Does there exist an irreducible polynomial $f(x)\in \mathbb{Z}[x]$ with degree greater than one such that for each $n>1$, $f(x^n)$ is reducible (over $\mathbb{Z}[x]$)?

6$\begingroup$ The polynomial $f(x) = x^2 + 1$ has the property that $f(x^n)$ is reducible except when $n$ is a power of 2, so it misses a very thin set. $\endgroup$– Stanley Yao XiaoAug 25, 2014 at 16:23

$\begingroup$ I'm pretty sure that this question appeared on some issue of Mathematical Reflections (proposed by G. Dospinescu), but I can't seem to find it. By the way, the answer should be "no", if I remember correctly. $\endgroup$– user41593Aug 25, 2014 at 16:31

1$\begingroup$ A stupid comment is that the question is equivalent to the same question in $\mathbf{Q}$ instead of $\mathbf{Z}$, so there is no reason to bother with integrality issues and the polynomials can be assumed monic in $\mathbf{Q}[x]$. $\endgroup$– YCorAug 25, 2014 at 17:41
1 Answer
There is no such polynomial.
It is clear that $f$ cannot be a cyclotomic polynomial (your condition $\deg{f} > 1$ excludes $x1$). So suppose $f$ is noncyclotomic and irreducible, of degree $d$, and consider a prime $p$ for which $f(x^p)$ is reducible. For $\alpha$ a root of $f(x^p)$, reducibility means that $[\mathbb{Q}(\alpha):\mathbb{Q}] < pd$. On the other hand $[\mathbb{Q}(\alpha^p):\mathbb{Q}] = d$ by the irreducibility of $f$, and we conclude that $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^p)] < p$; this means that the polynomial $x^p  \alpha^p$ is reducible over $\mathbb{Q}(\alpha^p)$. By the VahlenCapelli theorem, this is only possible if $\alpha^p = \eta^p$ with $\eta \in \mathbb{Q}(\alpha^p)$. We conclude that for every prime $p$, either $f(x^p)$ is irreducible, or else each root $\xi$ of $f$ has the property that $\xi$ is a $p$th power from $K := \mathbb{Q}(\xi)$.
It remains to note that for any number field $K$, and any $\xi \in K^{\times} \setminus \mu_{\infty}$ not a root of unity, $\xi \in (K^{\times})^p$ for only finitely many primes $p$. This follows for instance from Dirichlet's unit theorem.

$\begingroup$ Very nice. It also works for degree 1 polynomials if we exclude multiples of $x$ and $x1$. $\endgroup$– YCorAug 25, 2014 at 20:48

4$\begingroup$ incidentally this shows that if $f$ (any monic irreducible polynomial) is cyclotomic , then $f(X^p)$ is reducible for all but finitely many primes $p$, while if $f$ is not cyclotomic (and $\neq X$), then $f(X^p)$ is irreducible for all but finitely many primes $p$. $\endgroup$– YCorAug 26, 2014 at 0:04