Let C be a geometrically integral curve over a number field K and let K' be a number field containing K. Does there exist a number field L containing K such that

  • $L \cap K' = K$, and
  • $C(L) \neq \emptyset$?

Note that the hypotheses on C are necessary -- the curve x^2 + y^2 = 0, with the origin removed, is not geometrically integral, but gives a counterexample for K = Q and K' = Q(i).

Also, I can prove that this is true when C has prime gonality. It would be odd, though, for this to be a necessary hypothesis.

  • $\begingroup$ Are you sure your proof only works for prime gonality, and not when there exists nonconstant map of prime degree to $\mathbb P^1$. All curves have a nonconstant map of prime degree to $\mathbb P^1$, because they have a nonconstant map of prime degree to $\mathbb P^n$ and you can project down to $\mathbb P^1$. $\endgroup$
    – Will Sawin
    Jun 11, 2012 at 23:36
  • $\begingroup$ @Will: That would do it. What is the map to P^n? $\endgroup$ Jun 12, 2012 at 0:39
  • $\begingroup$ One can find a map to $\mathbb P^n$ by choosing a divisor of sufficient degree and applying Riemann-Roch. But of course you really need a $K$-rational map, which requires a $K$-rational divisor of prime degree, which need not exist. So gonality isn't really the issue with this argument. $\endgroup$
    – Will Sawin
    Jun 12, 2012 at 1:19
  • $\begingroup$ The following MO thread might be relevant: mathoverflow.net/questions/77644/… $\endgroup$
    – Harry
    Jun 12, 2012 at 6:47

2 Answers 2


Yes, this follows from a Theorem of Moret-Bailly, see for example Corollary 1.5


Roughly speaking, given a finite set $S$ of primes with $C(K_v)$ is non-empty, this produces a field $L$ with $C(L) \neq \emptyset$ and $L_v = K_v$ for all $v \in S$.

To guarantee that $L \cap K' = K$, one may as well assume that $K'/K$ is Galois with Galois group $G$. Then for every conjugacy class $g \in G = \mathrm{Gal}(K'/K)$, let $v$ be a prime such that $\langle \mathrm{Frob}_v \rangle = \langle g \rangle \in G$ and $C(K_v) \ne \emptyset$. (The existence of such $v$ follows from Cebotarev, the Weil conjectures, and Hensel's Lemma.) If $S$ is the resulting set, then one may find $L$ with $C(L)$ non-empty and $L_v = K_v$ for all $v \in S$, and so (by Cebotarev) that $L \cap K' = K$.

This theorem gets used all the time in "potential modularity" theorems.


I think this is a consequence of (variants of) Hilbert's irreducibility theorem. Let me explain why. Suppose that $C$ is a geometrically integral curve defined over a number field $K$. Let $K'/K$ be a normal finite extension. We will show that infinitely many points of $C$ are defined over a field disjoint from $K'$.

Since both the hypotheses and the conclusion are birational invariants, we may suppose that $C$ is a closed subset of $\mathbb{A}_K^2$ (take an affine open of $C$, embed it in $\mathbb{A}^N$ for some $N$ and take a generic projection to $\mathbb{A}^2$). Choose a generic projection $p:C\to\mathbb{A}^1_K$. The curve $C$ is described by an equation $F(t,x)=0$, where $t$ is the coordinate of $\mathbb{A}^1_K$, and $F$ is an irreducible polynomial.

Now, since $C$ is geometrically integral, $F_{K'}$ is still irreducible. By [Serre, Topics in Galois theory, Proposition 3.3.1], $x\mapsto F_{K'}(\lambda',x)$ is irreducible for every $\lambda'\in K'$ outside of a thin set. Hence, by [Serre, Topics in Galois theory, Proposition 3.2.1], $x\mapsto F_{K'}(\lambda,x)$ is irreducible for every $\lambda\in K$ outside of a thin set. Since $K$ is Hilbertian, this holds for infinitely many $\lambda\in K$.

Let us fix such a $\lambda$. We denote by $q$ and $q'$, the points of $\mathbb{A}^1_K$ and $\mathbb{A}^1_{K'}$ with coordinate $\lambda$. By choice of $\lambda$, $x\mapsto F_{K'}(\lambda,x)$ hence also $x\mapsto F(\lambda,x)$ are irreducible polynomials. Hence $p_{K'}^{-1}(q')\subset C_{K'}$ (resp. $p^{-1}(q)\subset C$) consists of a unique (reduced) point $p'\in C_{K'}$ (resp. $p\in C$). Let $L$ and $L'$ be the residual fields of $p$ and $p'$. By construction, $p'=p\times_{q} q'$ so that $L'=L\otimes_K K'$. This implies that $L$ is disjoint from $K'$.

  • $\begingroup$ @Olivier: Thanks for the response! One question: why is L' = L \otimes_K K'? It seems like for K = Q, K' = Q(i), C = x^2 + y^2 + 1 = 0, and p' = (0,i), that L = L'. $\endgroup$ Jun 12, 2012 at 1:01
  • $\begingroup$ I edited my answer ; it should be easier to read. The equality L'=L\otimes_K K' comes by base change from the cartesian square involving \mathbb{A}^1_K, C,\mathbb{A}^1_{K'} and C_{K'}$. $\endgroup$ Jun 12, 2012 at 7:32

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