If a monic polynomial $p(x)$ is irreducible over rationals and has $\alpha$ as root. Is it possible that $\alpha$ is root of some other irreducible monic polynomial $q(x)$ such that $\deg(p(x))\neq \deg(q(x))$?
This question arose when I was thinking about minimal polynomials of certain elements.
No if $p$ is the minimal polynomial of $\alpha$ in $\mathbb{Q}$, $p$ will divide every other polynomial which has $\alpha$ as a root. This is because the ring of polynomials over a field is a principal ideal domain. So the ideal $I$ of all polynomials which has a root $\alpha$ will be principal say equal to some $\langle h \rangle$, $h$ - monic. Now if $p, q \in \langle h \rangle$ then $h$ divides $p$ and $q$ hence $h = p = q$, because $p$ and $q$ are irredicible.
Suppose you have two such polynomials $p(x)$ and $q(x)$ which have the same root $\alpha$. Then, unless one of $p, q$ is a factor of the other (including the case where they are equal), the division algorithm will produce a further polynomial $r(x)$ of lower degree than both $p$ and $q$ and with $\alpha$ as a root.
The division algorithm can be carried out without leaving the original ground field. So this applies more widely than simply $\mathbb Q$.
