This question came to my mind when I was analyzing the following situation. Suppose we have a finite field $\mathbb{F}_q$ and two monic irreducible polynomials $f(x)$ and $g(y)$ of degree $n$. We know that the fields $\mathbb{K}_1 = \mathbb{F}_q[x]/(f(x))$ and $\mathbb{K}_2 = \mathbb{F}_q[y]/(g(y))$ are isomorphic, but...
is it true that the map $x\mapsto y$ is always an isomorphism?
I think this is (trivially) true: the map is injective because $x\mapsto y$ is not the zero map ($y$ is not zero in $\mathbb{K}_2$) and it's clearly surjective because every element of $\mathbb{K}_2$ can be written as a polynomial of degree at most $n-1$ in $y$, which is the image of exactly the same polynomial in $x$.
I think the same argument generalizes easily to $\mathbb{F}(\alpha) \cong \mathbb{F}(\beta)$.
However, something that makes me feel a bit uncomfortable is that if $x\mapsto y$ is an isomorphism, then $x$ and $y$ must have the same minimal polynomial $h(t)$ over $\mathbb{F}_q$ and therefore, since $f(x)=0$ and $g(y) = 0$ in $\mathbb{K}_1$ and $\mathbb{K}_2$ respectively, $h(t)$ must divide both $f(t)$ and $g(t)$...
but how is this possible, if both $f(t)$ and $g(t)$ are irreducible?