Intermediate fields between $\mathbb{Z}_2 (\sqrt{x},\sqrt{y})$ and $\mathbb{Z}_2 (x,y)$ Let $K=\mathbb{Z}_2 (x,y)$, where $x,y$ are independent, and $L$ be a splitting field extension of $(X^2 - x) (X^2 - y)$, then $[L:K] = 4$ and $L = K(\sqrt{x},\sqrt{y})$ where $\sqrt{x},\sqrt{y}$ are roots of $X^2-x$, $X^2 - y$ respectively. What are the subextensions of $L:K$?
I know all elements in $L$ square to something in $K$, so all the intermediate fields are $K(\sqrt{k})$ for some $k\in K$, but some of them are the same, say $K(\sqrt{x/y}) = K(\sqrt{xy})$... 
Note: $\mathbb{Z}_2$ means $\mathbb{Z} / 2\mathbb{Z}$

I don't like writing square roots, so let's just pick $K = \Bbb F_2(X^2,Y^2)$ and $L = F_2(X,Y)$.
Since $K \subset L$ is of degree $4$, any intermediate field is of degree $2$. As you said, every element of $L$ square to something in $K$ (in fact, the map $x \in L \mapsto x^2 \in K$ is an isomorphism of fields), so those fields are the $K(a)$ for $a \in L \setminus K$ .
The intermediate fields are the $2$-dimensional $K$-vector spaces containing $K$.
One direction is obvious, and if $F = \langle 1,a \rangle$ is such a vector space, then $F$ is stable by multiplication (because $1 \cdot a = a \in F$, and $a \cdot a = a^2 \in K \subset F$), so it is the field $K(a)$. 
Those correspond to $1$-dimensional vector spaces in the ($3$-dimensional) quotient $L/K$, and so to elements of $(L/K)^* / K^* \simeq \Bbb P^2(K)$.
Given an element $[x:y:z] \in \Bbb P^2(K)$ we can associate $a = xX+yY+zXY$ and the intermediate field $K(a)$. 
Anyway this means that there are infinitely many (because $K$ is an infinite field) intermediate fields.
