Polynomial basis representation of elements of a finite field

Question

Confusing text:

I thought different elements are represented by different coefficients, $g_i$. Can someone explain me the second part of the text, one about roots of the polynomial.

Answer 1

The general representation of $\Bbb F_{p^m}$ is $\Bbb F _p [X] / (F)$ where $F$ is any irreducible polynomial in $\Bbb F_p [X]$ of degree $m$ (it turns out that all these polynomials lead to isomorphic quotient rings).

Now, since $F$ is irreducible, in particular it cannot have roots in $F_p$. Let $\alpha$ be any root of $F$ in the algebraic closure $\overline {\Bbb F_p}$ of $\Bbb F_p$. Consider the ring morphism $i : \Bbb F _p [X] \to \Bbb F _p (\alpha)$ given by $i(f) = f(\alpha)$ (where one naturally views $f \in \Bbb F _p [X]$ as an element of $\Bbb F _p (\alpha) [X]$).

Let us prove that $i$ is surjective. If $\beta \in \Bbb F_p (\alpha) = \Bbb F_p [\alpha]$ (the last equality being true because $\alpha$ is algebraic over $\Bbb F_p$), then $\beta = \beta_0 + \beta_1 \alpha + \dots + \beta_m \alpha^m$ with $\beta_i \in \Bbb F_p \ \forall i$. Then, if $f = \beta_0 + \beta_1 X + \dots + \beta_m X^m \in \Bbb F_p [x]$, it follows that $i(f) = \beta$, so $i$ is indeed surjective.

Concerning the kernel, notice that $i(f) = 0 \iff f(\alpha) = 0 \iff F \mid f$, so $\ker i \simeq (F)$.

By the fundamental isomorphism theorem, then, you get that $\Bbb F _p [X] / (F) \simeq \Bbb F_p (\alpha)$ and since we already knew that $\Bbb F_{p^m} \simeq \Bbb F _p [X] / (F)$, it follows that $\Bbb F_p (\alpha) = \Bbb F_p [\alpha]$ is another representation of $\Bbb F_{p^m}$.