# Geometric generic fibre

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!

• Think of $j$-function for elliptic curves. In particular, cubic curves in $\mathbb{P}^2_{\mathbb{C}}$. – Mohan Jul 5 '15 at 15:52
• 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. – Donu Arapura Jul 5 '15 at 16:26
• This is true only for one-dimensional families. For higher dimensions, you must use countable many hypersurfaces. – Will Sawin Jul 5 '15 at 17:43
• @WillSawin: of course, sorry for the sloppiness. I fixed it. – potentially dense Jul 5 '15 at 19:05
• @DonuArapura: right, that's the kind of situation I had in mind to convince myself this isn't completely crazy! – potentially dense Jul 5 '15 at 19:09