1
$\begingroup$

Let $V \subset \mathbb{A}_{\mathbb{C}}^2 = \operatorname{Spec}\mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $\pi \colon V \to \mathbb{A}_{\mathbb{C}}^1 = \operatorname{Spec} \mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U \subset \mathbb{A}_{\mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U \cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' \subset \mathbb{A}_{\mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U \cap \pi^{-1}(p)$ is independent of the choice of $p \in U'$?

What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $\mathbb{A}_{\mathbb{C}}^2$. Indeed, take $$V = \operatorname{Spec} \mathbb{C}[[x]][t]/(x(x-t))$$ and take $\pi$ to be the morphism of affine schemes induced by the obvious ring map $\mathbb{C}[t] \to \mathbb{C}[[x]][t]/(x(x-t))$. Then $\pi$ is quasifinite because it has finite fibers (multiplicity $1$ for $t\neq 0$ and multiplicity $2$ for $t =0$), and it is also flat, because it is obtained by precomposing the flat morphism $\operatorname{Spec} \mathbb{C}[x][t]/(x(x-t)) \to \operatorname{Spec} \mathbb{C}[t]$ with the morphism of affine schemes induced by the localization $\mathbb{C}[t] \to \mathbb{C}[[t]]$. However, the fiber multiplicity jumps from $1$ to $2$ at $t=0$.

What appears to go wrong in the above example is that the subscheme $V$ picks up some additional fuzz in the $x$-direction at $t=0$, but intuitively, it seems like this can't happen if $V$ is a closed subscheme to begin with.

$\endgroup$
  • 2
    $\begingroup$ What about $V=\operatorname{Spec}\mathbb{C}[x,t]/(tx^2-x)\to\operatorname{Spec}\mathbb{C}[t]$. Then $V$ is the union of two irreducible components : $x=0$ and $xt=1$. Both are flat. The first one is onto, hence so is $V$. The second one is not finite, hence neither is $V$. $\endgroup$ – Roland Dec 16 '18 at 10:12
  • $\begingroup$ @Roland Ah, that's a great counterexample! Now suppose we stipulate that no irreducible components of $V$ go off to infinity as $t \to 0$. More precisely, suppose the closure of $V$ in $\mathbb{P}_{\mathbb{C}}^2$ is not supported at the point defined by $t = 0$ in the line at infinity. Is it then true that $\pi$ is finite? $\endgroup$ – Ashvin Swaminathan Dec 16 '18 at 17:59
  • 2
    $\begingroup$ Well the problem does not happen only at $t=0$. Take $V=\operatorname{Spec}\mathbb{C}[x,t]/((t-1)x^2-x)$. Then there is a problem at $t=1$... $\endgroup$ – Roland Dec 16 '18 at 18:49
  • $\begingroup$ @Roland Right, this problem could indeed occur at any $t$. I guess I'm more interested in what happens locally near a point of $V$, and I've rewritten my question in that context. $\endgroup$ – Ashvin Swaminathan Dec 16 '18 at 19:50
  • $\begingroup$ I'm still thinking about it. I honestly didn't understand the example with $xy-x-t = 0$ mapping to $\operatorname{Spec} \mathbb{C}[t]$ via $(x,y,t) \mapsto t^2 - t$. Why is the fiber above $0$ one point? I also didn't understand what condition you're saying is insufficient. $\endgroup$ – Ashvin Swaminathan Jan 10 at 5:17
0
$\begingroup$

Let $V \subset \mathbb{A}_{\mathbb{C}}^2 = \operatorname{Spec}\mathbb{C}[x,t]$ be a closed subscheme containing the point $x = t = 0$, and suppose we have a quasifinite flat surjective morphism $\pi \colon V \to \mathbb{A}_{\mathbb{C}}^1 = \operatorname{Spec} \mathbb{C}[t]$ that sends the point $x = t = 0$ to the point $t = 0$. Choose a sufficiently small analytic neighborhood $U \subset \mathbb{A}_{\mathbb{C}}^2$ containing the point $x = t = 0$ such that every component of $U \cap V$ passes through the point $x = t = 0$. Does there exist an analytic neighborhood $U' \subset \mathbb{A}_{\mathbb{C}}^1$ of the point $t = 0$ such that the multiplicity of the fiber $U \cap \pi^{-1}(p)$ is independent of the choice of $p \in U'$?

I think that this holds for sufficiently small $U$, but the condition you impose is not sufficient, e.g. let $U = V$ be defined by $xy = x + t$ and $\pi$ be defined by $t^2 - t$. Then the fiber above $0$ is one point---reduced---and the other fibers above closed points consist of $2$ points.

By standard arguments, for some open neighborhoods $U$ and $U'$ we have an induced finite flat map $U \to U'$---cf. Lemma in Fischer's Complex Analytic Geometry, page 132---such that the point $x = t = 0$ is the only point mapping to the point $t = 0$. Then for $U'$ connected the direct image of the structure sheaf is a locally free coherent sheaf of some constant rank $d$, and $U_1 \subset U$ is an open neighborhood of the origin let $U'_1$ be the complement in $U'$ of the image of $U - U_1$, then for $p$ in $U'_1$ the fiber of $U_1 \to U'$ above $p$ is of degree $d$.

What I know so far: it is certainly possible for this to fail if we drop the condition that $V$ is a closed subscheme of $\mathbb{A}_{\mathbb{C}}^2$. Indeed, take $$V = \operatorname{Spec} \mathbb{C}[[x]][t]/(x(x-t))$$ and take $\pi$ to be the morphism of affine schemes induced by the obvious ring map $\mathbb{C}[t] \to \mathbb{C}[[x]][t]/(x(x-t))$. Then $\pi$ is quasifinite$\ldots$

It is not of finite type, so not quasifinite in the sense of EGA.

$\endgroup$

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.