Let $p(x),q(x)\in \mathbb{Q}[x]$ be irreducible polynomials, $p(a)=0$, for $a\in\mathbb{C}$ and $q(b)=0$, for $b\in\mathbb{C}$. How can we show that $p(x)$ is irreducible in $\mathbb{Q}(b)[x]$ if and only if $q(x)$ is irreducible in $\mathbb{Q}(a)[x]$?
Im using the field extension $\mathbb{Q}(a,b)$ and the fact that $Irr(a,\mathbb{Q})|p(x)$ and $Irr(b,\mathbb{Q})|q(x)$ to show that $[\mathbb{Q}(a,b):\mathbb{Q}]\leq nm$ where $n=\deg p(x)$ and $m=\deg q(x)$ but im kinda stuck on how to follow next. My guess is that i need to show that there is an isomorphism between $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$.
Any hint would be appreciated!
We may assume that $p,q$ are monic.
Since $q$ is irreducible over $\mathbb Q$ it is the minimal polynomial of $b$ over $\mathbb Q$. This implies that $[\mathbb Q(b):\mathbb Q]=\deg q$.
Moreover, if $p$ is irreducible over $\mathbb Q(b)$, from $p(a)=0$ we deduce that it is the minimal polynomial of $a$ over $\mathbb Q(b)$. It follows that $[\mathbb Q(a,b):\mathbb Q(b)]=\deg p$. Thus we get $[\mathbb Q(a,b):\mathbb Q]=\deg p\deg q$.
Since $p$ is irreducible over $\mathbb Q$ it is the minimal polynomial of $a$ over $\mathbb Q$. In particular, $[\mathbb Q(a):\mathbb Q]=\deg p$.
Now we get $[\mathbb Q(a,b):\mathbb Q(a)]=\deg q$. Since $q(b)=0$ we get that $q$ is the minimal polynomial of $b$ over $\mathbb Q(a)$, hence $q$ is irreducible over $\mathbb Q(a)$.
The converse is similar.