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Let $X \subset \mathbb{P}^3$ be a non-singular surface defined over $\mathbb{Q}$ of degree $d \geq 3$. It is a theorem of Colliot-Thelene (see the appendix to this paper: http://www.jstor.org/stable/3062125?seq=1#page_scan_tab_contents) that there are finitely many curves of degree $\delta \leq d-2$ contained in $X$. Let $\mathcal{C}(X)$ denote this finite set of curves. Since each curve is defined over some number field whose degree over $\mathbb{Q}$ is finite, it follows that there is a unique largest finite extension of $\mathbb{Q}$, which we denote by $K = K_{\mathcal{C}(X)}$, for which all curves in $\mathcal{C}(X)$ are defined over $K$.

Do we know anything about this field $K$? For example, can we bound $[K:\mathbb{Q}]$ in terms of $d$, or maybe the coefficients of the polynomial defining $X$? A quick literature review turned up nothing. Any help would be appreciated.

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    $\begingroup$ Where is this question coming from? We do not even know the maximum number of degree $1$ curves that can be contained in a smooth degree $d$ surface. The number is bounded below by $3d^2$, and it is bounded above by something like $11d^2 + O(d)$, but the precise value remains open (except for $d=1,2,3,4$). $\endgroup$
    – Jason Starr
    Jan 12, 2016 at 0:57
  • $\begingroup$ I am concerned with how atypical it is for such curves to be defined over $\mathbb{Q}$, a very rough measure would be the degree of $K$ $\endgroup$
    – Stanley Yao Xiao
    Jan 12, 2016 at 1:45
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    $\begingroup$ For $m\leq (d+3)(d+2)/6$, for a closed subscheme $Z\subset \mathbb{P}^3_{\mathbb{Q}}$ that is geometrically a union of $m$ lines, there exists a degree $d$ hypersurface containing $Z$. I am certain that we can arrange the "splitting field" of $Z$ to have degree $m!$ over $\mathbb{Q}$. This is for general configuations of $m$ lines. If you use special configurations, you might make the degree much larger. $\endgroup$
    – Jason Starr
    Jan 12, 2016 at 2:15

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I will explain what happens in the case of cubic surfaces. As is well known, any smooth cubic surface over an algebraically closed field contains $27$ lines. For the generic cubic surface over $\mathbb{Q}$, its configuration space of lines has Galois group $W(E_6)$ (Weyl group of the $E_6$ root system). Hence an application of Hilbert's irreducibility theorem shows that for "most" cubic surfaces over $\mathbb{Q}$, the Galois group of its splitting field is $W(E_6)$, hence has degree $\#W(E_6)=51840$.

An explicit example of such a surface is $$11w^3 + 56w^2x + 33x^3 + 28x^2y + 11y^3 + 14y^2z + 28wz^2 + 11z^3 = 0,$$ as found by Ekedahl in his paper "An effective version of Hilbert's irreducibility theorem".

Of course the case of cubic surfaces is very special. In general I don't see how one can do any better than what Jason Starr suggests.

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