An affine singular surface 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?
 A: 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).
A: 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 $\mathbb C$.
