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.

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    $\begingroup$ Some confusion in your argument. Because $\gcd(q,p)=1$ there exists integers $u,v$ such that $uq+vp=1$. This implies that $$x=x^{uq+vp}=(x^q)^u\cdot(x^p)^v$$ is contained in all the fields $\Bbb{F}_p(x^p,y^p)(x^q)$. In fact, this argument shows that $$\Bbb{F}_p(x^p,y^p)(x^q)=\Bbb{F}_p(x,y^p)$$ for all primes $q\neq p$. $\endgroup$ – Jyrki Lahtonen Jun 13 '16 at 6:41
  • $\begingroup$ @JyrkiLahtonen Thanks for pointing this! I'll edit the question. So, essentially we have two proper intermediate fields $\mathbb{F}_p(x,y^p)$ and $\mathbb{F}_p(x^p,y)$. $\endgroup$ – Larara Jun 13 '16 at 6:50

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 $[K(u):K]\le p<[L:K]$.

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    $\begingroup$ The infinite collection of intermediate fields above is not comprehensive. We also have fields like $K(x+y^2)$ when $p>2$, and $K(xy)$. In the case $p=2$ it is not too difficult to describe all the intermediate fields, because the elements $z_1 x+z_2y+z_3xy$ and $z_1' x+z_2'y+z_3'xy$, $z_i,z_i'\in K$ generate the same field iff the lists of coordinates are scalar multiples of each other. $\endgroup$ – Jyrki Lahtonen Jun 13 '16 at 7:04
  • $\begingroup$ What if we take $w=zx+y$ instead. Will $K(w)\neq K(w^{\prime})$, for every $w^{\prime}=x+zy$, $z\in K$? $\endgroup$ – Larara Jun 13 '16 at 7:05
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    $\begingroup$ @Larara $zx+y$ and $x+z^{-1}y$ generate the same field because $z\in K$. $\endgroup$ – Jyrki Lahtonen Jun 13 '16 at 7:08
  • $\begingroup$ Why can we choose infinitely many elements $z\in K$ if $K$ is finite? $\endgroup$ – yung_Pabs Apr 27 '17 at 21:27
  • $\begingroup$ @yung_Pabs $K$ is infinite. For example all the monomials $x^{mp}y^{np}$, $m,n$ arbitrary integers, are distinct elements of $K$. $\endgroup$ – Jyrki Lahtonen Apr 27 '17 at 21:36

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