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 $[\mathbb A/\mu_2]$ 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 $f$.

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 $f$.