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Let $X$ be an algebraic variety defined over a field $k$ of characteristic zero. Suppose $X(K)$ is non-empty for some extension $K/k$. Is it true that $X(L)$ is non-empty where $L$ is the algebraic closure of $k$ in $K$? An equivalent formulation is: suppose $X(k)$ is empty; does that force $X(K)$ to be empty for every purely transcendental extension $K/k$?

In my particular application $k$ is a number field with a real place and $K=\mathbb{R}$ and then claim is true since the theory of real closed fields eliminates quantifiers, but I'm wondering about the general case. Relatedly Chevalley's Theorem (elimination of quantifiers for algebraically closed fields) does produce $\bar{k}$-points, but I don't see why they have to be $\bar{k}\cap K$-points.

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    $\begingroup$ Let me first assume $K/k$ has transcendence degree $1$. Then, an element of $X(K)$ extends to a morphism $U\to X$ over $k$, where $U$ is a dense open of $\mathbb{P}^1_k$. Since $X(k)$ is empty, the set $U(k)$ is empty. But this is absurd, as $U(k)$ is dense in $U$. So, we conclude that $X(K)$ is empty. $\endgroup$
    – Ariyan Javanpeykar
    May 11, 2022 at 6:19
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    $\begingroup$ By induction on the transcendence degree, we obtain that $X(K)$ is empty for every $K/k$ pure of finite transcendence degree. $\endgroup$
    – Ariyan Javanpeykar
    May 11, 2022 at 6:20
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    $\begingroup$ If $K/k$ is purely transcendental, then any point in $X(K)$ will "come from" a point in $X(K_0)$ with $k\subset K_0\subset K$ a purely transcendental extension of finite transcendence degree. Since $X(K_0)$ is empty, we conclude that $X(K)$ is empty, as desired. $\endgroup$
    – Ariyan Javanpeykar
    May 11, 2022 at 6:22
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    $\begingroup$ I just realized that comment 1 and 2 can be merged by simply saying that, for $K/k$ of transcendence degree $n\in \mathbb{N}$, any element of $X(K)$ extends to a morphism $U\to X$ with $U\subset \mathbb{P}^n_k$ some dense open. Since $U(k)$ is dense, this contradicts $X(k)$ empty. Thus $X(K)$ is empty. $\endgroup$
    – Ariyan Javanpeykar
    May 11, 2022 at 8:56
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    $\begingroup$ The equivalence stated at the beginning is not true. If, say, $X$ is smooth and geometrically connected over $k$, with function field $K$, then $k$ is algebraically closed in $K$ and, obviously, $X(K)\neq\emptyset$ while $X(k)$ may be empty (but isn't if $K/k$ is purely transcendental). $\endgroup$
    – Laurent Moret-Bailly
    May 11, 2022 at 9:22

1 Answer 1

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The following answers your second question.

Let $X$ be a finite type separated scheme over an infinite field $k$. Let $K/k$ be a purely transcendental extension. If $X(k)$ is empty, then $X(K)$ is empty.

Proof. We may assume that $K$ has finite transcendence degree over $k$. Then, any element of $X(K)$ extends to a morphism $U\to X$, where $U$ is a dense open of some $\mathbb{P}^n$. Since $k$ is infinite, the set $U(k)$ is dense (hence non-empty). This contradicts $X(k)$ being empty.

Your first question is answered by Laurent Moret-Bailly.

Let $X$ be a geometrically connected finite type scheme over $k$ (and thus non-empty). Let $K=K(X)$ be the function field of $X$. Then $X(K)$ is non-empty. However, there is no reason for $X(k)$ to be non-empty. Take $X$ to be a smooth proper conic over $\mathbb{Q}$ with no $\mathbb{Q}$-points, for example.

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  • $\begingroup$ Your answer prompts me to ask this new question. $\endgroup$
    – Gro-Tsen
    May 11, 2022 at 12:56
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    $\begingroup$ @LaurentMoret-Bailly Of course! I made the necessary correction. $\endgroup$
    – Ariyan Javanpeykar
    May 11, 2022 at 19:44

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