ag.algebraic geometry - Is $mathbb{A}²$ the universal smooth scheme which is a finite cover of $mathbb{A}²/μ₂$?

One thing that confused me about Francesco's answer was how to actually construct the branch covers $f_k:Y_kto S$ 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 $S=Spec(k[a^2,ab,b^2])$:

• $mathbb A^2to S$, 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 $sqrt gmapsto -sqrt g$. 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 $mu_2times mu_2$. 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 $tilde Yto Y$ 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 $k[a^2,ab,b^2,asqrt g,bsqrt g]$. The $mu_2$ action on $Y$ is $(a^2,ab,b^2,asqrt g,bsqrt g)mapsto (a^2,ab,b^2,-asqrt g,-bsqrt g)$.

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