The Stacks project

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 33.41. 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 33.44.10). 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 (5)

Comment #5779 by Jonas Ehrhard on

If is algebraically closed, there are no nontrivial extensions .

Comment #5782 by on

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

Comment #7709 by Cristian D. Gonzalez-Aviles on

Hi, I'm a bit confused about this Lemma. It seems to me that is (or should be) the normalization of , so the diagram displayed above is a diagram that is composed of two smaller diagrams: the top one would be the pushout that produces from its normalization (this is a square of k'-schemes) and the bottom half would be the diagram that one obtains by base changing to the line . Am I right? Thanks

Comment #7712 by on

Whenever one has a commutative square, there is a morphism from one corner to the fibre product of the rest of the diagram. In the first line of the proof we see that is the normalization of . But in general is not a variety (eg it could have more than irreducible component and/or could be nonreduced) so what you said will not be true.

As an example consider given by in over the field .

Comment #7724 by Cristian D. Gonzalez-Aviles on

Aha! I overlooked the fact that X need not be geometrically irreducible (currently I'm only working with geometrically integral k-schemes). Thanks for taking the time to answer my question.

There are also:

  • 2 comment(s) on Section 53.10: Curves of genus zero

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DJB. Beware of the difference between the letter 'O' and the digit '0'.