4
$\begingroup$

In this question I ask whether ambient spaces descend to models of varieties.

Let $k\subset K$ be a non-trivial extension of algebraically closed fields, e.g., $\overline{\mathbb Q}\subset \mathbb C$.

Let $X$ be a projective variety over $K$ which can be defined over $k$ (as an abstract scheme).

Assume that $X$ can be embedded in $\mathbb P^n_K$, i.e., there is a closed immersion from $X$ into $\mathbb P^n_K$. Is there a model $X_0$ of $X$ over $k$ such that $X_0$ can be embedded in $\mathbb P^n_k$?

What if $k$ is of characteristic zero?

$\endgroup$

1 Answer 1

1
$\begingroup$

Write $X=Y\times_k K$ for some variety $Y$ over $k$. The closed immersion $i: Y\times_k K\to\mathbb P^n_K$ is defined by using finitely many coefficients in $K$, so it is defined over a finitely generated $k$-algebra $A$: $$i_S: Y\times_k S\to \mathbb P^n_S=\mathbb P^n_k \times_k S $$ where $S$ is the affine variety associated to $A$, and $i_S$ is an $S$-morphism. Take a closed point $s\in S$ and you will have a closed immersion $$ Y=Y\times \{ s \} \to \mathbb P^n_k \times \{ s\}=\mathbb P^n_k.$$

$\endgroup$
2
  • $\begingroup$ This works indeed and answers my question. (I don't think the base-change to $K$ of Y embedded in $\mathbb P^n_k$ will be X with its chosen embedding in $\mathbb P^n_k$ because we're "specializing". Nevertheless, it answers the question.) $\endgroup$
    – Andrew
    May 19, 2014 at 16:47
  • $\begingroup$ @Andrew: of course not, otherwise the initial embedding would be a "constant" one and there is no reason for that. $\endgroup$
    – Cantlog
    May 19, 2014 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.