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Let $K=F_p(X,Y)$, where $F_p$ is a finite field of characteristic $p$, and $F=F_p(X^p,Y^p)$. I have been given the following problem:

Determine the degree of extension $[K:F]$.

My experience with problems regarding the degree of field extensions is limited to the case where the field extension is generated by a finite number of elements, such as the prototypical example $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. In general, how does one proceed on a problem such as this, where one does not have a field extension generated by the adjunction of elements to a base field?

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Consider the intermediate extension $F_p(X^p, Y)$. – user18119 Apr 18 '12 at 16:21
In this context, the base field is $F$, not $F_p$. – Lubin Apr 18 '12 at 16:47
In fact, there are infinitely many subextensions of degree $p$. – i707107 Jul 31 '15 at 23:27
up vote 8 down vote accepted

Ok, so the basic idea whenever you have an extension which is the appendation of multiple elements is to take things one at a time. Namely, let's first find $n$ and $m$ in the following diagram

$$\begin{array}{c}F_p(X,Y)\\ \vert n\\ F_p(X^p,Y)\\ \vert m\\ F_p(X^p,Y^p)\end{array}$$

The basic idea now is that going from $F_p(X^p,Y^p)$ to $F_p(X^p,Y)$ is just appending one element since $F_p(X^p,Y)=F_p(X^p,Y^p)(Y)$. So, the question is what is the minimal polynomial of $Y$ over $F_p(X^p,Y^p)$? Well, we clearly have a candidate since $f(t)=t^p-Y^p\in F_p(X^p,Y^p)[t]$ and $f(Y)=0$. The question then is why $f(t)$ is irreducible in $F_p(X^p,Y^p)[t]$. The reason why would expect it to be is that any factorization should be one involving the fact that $t^p-Y^p=(t-Y)^p$--but of course this won't work since $t-Y\notin F_p(X^p,Y^p)[t]$. To make this more concrete suppose that $f(t)=p(t)q(t)\in F_p(X^p,Y^p)[t]$ is a factorization into irreducibles. Note then that we have factorized $f(t)$ in $F_p(X,Y)[t]$ and since $F_p(X,Y)$ is a field we know that $F_p(X,Y)[t]$ is a UFD and so use this to conclude that, up to constants, $p(t),q(t)$ are powers of $(t-Y)^k$. Conclude then that (from the primality of $p$ which tells you that $Y^k$ is not in $F_p(X^p,Y^p)$ for $k<p$) the one factor must be $(t-Y)^p$ and so the other is a unit. Conclude that $f$ is irreducible and so the minimal polynomial of $Y$ over $F_p(X^p,Y^p)$. Conclude that $m=p$.

Now you try for $n$ and then use the multiplicative property of field towers.

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+1 For this concise answer. – user38268 Jan 26 '13 at 9:46

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