Rings that are generated as an Algebra over a field by an arbitrary amount of algebraic elements In an introduction course in algebra, you learn, that if you take a field $F$ and an element $x$, which is algebraic over $F $, then the smallest generated Ring by $F$ and $x$, mostly called $F[x]$ is already a field. By induction you can see that this also holds true for finitely many elements $x_1, x_2,\cdots ,x_n$, But what if I want to adjoint an arbitrary amount of elements to $F$. Is it still true?
If yes, do you prove it by transfinite induction, or is there a more obvious argument that I'm missing?

Let $\Omega$ be a field containing $F$ and let $S \subseteq \Omega$ such that all the elements of $S$ are algebraic over $F$. Let $K=F[S]$, the smallest subring of $\Omega$ containing $F$ and $S$. Take a non-zero $\alpha \in K$. Then $\alpha$ is a polynomial in finitely many elements $s_1, s_2, \dots, s_n \in S$. This means that $\alpha \in F[s_1, s_2, \dots, s_n]$, which is a field because the $s_i$ are algebraic. So, $\alpha$ has an inverse in $F[s_1, s_2, \dots, s_n]\subseteq K$.
