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Fix a prime number $p$. Consider the affine curve $C$ in $\mathbf{A}^2$ over a number field $K$ given by the equation $x^p+y^p - (x+y)^p =0$.

Its Jacobi matrix is $(px^{p-1} -p(x+y)^{p-1} \ py^{p-1} - p(x+y)^{p-1})$. This has rank zero if and only if $(x,y) = (0,0)$. So we have a singularity in the origin.

What is the normalization of $C$?

The reason I find this curve peculiar is because it seems that when I consider $C$ as a projective curve in $\mathbf{P}^2$ it has not singularities...

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Do you mean as a set of points in $\mathbb{P}^1$? Of course a set of points won't have singularities... –  Zhen Lin Jan 28 '12 at 16:06
No I meant as a subset of $\mathbf{P}^2$. The equation is homogeneous and the variable $z$ is "free". –  Ali Jan 28 '12 at 17:17

1 Answer 1

up vote 2 down vote accepted

First of all, you are not correct that the Jacobian is nonzero away from $(x,y) = (0,0)$. If $p \equiv 1 \mod 6$ and $y = \omega x$ with $\omega$ a primitive third root of unity, then $x^{p-1} - (x+y)^{p-1} = y^{p-1} - (x+y)^{p-1} =0$. So, when $p \equiv 1 \mod 6$, your curve is singular along the entire pairs of lines $x = \omega y$ and $x = \omega^{-1} y$. More precisely, your curve is non-reduced along these lines. If $K$ does not contain cube roots of unity, then the right way to formulate this statement is that your curve is nonreduced along $x^2+xy+y^2=0$.

Since normalization usually isn't defined for nonreduced schemes, I'll stick to the case that $p \equiv 5 \mod 6$. That said, the case that $p \equiv 1 \mod 6$ isn't much more complicated; you just need to keep track of those two nonreduced lines.

It's easiest to describe the situation when $K$ is algebraically closed. The equation $x^p+y^p-(x+y)^p$ is homogenous. That means its zero scheme will be a union of lines through the origin. Normalization breaks those lines apart. So the normalization of $C$ is a union of $p$ copies of $\mathbb{A}_1^{K}$, one for each root of $z^p+1-(z+1)^p$, plus one for the "root at $\infty$". Explicitly, you can write down the normalization by blowing up the origin: There are two charts, one with coordinates $(x,z)$ and the other with coordinates $(y,w)$. The idea is that $xz=y$ and $x=yw$. So, in the first chart, the blow up is $K[x,z]/(z^p+1-(1+z)^p)$.

You wanted $K$ to be a number field. In that case, $z^p+1- (z+1)^p$ may not factor into linear factors over $K$. Rather, for each irreducible factor $g$, we get a component of the normalization which looks like $K[z]/g(z) \times_K \mathbb{A}^1_{K}$.

To complete this answer, I should write out a proof that the Jacobian only vanishes at the claimed points, but I'll leave that to you.

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Wow this is an amazing answer. So much for me to learn! –  Ali Jan 28 '12 at 17:17

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