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.

  • $\begingroup$ 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). $\endgroup$
    – Pop
    Oct 4, 2018 at 19:52
  • $\begingroup$ That is a good point, and I have tried to make that intuition into a complete argument, but haven't been successful. $\endgroup$ Oct 4, 2018 at 20:35
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    $\begingroup$ 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. $\endgroup$ Oct 4, 2018 at 20:46
  • $\begingroup$ 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? $\endgroup$ Oct 4, 2018 at 21:53
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    $\begingroup$ 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. $\endgroup$ Oct 17, 2018 at 16:10

2 Answers 2


The flatness statement you would like follows from Theorem 3.3.27 of Schoutens's book on ultraproducts.


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.


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