Subfields of a transcendental extension

Question

Is there some nonprime subfield of $\mathbb Q(\pi)$ that is finite dimensional over $\mathbb Q$? Or more generally let $\alpha $ be transcendental over the rationals, is there some subfield $\mathbb L\neq\mathbb Q$ of $\mathbb Q(\alpha)$ that is finite dimensional over $\mathbb Q$?

Answer 1

No. Assume that such a field $L$ exists; then by the primitive element theorem, you can write $L=\mathbf Q(\beta)$ for some $\beta\in L$. But then $\beta\in\mathbf Q(\alpha)$, so there is a (non-zero) rational function $F(T)\in\mathbf Q(T)$ such that $\beta = F(\alpha)$. Now $\beta$ is algebraic over $\mathbf Q$, so there is a non-zero polynomial $G(T)\in\mathbf Q[T]$ such that $G(\beta)=0$, and so the rational function $G\circ F$ vanishes at $\alpha$. Its numerator (a polynomial) does as well, which is absurd if $\alpha$ is transcendental.

Answer 2

No such subfield exists.

$K(\alpha),$ where $K$ is a field and $\alpha$ is transcendental over $K$, is basically a field of rational functions. In other words, one can treat $\alpha$ as an indeterminate (there are simply no non-trivial relations involving $\alpha$, so basically it is an indeterminate).

So let us prove that $K(X),$ the field of rational functions over $K$, has no algebraic elements that are not in $K$. Suppose $\frac{f(X)}{g(X)}$ is a rational function, $f$ and $g$ coprime. Suppose $\frac{f(X)}{g(X)}$ is algebraic over $K$, i.e. there is a relation of the form

$$\sum_{i=0}^n a_i\frac{f^i(X)}{g^i(X)}=0, \;\; a_i \in k,\;\; a_n \neq 0.$$

Denote the degrees $\deg f=k, \; \deg g=l$. To get a non-trivial rational function, we may assume $k>0$ or $l>0$. Without loss of generality assume that $k\geq l$ (if $\frac{f(X)}{g(X)}$ is algebraic, so is $\frac{g(X)}{f(X)}$) - in particular, $k>0$.

Multiplying by $g^n(X)$, we have

$$\sum_{i=0}^n a_if^i(X)g^{n-i}(X)=0.$$

Rewrite this as $$a_n f^n(X)=-g(X)\cdot \left(\sum_{i=0}^{n-1} a_if^i(X)g^{n-i-1}(X)\right)$$.

Denote the polynomial on the RHS $h(X)$. Since $f, g$ were taken coprime and of positive degree, we see that the polynomials $f^n(X), h(X)$ have different factorizations into irreducibles (g(X) divides h(X), but not $f^n(X)$). Thus, this can happen only if $a_n=0$. This is a contradiction, since we assumed (again WLOG) that $a_n \neq 0$.