5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '18 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. $\endgroup$ – WSL Oct 4 '18 at 20:35
  • 1
    $\begingroup$ 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. $\endgroup$ – Sándor Kovács Oct 4 '18 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? $\endgroup$ – WSL Oct 4 '18 at 21:53
  • 1
    $\begingroup$ 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. $\endgroup$ – Sándor Kovács Oct 17 '18 at 16:10
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.