Irreducible polynomial in finite-fields Let $\overline{\mathbb{F}}_2$ be an algebraic closure of $\mathbb{F}_2$, and let $\alpha\in\overline{\mathbb{F}}_2$ be such that $\alpha^2+\alpha+1=0$. Prove that if a polynomial $P$ is irreducible in $\mathbb{F}_2$ but $P$ is reducible in $\mathbb{F}_2[\alpha]$, the deg $P$ is even.
 A: 
Lemma: Let $q$ be a power of a prime natural number.  Write $\bar{\mathbb{F}}_q$ for the algebraic closure of the field $\mathbb{F}_q$ with $q$ elements.  Fix $\alpha\in\bar{\mathbb{F}}_q$ such that $\big[\mathbb{F}_q[\alpha]:\mathbb{F}_q\big]=m$ for some positive integer $m$.  Suppose that $P(x)\in\mathbb{F}_q[x]$ is irreducible over $\mathbb{F}_q$ but is reducible over $\mathbb{F}_q[\alpha]$.  Then, $$\gcd(m,n)>1\,,$$ where $n$ is the degree of $P(x)$.

Let $\beta\in\bar{\mathbb{F}}_q$ be a root of $P(x)$. Then, $\mathbb{F}_q[\alpha,\beta]=\mathbb{F}_q[\gamma]$ for some $\gamma\in\bar{\mathbb{F}}_q$ (every finite-degree extension of a field is a simple extension, that is, it can be generated by a single generator due to the Primitive Element Theorem).  Let $Q(x)\in\mathbb{F}_q[x]$ be a minimal polynomial over $\mathbb{F}_q$ with root $\gamma$.  Hence, $$\deg\big(Q(x)\big)=\big[\mathbb{F}_q[\alpha,\beta],\mathbb{F}_q\big]=\big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\beta]\big]\,\big[\mathbb{F}_q[\beta]:\mathbb{F}_q\big]=kn\,,$$
where $k:=\big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\beta]\big]=\big[\mathbb{F}_q[\beta][\alpha]:\mathbb{F}_q[\beta]\big]$.  Note that $k$ is a divisor of $m$ (see the hidden portion below). On the other hand,
$$\deg\big(Q(x)\big)=\big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\alpha]\big]\,\big[\mathbb{F}_q[\alpha]:\mathbb{F}_q\big]=m\, \big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\alpha]\big]\,.$$
If $\gcd(m,n)=1$, then $m\mid k$, whence $k=m$.  Thus,
$$\big[\mathbb{F}_q[\alpha][\beta]:\mathbb{F}_q[\alpha]\big]=\big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\alpha]\big]=n\,.$$
That is, the minimal polynomial of $\beta$ over $\mathbb{F}_q[\alpha]$ must have degree $n$, but this contradicts the assumption that $P(x)$ is reducible over $\mathbb{F}_q[\alpha]$.  Hence, $\gcd(m,n)=1$. In particular, if $m=p$ for some prime natural number $p$, then $p\mid n$, $k=1$, $Q(x)=P(x)$, and $\alpha\in\mathbb{F}_q[\beta]$.

  A proof that $k\mid m$ is given here.  It is clear that, if $F(x)\in\mathbb{F}_q[x]$ is a minimal polynomial of $\alpha$ over $\mathbb{F}_q$ and $G(x)\in\mathbb{F}_q[\beta][x]$ is a minimal polynomial of $\alpha$ over $\mathbb{F}_q[\beta]$, then $G(x)$ divides $F(x)$.  Now, since $F(x)$ splits over $\mathbb{F}_q[\alpha]$, $G(x)$ splits into linear factors $x-\tau$ with $\tau\in\mathbb{F}_q[\alpha]$.  Therefore, $G(x)\in\mathbb{F}_q[\alpha][x]$.  Hence, $G(x)\in K[x]$, where $K:=\mathbb{F}_q[\alpha]\cap\mathbb{F}_q[\beta]$.  Note that $G(x)$ is irreducible over $K$.  Let $L$ be the splitting field of $G(x)$ over $K$, then $L\subseteq \mathbb{F}_q[\alpha]$ (in fact, $L=\mathbb{F}_q[\alpha]$), so that  $$\deg\big(G(x)\big)=[L:K]\;\Big|\;\big[\mathbb{F}_q[\alpha]:L\big]\,[L:K]\,\big[K:\mathbb{F}_q\big]=\big[\mathbb{F}_q[\alpha]:\mathbb{F}_q\big]=m\,.$$  Since $k=\big[\mathbb{F}_q[\alpha,\beta]:\mathbb{F}_q[\beta]\big]=\deg\big(G(x)\big)$, the claim follows.

Note that the converse also holds.  The proof is omitted here.

Converse: Let $q$ be a power of a prime natural number.  Fix $\alpha\in\bar{\mathbb{F}}_q$ such that $\big[\mathbb{F}_q[\alpha]:\mathbb{F}_q\big]=m$ for some positive integer $m$.  Suppose that $P(x)\in\mathbb{F}_q[x]$ is irreducible over $\mathbb{F}_q[\alpha]$.  Then, $$\gcd(m,n)=1\,,$$ where $n$ is the degree of $P(x)$.

A: Here’s a shorter argument:
Your $\alpha$ is a cube root of unity, and $\Bbb F(\alpha)=\Bbb F_4$. The extension os Galois of course, and the nontrivial automorphism exchanges $\alpha$ with $\bar\alpha=\alpha^2=\alpha+1$.
A monic polynomial $g\in\Bbb F_4[X]$ either has coefficients involving $\alpha$ or it doesn’t, in which latter case, it’s in $\Bbb F_2[X]$. We can denote by $\bar g$ the polynomial gotten from $g$ by applying the conjugation to every coefficient, and for $g\in\Bbb F_4[X]$, you see that $g\in\Bbb F_2[X]$ if and only if $\bar g=g$.
Now take your $\Bbb F_2$-irreducible polynomial $f(X)$, and factor it into irreducibles as an $\Bbb F_4$-polynomial. If any factors had no $\alpha$, they would have been $\Bbb F_2$-factors of the original, so all factors of $f$ must have $\alpha$ in ’em, and since the product is an $\Bbb F_2$-polynomial, they occur in conjugate pairs. So $f$ had even degree.
