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.

  • 2
    $\begingroup$ 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 $q\neq p$. $\endgroup$ Jun 13, 2016 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, 2016 at 6:50

1 Answer 1


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

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 13, 2016 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, 2016 at 7:05
  • 1
    $\begingroup$ @Larara $zx+y$ and $x+z^{-1}y$ generate the same field because $z\in K$. $\endgroup$ Jun 13, 2016 at 7:08
  • $\begingroup$ Why can we choose infinitely many elements $z\in K$ if $K$ is finite? $\endgroup$
    – yung_Pabs
    Apr 27, 2017 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 $K$. $\endgroup$ Apr 27, 2017 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.