Let α and β be algebraic over $F$ with minimal polynomials $f(x)$ and $g(x)$ respectively. Show that $f(x)$ is irreducible over $F(β)$ if and only if $g(x)$ is irreducible over $F(α)$.
My method is to consider the extension $F(a,b)$ so let $m=deg (f)$, $n=deg (g)$. Let's just show the the forward implication. Suppose $f$ irreducible over $F(b)$, this implies $[F(a,b):F(b)] =m$. Then $[f(a,b):F]=mn$. Is this true? Here I am not sure why $[F(a,b):F]=mn$ though?
Then by the tower theorem,$ [F(a,b):F]= [F(a,b):F(a)][F(a):F] =[F(a,b):F(a)]m$. Then $[F(a,b):F(a)]=n$ by the prev paragraph and this implies $g(x)$ irreducible over $F(a)$. Is this correct?
It seems like what you need is a quick explanation of why the degree of the minimal polynomial of an algebraic element is the same as the degree of the field extension.
Let $f(x) = \sum\limits_{i=0}^m f_ix^i$ be the minimal polynomial for $\alpha$ over $F$. It's clear that the first $m$ powers of $\alpha$ form a vector-space basis for $F(\alpha)$ over $F$, so that $[F(\alpha) : F] \leq m$. Conversely, suppose that $[F(\alpha) : F] = n$, i.e. $F(\alpha)$ has a basis $b_1, b_2, \ldots b_n$. Then any $n+1$ elements of $F(\alpha)$ satisfy a linear dependence, and in particular so do the first $n$ powers of $\alpha$, i.e. there exists $g(x) = \sum\limits_{i=0}^n g_ix^i$ in $F[x]$ such that $g(\alpha) = 0$. Since $f$ is the minimal polynomial of $\alpha$, it divides $g$ and a fortiori $[F(\alpha) : F] = n \geq m$. Thus we have $\deg(f) = [F(\alpha) : F]$
Note also that $f$ is irreducible over $F(\beta)$ iff $f$ is the minimal polynomial for $\alpha$ over $F(\beta)$, so the if $f$ is irreducible over $F(\beta)$ then the above shows that $[F(\alpha,\beta) : F(\beta)] = [F(\alpha) : F]$