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I have a flat morphism of relative dimension $d$. Is it true that a base change of this morphism is a flat morphism of the same relative dimension? We can assume that all schemes are of finite type over some field and equidimensional.

Galois group
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  • Yes, you show this using that if $X$ is a finite type $k$-scheme of pure dimension $n$ (where $k$ is a field), and if $K/k$ is a field extension, then the base change $X_K=X\times_k K$ also has pure dimension $n$. See Vakil's "Foundations of Algebraic Geometry'', Exercise 11.2.J for a series of useful hints if you want to prove this. If you know this statement already, then your statement follows as flatness is also preserved by base change. – Daniël Apol Nov 15 '23 at 09:39
  • I know the fact about extension of scalars. Can you please elaborate how the statement follows? – Galois group Nov 15 '23 at 09:58
  • I will put it in an answer. – Daniël Apol Nov 15 '23 at 10:14

1 Answers1

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Let $\pi\colon X\to Y$ be a flat morphism of finite type $k$-schemes, which has relative dimension $d$. Let $f\colon Y'\to Y$ be another morphism of schemes over $k$. We have to show that the pullback $\pi'\colon X'=X\times_Y Y'\to Y'$ is flat of relative dimension $d$. We know that $\pi'$ is flat, so now we have to show that for an arbitrary point $y'\in Y$, the fiber $\pi'^{-1}(y')$ has dimension $d$ over $k(y')$. We have the following diagram of pullbacks: enter image description here

(I'm sorry for its comical size.) By pullback pasting, the large rectangle is also a pullback diagram. However, the bottom composite equals $\mathrm{Spec}\,k(y')\to\mathrm{Spec}\,k(f(y'))\to Y$, and hence $\pi'^{-1}(y')$ is isomorphic to the base change $\pi^{-1}(f(y'))\times_{k(f(y'))} k(y')$. We know that $\pi^{-1}(f(y'))$ is of dimension $d$ over $k(f(y'))$ as $\pi$ is flat of relative dimension $d$, so $\pi'^{-1}(y')$ is also $d$-dimensional over $k(y')$, because the pure dimension of finite type schemes over a field is preserved under any field extension.

Daniël Apol
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