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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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If it is a problem in the first chapter why don't you try to do it from the definitions first? – Phira May 23 '11 at 5:43
yes, taking the partial derivatives it's ok. You should be warned, though, that since you're not working in the projective plane, you should also add to you system of equation with the partial derivatives also the condition that the points you find lie on the cruve. – Andy May 23 '11 at 10:18
also, when you're taking partial derivatives, you're only finding the coefficients of the Taylor expansion of your polynomial (which is, actually, the polynomial itself) – Andy May 23 '11 at 10:22
up vote 6 down vote accepted

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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Oh, I think I understand. Simply reducing it to one of two possible forms by some diagonaliation etc.? – Dedalus May 23 '11 at 6:25
@Dedalus: actually, that's just the way you classify singular points. Do you know how to classify double points? Try that with you second degree polynomial, what happens? – Andy May 23 '11 at 7:41
Andy: No, I don't. From what I now know is that a curve is singular at a point if the partial derivatives are zero there. Would you mind expanding? – Dedalus May 23 '11 at 8:06
@Dedalus: if a point is singular, it means it does not have a unique tangent line (actually it has as many principal tangents as its degree, i.e. a double point will have two tangent lines). What does this mean in terms of the polynomial expansion? (Mind that $f_0 =0$ because the origin is on the curve now) If some more explanation is needed I'll add an answer :) – Andy May 23 '11 at 9:43
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). – Dedalus May 23 '11 at 10:10

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