Let $\mathbb{F}_p$ be the finite field with $p$ elements, where $p$ is a prime number. Let $x$ and $y$ be transcendental and algebraically independent over $\mathbb{F}_p$. The extension $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite of degree $p^2$. To see this consider the polynomials $T^p-x^p$ and $T^p-y^p\in\mathbb{F}_p(x^p,y^p)[T]$.
So, if $K$ is any (proper) intermediate field of this extension, we must have $[K:\mathbb{F}_p(x^p,y^p)]=p$. For each prime $q\neq p$, consider the polynomial $T^p-x^{qp}\in\mathbb{F}_p(x^p,y^p)[T]$. It is an irreducible polynomial of degree $p$ with $x^q$ as root. So the extension $\mathbb{F}_p(x^p,y^p)(x^q)/\mathbb{F}_p(x^p,y^p)$ has degree $p$ and hence, $\mathbb{F}_p(x^p,y^p)(x^q)$ is a proper intermediate field of the original extension. Note that we can do the same with $y$ instead of $x$. Since there are infinite primes $\neq p$, there will be infinite (proper) intermediate fields of the original extension. These fields are distinct, because they are obtained by adjoining a root of an irreducible polynomial, and their irreducible polynomials are distinct.
This shows that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is finite but not simple, hence, it is not separable.
My question is: are there other intermediate fields other than that pointed out previously?
I first tried to show that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by supposing the existence of a primitive element then trying to get a contradiction, specifically, that $x$ and $y$ are algebraically dependent. The problem is, they are algebraically dependent over $\mathbb{F}_p(x^p,y^p)$, because they vanish this (non-zero) polynomial $T_1^p-x^p+T_2^p-y^p\in\mathbb{F}_p(x^p,y^p)[T_1,T_2]$. If you know someway of proving that $\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)$ is not simple by contradiction, please answer it too.
Best Answer
To see why there are infinitely many intermediate fields between $K=\Bbb{F}_p(x^p,y^p)$ and $L=\Bbb{F}_p(x,y)$ you can do the following.
Let $z$ be any element of $K$. Consider $w=x+zy$. We see that $w^p=x^p+z^py^p\in K$, so $K(w)$ is a degree $p$ extension of $K$. If $w'=x+z'y$ for some $z'\in K$, $z'\neq z$, then we easily see that $K(w,w')=L$. Therefore different choices of $z$ lead to different intermediate fields. The element $z$ can be chosen in infinitely many ways, and the claim follows.
IMHO the standard way of showing that $L/K$ is not simple is to observe that any element $u\in L$ has the property $u^p\in K$. Therefore $[K(u):K]\le p<[L:K]$.