Chebyshev polynomials factoring uniformly modulo all primes

Question

Consider the Chebyshev polynomial of the first kind $T_n(x)$ and its factorization in $\mathbb F_p$ for a given prime $p$. Most often, this factorization is not uniform (meaning that the irreducible factors have not all the same degree, as in this thread). But the following seems to hold:

For given $n$, $T_n(x)$ factors uniformly for all $p$ iff $n$ is a power of $2$.

More precisely: define $\nu(p):=\max( \nu_2(p-1),\nu_2(p+1))=\nu_2(p^2-1)-1$ (where as usual, $\nu_2(\cdot)$ is the multiplicity of the prime factor $2$).
Then it appears that each irreducible factor of $T_{2^k}(x)$ over $\mathbb F_p$ has degree $2^{max(k+2-\nu(p),0)}$.

How to prove this?

Answer 1

Let $\zeta$ be a primitive $4n$th root of unity. Then $\alpha = (\zeta+\zeta^{-1})/2$ is a root of $T_n$. $\alpha$ generates the real subfield of the $4n$th cyclotomic field, of degree $\varphi(4n)/2$ over $\mathbb Q$ (where $\varphi$ is the Euler phi function). Now $\varphi(4n) \le 2n$, with equality if and only if $n$ is a power of 2. This implies that $T_n$ is irreducible over $\mathbb Q$ if and only if $n$ is a power of 2. The uniformity follows from that (since the field generated by a root is Galois, even cyclic).

Let $n = 2^k$. The degree of an irreducible factor mod $p$ is the smallest $e$ such that $p^e \equiv \pm 1 \bmod 2^{k+2}$. If $l = v(p)$, then this is 1 for $k \le l-2$ and $2^(k+2-l)$ for $k \ge l-2$ as claimed.