Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:18:29Z http://mathoverflow.net/feeds/question/35699 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35699/is-mathbba-the-universal-smooth-scheme-which-is-a-finite-cover-of-mathbb Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? Anton Geraschenko 2010-08-16T01:02:54Z 2011-06-23T04:02:03Z <p>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.</p> <p>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.</p> <blockquote> <p>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 $f:\mathbb A^2\to Spec(k[x^2,xy,y^2])$?</p> </blockquote> <p>A couple of remarks:</p> <ul> <li>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.</li> <li>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 $f$.</li> </ul> http://mathoverflow.net/questions/35699/is-mathbba-the-universal-smooth-scheme-which-is-a-finite-cover-of-mathbb/35740#35740 Answer by Francesco Polizzi for Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? Francesco Polizzi 2010-08-16T09:49:12Z 2010-08-16T17:27:36Z <p>It seems to me that in the global case the answer should be $no$ because of the following argument.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>When $k=1$ we have $Y_1=\mathbb{A}^2$.</p> <p>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.</p> <p>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 $\mathbb{A}^2$.</p> <p>In fact, $f_k$ does not factor through $f$ except for $k=1$.</p> http://mathoverflow.net/questions/35699/is-mathbba-the-universal-smooth-scheme-which-is-a-finite-cover-of-mathbb/36699#36699 Answer by inkspot for Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? inkspot 2010-08-25T22:01:43Z 2010-08-25T22:01:43Z <p>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 <code>$V=Spec\ R$</code> outside the origin. Put $X=V/G$, and suppose $S$ is another complete regular local ring with a finite surjective morphism <code>$Y=Spec\ S\to X$</code>. </p> <p>Claim: $Y\to X$ factors through $V$.</p> <p>Proof: Denote the punctured spectra by asterisks. Take the fiber product <code>$W^*=V^*\times_{X^*}Y^*$</code>. Since <code>$V^*\to X^*$</code> is finite and etale, so is <code>$W^*\to Y^*$</code>. But <code>$Y^*$</code> is simply connected ("purity of the branch locus"), so <code>$W^*\to Y^*$</code> has a section. Compose this with the projection <code>$W^*\to V^*$</code> to get <code>$Y^*\to V^*$</code>, and extend this across the punctures to get <code>$Y\to V$</code>.</p> http://mathoverflow.net/questions/35699/is-mathbba-the-universal-smooth-scheme-which-is-a-finite-cover-of-mathbb/68584#68584 Answer by Anton Geraschenko for Is $\mathbb{A}²$ the universal smooth scheme which is a finite cover of $\mathbb{A}²/μ₂$? Anton Geraschenko 2011-06-23T04:02:03Z 2011-06-23T04:02:03Z <p>One thing that confused me about <a href="http://mathoverflow.net/questions/35699/is-mathbba-the-universal-smooth-scheme-which-is-a-finite-cover-of-mathbb/35740#35740" rel="nofollow">Francesco's answer</a> was how to <em>actually construct</em> 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.</p> <p>Let $g(x,y,z)$ be a polynomial which does not vanish at the origin. We then have two interesting degree 2 maps to $S=Spec(k[a^2,ab,b^2])$:</p> <ul> <li>$\mathbb A^2\to S$, 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.</li> <li>$S[\sqrt{g}]\to S$ (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.</li> </ul> <p>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 <em>diagonal</em> $\mu_2$ action, $(a,b,\sqrt g)\mapsto (-a,-b,-\sqrt g)$.<sup>&dagger;</sup> 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 $Y/\mu_2 = S$.</p> <p><code>$$\begin{array}{cccccc} &amp; &amp; \tilde Y\\ &amp; \swarrow &amp; \downarrow &amp; \searrow\\ \mathbb A^2 &amp; &amp; Y &amp; &amp; S[\sqrt g]\\ &amp; \searrow &amp; \downarrow &amp; \swarrow \\ &amp; &amp; S \end{array}$$</code></p> <p><sup>&dagger;</sup> 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 $(a^2,ab,b^2,a\sqrt g,b\sqrt g)\mapsto (a^2,ab,b^2,-a\sqrt g,-b\sqrt g)$.</p>