# 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.

• Yes $M$ exists, it is just the set of algebraic elements in $F(S)$. And if $S$ is finite then $M$ is finite dimensional over $F$. – Gregory Grant Apr 19 '15 at 16:14
• @GregoryGrant: Why? I know that the algebraic elements of $F(S)$ over $F$ form a field, but why is it finite dimensional over $F$? It is easy if $S$ is a set of algebraic elements over $F$, since $[F(S):F]$ is finite, but this does not give anything if $S$ has transcendental elements over $F$. – user21820 Apr 20 '15 at 1:33
• It's true in general that if $F=K(u_1,\dots,u_n)$ is an arbitrary finitely generated field extension of $K$ and $E$ is any intermediate field, then $E$ is a finitely generated extension of $K$. This is exercise 5 of Section VI.1 of Hungerford's Algebra. It follows that the field of algebraic elements is finitely generated over $K$ (as a field) and is therefore finite dimensional over $K$. – Gregory Grant Apr 20 '15 at 3:02
• @GregoryGrant: Would you mind posting a brief sketch of the proof in an answer? As I said I don't think I know enough to figure it all out myself. – user21820 Apr 20 '15 at 13:00
• I found this same question a bunch of times here. Take a look at these pages and if you still don't understand the proof let me know and I'll try to help. math.stackexchange.com/questions/125815/… math.stackexchange.com/questions/34424/… math.stackexchange.com/questions/138094/… math.stackexchange.com/questions/125815/… – Gregory Grant Apr 20 '15 at 15:35

## 1 Answer

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 $F(B)$.

• I'm having trouble following this proof. Which of the links did you use to base it on? – Gregory Grant Apr 20 '15 at 19:51
• @GregoryGrant: There were too many gaps I couldn't fill in some of the links, so I looked mostly at Pete L. Clark's pdf notes (Section 12.4). It was too dense so I distilled out the main idea. Which part of my proof does not make sense? – user21820 Apr 21 '15 at 5:08
• @GregoryGrant: I've added more details, so if there are mistakes could you point them out? – user21820 Apr 21 '15 at 6:26
• @GregoryGrant: So do you see any mistake in my proof? – user21820 Apr 23 '15 at 7:05