3
$\begingroup$

Let $f:X \to Y$ be a surjective, smooth projective morphism of noetherian schemes. Let $\mathcal{L}$ be an inverible sheaf on $X$ satisfying $f_*\mathcal{L}$ is locally free and $s \in H^0(\mathcal{L})$ be a global section of $\mathcal{L}$. Is the zero locus of $s$ (in $X$), flat over $Y$?

N.B. If necessary, one can assume the fibers of $f$ are irreducible and $Y$ is affine. Also can assume $\pi_*\mathcal{L}$ is free.

$\endgroup$
7
$\begingroup$

You already have trivial counterexamples for your statement, but perhaps you were thinking of a section whose zero locus is irreducible and dominates $Y$. It is false even with that additional assumption:

Let $Y$ be an arbitrary non-singular surface and $Z$ the blow-up up of a (closed) point on $Y$. Then by construction/definition $Z\subseteq X=Y\times \mathbb P^1$. Since $X$ is non-singular, $Z$ is a Cartier divisor and hence there is a line bundle $\mathscr L$ on $X$ such that $Z$ is the zero locus of a global section of $\mathscr L$. Clearly the projection $f:X\to Y$ is flat, so this is a counterexample with the additional property that the zero locus of the section is irreducible and surjects onto $Y$.

EDIT: Oops, I've just realized that there is one more assumption to check, namely that $f_*\mathscr L$ is locally free, but it is easy to check that that also holds.

$\endgroup$
7
$\begingroup$

No. Take $X=\mathbb{P}^1\times \mathbb{P}^1$, $Y=\mathbb{P}^1$, $f$ the first projection, $\mathcal{L}=\mathcal{O}_{\mathbb{P}^1}(1)\boxtimes \mathcal{O}_{\mathbb{P}^1}(1)$, $s=X\otimes X'$, where $(X,Y)$ are the coordinates of the first factor $\mathbb{P}^1$ and $(X',Y')$ the coordinates of the second factor. The zero locus of $s$ is $\mathbb{P}^1\times \{0\}\cup \{0\}\times \mathbb{P}^1 $, obviously not $f$-flat.

$\endgroup$
1
  • 11
    $\begingroup$ More trivially, you can take $X=Y=\mathbb{A}^1$, $\mathcal{L}=\mathcal{O}_X$, $s=$ the standard coordinate on $X$. $\endgroup$ – Laurent Moret-Bailly Jun 4 '16 at 6:05

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.