# Why generator polynomial of $GF(2^m)$ are irreducible?

Why generator polynomial of the cyclic group $GF(2^m)$ are irreducible?

• Galois fields are not just groups; they are finite fields (en.wikipedia.org/wiki/Finite_field). The generator polynomial needs to be irreducible to ensure that you actually get a field. May 13 '15 at 17:27

It looks like you're confusing two different meanings of "generator polynomial".

First, $GF(2^m)$ is the field $\mathbb F_2[X]/ \langle p\rangle$ for some irreducible polynomial $p$ of degree $m$ over $\mathbb F_2$. This $p$ is sometimes know as the generator polynomial for the field (representation) in question. It needs to be irreducible, because otherwise we're not getting a field, just a ring with zero divisors.

(It turns out that all irreducible polynomials of the same degree give the same field up to isomorphism, so for high-level purposes we don't need to specify which).

Second, however, the multiplicative group of any finite field such as $GF(2^m)$ is always cyclic and so has a generator. Concretely, this generator can be written down as a polynomial of degree $<m$, and that polynomial does not have to be irreducible.

For example, $GF(32)$ can be represented as $\mathbb F_2/\langle X^5+X^2+1\rangle$, because $X^5+X^2+1$ is irreducible.

The multiplicative group of $GF(32)$ is a cyclic group with 31 elements, each of which is a generator (because 31 is prime). So, for example $X^2$ is a generator of the multiplicative group as a cyclic group, but it is plainly not irreducible as a polynomial.

(On the other hand, $X^5+X^2+1$ is not a member of the multiplicative group at all, and so in particular not a generator).