0

Why is $F'(a,z_i) \ne 0$?

An algebraic function $y=f(x)$ is defined by the algebraic equation $$ F(x,y) := g_n(x)y^n + g_{n-1}y^{n-1} + \cdots + g_0(x) = 0 $$ where $g_j$ are polynomials. In what follows for simplicity we will assume that $g_n(x)\equiv 1$ (then the roots will not escape to infinity).

Let a point $n\in \mathbb{C}$ be such that the equation $F(x,y)=0$ has $n$ different roots $y=z_1,\dots,z_n$. then $F'_y(a,z_i)\ne 0$ and Implicit Function Theorem asserts that for any $x$ from some ...

It looks to me like it should be something obvious but I’ve been trying to figure it out a while now. Have I missed something or is this from some deeper result a second year undergraduate such as myself may not have seen?

Original image

Somos
  • 35,251
  • 3
  • 30
  • 76
oliver maclean
  • 3

1 Answers1

0

I think it means that fixing $x=a$, the polynomial (thought of as a variable in $y$) $F(a,y)=c(y-z_1(a))...(y-z_n(a))$ where $z_1(a),...,z_n(a)$ are distinct complex numbers. Then you can check that $F'(a,z_i(a)) \neq 0$. (It's because the roots are distinct.)

By allowing $x$ to vary, the roots will depend on $x$, so you have in general functions $z_1(x),...,z_n(x)$.

zoidberg
  • 1,305