Sorry, misspoke. Thanks for the correction

Thanks, this is very nice

I emailed Johan de Jong about this question and he gave an example which you can find here: github.com/stacks/stacks-project/commit/….

@JasonStarr probably I'm miscalculating, but won't the image of that set be (naively) constructible? Let A be the locally closed locus cut out by $(x_i - x_j)$ for all i, and $x_1 \neq 0$. Let $B$ be the closed locus cut out by $(x_i - x_j)(x_j - x_k)(x_i - x_k)$ for all triples $i,j,k$ of distinct indices. Let $C$ be the open locus which is the union, over all $i \neq j$, of the locus where $x_i \neq x_j, x_i \neq 0, x_j \neq 0$. Then I think your image set is $A \cup (B \cap C)$.

@EmilJeřábek yes, that's what I had in mind (and as you said, it extends to subfields of $\mathbb{R}$ but maybe not further).