# smooth equidimensional fibers over a smooth base

This is probably a simple-minded question, but I haven't been able to prove it or find a counterexample. This old question seems to dance around my question, but I don't think any of the answers address exactly my situation. (Please tell me if I am wrong!)

Suppose $$\varphi: X\to Y$$ is a surjective morphism of algebraic varieties (reduced, irreducible, separated schemes, finite type over an algebraically closed field), and furthermore assume that $$Y$$ is smooth. Also, I can assume that $$\varphi$$ is finitely presented.

If the fiber $$X_y$$ is smooth and equidimensional of dim $$n$$ for any $$y\in Y$$, is the morphism flat? I know that equidimensionality alone does not mean flat, but I wonder if the smoothness assumptions are enough. Obviously, I want to conclude that $$X$$ is smooth and this is enough.

The equidimensionality assumption rules out the blow-up examples in the above cite problem, and the normalization of a node on a curve is ruled out by smoothness of $$Y$$. I would be happy with a counterexample though.

• By "miracle flatness", if f is not flat then X is not Cohen--Macaulay. But if x∈X is a non-CM point then I am very doubtful that the fibre through x can really be smooth (e.g. because locally near X the fibre is cut out by a regular sequence). – Pop Oct 4 '18 at 19:52
• That is a good point, and I have tried to make that intuition into a complete argument, but haven't been successful. – Spencer Leslie Oct 4 '18 at 20:35
• If $Y$ is smooth, then every $y\in Y$ is a complete intersection in $Y$ and hence $X_y$ is a complete intersection in $X$. If $X_y$ is CM, then this implies that $X$ is CM...It seems though that you don't need $Y$ to be smooth. See the link I posted below. – Sándor Kovács Oct 4 '18 at 20:46
• I will take a look at the book! Is your point with not needing $Y$ to be smooth that you only need every $y\in Y$ a complete intersection? – Spencer Leslie Oct 4 '18 at 21:53
• I should have said "if $X_y$ is CM, then $X$ is CM in a neighborhood of $X_y$", but if this holds for all $y$, then $X$ is CM. – Sándor Kovács Oct 17 '18 at 16:10

One can also show directly that $$X$$ is smooth (assuming it is irreducible or even just equidimensional). The problem is local on $$X$$ and $$Y$$ so we may assume that $$X\subset \mathbb{C}^N$$ is affine of codimension $$k=N-n$$ and $$Y=\mathbb{C}^d$$. Choose polynomials $$f_1,...,f_m, t_1,...,t_d\in \mathbb{C}[X_1,...,X_N]$$ such that $$X\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots = f_m(x)=0\}$$ and $$\varphi(x)=(t_1(x),...,t_d(x))$$ for all $$x\in X$$. Fix $$p\in X$$ and assume $$0=f(p)\in\mathbb{C}^d$$. Then
$$X_0\ = \ \{x\in\mathbb{C}^N\,:\, f_1(x)=\cdots =f_m(x)=t_1(x)=\cdots =t_d(x)=0\ \}$$ Since $$p$$ is a smooth point of $$X_0$$, after reordering the indices if necessary, there exist $$r\leq m$$ and $$s\leq d$$ such that $$r+s=k+d$$ and the following set is linearly independent.
$$\{df_1(p),...,df_r(p),dt_1(p),...,dt_s(p)\}$$ Since $$s\leq d$$ we see $$r\geq k$$. But $$X$$ has pure codimension $$k$$ so $$r\leq k$$. Thus $$r=k$$ and $$\{df_1(p),...,df_k(p)\}$$ is linearly independent so $$p\in X$$ is a smooth point.