Irreducibility at compositions of polynomial expressions in general?

Question

Let $P(X),Q(X),p(X)\in\overline{\mathbb{Q}}[X]$ (or $\mathbb{Q}[X]$) and let $P(X),Q(X),p(X)$ be not constant.
Let $P(p(x))=Q(x)$.

Let's consider irreducibility over $\overline{\mathbb{Q}}$ (or $\mathbb{Q}$).

1. In which cases can the irreducibility of $Q(x)$ be concluded from the irreducibility of $P(x)$?

2. In which cases can the irreducibility of $P(x)$ be concluded from the irreducibility of $Q(x)$?

I don't know how I could start to derive an answer.

I need the answer because I want to try to prove some of the conjectures in Closed-form solvability of elementary transcendental equations?.

Answer 1

As for question 2, the irreducibility of P(x) follows from the irreducibility of Q(x) as long as p(x) isn’t constant . This can easily be proven by contradiction: Let p(x) be any non-constant polynomial and suppose P(x) is some reducible polynomial such that Q(x) = P( p(x) ) is irreducible. Now since P(x) is reducible, it has some factorization into two non-constant polynomials which we will call R(x) and S(x), so P(x)=R(x)•S(x). Substituting this in, we get that Q(x) = P( p(x) ) = R( p(x) )•S( p(x) ). And since p(x), R(x), and S(x) are all non-constant, the compositions R( p(x) ) and S( p(x) ) are both non-constant. This means that the equation Q(x) = R( p(x) )•S( p(x) ) Is a non-trivial factorization of Q, meaning Q is reducible, contradicting our assumption that it was not. Thus for any p(x), the irreducibility of Q implies the irreducibility of P

If p(x) were constant, Q(x) would also be constant so it would obviously not be irreducible.

The answer to question 1, is that the irreducibility of P implies the irreducibility of Q only when p(x) is linear. With algebraic polynomials it’s easy to see that the only irreducible polynomials are the linear ones since algebraic polynomials always have algebraic zeroes and thus divide the linear polynomials with those zeroes. Returning to the rationals, I can tell you that for any quadratic p(x), there is a linear P(x) which essentially “completes the square” so that Q(x) is a square polynomial (so P is irreducible but Q is reducible).

This can be improved upon because for any polynomial p(x) with degree grater than one, you can let P(x) be the linear function that subtracts the constant term of p(x). As a result, Q(x) will have no constant term, and must divide the polynomial f(x)=x.