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Let $X\subset \mathbb{A}^2$ be a curve defined by an equation $f_{n-1}(x,y)+f_{n}(x,y)$ where $f_{n-1},f_n$ are homogeneous polynomials of degree $n-1$ and $n$ respectively. Then $X$ has a parametrization by $\mathbb{A}^1$.

In this problem we consider the curve $y=tx$ and intersect this curve with the curve $X$. Then:

$f_{n-1}(x,tx)+f_n(x,tx)=x^{n-1}f_{n-1}(1,t)+x^n f_n(1,t)$

Deleting the obvius solution at $x=0$ we get the equation

$x f_n(1,t) + f_{n-1}(1,t) =0$

Therefore: $$x=-\frac{f_{n-1}(1,t)}{f_{n}(1,t)}$$

Is this ok?

Miguel
  • 339
Arthur Levip
  • 93

1 Answers1

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Put $T_2=t_2T_1, \: T_3=t_3T_1,\: \dots,\: T_n = t_nT_1$. Then

$$f_{n-1}(T_1, T_2, \dots, T_n) + f_{n}(T_1, T_2, \dots, T_n)$$

becomes

$$T_1^{n-1}f_{(n-1)}(1, t_2, \dots, t_n) + T_1^n f_n(1_, t_2, \dots, t_n)$$

thus, assuming $T_1 \neq 0$,

$$T_1 = \frac{f_{(n-1)}(1, t_2, \dots, t_n)}{f_n(1_, t_2, \dots, t_n)},$$

$$T_j = t_j T_1 = t_j\frac{f_{(n-1)}(1, t_2, \dots, t_n)}{f_n(1_, t_2, \dots, t_n)}.$$

Bruno Joyal
  • 54,711