Splitting field of a separable polynomial in a finite field

Question

Let $F_2$ be the field with 2 elements.

(a) Factor the polynomial $ f(x) = x^4+x^2+x+1 \in F_2[x]$

into irreducible factors in the ring $F_2[x]$.

(b) Let $E$ be the splitting field of f over $F_2$. How many elements does $E$ have?

I calculated the solution for (a) as

$x^4+x^2+x+1 = (x^3+x^2+x+1)*(x+1)$

For (b) I figured out that (x+1) is redundant because it already splits in $F_2$

I want to find the splitting field $E$ explicitly. I think it is given by $F_{2^3}$ but I don't know how to prove that. Can someone help me?

I think the splitting field of an irreducible polynomial $f \in F_p[X]$ with $deg(f) = n$ is always isomorphic to $F_{p^n}$ but I can't find that statement anywhere. Is it true?

Answer 1

Your factorization of $p(x) =x^4+x^2+x+1$ is wrong.

You have in fact

$$p(x) = (x^3+x^2+1)(x+1)$$

$q(x) = x^3+x^2+1$ is irreducible as it is of degree $3$ and neither $0$ nor $1$ are roots of $q$.

Indeed, the splitting field of an irreducible polynomial $f \in \mathbb F_p[X]$ with $deg(f) = n$ is always isomorphic to $\mathbb F_{p^n}$. Why?

A rupture field $\mathbb F_p \subset \mathbb F_p(a) \cong \mathbb F_p[x]/(f)$ is a field extension that can be seen as a vector space of dimension $\deg f = n$ over $\mathbb F_p$. Therefore it has $p^n$ elements and is isomorphic to $\mathbb F_{p^n}$. $a$ is a root of the polynomial $x^{p^n} -x$ and $f$ divides $x^{p^n} -x$. The splitting field of $x^{p^n} -x$, namely $\mathbb F_{p^n}$, is an algebraic extension of $\mathbb F_p(a)$. As both have $p^n$ elements, they are equal.