Question about irreducible polynomials over a finite field. If a polynomial $f(x)$ is irreducible over a finite field, does that mean the only factors are $\{1, f(x)\}$?
How would I go about proving a polynomial $f(x)$ is irreducible over a finite field? A bit of searching on StackExchange showed me this:

Irreducibility criterion: A polynomial $P\in\mathbf F_q[X]$ with degree $n$ is irreducible if  and only if

*

*$P$ divides $X^{q^n}-X$;


*$P$ is coprime with all  $X^{q^r}-X$, $\;r=\dfrac nd$, where $d$ is a prime divisor of $n$.

How would I apply this to $f(x) = x^8 + x^4 + x^3 + x + 1$ over $GF(2^8)$?

By definition a non-zero non-unit element $r$ in a ring $R$ is irreducible if whenever $d$ divides $r$, $d$ is either a unit, or $d$ is the product of $r$ by a unit. So the set of divisors may be considerably larger and contains all elements of the field.
