Irreducibility of iterations of a polynomial

Question

I am look for some nice examples or conditions that force $f^{(n)}(x)$ to be irreducible (this is composition). I looked at things online and it seemed that topic is quite complex and involves calculating discriminants. I would like to do my own little exposition/project on perhaps an example where I could prove $f^{(n)}(x)$ is irreducible using elementary methods. I know a little bit of dynamical systems and field theory up until basics of Galois.

Does such an example exist? Perhaps something akin to that? In general I am looking for something I could write extensively about that relates to iterations of polynomials, irreducibility is the first thing that came to mind.

Edit: I ran some tests with $x^2+1$ and it seems the first 10 compositions are irreducible so I think considering this polynomial could be a good start. I am just not sure what to do as I never actually considered reducibility of compositions.

The case of first degree polynomials is easily discussed, so your exposition might quickly segue from those to iterations of quadratic polynomials. I'll be interested to see what comes of your Question. — hardmath, Nov 05 '20 at 16:44
Rafe Jones's 2012 paper An iterative construction of irreducible polynomials reducible modulo every prime develops sequences of irreducible-over-$\mathbb Q$ iterates of certain monic quadratic polynomials, e.g. $f(x) = (x-2)^2 + 3$, which are all reducible wrt to every prime modulus after sufficiently many iterations. — hardmath, Nov 06 '20 at 00:36
See also the 2012 paper Newly reducible iterates in families of quadratic polynomials, where "newly reducible" means $f^{(n)}$ is irreducible but $f^{(n+1)}$ is reducible. — hardmath, Nov 06 '20 at 00:45

score 1 · Accepted Answer · answered Nov 06 '20 at 07:11

Let $f(x) = (x-1)^2 + 1$. Then $f^{(n)}(x)$ is irreducible over $\mathbb Q$ for all $n\ge 1$, and we can show this using just Eisenstein's criterion.

First we deduce the closed form:

$$ f^{(n)}(x) = (x-1)^{2^n} + 1 $$

This can be shown easily by induction, but it can be seen essentially by inspection if we write:

$$ f(x) = \tau^{-1} \circ g \circ \tau(x) $$

where $g(x) = x^2$ and $\tau(x) = x-1$ (so that $\tau^{-1}(x) = x+1$).

Thus $f^{(n)}(x) = \tau^{-1} \circ g^{(n)} \circ \tau(x)$, in agreement with the closed form above, since $g^{(n)}(x) = x^{2^n}$.

Now apply Eisenstein's criterion with $p=2$:

$$ f^{(n)}(x) = (x-1)^{2^n} + 1 = x^{2^n} + \left[ \sum_{k=1}^{2^n -1} \binom{2^n}{k} (-x)^k \right] + 2 $$

For proofs that $p=2$ divides each binomial coefficient $\binom{2^n}{k}$ for $1\le k \le 2^n -1$ see the previous Math.SE Question Prime dividing the binomial coefficients.

Irreducibility of iterations of a polynomial

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