Algebraic Field Extensions This is exercise $6.34$ from Lang's book:

Give an example of a field $K$ which is of degree $2$ over two distinct subfields $E$ and $F$, respectively, but such that $K$ is not algebraic over $E\cap F$.

So, I've nearly proven the existence of such a field, but there is one part I am struggling with, and if anyone could provide an actual example, I'd greatly appreciate it! My proof goes as follows:
Consider the diagram consisting of $k(s,t)$ at the top, $k(t+t^{-1},s)$ on the lower middle left, and $F$ on the lower middle right, where $F$ is the fixed field of the map $\sigma$ with $\sigma(s)=st$ and $\sigma(t)=t^{-1}$. It's easy to see that $x^2-(t+t^{-1})x+1$ is an irreducible polynomial over the left field, hence it is of degree $2$. Now, I've made the claim that $s\notin F$ since $\sigma^n(s)=st^n$ is an infinite chain, but I'm not sure why this works, nor am I sure why $F$ is of degree $2$. After these two steps, then it's clear that the intersection can't contain $s$, hence $k(s,t)$ can't be algebraic over $k(t+t^{-1},s)\cap F$. Can anyone help me fill in these two gaps?
 A: A simpler example: $K=\mathbf{Q}(x)$ has order-$2$ automorphisms $\sigma:x\mapsto -x$ and $\tau:x\mapsto 1-x$ for which $\sigma\tau$ has infinite order (since it maps $x\mapsto x+1$).  I leave it to you to determine $E$ and $F$.
A: I would fill in the details along the following lines. Essentially you have introduced two automorphisms of the field $K=k(s,t)$. The first automorphism
$\tau$ is defined by the formulas $\tau(t)=t^{-1}$, $\tau(s)=s$. The other automorphism $\sigma$ is similarly defined by setting $\sigma(s)=st$, $\sigma(t)=t^{-1}$.
It is easy to see that both $\tau$ and $\sigma$ are of order two. Therefore Galois theory tells us their respective fixed fields $E$ and $F$ have the property 
$$[K:E]=\operatorname{ord}(\tau)=2=\operatorname{ord}(\sigma)=[K:F].$$
As you have observed, $E=k(s,t+t^{-1})$. Unless I made a mistake, we also get that
$F=k(s+st,t+t^{-1})$, as $t$ is algebraic of degree two over $k(s+st,t+t^{-1})$
and $K=k(s+st,t+t^{-1})(t)$.
Anyway, the remaining task is to show that $[K:E\cap F]=\infty.$ Anything in $E\cap F$ is fixed under both $\sigma$ and $\tau$. Conversely if an element of $K$ is fixed by the group of automorphisms $G=\langle\sigma,\tau\rangle$, then
it is in the intersection $E\cap F$. If the extension $[K:E\cap F]$ were finite, then by basic Galois theory, the group $G$ would have to be finite. But this is not the case, as the automorphism $\alpha=\sigma\circ\tau$ maps the generators like $\alpha(s)=st$, $\alpha(t)=t$, and hence has infinite order. This settles the claim.
I would guess that the fixed field of $G$ is actually $E\cap F=k(t+t^{-1})$. I haven't checked that this is the case though.
