**Proof.**
Let $\nu : X' \to X$ be the normalization of $X$, see Varieties, Sections 33.27 and 33.40. Since $X$ is singular $\nu $ is not an isomorphism. Then $k' = H^0(X', \mathcal{O}_{X'})$ is a finite extension of $k$ (Varieties, Lemma 33.26.2). The short exact sequence

\[ 0 \to \mathcal{O}_ X \to \nu _*\mathcal{O}_{X'} \to \mathcal{Q} \to 0 \]

and the fact that $\mathcal{Q}$ is supported in finitely many closed points give us that

$H^1(X', \mathcal{O}_{X'}) = 0$, i.e., $X'$ has genus $0$ as a curve over $k'$,

there is a short exact sequence $0 \to k \to k' \to H^0(X, \mathcal{Q}) \to 0$.

In particular $k'/k$ is a nontrivial extension.

Next, we consider what is often called the *conductor ideal*

\[ \mathcal{I} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\nu _*\mathcal{O}_{X'}, \mathcal{O}_ X) \]

This is a quasi-coherent $\mathcal{O}_ X$-module. We view $\mathcal{I}$ as an ideal in $\mathcal{O}_ X$ via the map $\varphi \mapsto \varphi (1)$. Thus $\mathcal{I}(U)$ is the set of $f \in \mathcal{O}_ X(U)$ such that $f \left(\nu _*\mathcal{O}_{X'}(U)\right) \subset \mathcal{O}_ X(U)$. In other words, the condition is that $f$ annihilates $\mathcal{Q}$. In other words, there is a defining exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{Q}, \mathcal{Q}) \]

Let $U \subset X$ be an affine open containing the support of $\mathcal{Q}$. Then $V = \mathcal{Q}(U) = H^0(X, \mathcal{Q})$ is a $k$-vector space of dimension $n - 1$. The image of $\mathcal{O}_ X(U) \to \mathop{\mathrm{Hom}}\nolimits _ k(V, V)$ is a commutative subalgebra, hence has dimension $\leq n - 1$ over $k$ (this is a property of commutative subalgebras of matrix algebras; details omitted). We conclude that we have a short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_ X \to \mathcal{A} \to 0 \]

where $\text{Supp}(\mathcal{A}) = \text{Supp}(\mathcal{Q})$ and $\dim _ k H^0(X, \mathcal{A}) \leq n - 1$. On the other hand, the description $\mathcal{I} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\nu _*\mathcal{O}_{X'}, \mathcal{O}_ X)$ provides $\mathcal{I}$ with a $\nu _*\mathcal{O}_{X'}$-module structure such that the inclusion map $\mathcal{I} \to \nu _*\mathcal{O}_{X'}$ is a $\nu _*\mathcal{O}_{X'}$-module map. We conclude that $\mathcal{I} = \nu _*\mathcal{I}'$ for some quasi-coherent sheaf of ideals $\mathcal{I}' \subset \mathcal{O}_{X'}$, see Morphisms, Lemma 29.11.6. Define $\mathcal{A}'$ as the cokernel:

\[ 0 \to \mathcal{I}' \to \mathcal{O}_{X'} \to \mathcal{A}' \to 0 \]

Combining the exact sequences so far we obtain a short exact sequence $0 \to \mathcal{A} \to \nu _*\mathcal{A}' \to \mathcal{Q} \to 0$. Using the estimate above, combined with $\dim _ k H^0(X, \mathcal{Q}) = n - 1$, gives

\[ \dim _ k H^0(X', \mathcal{A}') = \dim _ k H^0(X, \mathcal{A}) + \dim _ k H^0(X, \mathcal{Q}) \leq 2 n - 2 \]

However, since $X'$ is a curve over $k'$ we see that the left hand side is divisible by $n$ (Varieties, Lemma 33.43.10). As $\mathcal{A}$ and $\mathcal{A}'$ cannot be zero, we conclude that $\dim _ k H^0(X', \mathcal{A}') = n$ which means that $\mathcal{I}'$ is the ideal sheaf of a $k'$-rational point $x'$. By Proposition 53.10.4 we find $X' \cong \mathbf{P}^1_{k'}$. Going back to the equalities above, we conclude that $\dim _ k H^0(X, \mathcal{A}) = 1$. This means that $\mathcal{I}$ is the ideal sheaf of a $k$-rational point $x$. Then $\mathcal{A} = \kappa (x) = k$ and $\mathcal{A}' = \kappa (x') = k'$ as skyscraper sheaves. Comparing the exact sequences given above, this immediately implies the result on structure sheaves as stated in the lemma.
$\square$

## Comments (2)

Comment #5779 by Jonas Ehrhard on

Comment #5782 by Johan on

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