PAC and totally real fields A field $K$ is called pseudo-algebraically closed (PAC) if every absolutely irreducible variety over $K$ has a $K$-point. Let $L$ be the maximal totally real subfield of $\overline{\mathbb Q}$. A few places claim that $L(i)$ is PAC, but I can't find any proofs of this. Does anyone have a reference, or know why this is true?
 A: The reason why $L(i)$ is PAC is the following theorem.

Theorem. Let $K$ be a global field. Let $S$ be a finite set of places of $K$. Let $X$ be a smooth, geometrically integral $K$-variety. For each $v\in S$, let $\Omega_v\subset X(K_v)$ be open (for the $v$-adic topology) and nonempty. Then there exist:

*

*a finite extension $M$ of $K$, totally split over $S$ (i.e. for each $v\in S$ and each place $w$ of $M$ above $v$, we have $M_w \cong K_v$),

*a point $x\in X(M)$ such that for all $w$ and $v$ as above, $x$ "is" in $\Omega_v$ (via the map $X(M)\subset X(M_w)\cong X(K_v)$).


The theorem follows from Theorem 1.3 in this paper, which is stated in terms of rings of integers of global fields but also works for localizations, including the field itself: see remark 1.7 in the paper.
Now, if $L\subset\overline{Q}$ denotes the field of totally real numbers, we immediately derive the following:

Corollary 1. Let $K$ be a totally real number field, and let $X$ be a smooth, geometrically integral $K$-variety. If $X(K_v)\neq\emptyset$ for each real place $v$ of $K$, then $X(L)\neq\emptyset$.

To prove that $L(i)$ is PAC amounts to proving:

Corollary 2. For $K$ as in Cor. 1, let $Y$ be a geometrically integral $K(i)$-variety. Then $Y(L(i))\neq\emptyset$.

Proof: we may assume $Y$ smooth by taking its smooth locus. Let $X$ be the $K(i)/K$-Weil restriction of $Y$. [EDIT: in general $X$ is an algebraic space, not necessarily a scheme. But it is an affine $K$-variety if $Y$ is affine, which we may always assume.] We need to see that $X(K)\neq\emptyset$. For each real place $v$ of $K$, $X(K_v)$ is the same thing as $Y(L\otimes_K K_v)$ which is nonempty since $L\otimes_K K_v\cong \mathbb{C}$. So the claim follows from Cor. 1.
Unfortunately, when I wrote the paper I was unaware of this nice consequence. I remember that a few years later I saw it mentioned somewhere (but where?), and it was attributed to Pop. I could not find an explicit reference, but it can be derived somewhat more directly from "Theorem $\mathfrak{S}$" here.
