Let $p,q \in \mathbb{Q}[x]$ two relatively prime polynomials. Let $h\in \mathbb{R}$ any number and let $F_h(x) = p(x) + h \cdot q(x)$.

What can be said about the irreducibility of the polynomial $F_h(x)$ over the field $\mathbb{Q}(h)$?

More precisely, let $H = \{h : F_h \mbox{ is reducible over }\mathbb{Q}(h) \}$. How big can $H$ be?

Is there some reasonable condition on $p$ and $q$ that guarantees that $H$ is empty? Or finite? Or at least discrete in some sense?

I understand that $H$ can only contain algebraic numbers. The linearity of $F$ in $h$ forces $F$ to be irreducible for transcendental $h$, because one of the factors would have to lie in $\mathbb{Q}(x)$ and that would contradict that $p$ and $q$ are coprime.

I would guess that the same is true when $h$ is algebraic but its degree is large relative to the degrees of $p$ and $q$, though I don't know how to approach this.


1 Answer 1


Surely, it is worth assuming $\max(\deg p,\deg q)>1$ (and $\deg q\geq \deg p$).

In this case, there is no hope for $H$ to be finite or discrete. Take any positive integer $n$ and any nonzero rational $r$. Let $h$ be a root of $G_{r,n}(x)=p(rx^{2n})+xq(rx^{2n})$ (this polynomial is of odd degree!). Then $h\in H$, since $x-rh^{2n}\mid F_h(x)$.

Fixing $n$ and varying $r$, you get that $H$ is dense in the set of value of $(-p/q)$.

Moreover, I would expect the polynomial $G_{r,n}(x)$ to be irreducible infinitely often; in this case $h$ can have arbitrarily high degree.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.