Let $m(\mu)$ be an irreducible polynomial of degree $d$ over $\mathbb{F}_2$, $F_{2^d} = \mathbb{F}_2[x]/(m(\mu))$ by a field extension given by that polynomial and let $d: \mathbb{F}_2[x] \to F_{2^d}[x]$ be the map $f \mapsto f(x + \mu) - f(x)$.
For the case $d=2$, there's only one choice for $m$, and the kernel of $d$ is $\mathbb{F}_2[x^4 + x]$, as discussed in this question: Is there an analogue of the jordan normal form of an nilpotent linear transform of a polynomial ring?
For $d=3$, we can have $m(\mu) = 1 + \mu^2 + \mu^3$, which gives a kernel of $\mathbb{F}_2[x^8+x]$, and $m(\mu) = 1 + \mu + \mu^3$ gives $\mathbb{F}_2[x^4+x^2+x]$.
We have a similar situation for $d=4$: $1+\mu + \mu^4$ gives $\mathbb{F}_2[x^8 + x^4+x^2+x]$, while $1 + \mu^3 + \mu^4$ and $1+\mu + \mu^2+\mu^3+\mu^4$ take us back to $\mathbb{F}_2[x^{16}+x]$
Of course, $x^{2^d} + x$ is the product of all irreducible polynomials of degree $k$, where $k|d$.
Essentially, I've got a map from the set of irreducible polynomials over $\mathbb{F}_2[x]$ to the set of subalgebras of $\mathbb{F}_2$ by $m(\mu) \mapsto \text{ker}(d)$ (this could be a functor of categories, but it's not obvious to me how an isomorphism of the fields turns into algebra homomorphism.)
My question is: why do some irreducible polynomials land on the algebra generated by $x^{2^d} - x$ and others land in an algebra generated by a factor of that polynomial? Can I characterize the polynomials that hit the former?
It's worth noting that if $\alpha: F_{2^d,m_1} \to F_{2^d, m_2}$ is a field isomorphism, then $\text{ker}(f \mapsto f(x +\mu) - f(x)) = \text{ker}(f \mapsto f(x + \alpha(\mu)) - f(x))$. But that doesn't answer my question.