# If $\mathbb{F}(\alpha)$ and $\mathbb{F}(\beta)$ are isomorphic, then $\alpha\mapsto\beta$ is necessarily an isomorphism?

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?

• The big problem is to show that $x\mapsto y$ is a homomorphism in the first place. Compare the infinite field case of $\Bbb Q[\sqrt 2]\cong \Bbb Q[\sqrt 8]$, but $\sqrt 2\mapsto\sqrt 8$ does not define a field homomoprhism. – Hagen von Eitzen Aug 8 '17 at 17:36
• Ok, thanks for the counter example. However, in the particular case I shown the map is an homomorphism, right? it's just polynomial "evaluation" – Daniel Aug 8 '17 at 17:45

Call $\Phi$ the vector space isomorphism sending $x$ to $y$ (abusing notation, I will write $x$ and $y$ also when I mean their equivalence classes).
Since $f(x)=0$, we have $\Phi((f(x))=0$.
On the other hand, if $\Phi$ is a field homomorphism we have $\Phi(f(x))=f(\Phi(x))=f(y)$, so $y$ is a root of $f$. By the uniqueness of the minimal polynomial, this implies $f=g$.
• Ok, so it basically shows that $\Phi$ is a field homomorphism if and only if the polynomials are the same, correct? – Daniel Aug 8 '17 at 18:03