20
$\begingroup$

This is a pretty elementary question about schemes, but it came up in the course of research, so let's try it here rather than MSE.

Question 1: Are the fibres of a family of complex varieties isomorphic as schemes to the geometric generic fibre, outside of a union of countably many subfamilies?

That seems outlandish, so let me explain below the reasoning that led me to ask. If my argument is flawed I would be glad to know why. If, by contrast, this is well-known, I would be glad of a reference, so I ask:

Question 2: Does anyone know of a published reference for this fact, if it is correct?

I fix the following simple setup for concreteness.

Let $f: X \rightarrow \mathbf A^1$ be a family of varieties over $\mathbf C$.

Let $k = \mathbf C(t)$, the function field of the base, and $K=\overline{k}$. Let $G$ be the geometric generic fibre of $f$, that is, the $K$-variety$ X \times_{\mathbf A^1} \operatorname{Spec K}$.

Now $K$ is algebraically closed, has characteristic zero, and has the same cardinality as $\mathbf C$. So there is a field isomorphism $\alpha: \mathbf C \simeq K$. (As I understand it, this depends on the axiom of choice, but that's alright.) So base change by $\alpha$ turns $G$ into a variety $G_\alpha$ over $\mathbf C$, isomorphic to $G$ as a scheme. (OK so far?)

Now suppose for concreteness that $f$ is a family of hypersurfaces in projective space $\mathbf P^n$, so it is given by a form $F(x_0,\ldots,x_n;t)$ where the $x_i$ are coordinates coordinates on $\mathbf P^n$ and $t$ is the coordinate on $\mathbf A^1$.

Now pick any number $z \in \mathbf C$ which is algebraically independent from all the coefficients of $F$. Then we can choose our field isomorphism so that $\alpha^{-1}$ fixes all the coefficients of $F$, and $\alpha^{-1}(t)=z$. Then base change by $\alpha$ just has the effect of substituting $z$ in place of $t$ in the form $F$: in other words, $G \simeq G_\alpha \simeq G_z$, the fibre of $f$ over $z \in \mathbf A^1$.

This argument has the disturbing (to me) consequence that all but countably many fibres of $f$ are isomorphic, albeit in a weird way, as schemes. (Of course, it doesn't claim that they are isomorphic as varieties over $\mathbf C$, which would be absurd.) But maybe this just shows that my intuition about scheme isomorphism is lacking. Either way, I would be glad to know!

Improve this question
$\endgroup$
4
  • $\begingroup$ Think of $j$-function for elliptic curves. In particular, cubic curves in $\mathbb{P}^2_{\mathbb{C}}$. $\endgroup$
    – Mohan
    Commented Jul 5, 2015 at 15:52
  • 9
    $\begingroup$ This looks right, even if counterintuitive. It may help to consider the following example. Let $E_t= \{y^2=x(x-1)(x-t)\}$. Then $E_\pi$ and $E_e$ are not isomorphic as $\mathbb{C}$-schemes. However, there exists a field automorphism of $\mathbb{C}$ taking $\pi$ to $e$, and thus the first curve to the second. $\endgroup$
    – Donu Arapura
    Commented Jul 5, 2015 at 16:26
  • $\begingroup$ @DonuArapura: right, that's the kind of situation I had in mind to convince myself this isn't completely crazy! $\endgroup$
    – Lazzaro Campeotti
    Commented Jul 5, 2015 at 19:09
  • $\begingroup$ @Mohan: as I mentioned in the question, I am not asking about isomorphism as varieties over $\mathbf C$. Donu's example shows what can happen when we allow this wider class of morphism. $\endgroup$
    – Lazzaro Campeotti
    Commented Jul 6, 2015 at 11:35

1 Answer 1

Reset to default
8
$\begingroup$

This statement is indeed pretty remarkable. A reference is Lemma 2.1 in C. Vial. Algebraic cycles and fibrations. Documenta Math. 18 (2013). The statement there is given for varieties, but it seems likely that it holds more generally.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .