I've found many exercises in which, given the minimal polynomial $p(x)\in \mathbb {F}[x] $ of $a\in \mathbb {F} $, the minimal polynomial of another expression related to that $a $ is requested. They usually adoperate easy expressions, but I was wondering if it is always possible to find the minimal polynomial of $a'=f(a)/g(a)$ where $f(a), g(a)\in \mathbb {F}[a] $.
I think that the answer is yes, but I can't really find a solid algorithm that will always work. The general idea could be to express $a $ in terms of $a'$, this is not always that easy though, and I haven't even managed to prove that it is always possible.
For a partial answer, the polynomials $\,p(x)\,$ and $\,a' g(x) - f(x)\,$ have the common root $\,x=a\,$, therefore the resultant is zero $\,\operatorname{Res}_x\big(p, a'g-f\big)\,$, which is a polynomial in $\,a'\,$ with coefficients in $\mathbb{F}$.
The resultant, however, is not necessarily the minimal polynomial, for example if $\,f \equiv g\,$ then the resultant is $\,(a'-1)^{\operatorname{deg} p}\,$ but the minimal polynomial is $\,a'-1\,$.
[ EDIT ] For a simple example, consider the case where $a' = \dfrac{\sqrt[3]{2}+1}{\sqrt[3]{4}+1}\,$, which corresponds to $\,a=\sqrt[3]{2}\,$, $\,p(x) = x^3-2\,$, $\,f(x) = x+1\,$, $\,g(x)=x^2+1\,$.
Then
resultant[ x^3-2, a'(x^2+1)-(x+1), x ] = 5 a'^3 + 3 a'^2 - 3 a' - 3, so the polynomial $\,5x^3 +3x^2-3x-3\,$ has $a'$ as a root.In this particular case, the resultant is in fact the minimal polynomial of $\,a'\,$, though that's not necessarily true in the general case.