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I'm currently working through Rick Miranda's book on Algebraic Geometry and Riemann Surfaces, and I've been stuck on a problem in the first chapter, and I can't seem to get anywhere. I think that for example Bezout's theorem would solve it, but I would want something more elementary, which I think there is.

Let X be an affine plane curve of degree 2, that is, defined by a quadratic polynomial f(z,w). Suppose that f is singular. Show that f(z,w) then factors as the product of linear factors.

UPDATE So far I've done the following, set $f(x,y) = ax^2+bxy+cx+dy+ey^2+f$. Say that $p=(m,n)$ is a root, and that it is singular. Set $z=(x+m)$, $w=(y+n)$. Then we have a polynomial: f(z,w) which will have a singular point at (0,0). Taking the partial derivatives, and further, we solve for some coefficients, and at the end we get that a conic, singular polynomial should be (after some transformations) of the form: $az^2+bzw+cw^2$, which is reducible into linear factors. However, I'm not completly sure this method is correct, so any tips would be helpful, as to whether I'm on the right path or not.

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    also, when you're taking $p$artial derivatives, you're only finding the coefficients of the Taylor expansion of your polynomial (which is, actually, the polynomial itself)2011-05-23

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Hint: Make a linear change of coordinates so that (one of) the singular point(s) is located at $(0,0)$. (Check that this is okay in the context of this particular problem.) Now consider the Taylor series expansion of $f(z,w)$, i.e. write $f = f_0 + f_1 + f_2 + ...$, where $f_n$ is homogeneous of degree $n$ in $z$ and $w$. For which degrees $n$ is $f_n$ non-zero? What does this tell you?

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    Andy: I edited my original up there with some new ideas. What I cam up with sounds basically somewhat akin to that you're saying. After shifting to the origin, the constant is 0, thus, all terms must be of degree 2 (since otherwise the partials couldn't both be 0).2011-05-23