Is $x^{100} - x^2 + 1$ separable in an algebraic closure of $\mathbb{F}_2$

Question

My approach: $f'(X) = 100x^{99} - 2x = 0x^{99} - 0x = 0$ since in $\mathbb{F}_2$. So the $\gcd(f,f') = f > 1$, thus not separable.

On the other hand, $f(0) \neq 0 \neq f(1)$, so irreducible. But irreducible, implies separable in a finite field, so no roots.

So is $f$ actuallly separable or not? What am I missing here?

Answer 1

The fact that your polynomial has no root does NOT imply that it is irreducible (this kind of argument works only for polynomials of degree $2$ or $3$. Think of $(x^2+1)^2\in\mathbb{R}[x])$.

(And your polynomial is indeed NOT separable, by the way .)