Rings that are generated as an Algebra over a field by an arbitrary amount of algebraic elements

Question

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?

Answer 1

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 $F[s_1, s_2, \dots, s_n]\subseteq K$.

Answer 2

Every $a$ element of an $F$-algebra $A$ generated by an infinite set of algebraic elements $x_i$ is already in the $F$-algebra $A'$ generated by a finite subset of the $x_i$; this is what the $F$-algebra generated by an infinite set means. But you know that $A'$ is a field, so that if $a\neq0$, it has an inverse in $A'$, and a fortiori in $A$.

Answer 3

If $L$ is a field, $K \subseteq L$ a subfield and $(x_i)_{i \in I}$ a familiy of elements of $L$ which are algebric over $K$, then we have $$ K[x_i \mid i \in I] = \mathrm{span}_K \{ x_{i_1} \dotsm x_{i_n} \mid n \in \mathbb{N}, i_1, \dotsc, i_n \in I\} = \bigcup_{\substack{J \subseteq I \\ |J| < \infty}} K[x_j \mid j \in J]. $$ So $K[x_i \mid i \in I]$ is the union of subfields of $L$, such that every two subfields are contained in another one. Because every finite subset of $K[x_i \mid i \in I]$ is therefore contained is a subfield of $L$ it follows that $K[x_i \mid i \in I]$ itself is also a field.