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Let $k$ be a field and let $V$ be a variety over $k$ (i.e. a reduced separated finite type $k$-scheme).

Let $\overline V$ be the base change $V\times_k \overline k$ to the separable closure $\overline k$ of $k$. We have a canonical fiber product map $\pi:\overline V\rightarrow V$ of $k$-schemes.

As a point set, $V$ is the quotient of $\overline V$ by the action of the absolute galois group $G = \operatorname{Gal}(\overline k / k)$, and $\pi$ is the quotient map. Let $x\in \overline V$ be such that $\pi(x)$ has residue field $k$. Then the galois action on points of $\overline V$ fixes $x$, so that $\pi$ (as a set map) is injective when restricted to such points. Thus $\pi^{-1}$ maps a copy of the $k$-points of $V$ back into $\overline V$, and, with some imprecision, we can call its image "the $k$-points of $\overline V$."

(As an example, let $k=\mathbb{R}$ and let $V=\mathbb{A}^1$. Then $\overline V$ is the complex line aka the complex plane, and the $\mathbb{R}$-points are the real line in the complex plane.)

It seems to me that the $k$-points are Zariski dense in $V$ if and only if the $k$-points (in my sense) are Zariski dense in $\overline V$. Is this true?

One direction is clear to me. If there is a proper closed subset of $V$, say $W$, containing all the $k$-points, then $\pi^{-1}(W)$ is a proper closed subset of $\overline V$ containing all the $k$-points. Thus if the $k$-points are dense in $\overline V$, they are dense in $V$.

The other direction also seems surely true to me, but the argument is less clear. Suppose $\overline W$ is a proper closed subset of $\overline V$ containing all the $k$-points. What I'd like to do is to say that the union of its conjugates under the galois action is also a proper closed subset of $\overline V$, so that then its image under $\pi$ is a proper closed subset of $V$ containing all the $k$-points.

I think I see why it has to be closed. Since $\overline V$ is a variety, it has a finite cover by affine $\overline k$-schemes and in each of them, $\overline W$ is given by a specific ideal in the coordinate ring, which is finitely generated because finite type $\overline k$-schemes are noetherian. The generators are contained in a finite extension of $k$, so they only have finitely many galois conjugates. Conclusion: $\overline W$ only has finitely many galois conjugates, and the union of finitely many closed sets is closed.

However, I don't see why it has to be proper unless I assume $V$ is geometrically irreducible. Can the claimed result fail if $V$ fails to be geometrically irreducible?

Ben Blum-Smith
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    Just wondering: how about taking the intersection of the conjugates instead of the union? This would be closed and would still contain the fixed points of the Galois action. It might not help with the geometric irreducibility question though. – Dap Dec 27 '17 at 15:56
  • @Dap No that completely solves it! The result holds irrespective of geometric irreducibility since a galois-invariant closed subset of $\overline V$ is the full preimage of its image in $V$, thus: if $\overline W$ is a proper closed subset of $\overline V$, the intersection of its conjugates is too, and because of what I just said, its galois-invariance implies its image in $V$ is also closed and proper. Want to make this an answer so I can accept something? – Ben Blum-Smith Dec 28 '17 at 15:27

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In the argument for the forward direction, take the intersection of the conjugates of $\overline W$ instead of their union. This intersection is contained in $\overline W$ so must be a proper subset, no extra assumption needed. And it will still be closed and contain the fixed points of the Galois action.

Dap
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  • Adding the point that because this intersection is fixed by the Galois action, it is the full preimage of its image in $V$, which implies its image in $V$ is proper since it itself is proper. – Ben Blum-Smith Dec 28 '17 at 18:59