# Algebraic extensions $F(\alpha), F(\beta)$ with a $F$-isomorphism yield the same minimal polynomial for $\alpha, \beta$

Suppose we are given two simple algebraic extensions $$F(\alpha), F(\beta)$$ (over a field $$F$$) with an $$F$$-isomorphism $$\phi$$ of extensions (i.e. fixing $$F$$) such that $$\phi(\alpha)=\beta$$.

Then we want to prove that $$\alpha, \beta$$ have the same minimal polynomial.

I'd like to argue that $$F(\alpha) \simeq F[x]/(f(x))$$ and $$F(\beta) \simeq F[y]/(g(y))$$ where $$f,g$$ are the minimal polys of $$\alpha, \beta$$ and from this I should get an isomorphism $$F[x] \simeq F[y]$$ sending $$x$$ to $$y$$ that is induced by $$\phi$$. But I'm not sure if this argument is actually beating around the bush, or if there is a more direct or obvious argument?

We have $$0 = \phi(0) = \phi(f(\alpha)) = f(\phi(\alpha)) = f(\beta)$$ implies that $$g$$ divides $$f$$.
By symmetry, $$f$$ divides $$g$$ and so $$f=g$$ (since both are monic).