I know how to prove that if $\alpha_1, ..., \alpha_n$ are algebraic over $K$, then $K[\alpha_1, ..., \alpha_n]$ is a field (i.e., $K[\alpha_1, ..., \alpha_n]=K(\alpha_1, ..., \alpha_n)$).
I also know the converse is true for $n=1$ and also know how to prove it.
However, I'm having real trouble to deal with the case $n\geq 2$.
I've tried to use the same strategy with $n=1$, which envolves the surjective homomorphism $\psi:K[X]\to K[\alpha]$ with $F \mapsto F(\alpha)$ and the fact that $K[X]$ is a principal ideal domain. But that doesn't work with a similar map $\psi:K[X_1, ..., X_n]\to K[\alpha_1, ..., \alpha_n]$ since $K[X_1, ..., X_n]$ is not a principal domain.
I couldn't disprove it either. I tried to find a small example with $K=\mathbb{R}$ and $n=2$, but it also couldn't figure it out.
Any ideas? Thanks!
Yes. This follows from (or is essentially the content of) the Zariski lemma: a finitely generated algebra over a field $k$ that is itself a field is finite over $k$.
You can find several proofs in these notes by Pete Clark, around page 206.