Finding singular points of a curve in affine space

Quote:

Originally Posted by Siknature

How do i find the singular points of the following curve in A^2(C) OR prove that there are no singular points? (where A is affine space and C is complex numbers)

The curve is

$x^2y^2-6x^2+2xy-6y^2+25=0$

Taking partial derivatives i get the jacobian matrix

$J = \begin{pmatrix}2xy^2-12x+2y & 2x^2y+2x-12y\end{pmatrix}<br />$
I do not know how to proceed from here.

The condition for a singular point is that the Jacobian should have less than full rank. The full rank of a 1×2 matrix is 1. So at a singular point, the matrix must have rank 0, which is the same as saying that both entries should be 0.

Therefore you need to solve the equations

$\left.\begin{aligned}2xy^2-12x+2y &= 0, \\ 2x^2y+2x-12y &= 0.\end{aligned}\right\}$

An efficient way to do that is to add the equations, getting $(xy-5)(x+y)=0$ . So either $xy=5$ , leading to the solutions $x=y=\pm\sqrt5$ ; or $x+y=0$ , leading to the solutions $x=-y=\pm\sqrt7$ and also the solution $x=y=0$ .

(You then have to check whether any of those points actually lie on the curve. If they don't, they obviously don't yield singular points.)