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.



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> **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?
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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:



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> **Question 2:** Does anyone know of a published reference for this fact, if it is correct? 
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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 $z \in \mathbf A^1$.




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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!



Geometric generic fibre
This statement is indeed pretty remarkable. A reference is Lemma 2.1 in C. Vial. [Algebraic cycles and fibrations](https://www.dpmms.cam.ac.uk/~cv248/algcycles-fibrations.pdf). Documenta Math. 18 (2013). The statement there is given for varieties, but it seems likely that it holds more generally.


