TITLE: Is $x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?
QUESTION [6 upvotes]: Problem:
Is $f(x) = x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?
My thought:
$f(x)$ is irreducible in $\mathbb{F}_2$ and has degree $5$. So we can conclude that $\mathbb{F}_{32} \simeq \mathbb{F}_2[x]/f(x)$. Then apparently $f$ is not irreducible in $\mathbb{F}_{32}$.
But I don't know how to work on the case $\mathbb{F}_8$. I know that $\mathbb{F}_8 \simeq \mathbb{F}[x]/g(x)$ where $g(x)$ is some irreducible polynomial of degree $3$ in $\mathbb{F}_2 [x]$. For example, it can be $g(x) = x^3 + x + 1$. But how would that help me?

REPLY [6 votes]: Here’s another approach, based on your good start. You’ve observed that $f$ is $\Bbb F_2$-irreducible, and that indeed $\Bbb F_{32}$ is its splitting field. In other words, $f(x)=\prod_i(x-\alpha_i)$, where the alpha’s are five different elements of $\Bbb F_{32}$. If you could take fewer than five of the factors above and multiply them together to get a polynomial in $\Bbb F_8[x]$, then its coefficients would be in $\Bbb F_{32}\cap\Bbb F_8=\Bbb F_2$, in other words a proper $\Bbb F_2$-divisor of $f$. Thus the only way you can multiply some of those factors together to get an $\Bbb F_8$-polynomial is to use all of them. So $f$ is $\Bbb F_8$-irreducible.