Primitive element of an irreducible polynomial

Question

Let's consider an irriducible polynomial $f(x)$ over $GF(p^{m})$; i've understood that

$f(x) \equiv 0 \space (mod \space f(x))$,

however why (intuitively) the polynomial $x$ is considered the primive element of the field ? What's special about the polynomial $x$ ?

And moreover, why (intuitively) the powers of $x$ construct the whole finite field ?

The polynomial $x+(f(x))$ is a solution of $f(x)=0$ in the field $F[x]/(f(x))$ — fantasie, May 26 '19 at 07:41

score 1 · Answer 1 · answered May 26 '19 at 07:34

Well, the elements of $GF(p^n)$ are exactly the roots of the polynomial $x^{p^n}-x$ over $GF(p)$.

Moreover, the irreducible polynomials of degree $d$ over $GF(p)$ with $d\mid n$ are exactly the divisors of $x^{p^n}-x$.

For instance, the irreducible polynomials of degree 4 over $GF(2)$ are $x^4+x+1$, $x^4+x^3+1$ and $x^4+x^3+x^2+x+1$. The first two of them are primitive, while the last is not, since it divides $x^5-1$.

Primitive element of an irreducible polynomial

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