A morphism that is defined over $\mathbb{Q}$ and is an isomorphism over $\mathbb{C}$ may not be an isomorphism over $\mathbb{Q}$.

Question

Is there an example of two algebraic varieties $X,Y$ over $\mathbb{Q}$ and a morphism $f:X\rightarrow Y$ defined over $\mathbb{Q}$, that is an isomorphism over $\mathbb{C}$ but not over $\mathbb{Q}$? That is, the inverse $f^{-1}$ doesn't have rational coefficients.

It sounds like an easy problem but I have been doing computations (with $X=\mathbb{A}^1$ and $Y$ a plane curve) and I couldn't find any.

score 5 · Accepted Answer · answered Apr 12 '20 at 20:37

5

No, there is no such morphism. This is a basic case of faithfully flat descent: the map $\Bbb Q\to \Bbb C$ is faithfully flat and quasicompact, so if $f:X\to Y$ is a $\Bbb Q$-morphism of $\Bbb Q$-schemes, then $f_{\Bbb C}: X_\Bbb C\to Y_\Bbb C$ is an isomorphism iff $f:X\to Y$ is an isomorphism. See EGA IV2, 2.7.1 (viii) for the original proof.

answered Apr 12 '20 at 20:37

KReiser

65,137

Wonderful! I didn't expect that. Thanks! – nowhere dense Apr 12 '20 at 20:42
This is just beautiful. – QU Binggang Apr 14 '20 at 08:31

A morphism that is defined over $\mathbb{Q}$ and is an isomorphism over $\mathbb{C}$ may not be an isomorphism over $\mathbb{Q}$.

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