3
$\begingroup$

Let $j: U \to X$ be a quasi-affine open embedding between schemes and $M$ be a flat quasi-coherent $O_U$-module. Is $j_*M$ flat as an $O_X$-module?

I think the answer is no in general, but: do we have a counterexample when $X$ is a smooth variety and $U$ is the complement of a smooth closed subvariety $Z$ (of codimension $\ge 2$)?

Improve this question
$\endgroup$
3
  • $\begingroup$ No, this is false, even with codimension assumptions. A good place to look for counterexamples is $U$ is the normal locus in $X$. If $X-U$ has codimension at least $2$ then $j_\ast\mathcal{O}_U=f_\ast\mathcal{O}_{X^N}$, where $X^N$ is the normalization of $X$ (e.g., use the fact that the structure sheaf doesn't change values on a normal scheme if you remove a subset of codimension at least 2). But, this won't be flat if $X$ is normal, else $f$ would be finite flat and so degree makes sense, but then it would have to be degree $1$ (as it's birational), but that implies it's an isomorphism. $\endgroup$
    – Alex Youcis
    Commented May 2 at 3:00
  • $\begingroup$ (cont.) I will have to think whether $X$ being smooth saves you somehow. $\endgroup$
    – Alex Youcis
    Commented May 2 at 3:01
  • $\begingroup$ Thank you! My main interest is in the smooth case. I knew in the general case there are stupid counterexamples when $X$ is not irreducible, but didn’t realize this good normalization construction! $\endgroup$
    – Lin Chen
    Commented May 2 at 3:05

1 Answer 1

Reset to default
5
$\begingroup$

Take $X=\mathbb{A}^n$ with $n\geq 3$, $Z=\{0\} $, and for $M$ the kernel of the homomorphism $\mathscr{O}_U^n\rightarrow\mathscr{O}_U$ defined by $(x_1,\ldots ,x_n)$. Then $M$ is locally free, but $(j_*M)_{0}=\operatorname{Ker} (\mathscr{O}_{0}^n\rightarrow \mathfrak{m}_0)$ has projective dimension $n-2$, hence $j_*M$ is not flat.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .