Can a primitive root of a polynomial over $GF(2)$ ever not generate a multiplicative group?
I have some notes from my review of finite field extensions a while ago that I've been rereading. It's the last statement that's throwing me. I've included some preceding notes for context.
If $p(x)$ is an irreducible polynomial of degree $n$, then adjoining a root of $p$ to $GF(2)$ generates an extension of degree $n$, which is necessarily a field, $E$, with $2^n$ elements.
The multiplicative group of nonzero elements in $E$ has order $2^n - 1$.
Thus by Lagrange's Theorem, every nonzero element $a$ of $E$ satisfies $a^{2^n - 1} = 1$. Thus every element $a$ in $E$ is a root of $g(X) = > X^{2^n} - X$.In other words, $E$ is exactly the set of all roots of $g(X)$. Now the roots of the original $p(x)$ are also roots of $g(X)$, and so $p(x)$ divides $g(X)$ (after making the variables the same).
Conversely, if $f(x)$ is any polynomial that divides $g(x)$, then the roots of $f(x)$ lie in $E$, so they generate a subfield of $E$. If they generate all of $E$, and if $f(x)$ is irreducible, then they must have degree $n$.
Now, let $\rho$ be a primitive root of $f$, where f is irreducible. So $\rho$ will generate $E$ as a field, but not necessarily generate $E-\{0\}$ as a multiplicative group.
I think this is an example of what you're after. The polynomial $x^4+x^3+x^2+x+1$ is irreducible over the field of 2 elements, so any root $\alpha$ generates the field of 16 elements. But $\alpha^5=1$ (note that the given polynomial is a factor of $x^5-1$), so $\alpha$ doesn't generate the 15-element multiplicative subgroup as a group.