# Transcendental extension not isomorphic to its closure

Suppose I'm given a field extension $K/F$ with $\alpha\in K$ transcendental over $F$, the claim is that $F(x)\cong F(\alpha)$. It's a statement without proof in our class notes, and the remarks preceding it are that we can define an evaluation map $E_\alpha :F[x]\to F[\alpha]$, $f(x)\mapsto f(\alpha)$, and that this map is injective iff $\alpha$ is transcendental (understood) and that $F[x]\cong F[\alpha]$ (1st isomorphism theorem if I'm not mistaken). I believe the claim can be proven from the fact that the fields of fractions of two isomorphic integral domains are isomorphic, but I feel as though there should be some other explanation following more the approach above. Thanks

• Following which approach? – Ravi Mar 30 '16 at 23:37
• What is the problem with this proof? It is so simple and clear. – Crostul Mar 31 '16 at 10:45