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Let $n$ be a positive integer and let $A$ be the subring of ${\mathbb C}[x,y]$ generated by $x,xy,...,xy^n$. Let $S=Spec(A)$. This is an affine surface, which is clearly singular if $n\neq 1$. Is this some sort of familiar surface? For example, is it normal? Does it have rational singularities? Can one construct an explicit resolution of singularities for it?

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It is a toric surface; indeed, $A$ is the semigroup algebra of the semigroup generated by those monomials, which is the set of integer points of a plane cone, the dual of which is the defining fan of the surface —among other things, this implies normality. You can construct a desingulatization by subdividing a plane cone, using Hirzebruch-Jung continued fractions; this is explained in Fulton's book on toric varieties. – Mariano Suárez-Alvarez Apr 2 '14 at 7:26
Yes, I understand, but I wonder if someone has already performed that. – Alexander Braverman Apr 2 '14 at 7:41
up vote 9 down vote accepted

Let $\mathbb{F}_n$ be the Hirzebruch surface of index $n$. You have two open subsets on it isomorphic to $\mathbb{A}^1\times \mathbb{P}^1$, with a glueing map given by $$(t,[u:v])\mapsto (1/t,[ut^n:v])$$ The section $u=0$ has self-intersection $-n$ and the section $v=0$ has self-intersection $n$. We remove the section $v=0$ and obtain an affine surface if $n>0$ (because the section is ample). The morphism to your surface given by $$(t,[u:v])\mapsto (\frac{u}{v},\frac{u}{v}t,\dots,\frac{u}{v}t^n)$$ on the first chart corresponds to $$(t,[u:v])\mapsto (\frac{u}{v}t^n,\frac{u}{v}t^{n-1},\dots,\frac{u}{v}t,\frac{u}{v})$$ on the other chart. Hence the resolution of your singularity is just the complement in $\mathbb{F}_n$ of one section of self-intersection $n$, and the exceptional divisor is the section of self-intersection $-n$.

The surface obtained is called a Gizatullin surface, because of the very nice theorem of Gizatullin that asserts that the isomorphism class of the complement of an ample section in a Hirzebruch surface only depends on the square of the section (and not of the section or the Hirzebruch surface).

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Here is an alternative, perhaps more familiar way to recognize this surface.

First consider the $\mathbb C$-algebra embedding $$ \mathbb C[x,y]\hookrightarrow \mathbb C[z,w,z^{-1}] $$ given by $$ x\mapsto z^n \qquad y\mapsto wz^{-1}. $$ Since $z,w$ are algebraically independent, this is clearly an embedding.

It is easy to see that via this embedding the image of the subring $\mathbb C[x,xy,\dots,xy^n]\subset \mathbb C[x,y]$ maps isomorphically to the subring $\mathbb C[z^n,z^{n-1}w,\dots,zw^{n-1},w^n]$. Spec of this latter ring and hence the surface $S$ in the question is the (affine) cone over the (projective) rational normal curve of degree $n$ in $\mathbb P^n$.

From this description it is easy to answer the additional questions and even more about the singularity:

  • Since the rational normal curve is projectively normal, the cone is normal
  • Blowing up the vertex of the cone resolves the singularity and the exceptional curve is a smooth rational curve with self-intersection $-n$. (And this implies that the resolution of the projectivized cone is $\mathbb F_n$ as in Jérémy's answer).
  • Then by Artin's criterion (or direct computation) it is easy to see that the singularity is rational.
  • Or one could observe that this is a quotient singularity which implies that it is rational.
  • One may ask if the singularity is Gorenstein, canonical, log terminal, etc. This is also easy:
  • From the explicit resolution it is easy to compute the discrepancy of the exceptional divisor, which is $-1 + 2/n$ showing that the singularity is canonical if and only if $n\leq 2$ and log terminal, but neither canonical, nor Gorenstein if $n>2$. Finally, for $n=2$ this cone is naturally embedded into $\mathbb A^3$ as it is a cone over a projective plane curve, so it is Gorenstein. And of course for $n=1$ it is just a plane, so it is smooth. In other words, this singularity is always log terminal which also implies that it is rational.
  • Since it is not Gorenstein, it is not factorial, but several of the above properties imply that it is $\mathbb Q$-factorial.

... and the same argument works over any algebraically closed field, not just $\mathbb C$.

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Nice way of seeing it. For me it is "less familiar" but certainly "more familiar" for others. Hence it is good to have both ways. – Jérémy Blanc Apr 3 '14 at 7:57
Indeed, I hesitated writing that about being "more familiar" as it surely depends on one's point of view. Also, at the end it is the same thing. From the fact that the exceptional curve is a smooth rational curve with self-intersection $-n$ it follows easily that the resolution of the projectivized cone is actually $\mathbb F_n$. – Sándor Kovács Apr 3 '14 at 15:30

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