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One very handy (counter)example I often think about is the scheme $Spec(k[a,b,c]/(ab-c^2))$ (where $k$ is a field), which you may also know as $Spec(k[x^2,xy,y^2])$, as $\mathbb A^2/\mu_2$, or as the $A_1$ singularity. As with other (counter)examples, I'd like to be able to say as much as possible about it.

There is a finite surjection $f:\mathbb A^2\to Spec(k[x^2,xy,y^2])$ corresponding to the inclusion $k[x^2,xy,y^2]\subseteq k[x,y]$. The question is whether this surjection is in some sense universal.

Suppose $g:Y\to Spec(k[x^2,xy,y^2])$ is finite, surjective, and $Y$ is a smooth $k$-scheme. Must $g$ factor through $f:\mathbb A^2\to Spec(k[x^2,xy,y^2])$?

A couple of remarks:

  • The finiteness hypothesis on $g$ is definitely necessary. Otherwise we could take $Y$ to be a resolution of the singularity (by a blow-up). If such a resolution factored through $\mathbb A^2$, you'd get a section of $f$ defined away from the singularity, which would imply that $f$ is a birational equivalence, which it isn't.
  • The assumption that $Y$ is a scheme is important. The couple of people I've talked to have pointed out that the smooth stack $[\mathbb A^2/\mu_2]$ is a finite cover of $X$. If $[\mathbb A^2/\mu_2]$ factored through $\mathbb A^2$, you'd again get a rational section of $f$.
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Anton, I assume you intend for $Y$ to also be connected. But then it's an application of the universal property of normalization (so you can get by with much weaker hypotheses on $g$). – BCnrd Aug 16 '10 at 2:01
@BCnrd: I hadn't intended for $Y$ to be connected (so I guess it boils down to asking if any finite map from a smooth scheme to $Spec(k[x^2,xy,y^2])$ factors through $\mathbb A^2$), but I'm happy to assume it. I don't understand how you apply the universal property of normalization ... $Spec(k[x^2,xy,y^2])$ is already normal. – Anton Geraschenko Aug 16 '10 at 2:53
Anton, mea culpa, I was disoriented and my suggestion about normalization is wrong. But you need to assume connectedness, or at least that $Y$ has pure dimension 2, to avoid some silliness with spurious connected components of dimension 1 or 0. Also, letting $X$ be the quotient you use, it is not right to say that $[\mathbf{A}^2/\mu_2] \rightarrow X$ is finite. It is proper and quasi-finite, but finite morphisms are (by definition) relatively representable by finite morphisms of schemes, so a morphism that is not affine should not be called finite – BCnrd Aug 16 '10 at 3:11
Anton, essential case is formal version over alg. closed $k$. Here's sol'n to that. Since $k[[u,v]]$ is UFD, need to consider subring $k[[fh,gh,\sqrt{fgh}]]$ for coprime $f, g$ over which $k[[u,v]]$ is module-finite. Subring contains some $u^n, v^n$, so $h$ is unit, so wlog $h = 1$. Again using UFD property, subring must be $k[[F^2, G^2, FG]]$, contained in $k[[F,G]]$ for coprime $F, G$ in $(u,v)$. Just need $F$ and $G$ formally independent. Map $k[[X,Y]] \rightarrow k[[u,v]]$ via $X \mapsto F$ and $Y \mapsto G$ has image over which $k[[u,v]]$ is finite, hence of dim. 2, so ker is 0. QED – BCnrd Aug 16 '10 at 3:52
Anton, here's why formal case is what matters. By normality, problem can be Zariski-localized to singularity on base ring, and we can embed the two semi-local normal finite extensions over it into common one. Task is to check containment within common such normal extension. By faithful flatness, enough to check after extending scalars to alg. closure of $k$ (& then normalize upstairs if necessary), & again by f.flatness can base change to completion on base local ring (preserves normality on top, by excellence). Various rings bust into product of locals domains, and you can do the rest. – BCnrd Aug 16 '10 at 4:23
up vote 6 down vote accepted

It seems to me that in the global case the answer should be $no$ because of the following argument.

Set $S:=Spec$ $k[x,y,z]/(z^2-xy)$. Then $S$ is isomorphic to a quadric cone in $\mathbb{A}^3$. The point is that there are plenty of smooth double covers of $S$, which are pairwise non-isomorphic.

To see this, notice first that the morphism $f \colon \mathbb{A}^2 \to S$ corresponds to the restriction of a double cover $\mathbb{P}^2 \to$ (Cone $\subset \mathbb{P}^3$) branched on the vertex of the cone and on a smooth conic contained in the hyperplane at infinity.

Now one can generalize this construction by taking a double cover $f_k \colon Y_k \to S$ which is the restriction to $S$ of the projective cover branched on the vertex and on a smooth curve of $even$ degree $2k$ not passing through the vertex. The fact that $f_k$ is branched at the vertex ensures that $Y_k$ is smooth.

When $k=1$ we have $Y_1=\mathbb{A}^2$.

When $k=2$, $Y_2$ is an affine, open subset of a smooth surface of general type with $p_g=4, q=0, K^2=5$. These surfaces were studied by Horikawa in his famous paper "On deformations of quintic surfaces"; it turns out that the projective double cover is actually the canonical map.

Of course $f_2$ does not factor through $f$, since they are covering of the same degree but $Y_2$, being of general type, is not isomorphic to $\mathbb{A}^2$.

In fact, $f_k$ does not factor through $f$ except for $k=1$.

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"the morphism f \colon \mathbb{A}^2 \to S corresponds to a double cover branched on the vertex of the cone and on a smooth conic." I don't understand. Working with k algebraically closed and of characteristic not 2, every closed point has two preimages except the vertex. So where is the smooth conic you speak of? – David Speyer Aug 16 '10 at 16:54
@David You are right. I wrote the answer thinking about the projective situation, when you compactify and obtain a map $\mathbb{P}^2 \to$ Cone in $\mathbb{P}^3$. So the map $f_k$ must be considered as the restriction to the affine cone of the double covering defined as before. The map $f$ in particular corresponds to the case $k=1$ when the conic is contained in the hyperplane at infinity. But when $k \geq 2$ you always have a divisorial component in the branch locus also in the affine case. – Francesco Polizzi Aug 16 '10 at 17:10
OK, but I think this makes your answer wrong. Let $Y$ be the projective closure of $S$. What you are pointing out is that there are many curves $C$ in $Y$ for which there are double covers of $Y \setminus C$. However, it is only when you take $C$ to be the boundary conic that you get a double cover of $S$, as originally requested. – David Speyer Aug 16 '10 at 17:34
I pointed out that there are many curves $C$ in $Y$ such that there is a double cover of $Y$ branched on the vertex and $over$ $C$. So it seems to me that it is only when I take $C$ to be the boundary conic that I get a double cover of $S$ which is branched only at the vertex. In the other cases, I still obtain a double cover of $S$, which is branched at the vertex and over a smooth curve $C'$, obtained by removing from $C$ the intersection points with the boundary conic. I think this is ok, since the original question only asked for a finite morphism. But if I'm wrong, please correct me. – Francesco Polizzi Aug 16 '10 at 18:40
I see, I was misreading the original question. Still digesting your answer, but it is starting to look right to me. – David Speyer Aug 16 '10 at 19:07

At least in the complete (or henselian) case, this generalizes. Suppose $R$ is a complete regular local ring of dimension at least $2$ and $G$ a finite group of automorphisms of $R$, acting freely on $V=Spec\ R$ outside the origin. Put $X=V/G$, and suppose $S$ is another complete regular local ring with a finite surjective morphism $Y=Spec\ S\to X$.

Claim: $Y\to X$ factors through $V$.

Proof: Denote the punctured spectra by asterisks. Take the fiber product $W^*=V^*\times_{X^*}Y^*$. Since $V^*\to X^*$ is finite and etale, so is $W^*\to Y^*$. But $Y^*$ is simply connected ("purity of the branch locus"), so $W^*\to Y^*$ has a section. Compose this with the projection $W^*\to V^*$ to get $Y^*\to V^*$, and extend this across the punctures to get $Y\to V$.

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One thing that confused me about Francesco's answer was how to actually construct the branch covers $f_k:Y_k\to S$ which are branched over the vertex and a given curve. Since I was sheepish enough not to ask, perhaps somebody else (maybe future me) will benefit from a description.

Let $g(x,y,z)$ be a polynomial which does not vanish at the origin. We then have two interesting degree 2 maps to $S=Spec(k[a^2,ab,b^2])$:

  • $\mathbb A^2\to S$, corresponding to the inclusion $k[a^2,ab,b^2]\to k[a,b]$. Think of $S$ as $\mathbb A^2/\mu_2$, where $\mu_2$ acts by $(a,b)\mapsto (-a,-b)$. This is branched only over the vertex, since $(0,0)$ is the only point with a non-trivial stabilizer.
  • $S[\sqrt{g}]\to S$ (almost certainly non-standard notation since I just made it up), corresponding to the inclusion of rings $k[a^2,ab,b^2]\to k[a^2,ab,b^2,\sqrt{g(a^2,ab,b^2)}]$. Think of $S$ as $S[\sqrt g]/\mu_2$, where $\mu_2$ acts by $\sqrt g\mapsto -\sqrt g$. This is branched over the vanishing locus of $g$, since that's exactly where you have non-trivial stabilizer.

We can then define a sort of common refinement, $\tilde Y=Spec(k[a,b,\sqrt{g(a^2,ab,b^2)}]$, which has an action of $\mu_2\times \mu_2$. Quotienting by the first $\mu_2$ gives us $S[\sqrt g]$. Quotienting by the second $\mu_2$ gives us $\mathbb A^2$. Quotienting by both gives you $S$. Define $Y$ as the quotient by the diagonal $\mu_2$ action, $(a,b,\sqrt g)\mapsto (-a,-b,-\sqrt g)$. This action is free since $g(0,0,0)\neq 0$, so $\tilde Y\to Y$ is actually an etale cover. If $V(g)\cap S$ is smooth, $\tilde Y$ is smooth, so $Y$ is smooth. We have a remaining $\mu_2$ action on $Y$ with $Y/\mu_2 = S$.

$$\begin{array}{cccccc} & & \tilde Y\\ & \swarrow & \downarrow & \searrow\\ \mathbb A^2 & & Y & & S[\sqrt g]\\ & \searrow & \downarrow & \swarrow \\ & & S \end{array}$$

You can very explicitly describe the ring of invariants under this action. $Y$ is the spectrum of $k[a^2,ab,b^2,a\sqrt g,b\sqrt g]$. The $\mu_2$ action on $Y$ is $(a^2,ab,b^2,a\sqrt g,b\sqrt g)\mapsto (a^2,ab,b^2,-a\sqrt g,-b\sqrt g)$.

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