2
$\begingroup$

Let $f\colon X\to D$ be a morphism of a complex analytic space $X$ into the 1-dimensional disk $D$. Assume for simplicity that $X$ has a single irreducible component which may not be reduced.

Question: is it true that if $f$ is onto then $f$ is flat?

(I am not a specialist, and the answer is not known to me even if $X$ is reduced.)

If the answer is no, I would be curious to know if there is a correct reformulation of this result and what is the natural generality. Of course, a reference will be most useful.

This question is an analytic version of the following well known algebraic criterion. Let $f\colon X\to Y$ be a morphism of schemes when $Y$ is integral regular 1-dimensional. Assume for simplicity that $X$ is reduced and irreducible (these conditions may be relaxed). Then $f$ is flat if and only if $X$ dominates $Y$. (See Hartshorne's book, Ch. III, Prop. 9.7.)

$\endgroup$

1 Answer 1

4
$\begingroup$

There are easy counterexamples if you allow "embedded components", e.g. $X\subset D\times\mathbb{C}$ given (in coordinates $z,t$) by the equations $zt=t^2=0$. In the local ring at $x=(0,0)$, the coordinate $z$ from $D$ is a zero divisor, so $\mathscr{O}_{X,x}$ is not flat over $\mathscr{O}_{D,0}$.
On the other hand, in the general case, flatness at $x\in X$ just means that $\mathscr{O}_{X,x}$ is a torsion-free $\mathscr{O}_{D,f(x)}$-module (because $\mathscr{O}_{D,f(x)}$ is a PID), or simply that the function $z-f(x)$ is not a zero divisor in $\mathscr{O}_{X,x}$. This will hold if you assume that $X$ has no embedded component at $x$, in the sense that every zero divisor in $\mathscr{O}_{X,x}$ is nilpotent (indeed, with your assumptions, $z-f(x)$ is certainly not nilpotent). In particular, $f$ is flat if $X$ is (irreducible and) reduced (and of course you may replace "onto" by "nonconstant").

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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