Algebraic subfield of transcendental extension I was recently thinking about whether it is possible to generate an infinite dimensional algebraic extension over a base field using just finitely many transcendental elements. Specifically, given a field $K$ and subfield $F$ and a finite set $S \subseteq K$, is it true that there is a finite extension $M$ over $F$ such that $M$ contains every algebraic element of $F(S)$ over $F$? If so, is there anything we can say about $[M:F]$? If not, what is a counter-example? And does the answer change if $F = \mathbb{Q}$ and $K = \mathbb{R}$?
I know a few things from an introductory Galois theory course, but nothing more, and hence have no idea how to approach this question. All I know is that adjoining one single transcendental element certainly gives no new algebraic elements, but can't even see what happens when there are two. For instance, given any algebraic $a$ over $\mathbb{Q}$ it is easy to make transcendental $x,y \in \mathbb{R}$ over $\mathbb{Q}$ such that $x-y = a$, but is it possible to get 'more' out of two transcendental elements?
This question is almost purely out of curiosity. I was wondering how much one can obtain from adjoining finitely many transcendental elements, since adjoining finitely many algebraic elements in contrast simply yields a finite extension.
 A: Based on the link that Gregory Grant gave, here is my solution.
Theorem
Take any field extension $K/F$ and finite set $S \subseteq K$.
Let $M = (F(S)/F)^{alg} = \{ r : r \in F(S) \land \text{$r$ is algebraic over $F$} \}$.
Then $M/F$ is finite.
Proof
Let $B \subseteq S$ be a transcendence basis for $F(S)$ over $F$.
Then $F(S)/F(B)$ is algebraic [by definition of $B$] and finitely generated [since $S$ is finite].
Thus $[F(S):F(B)] < \infty$ [by Tower law].
Given any field $M$ such that $F(S)/M/F$:
  If $[M:F] > [F(S):F(B)]$:
    If $[M:F] = \infty$:
      No finite chain of finite extensions of $F$ can contain $M$.
      Thus there is an infinite chain of nontrivial finite extensions of $F$ within $M$.
      Thus there are extensions of $F$ within $M$ of arbitrarily large finite degree.
    Let field $N$ be such that $M/N/F$ and $\infty > [N:F] > [F(S):F(B)]$.
    Then $F(S)/N(B)/F(B)$ and $[F(S):F(B)] \ge [N(B):F(B)] = [N:F]$ [by below lemma].
    Contradiction.
  Therefore $[M:F] \le [F(S):F(B)]$.
Lemma
[Here I define rational functions as the ratios of two polynomials with nonzero denominator.]
Given any field tower $K/N/F$ with finite $N/F$ and algebraically independent $B \subseteq K$ over $F$:
  $B$ is algebraically independent over $N$ otherwise:
    Let $b \in B$ and $B' = B \setminus \{b\}$ and $p$ be a nonzero polynomial over $N(B')$ such that $p(b) = 0$.
    Then $N(B)/F(B')$ is finite because $N(B)/N(B')$ and $N(B')/F(B')$ are both finite.
    Thus $F(B)/F(B')$ is finite because $F(B) \subseteq N(B)$, and hence $b$ is algebraic over $F(B')$.
    Thus $B$ is not algebraically independent over $F$, a contradiction.
  Thus $N(B)$ is isomorphic to the rational functions $R$ in $\#(B)$ variables over $N$ because:
    For any rational function $r = \frac{p}{q} \in R$:
      $q(B) \ne 0$ because $B$ is algebraically independent over $N$.
      Then $r(B) = \frac{p(B)}{q(B)} \in N(B)$.
    Let $φ : R \to N(B)$ be the evaluation map $r \mapsto r(B)$, which is a homomorphism.
    Then $φ$ is injective because $B$ is algebraically independent over $N$.
    Also $φ$ is surjective by definition of $N(B)$.
  Let $X = (x_k)$ be a basis for $N/F$.
  For any coefficients $(a_k)$ from $F(B)$ such that $\sum_k a_k x_k = 0$:
    Let polynomials $(p_k),(q_k)$ over $F$ such that $a_k = \frac{p_k(B)}{q_k(B)}$.
    Then $\sum_k ( p_k(B) \prod_{m \ne k} q_m(B) ) x_k = 0$.
    Thus $\sum_k ( p_k \prod_{m \ne k} q_m ) x_k$ is the zero polynomial over $N$ [by the isomorphism].
    Thus each coefficient of $p_k \prod_{m \ne k} q_m$ is $0$ because $X$ is independent over $F$.
    Thus $a_k = 0$.
  Therefore $X$ is independent over $F(B)$.
