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?


1 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.

  • $\begingroup$ 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? $\endgroup$
    – Godsbane
    Aug 29, 2019 at 17:57
  • $\begingroup$ @Godsbane I reworded my answer to develop elements about your question. $\endgroup$ Aug 30, 2019 at 8:28
  • 1
    $\begingroup$ @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.) $\endgroup$ Aug 30, 2019 at 8:54
  • 1
    $\begingroup$ @AlexWertheim You're entirely right, of course. I done goofed. It lies somewhere between $n$ and $n!$. $\endgroup$
    – Arthur
    Aug 30, 2019 at 8:55
  • $\begingroup$ 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. $\endgroup$
    – Kolja
    Aug 30, 2019 at 8:58

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