Why are $q$-polynomials commutative while linear transformations are not?

Question

There is a one-to-one correspondence between those $q$-polynomials over $\mathbb{F}_{q^n}$ of the form$$\sum_{i=0}^{n-1}a_ix^{q^i},$$ and those linear transformations of $\mathbb{F}_{q^n}$ over $\mathbb{F}_q$. However, for two $q$-polynomials $f,\,g$, $$f(g(x))=g(f(x)),\text{ for all }x\in\mathbb{F}_{q^n},$$ while linear transformations are not commutative, right? So the question is how to explain the seemingly contradiction.

Answer 1

Linearized polynomials don't commute either. Consider $$ f(x)=x^q\qquad\text{and}\qquad g(x)=x^q+ax, $$ where $a\in\Bbb{F}_{q^n}\setminus\Bbb{F}_q$. Then $$ f(g(x))=g(x)^q=x^{q^2}+a^qx^q $$ and $$ g(f(x))=f(x)^q+a f(x)=x^{q^2}+ax^q. $$ These are distinct for $a^q\neq a$ whenever $a\notin\Bbb{F}_q$.

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On the other hand, if you restrict yourself to linearized polynomials of the form $$\sum_{i=0}^{n-1}a_ix^{q^i}$$ with all the coefficients $a_i$ in the smaller field $\Bbb{F}_q$, call this set of linear transformations $S$, then any two transformations from $S$ do commute. This is because the elements of $\Bbb{F}_q$ are fixed under Frobenius.

However, $S$ is not the set of all $\Bbb{F}_q$ linear transformations. You immediately see that the cardinality is wrong – $S$ is only an $n$-dimensional space whereas the ring of linear transformations has dimension $n^2$ (isomorphic to the ring of square matrices).

An alternative and generalizable explanation is that $S$ is the ring of linear transformations generated by scalars matrices (=elements of the smaller field) and the Frobenius automorphism. Whenever you have a vector space $V$ over a field $k$, and a single $k$-linear transformation $T:V\to V$, the $k$-algebra generated by powers of $T$ is commutative. Basically because powers of $T$ commute with each other and the scalars. The ring $k(T)$ is isomorphic to a quotient ring $k[T]/\langle m(T)\rangle$, where $m(T)$ is the minimal polynomial of $T$. The latter ring inherits commutativity from the polynomial ring. In the case of $T=$ Frobenius of $\Bbb{F}_{q^n}$ we have $m(T)=T^n-1$.