Let $GF(2^m)$ be a binary extension field with constructing polynomial $f(z)$ be an irreducible, primitive polynomial over $GF(2)$.
Is there any possibility that two (or more) different $f(z)$ can produce exactly the same field (not simply isomorphic) for calculations with its polynomials as elements?
For example, for $GF(2^{233})$ is there any case that e.g. the following irreducible constructing polynomials $f_1(z): z^{233} + z^{74} + 1$ and $f_2(z): z^{233}+z^{159}+1$ to produce identical field traces for all elements $\in GF(2^{233})$?
Thank you for your time,
No, it is not possible. If $f(z)$ and $g(z)$ are two distinct irreducible primitive polynomials, then none of them is a multiple of the other one in $GF(2^m)[z]$. But the $0$ element of the field $GF(2^m)[z]/\langle f(z)\rangle$ is the set of all multiples of $f(z)$ and the $0$ element of the field $GF(2^m)[z]/\langle g(z)\rangle$ is the set of all multiples of $g(z)$. These are distinct sets. So, $g(z)$ does not belong to the first set, whereas $f(z)$ does not belong to the second one.