# Irreducible polynomial such that its roots satisfy a given relation $x_3 = f(x_1,x_2)$

My general question is:

What are the polynomials $f \in \Bbb Q[X,Y]$ such that there exists an irreducible polynomial $P \in \Bbb Q[X]$ having three distinct zeros $x_1,x_2,x_3$ that satisfy $x_3 = f(x_1,x_2)$ ?

It is probably too broad, so I'd like to begin with the case $f(X,Y)=aX+bY+c$ where $a,b,c$ are rational numbers.

Apparently, we can't have $ab=0$ (because $X \mapsto aX+c$ is an automorphism of $\Bbb Q[X]$ if $a\neq 0$, and clearly if $a=0$, we can't have $x_3=c \in \Bbb Q$). Moreover, the cases $a=1,b=0=c$ and $b=1,a=0=c$ are forbidden.

It is less trivial, but the case $c=0,a=b=1/2$ can't happen, see here. The main point of the answer is that the Galois group of the splitting field of $P$ over $\Bbb Q$ acts transitively on the finite set $R \subset \Bbb C$ of the roots of $P$, so that any root $x_j$ can be expressed as $x_j=f(a_j,b_j)$ where $a_j \neq b_j \in R$.

We can call this property the $f$-transitivity: a non-empty set $E \subset \Bbb C$ is $f$-transitive if any $x \in E$ can be written as $x=f(y,z)$ for distinct elements $y,z \in E$. Maybe for many $f \in \Bbb Q[X,Y]$, an $f$-transitive set must be infinite, as this is the case for $f(X,Y)=(X+Y)/2$. Even for $f(X,Y)=X+Y$, I wasn't able to prove it, but I believe it's true for $f(X,Y)=aX+bY+c$, and maybe for higher degree…

However, $f(X,Y)=XY$ is possible, using some cyclotomic polynomial.

Any hint is welcome. Thank you!