# Isomorphism between fields

Suppose that $F$ and $G$ are two fields and $F[x]$ and $G[x]$ the polynomial rings in $x$. We know that the equivalence classes of the fractions of elements of $F[x]$ is a field $F(x)$ and the same is for $G(x)$.

If $F[x]$ and $G[x]$ are isomorphic as rings and/or $F(x)$ and $G(x)$ are isomorphic as fields, what can we say about $F$ and $G$?

I remember that I've read that we cannot prove that $F$ and $G$ are isomorphic, but I don't remember why and I suppose that the proof is not simple.

• mathoverflow.net/questions/96777/… – user26857 Sep 23 '15 at 19:38
• @Emilio: this could be related math.stackexchange.com/questions/13504 – Watson Aug 18 '16 at 12:47
• @Watson: Thankyou. The question that remain open is if: $F(x)$ isomorphic to $G(x)$ implies $F$ isomorphic to $G$ but $F[x]$ not isomorphic to $G[x]$. I cannot figure such a situation....but I don't find a proof . – Emilio Novati Aug 18 '16 at 14:52
• I don't understand : $F$ isomorphic to $G$ but $F[x]$ not isomorphic to $G[x]$ seems impossible to me, even when $F,G$ are rings: you just send $x$ to $x$ and $F$ to $G$ via the given isomorphism. – Watson Aug 18 '16 at 15:02
• This could be related to Zariski's cancellation problem, see here. – Watson Nov 5 '16 at 14:18

## 1 Answer

Here's one part: Assume $\phi\colon F[x]\to G[x]$ is an isomorphism of rings with inverse $\psi$. For $a\in F^\times$, we must have $\phi(a)\in G[x]^\times =G^\times$. As also $\phi(0)=0$, we have that $\phi|_F$ is a ring homomorphism $F\to G$ with inverse $\psi|_G\colon G\to F$, in other words, $F$ and $G$ are isomorphic.

• Thank you , this prove the statement for the rings of polynomials (+1), but for the field of fractions? – Emilio Novati Sep 23 '15 at 16:09