# When does a quasifinite surjective flat morphism have constant fiber multiplicity near a point?

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.

• 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 $V$. – Roland Dec 16 '18 at 10:12
• @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? – Ashvin Swaminathan Dec 16 '18 at 17:59
• 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 $t=1$... – Roland Dec 16 '18 at 18:49
• @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. – Ashvin Swaminathan Dec 16 '18 at 19:50
• 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. – Ashvin Swaminathan Jan 10 at 5:17

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$$\ldots$$