I have precedently posted the same question on Math.Stackexchange (<https://math.stackexchange.com/questions/4277856/quasi-compact-surjective-morphism-of-smooth-k-schemes-is-flat>), but to no avail; I hope this is not too low-level for this site.


In the article "The Greenberg functor revisited'' (<https://doi.org/10.1007/s40879-017-0210-0>) by Bertapelle and González-Avilés, one can read, before the statement of Theorem A.12, that a given morphism, call it $f\colon X\to Y$, "is a quasi-compact [even affine] and surjective morphism of smooth $k$-schemes [$k$ is a field] and therefore faithfully flat and locally of finite presentation.'' This might be trivial, but I have no idea how to prove flatness: the most general result of this kind which I have been able to find in the literature is EGA IV.15.4.2, which however requires one of the following conditions (I am keeping the same notation as in EGA), none of which I am able to deduce from the information above:


* $f$ is universally open;
* for any $x\in X$, $f$ is open at the generic points of the irreducible components of $X\_{f(x)}$ containing $x$;
* for any $x\in X$, $\dim\mathcal{O}\_{X,x}=\dim\mathcal{O}\_{f(x)}+\dim(\mathcal{O}\_{X,x}\otimes\_{\mathcal{O}\_{f(x)}}\kappa(f(x)))$


I would be grateful for an explanation of why $f$ is flat.



Quasi-compact surjective morphism of smooth k-schemes is flat
The simplest blowup morphism $\mathrm{Bl}\_0(\mathbb{A}^2) \to \mathbb{A}^2$ (with center at a point) is not flat.


EDIT. Here is an example with affine morphism. Let
$$
X = \{ x\_1y\_1 + x\_2y\_2 + x\_3y\_3 = 0 \} \subset \mathbb{A}^4\_{x\_1,x\_2,x\_3,x\_4} \times \mathbb{A}^4\_{y\_1,y\_2,y\_3}
$$
and let $f \colon X \to \mathbb{A}^3$ be the projection to the second factor. This example, however, is **singular** at the point $(0,0)$.


EDIT 2. Consider the variety
$$
\bar{X} = \{x\_1y\_1 + x\_2y\_2 + x\_3y\_3 = 0\} \subset
\mathbb{P}^2\_{x\_1:x\_2:x\_3} \times 
\mathbb{A}^3\_{y\_1,y\_2,y\_3}.
$$
It is smooth, because the projection to $\mathbb{P}^2$ is a fibration with fiber $\mathbb{A}^2$. On the other hand, the projection $\bar{f} \colon \bar{X} \to \mathbb{A}^3$ is not flat, because the dimension of the fiber jumps at $0$.


Now let
$$
X = \bar{X} \cap 
((\mathbb{P}^2 \setminus C) \times \mathbb{A}^3),
$$
where $C$ is a smooth conic. Then


1. $X$ is smooth, because it is open in $\bar{X}$;
2. $X$ is affine over $\mathbb{A}^3$ because $\mathbb{P}^2 \setminus C$ is affine,
3. the map $f \colon X \to \mathbb{A}^3$ is surjective, because the smooth conic $C$ cannot contain a fiber of $\bar{f}$ (a line or the plane),
4. the map $f$ is not flat, because the dimension of the fiber still jumps at $0$.


