a set of pairwise commuting polynomials The set $\{1,x,x^2,x^3,\dots\}$ has the following properties:

*

*any two of these polynomials commute


*for any $n\ge 0$ there is exactly one polynomial of degree $n$ inside
Is this the only example? If not, is there a general form of such sets? I was thinking of this problem during a long flight, it seems very natural and might have been investigated, but I didn't found this result so far.

The above question has a negative answer - as flawr mentioned (see below), the Chebyshev polynomials of the first kind are another example.
Also, Chris Culter pointed out, that we can replace any polynomial $f$ in our "good" set with a conjugate of the form $\sigma f \sigma^{-1}$, where $\sigma(x)=ax+b$ with $a\neq 0$.

So now, a conjecture:
$\{1,x,x^2,x^3,\dots\}$ and $\{T_0,T_1,T_2,\dots\}$ and the conjugates among them are the only examples.


Is the above conjecture true? Is such a thing already known?


Edit after a clarification request:
The field is $\mathbb{R}$.
I consider polynomials of real coefficients and one real variable.
Polynomials $P$ and $Q$ commute, if $P(Q(x))=Q(P(x))$ for every $x\in\mathbb{R}$.
 A: the examples you mentioned are not the only examples: Consider the Chebyshev Polynomials $\{T_n \mid n \in \mathbb N\}$. These have the property that
$$T_n(T_m(x)) = T_{nm}(x)$$
and therefore they commute, but they are not repeated compositions of themselves.
A: Yes, every such polynomial chain is similar to (exactly one of) the power polynomials $\{x^1,x^2,\ldots\}$ or the Chebyshev polynomials $\{T_1, T_2,\ldots\}$.

Doing some internet digging, I found that the result was discovered by Block and Thielman; there's a nice overview of commuting polynomials, leading up to the result, in this article:
https://digitalworks.union.edu/cgi/viewcontent.cgi?article=1784&context=theses
The theorem itself appears on p33, as the  Block-Thielman theorem.  I present my own streamlined version of the results below.

In detail, you can establish that:

*

*For any quadratic polynomial $q(x)$, and for every degree $k$, there is at most one degree-$k$ polynomial that commutes with $q$.
It follows that a chain of polynomials (which mutually commute and have exactly one term of each degree) is uniquely determined by its quadratic term (!).


*Every quadratic polynomial is similar to a unique monic polynomial of the form $x^2+c$.


*In a chain, the value of $c$ must be equal to 0 or -2. In the first case, the chain is similar to the power chain. In the second case, it's similar to the Chebyshev chain.
Here's a sketch of how to prove #3 once you know #1 and #2. Take a chain $\{p_1, p_2, p_3, \ldots\}$, which by #1 we know is uniquely determined by its quadratic  $p_2$. By #2, $p_2$ is similar to a monic quadratic of the form $x^2+c$. Apply that similarity transform to every term in the chain, obtaining a new chain:
$$\{q_1, x^2+c, q_3, q_4, \ldots\}$$
and consider the degree-three term $q_3$. Because $q_3$ commutes with $x^2 + c$, we get a functional equation stating that the following two polynomials are equal everywhere:
$$q_3\circ q_2 = q_2\circ q_3$$
When two polynomials are equal, their coefficients must be equal. Hence we have a system of six equations relating $c$ to the coefficients of $q_3(x)$. ($q_m\circ q_n$ has degree $m+n$.)
The equations simplify. In fact, we can use the fact that $q_3$ commutes with $x^2+c$ to solve for all of $q_3$'s coefficients (see below).  The rest of the system tells us that $c$ was in fact highly constrained:
$$c(c+2) = 0$$
That is, $c=0$ or $c=-2$. At a higher level, this shows that a chain of polynomials can only exist when its quadratic term is similar to $x^2$ or to $x^2-2$— nothing else.
In the first case, the quadratic is similar to $x^2$, so the chain is similar to the power chain (by uniqueness #1). In the second case, the quadratic is similar to $x^2-2$. This is in turn similar to the second Chevyshev polynomial $T_2(x)=2x^2-1$, through the similarity $\alpha(x)=2x$, so the chain is similar to the Chebyshev chain.

We can derive the coefficients of $q_3$ from the fact that it commutes with $x^2+c$.  Take note of the following facts:


*For even polynomials $f(-x)=f(x)$, all the odd coefficients vanish. For odd polynomials $f(-x)=-f(x)$, all the even coefficients vanish. (N.B. Not all polynomials are odd or even functions.)


*If a polynomial commutes with a degree $n$ monic polynomial, then its leading coefficient must be a root of $x^{n-1}-1$.
Then here's what we know about $q_3$:

*

*It commutes with $(x^2+c)$, obeying the equation  $$q_3\circ(x^2+c) = (x^2+c)\circ q_3$$


*By #5, its leading coefficient must be 1 (the only root of $x-1$).


*By expanding out the equation, we find that $q_3(x)^2 \equiv q_3(-x)^2$. Hence $q_3(x)$ must be an odd or even function $=\pm q_3(-x)$. But it's cubic, so its degree-3 term can't vanish (see #4) — we conclude it's odd.


*It's an odd function, so its only nonzero coefficients are its leading coefficient and its linear coefficient. (See #4)


*The linear coefficient $b_1$ is the only unknown; our system lets us solve for it, finding $3c = 2b_1$.

A: You could replace each polynomial $f(x)$ in your set with the conjugate $f'(x)=f(x-1)+1$. This will yield $\{2,x,x^2-2x+2,x^3-3x^2+3x,\ldots\}$.
A bit more generally, let $\sigma(x)=ax+b, a\neq0$ be an invertible polynomial. Then you can conjugate like so: $f'(x)=\sigma(f(\sigma^{-1}(x)))$.
