)