Lemma 53.10.5. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. Assume $X$ is singular and has genus $0$. Then there exists a diagram

$\xymatrix{ x' \ar[d] \ar[r] & X' \ar[d]^\nu \ar[r] & \mathop{\mathrm{Spec}}(k') \ar[d] \\ x \ar[r] & X \ar[r] & \mathop{\mathrm{Spec}}(k) }$

where

1. $k'/k$ is a nontrivial finite extension,

2. $X' \cong \mathbf{P}^1_{k'}$,

3. $x'$ is a $k'$-rational point of $X'$,

4. $x$ is a $k$-rational point of $X$,

5. $X' \setminus \{ x'\} \to X \setminus \{ x\}$ is an isomorphism,

6. $0 \to \mathcal{O}_ X \to \nu _*\mathcal{O}_{X'} \to k'/k \to 0$ is a short exact sequence where $k'/k = \kappa (x')/\kappa (x)$ indicates the skyscraper sheaf on the point $x$.

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

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

2. 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. $\square$

Comment #5779 by Jonas Ehrhard on

If $k$ is algebraically closed, there are no nontrivial extensions $k'/k$.

Comment #5782 by on

@Jonas: Sure! Do you have a question/comment/suggetion about the lemma?

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