Does the generator of a finite field depend on the used irreducible polynomial?

Question

Let our finite field be $\operatorname{GF}(2^{8})$ with an irreducible polynomial: $$p_1=x^8+x^4+x^3 +x^2+1$$

I saw here The Laws of Cryptography: Generate Multiplication Tables that $$\alpha=x+1$$ is a generator of $\operatorname{GF}(2^{8})$. But when I used this finite field calculator, the order of $\alpha$ was displayed as:

$$\operatorname{ord}(\alpha)=51$$

They used a different irreducible polynomial on their site ($p_2=x^8 + x^4 + x^3 + x + 1$). So does the generator of a finite field depend on the used irreducible polynomial?

Is there something like an online calculator to find the generators over a finite field with a given irreducible polynomial?

Update

My Python 3 code to calculate roots and generators can be found in this gist. Enjoy!

Answer 1

Well, it does. For instance, the polynomials $x^4+x+1$ and its conjugate $x^4+x^3+1$ are irreducible over $GF(2)$ and primitive, i.e., each zero $\alpha$ of them is a primitive generator of the multiplicative group of the field $GF(16)$.

On the other hand, the polynomial $x^4+x^3+x^2+x+1$ is irreducible over $GF(2)$ but a divisor or $x^5-1$ and so not primitive, i.e., each zero $\alpha$ of it satisfies $\alpha^5=1$.

All in all, the polynomial $x^{16}-x$ factors as follows: $$x(x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1).$$