Let $K$ be a number field (or perhaps more generally a Hilbertian field). Let $X_K\rightarrow \mathbb{P}^1_K$ be a regular (i.e. without extension of scalars) $G$-Galois branched cover. Hilbert's Irreducibility implies that there are infinitely many $K$-rational points on $\mathbb{P}^1_K$ such that the fiber product $Spec(K)\times_{\mathbb{P}^1_K}X_K$ is connected. (This is frequently stated as: there are infinitely many $K$-rational points on $\mathbb{P}^1_K$ such that specializing to them gives a $G$-Galois extension of fields over $K$.)

My question is whether the other extreme of this is true. I.e., is there a $K$-rational point on $\mathbb{P}^1_K$ such that $Spec(K)\times_{\mathbb{P}^1_K}X_K$ has $|G|$ connected components (each one isomorphic to $Spec(K)$)? In other words, is there a $K$-rational point on $\mathbb{P}^1_K$ that "splits completely"?

This reminds me of the statement that in a Galois extension of number fields there are infinitely many primes that split. This is a far from trivial statement that comes from Class Field Theory. However the analogy isn't perfect, so I don't immediately see how the same methods can be used here.

  • $\begingroup$ I am reasonably certain that the statement that there are infinitely many split primes is actually quite elementary; the argument in mathoverflow.net/questions/15220/… should extend. $\endgroup$ Jan 15, 2012 at 2:30

1 Answer 1


Perhaps I've misunderstood what you're asking, but as I read you, there are only finitely many such points whenever there are only finitely many points of X(K), and there are no such points when there are no points of X(K) -- a situation which your conditions certainly don't rule out.

  • $\begingroup$ Your reading is correct. I've overlooked this because I was only interested in whether there existed any rather than infinitely many. I will edit the question. $\endgroup$ Jan 15, 2012 at 2:46
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    $\begingroup$ Even so -- why should X(K) have any points at all? There are lots of genus 2 curves / Q with no Q-rational points; let X be one such, and map X -> P^1 via the hyperelliptic quotient (so G = Z/2Z.) Then no points of P^1(Q) split completely. $\endgroup$
    – JSE
    Jan 15, 2012 at 3:38
  • $\begingroup$ Right. I guess that solves it! $\endgroup$ Jan 15, 2012 at 4:42

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