Proving that $x^4+x^3+1$ has no solution over $\mathrm{GF}(2^e)$, $e$ odd.

Question

I'm trying to find out whether $x^4+x^3+1$ has a solution over $\mathrm{GF}(2^e)$, $e$ odd. Some quick calculations for small values of $e$ indicate that it does not, but my background is not in number/finite field theory, so I don't know how to prove it in general.

I've tried using the Trace function as follows:

$\mathrm{Tr}(x^4+x^3+1) = \mathrm{Tr}(0) \\ \Rightarrow \mathrm{Tr}(x^4)+\mathrm{Tr}(x^3)=1 \\ \Rightarrow \mathrm{Tr}(x)+\mathrm{Tr}(x^3)=1$.

So I guess I need to show that $\mathrm{Tr}(x^3) = \mathrm{Tr}(x)$ in $\mathrm{GF}(2^e)$, $e$ odd. Any help is much appreciated!

Answer 1

Let $\mathbb{F}_{2^{e}}$ be the finite field with $2^{e}$ elements, $e$ odd. Note that $f(X) = X^{4}+X^{3}+1$ is irreducible over $\mathbb{F}_{2}$, since it does not have any roots in $\mathbb{F}_{2}$, and is not the square of the only irreducible quadratic over $\mathbb{F}_{2}$ (which is $X^{2}+X+1$). If $f$ were reducible over $\mathbb{F}_{2^{e}}$, then $\mathbb{F}_{2^{e}}$ would necessarily contain a subfield isomorphic to $\mathbb{F}_{2}[X]/\langle f(X) \rangle \cong \mathbb{F}_{2^{4}}$, namely $\mathbb{F}_{2}(\alpha)$, the subfield of $\mathbb{F}_{2^{e}}$ generated by the root $\alpha$ of $f$ over the prime subfield $\mathbb{F}_{2}$. But this is a contradiction, as $\mathbb{F}_{p^{k}}$ is a subfield of $\mathbb{F}_{p^{n}}$ if and only if $k \mid n$; this is clearly impossible, since $e$ is odd.

Answer 2

$x^4 + x^3 + 1$ is irreducible over $\mathbb{F}_2$. You can see directly that it has no zeros; it can't split into a product of degree-$2$ factors because it is coprime with $x^4 - x.$ (In general, all irreducible polynomials of degree $n$ divide $x^{2^n} - x.$)

Therefore each of its zeros is a generator of $\mathbb{F}_{16}.$ This is not contained in $\mathbb{F}_{2^e}$ for any odd $e$, since $\mathbb{F}_{2^k} \subseteq \mathbb{F}_{2^l}$ if and only if $k | l.$