If $\Bbb Q[x,y]$ is a field, are $x$ and $y$ algebraic over $\Bbb Q$?

Question

If $x,y$ are two complex numbers, such that $\Bbb Q[x,y]$ is a field, then is it true that $x$ and $y$ algebraic over $\Bbb Q$?

I know that the converse is true, see this question. This is true if I replace $\Bbb Q[x,y]$ by $\Bbb Q[x]$. This is wrong if I replace $\Bbb Q[x,y]$ by $\Bbb Q[x_1,x_2,\dots]$. For instance, similarly to this question, I can complete $\{e\}$ into a countable $\mathbb Q$-basis $A$ of $\Bbb Q(e)$. Therefore, $\Bbb Q[e,a_1,a_2, \dots] = \Bbb Q(e)$ is a field.

If $x$ was not algebraic, maybe we could have $y=x^{-1}$. Then I should show that $x+y$ would have no inverse, for instance, but I'm stuck.

Answer 1

This is precisely the statement of Zariski's lemma: if $ K $ is a finitely generated algebra over a field $ k $ and $ K $ is a field, then $ K $ is a finite extension of $ k $. Clearly $ \mathbf Q[x, y] $ is finitely generated over $ \mathbf Q $. The condition that it is a field forces it to be a finite extension, and therefore both $ x $ and $ y $ are algebraic over $ \mathbf Q $.

Answer 2

Yes. Every field extension $L|K$ with $L$ finitely generated as a $K$-algebra is algebraic. This is a consequence of the Noether normalization lemma.