2

$ f $ is monic , $[E:\mathbb{F}_q]=s$ , $ E $ is an extension of $\mathbb{F}_q$

$deg(f)| s$

The book states that:

$f(x)^\frac{s}{deg(f)} = x^s - c_{1}x^{s-1} ...- c_k$

I know $ f $ has deg (f) distinct roots in $ E $, but I do not see how to have a term of degree $ s-1 $

$f(x)= (x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_{deg(f)})$

In fact the book states that the term $ c_1 $ is exactly equal to $Tr (\alpha_i)$ where $ \alpha_i $ is a root of $ f $, but I do not even know why it has degree term $ s-1 $, and much less I know because this term is $ Tr(\alpha_i) $. I know only that the degree term $ deg(f) -1 $ in the polynomial $ f $ has coefficient $-Tr(\alpha_i) $

user489941
  • 133

1 Answers1

1

This notation in no way asserts that $c_1\neq 0$; it's just giving a name to the coefficient of $x^{s-1}$. It's entirely possible that $c_1=0$, in which case you could say "there is no $x^{s-1}$ term".

As for why $c_1=Tr(\alpha_i)$, that follows (for instance) from the fact that the Frobenius automorphism $F$ permutes the $\alpha_i$ cyclically. So, when you add up $Tr(\alpha_i)=\alpha_i+F(\alpha_i)+\dots+F^{s-1}(\alpha_i)$, each root of $f$ appears $s/\deg(f)$ times in the sum. The negative of that sum is therefore exactly what you get as the coefficient of $x^{s-1}$ when you expand $$f(x)^{s/\deg(f)}= (x-\alpha_1)^{s/\deg(f)}(x-\alpha_2)^{s/\deg(f)}\cdots (x-\alpha_{deg(f)})^{s/\deg(f)}.$$

(Incidentally, it is not true that the coefficient of degree $\deg(f)-1$ in $f(x)$ is $-Tr(\alpha_i)$. This would be true only if the field $E$ is generated by the $\alpha_i$, or equivalently in the case $\deg(f)=s$.)

Eric Wofsey
  • 342,377