$f,g \in \mathbb Q[x]$ , $f(a)=g(b)=0$ ; $f$ is irreducible in $\mathbb Q(b)[x]$ iff $g$ is irreducible in $\mathbb Q(a)[x]$?

Question

Let $a,b$ be complex roots of irreducible polynomials $f(x),g(x) \in \mathbb Q[x]$ . Let $F:=\mathbb Q(a) , K:=\mathbb Q(b)$ ; then is it true that $f(x)$ is irreducible in $K[x]$ if and only if $g(x)$ is irreducible in $F[x]$ ?

Answer 1

Yes. Let $n$ and $m$ be the degrees of $f(x)$ and $g(x)$, respectively. Since $f(x)$ and $g(x)$ irreducible, they are the minimal polynomials of $a$ and $b$, respectively, over $\mathbb{Q}$. Therefore $[\mathbb{Q}(a) : \mathbb{Q}] = n$ and $[\mathbb{Q}(b) : \mathbb{Q}] = m$.

Suppose $f(x)$ is irreducible over $\mathbb{Q}(b)[x]$. Then $f(x)$ is the minimal polynomial of $a$ over $\mathbb{Q}(b)$, and so $[\mathbb{Q}(a,b) : \mathbb{Q}(b)] = n$, and so $[\mathbb{Q}(a,b) : \mathbb{Q}] = mn$ by the tower law. We have $$[\mathbb{Q}(a,b) : \mathbb{Q}(a)] = [\mathbb{Q}(a,b) : \mathbb{Q}]/[\mathbb{Q}(a) : \mathbb{Q}] = mn/n = m.$$ It follows that the minimal polynomial of $b$ over $\mathbb{Q}(a)$ has degree $m$. Since $g(b) = 0$ and $g(x)$ has degree $m$, $g(x)$ is the minimal polynomial of $b$ over $\mathbb{Q}(a)$, and is thus irreducible over $\mathbb{Q}(a)[x]$.