Is every subextension of a simple field extension simple itself? Let $K \subseteq M \subseteq L$ be three fields such that $L=K(\alpha)$ with $\alpha\in \overline{K}$, i.e. $L$ is a simple extension of $K$.
Then (?) $M$ is a simple extension of $K$.
So I've spent some time thinking about this question. If $L$ is a finite extension of $K$, then of course the answer is yes - there are only finitely many intermediate fields between $K$ and $L$, therefore there must be finitely many intermediate fields between $K$ and $M$ and thus it is simple.
If $L$ is an infinite algebraic extension of $K$, then also this statement holds - because then $L$ cannot be a simple extension of $K$.
So this question becomes interesting for transcendental extensions: for example, let $M$ be some field s.t. $ \mathbb{Q} \subseteq M \subseteq \mathbb{Q}(\pi)$. Then (?) $M$ is a simple extension of $\mathbb{Q}$.
I have no idea how to tackle this in the transcendental case, or if it is generally correct. Any help is appreciated.
 A: A proof using some algebraic geometry with main tools: *function fields correspond to smooth projective curves *genus is zero iff the function field is isomorphic to the field of rational functions *Riemann-Hurwitz formula.
Let $K$ be a field and $K(\alpha)$ the field of rational functions in $\alpha$. Let $K\subsetneq M\subseteq K(\alpha)$ be a subextension. Then $K(\alpha)/M$ is algebraic (indeed, $M$ contains a non-constant rational function $u=f(\alpha)/g(\alpha)$ and so $\alpha$ satisfies the polynomial  $f(X) - g(X)u=0$ which has coefficients on $M$).
We can view $K(\alpha)$ as the function field of the $K$-projective line $\mathbb{P}^1_{K}$, $M$ as the function field of some smooth projective $K$-curve $C$. Then the extension $K(\alpha)/M$  corresponds to a branched covering $\phi\colon \mathbb{P}^1_K\to C$. 
Applying Riemann-Hurwitz formula with $n=\deg\phi=[K(\alpha):M]$, with $g=$ genus of $C$, and noting that the genus of $\mathbb{P}^1$ is $0$, we get that 

$ - 2 = n(2g-2) + \deg R$

where $R$ is the ramification divisor (the only relevant information on $R$ for us is that $\deg R\geq 0$). Since the left-hand-side is negative, we must have a negative summand in the right-hand-side of the equation, and this can happen only if $g=0$. Thus $C$ is of genus zero, and so $M=K(\beta)$ for some $\beta\in M$.
