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Let $X$ be the Fermat quartic $x^4+y^4+z^4+w^4=0$ in $\mathbb P^3$. It is known that $X$ contains infinitely many $(-2)$-curves, that is, smooth rational curves. (One way to obtain in infinitely many is to use the various elliptic fibrations on $X$, and use translations in the fibers.) Note however that these curves are typically not defined over $\mathbb Q$, even though $X$ is. My question is the following:

Is there a finite extension $K$ of $\mathbb Q$ such that all of the smooth rational curves are defined over $K$ and have a $K$-rational point?

I suspect that the answer is no, but I don't see how to prove it.

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    $\begingroup$ (Edit.) I suspect the opposite. The geometric Picard group $\text{Pic}(X\otimes_{\mathbb{Q}}\overline{\mathbb{Q}})$ is a finitely generated Abelian group. Thus there exists a finite extension $K$ of $\mathbb{Q}$ over which there are defined invertible sheaves on $X\otimes_{\mathbb{Q}}K$ whose images generate the geometric Picard group. Every invertible sheaf defined on $X\otimes_{\mathbb{Q}}\overline{\mathbb{Q}}$ is already defined on $X\otimes_{\mathbb{Q}}K$. In particular, those invertible sheaves with self-square $-2$ are defined over $K$. $\endgroup$ – Jason Starr Jun 30 '16 at 15:04
  • $\begingroup$ Interesting! I realize what I wanted to ask was that also the rational curves are all rational over this field extension (i.e., they have a rational point). Is this true? $\endgroup$ – byu Jun 30 '16 at 15:24
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    $\begingroup$ Please confer Corollary 4.6 of Chapter 8, p. 161, and Corollary 2.4 of Chapter 15, p. 315, of Daniel Huybrechts, "Lectures on K3 Surfaces", math.uni-bonn.de/people/huybrech/K3Global.pdf There is a finite field extension over which all rational curves have a rational point. Did I understand correctly your question? $\endgroup$ – Jason Starr Jun 30 '16 at 18:13
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I am posting my comments as an answer. I am concerned that I misunderstand the OP, so let me state first the result. There exists a finite field extension $K/\mathbb{Q}$ such that for every closed immersion $\mathbb{P}^1_\mathbb{C} \hookrightarrow X\otimes_{\mathbb{Q}}\mathbb{C}$, there exists a closed immersion $\mathbb{P}^1_K \hookrightarrow X \otimes_{\mathbb{Q}} K$ with the same image after base change to $X\otimes_{\mathbb{Q}}\mathbb{C}$.

To prove this, first observe that the geometric Picard group $\text{Pic}(X\otimes_{\mathbb{Q}}\mathbb{C})$ is finitely generated. Thus, there exists a finite field extension after which a finite generating set of the geometric Picard group is defined over that field.

Second, the geometric automorphism group of the complex scheme, $\text{Aut}_{\mathbb{C}}(X\otimes_{\mathbb{Q}}\mathbb{C})$, is a discrete group that is finitely generated by Corollary 2.4 of Chapter 15, p. 315 of the following; result due to H. Sterk -- Finiteness results for algebraic K3 surfaces. Math. Z., 189(4):507–513, 1985.

Daniel Huybrechts
Lectures on K3 Surfaces
Part of Cambridge Studies in Advanced Mathematics
September 2016
ISBN: 9781107153042
http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf

Thus, after a further finite field extension, there exists a finite subset of $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ whose images generate the geometric automorphism group. Thus, $\text{Aut}_K(X\otimes_{\mathbb{Q}} K)$ is already the full geometric automorphism group.

Third, by Corollary 4.6 of Chapter 8, p. 161 -- result again proved by Sterk in the article cited above -- the action of the geometric automorphism group on the set of smooth rational curves has only finitely many orbits. For each of those finitely many orbits, after a degree $2$ field extension, every curve in that orbit acquires a rational point. Thus, after a further finite field extension, every smooth, genus $0$ curve on $X$ acquires a rational point.

Please note, all of this is valid for every K3 surface over every characteristic $0$ field, not only for the Fermat quartic surface over $\mathbb{Q}$. Using extensions of Sterk's results by Lieblich and Maulik, this should also be true in positive characteristic.

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