Chebyshev polynomials and commuting chains.

The Chebyshev polynomials is a family of polynomials $T_n(x) \in {\bf Z}[x]$ such that $$ \cos(nx) = T_n(\cos(x)) $$ with ${\text deg} T_n = n$, for each positive integer $n$. In particular we have $$ T_1(x) = x\\ T_2(x) = 2x^2 - 2\\ T_3(x) = 4x^3 -3x \\ \vdots $$ Now an observation is that this Chebyshev family of polynomials ${\mathscr T}=\{T_n\}$ forms a commuting family of polynomials under function composition. Namely $T_n\circ T_m = T_m \circ T_n$. In fact we have $T_n\circ T_m = T_{nm}$. 

There are other commuting families. Say $\{x + 1, x+ 2, x+3\}$, or ${\mathscr P}=\{x, x^2, x^3, x^4, \ldots\}$. Now the families $\mathscr T$ and $\mathscr P$ are called chains, where there is exactly one polynomial in each family per positive degree. As it turns out, $\mathscr T$ and $\mathscr P$ are essentially the only two commuting chains, up to similarity. 

What do we mean by similarity? Note an invertible polynomial is of the form $\sigma(x) = ax + b$ for some $a\neq 0$. We say polynomial families $\mathscr A$ and $\mathscr B$ are similar if ${\mathscr A} = \sigma {\mathscr B} \sigma^{-1}$, for some invertible $\sigma$ whose inverse is also a polynomial.  

So we have:

Theorem. (Block-Thielman)

If $\mathscr A$ is a commuting chain of polynomials, then $\mathscr A$ is similar to $\mathscr T$, the Chebyshev polynomials, or $\mathscr A$ is similar to $\mathscr P$, the power monomials. 

To prove this one analyzes the degrees 1, 2, and 3 polynomials in this chain, under a suitable similarity transform. And use the fact that:

Lemma. 

For a fixed quadratic polynomial $f$, and a fixed $d$ there exists at most one polynomial $g$ of degree $d$ such that $f\circ g  = g\circ f$. 

(Try proving these!)