Is the structure morphism flat?

Question

Let $X$ be a smooth projective variety over $k$. Variety here meaning a separated geometrically integral $k$-scheme of finite type. Is the structure morphism $f : X \to \text{Spec}\;k$ flat?

I guess this means we should check that for any $x \in X$ the induced map $f_x : k \to \mathcal O_{X,x}$ is flat, so for any $x \in X$ the stalk of the structure sheaf should be a flat $k$-module. But then a $k$-module is just a $k$-vector space, which is always flat. Am I missing something?

Your line of reasoning is correct. "Everything" is flat over a field....Your $X$ does not even have to be finite type. — Ariyan Javanpeykar, Jul 04 '19 at 21:22

score 3 · Accepted Answer · answered Jul 04 '19 at 21:31

3

Your proof is correct (as was already confirmed by others in the comments).

A little picture: heuristically, a morphism is flat if all the fibers look the same, and for projective varieties this heuristic is a theorem (flat iff all the hilbert polynomials of all the fibers are the same).

Over a field, there's only one fiber, so they tautologically "all look the same."

answered Jul 04 '19 at 21:31

hunter

29,847

remark: "look the same" is not exactly the right thing to say informally, since e.g. non-singular varieties can deform to singular varieties in flat families. I couldn't think of a better thing to say though than "intuitively, flatness is like flatness" which isn't helpful :) – hunter Jul 04 '19 at 21:32
That's very insightful, thank you. – Ruben Jul 04 '19 at 21:32

Is the structure morphism flat?

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