Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$?

Question

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 $f$.

Answer 1

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 $k=1$.

Answer 2

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 $Y\to V$.

Answer 3

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 $(a^2,ab,b^2,a\sqrt g,b\sqrt g)\mapsto (a^2,ab,b^2,-a\sqrt g,-b\sqrt g)$.