Consider the field of bivariate rational functions $ \Bbb{C}(x,y) $. Let $$ a=\frac{4x(1-x)(1-2y)}{(1-2xy)^{2}} $$ $$ b=\frac{4y(1-y)(1-2x)}{(1-2xy)^{2}} $$ and $ \tau_{1}, \tau_{2} $ be automorphisms of $ \Bbb{C}(x,y) $ defined by $$ \tau_{1}: (x,y) \mapsto (1-x,y) $$ $$ \tau_{2}: (x,y) \mapsto (x,1-y) $$ Let $$ A=a+\tau_{1}(a)+\tau_{2}(a)+\tau_{1} \circ \tau_{2}(a) $$ and $$ L=\Bbb{C}(a,b) $$ $$ K=\Bbb{C}(\frac{a}{b},A) $$ I proved that $ A $ belongs to $ L $ so of course $ K $ is a subfield of $ L $.
Is it then true that $ [L:K] \leq 2 $?
This question rose while I was studying a certain field. If indeed $ [L:K]\leq 2 $ then I would have successfully computed my desired field but I don't know how to prove that this degree is at most $ 2 $ even more so, I don't know whether it is true or not.
I would appreciate any help. Thank you!
Experimentation in finite fields show that $[K : \Bbb C(x,y)] = 32$, that $\Bbb C(x,y)$ is not Galois over $K$ and that if $M$ is the Galois closure then $[K:M] = 512$ (though I'm not sure I went to fields large enough for this).
Since $[L:\Bbb C(x,y)]$ has order $8$, $[K :L] = 4$ and this is not what you want.
If you have explicit formulas expressing $A$ in terms of $a$ and $b$ you can implement them instead and look at $[K : L]$ directly.
Like I said in answer to the other questions, since $\Bbb C(x,y)$ is Galois over both $K_1$ and $K_2$, their intersection $K_3$ is the set of rational fractions that are invariant by both Galois groups $G_1$ and $G_2$. Those two groups generate a subgroup $G$ of $Aut_{\Bbb C}(\Bbb C(x,y))$ (also called the Cremona group).
Since $G$ is finite of order $32$ (I urge you to check this yourself), $\Bbb C(x,y)$ is Galois over $K_3$ and the group is $G$.
Looking more into $G$ we find that it is generated by five transformations $(x,y) \mapsto (1-x,y), (-\frac x{1-2x},y), (x,1-y), (x,-\frac y{1-2y}), (\frac 1 {2y},\frac 1{2x})$.
Letting $u = 1-2x$ and $v=1-2y$, the first four become
$(u,v) \mapsto (-u,v),(u^{-1},v),(u,-v),(u,v^{-1})$
Those generate a normal subgroup $H$of $G$, and it should be easy to show that if $(U,V) = (u^2+u^{-2},v^2+v^{-2})$ then $\Bbb C(x,y)^H = \Bbb C(u,v)^H = \Bbb C(U,V)$.
Then the Galois correspondance say that $\Bbb C(U,V)$ is a Galois extension of order $2$ of $K_3$, the Galois group being $G/H$ and generated by the fifth transformation $(x,y) \mapsto (\frac 1{2y},\frac 1{2x})$.
This transformation, translated in terms of $u,v$ is $(u,v) \mapsto (-\frac{1+v}{1-v},-\frac{1+u}{1-u})$.
When applying it to $U$ and $V$ the results should be expressible in terms of $U,V$ too :
$\frac{(1+v)^2}{(1-v)^2} + \frac {(1-v)^2}{(1+v)^2} = \frac{(1+v)^4+(1-v)^4}{(1-v^2)^2} = \frac{2+12v^2+2v^4}{1-2v^2+v^4} = \frac{2V+12}{V-2}$
so it is $(U,V) \mapsto (\frac{2V+12}{V-2},\frac{2U+12}{U-2})$
The map $U \mapsto \frac {2U+12}{U-2}$ is a Mobius transform whose fixpoints are $-2$ and $6$, so letting $(p,q) = (\frac{U-6}{U+2}, \frac {V-6}{V+2})$, you have
$\Bbb C(U,V) = \Bbb C(p,q)$, and the transformation becomes $(p,q) \mapsto (-q,-p)$
So this is simply switching $p$ with $-q$. Except with the sign flip, this is the archetypical example of an $S_2$ extension. Now it should be clear that $\Bbb C(p,q) = \Bbb C(p,-q)$ and that $\Bbb C(p,-q)^{G/H} = \Bbb C(p-q,-pq) = \Bbb C(p-q,pq)$
And so $K_3 = \Bbb C(x,y)^G = (\Bbb C(x,y)^H)^{G/H} = \Bbb C(U,V)^{G/H} = \Bbb C(p,-q)^{G/H} = \Bbb C(p-q,pq)$
Also note that there are many ways of going up/down the chain of subgroups/subextensions and of finding generators. Those are not "the" generators of $K_3$, they are merely one generating pair. Different procedures and different decisions will lead to other descriptions of $K_3$ with other pairs of generators.