I was wondering whether it is not very difficult to see the following: 


for a nonconstant irreducible polynomial $p(x) \in \mathbb{Q}[x]$ does there always exist a polynomial $q(x) \in \mathbb{Q}[x]$ of degree at least $2$ such that the composition $p(q(x))$ is irreducible in $\mathbb{Q}[x]$?


Thank you,
Albertas



Irreducibility of compositions of polynomials
Yes, there does. In fact, we will show that a polynomial $q(x) = x^2 + d$ works for some $d \in \mathbb{Q}$. The conclusion will follow from [Hilbert's Irreducibility Theorem](http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem) once we show that $p(x^2 + d)$ is irreducible (as a polynomial in two variables).


Suppose, therefore, that $p(x^2 + d) = (d^a f\_a(x) + \dotsb)(d^bg\_b(x) + \dotsb)$ splits. Both $a$ and $b$ are positive (else you could plug in some value of $x$ in $\mathbb{C}$ to make the RHS vanish regardless of $d$, which is impossible by starring at the LHS). But then $f\_a$ and $g\_b$ are nonzero and so there is a rational $x$ which is a zero of neither of them. Plug it in and get a nontrivial splitting of (a shift of) $p(x)$. Contradiction, as $p(x)$ was assumed to be irreducible.


