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 $f$.



Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$?
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 $k=1$.


