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Math Help - Finding singular points of a curve in affine space

  1. #1
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    Finding singular points of a curve in affine space

    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^2)(y^2)-6(x^2)+2xy-6(y^2)+25=0

    Taking partial derivatives i get the jacobian matrix

    J = (2x(y^2)-12x+2y 2(x^2)y+2x-12y)

    I do not know how to proceed from here.

    Any help is appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by Siknature View Post
    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 12 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.)
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