The field $\mathbb F_q$ is it $\cong \mathbb F_p[X]/(X^q-X)$?

I know that $\mathbb F_q$ where $q=p^n$ is a field with $q$ element and that is by definition the splitting field of $X^q-X$

Q1) Is it really by definition ?

Can we say that $\mathbb F_q\cong F_p[X]/(X^q-X)$ ?

I would say no since $X^q-X$ is not irreducible.

Q2) So how can we describe $\mathbb F_q$ ?

I know that $X^{q-1}-1$ is irreducible,

Q3) So what would be the field $\mathbb F_p[X]/(X^{q-1}-1)$ ? (despite of the splitting field of $X^{q-1}-1$). Would it be a field with $q-1$ elements ? (it looks to be impossible since every finite field has $p^n$ element with $p$ prime).

As you can see, I'm a little bit lost...

• $X^{q - 1}-1$ is not irreducible (except in the non-interesting case $q=2$), for example because it has 1 as a root. – Marc Paul Dec 19 '15 at 10:43
• You need an irreducible polynomial of degree $n$. – Morgan Rodgers Dec 19 '15 at 10:44

Q1: You're right. $\;\mathbf F_p[X]/(X^q-X)$ is not even an integral domain.
Q2: The simplest description is it's the splitting field (over $\mathbf F_p$) of $X^q-X$. Unless $q=p=2$, $X^{p-1}-1=\Bigl(X^{\tfrac{p-1}2}-1\Bigr)\Bigl(X^{\tfrac{p-1}2}+1\Bigr)$ is certainly not irreducible.
$\mathbf F_q$ is isomorphic to $\mathbf F_p[X]/(P(X))$, where $P(X)$ is any irreducible factor of the cyclotomic polynomial $\Phi_{q-1,\mathbf F_p}(X)$ – they all have degree $n$.
Q3: $\mathbf F_p[X]/(X^{q-1}-1)$ is only a product of fields.