# Why are singular conics reducible?

I'm presently resolving Rick Miranda is publication on Algebraic Geometry and also Riemann Surfaces, and also I've been stuck on a trouble in the first phase, and also I can not appear to get anywhere. I assume that as an example Bezout is theory would certainly address it, yet I would certainly desire something extra primary, which I assume there is.

Allow X be an affine aircraft contour of level 2, that is, specified by a square polynomial f (z, w). Intend that f is single. Show that f (z, w) after that variables as the item of straight variables.

UPGRADE So much I've done the adhering to, set $f(x,y) = ax^2+bxy+cx+dy+ey^2+f$. Claim that $p=(m,n)$ is an origin, which it is single. Set $z=(x+m)$, $w=(y+n)$. After that we have a polynomial: f (z, w) which will certainly have a single factor at (0,0). Taking the partial by-products, and also better, we address for some coefficients, and also at the end we get that a conic, single polynomial need to be (after some makeovers) of the kind: $az^2+bzw+cw^2$, which is reducible right into straight variables. Nonetheless, I'm not completly certain this method is proper, so any kind of pointers would certainly be handy, regarding whether I'm on the appropriate course or otherwise.

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2022-06-07 14:35:56
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Hint: Make a straight adjustment of works with to make sure that (among) the single factor (s) lies at $(0,0)$. (Check that this is all right in the context of this certain trouble.) Currently take into consideration the Taylor collection development of $f(z,w)$, i.e. write $f = f_0 + f_1 + f_2 + ...$, where $f_n$ is uniform of level $n$ in $z$ and also $w$. For which levels $n$ is $f_n$ non - absolutely no? What does this inform you?