# Splitting field of a separable polynomial in a finite field

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?

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.

• I agree with your statement but I don't quite see why the splitting field of $f$ necessarily has $p^n$ elements. Couldn't it be even bigger? Aug 29, 2019 at 17:57
• @Godsbane I reworded my answer to develop elements about your question. Aug 30, 2019 at 8:28
• @Arthur: I don't think that's true, is it? Take $F = \mathbb{Q}$ and $f(X) = X^{3}+2$, for example; then if $K$ is a splitting field of $f$ over $F$, we have $[K:F] = 6$, not $3$. (The fact that $F$ is finite here plays an essential role.) Aug 30, 2019 at 8:54
• @AlexWertheim You're entirely right, of course. I done goofed. It lies somewhere between $n$ and $n!$. Aug 30, 2019 at 8:55
• You can prove that the splitting field of an irreducible $f$ of degree $n$ is $\mathbb{F}_{p^n}$ by trying to find the roots of $f$. If $\alpha$ is one root then $\alpha^p$ is another root (why?). Think about it for a little while, and show that all the roots of $f$ are in $\mathbb{F}_{p^n}$, it's a nice exercise. Aug 30, 2019 at 8:58