Degree of field extension $F_{p}(X^p,Y^p) \subset F_{p}(X,Y) $ What is the degree of the following field extension?
$$ K=F_{p}(X^p,Y^p) \subset  F_{p}(X,Y)=L  $$
Because of $ F_{p}(X^p,Y^p) \subset_{(1)} F_{p}(X,Y^p) \subset_{(2)}  F_{p}(X,Y)  $
and $(1)$ the minimal polynomial $T^p-X^p$ and $(2)$ the polynomial $T^p - Y^p$ it follows $ [L:K] \leq p^2$
So why is $T^p-X^p$ irreducible and how can I proof that $[L:K] = p^2$?
 A: Hint: You could take a look at the basis of the extension. Try to find it!
If you can't find the solution, I can elaborate and provide details. 
You can also argue that due to the irreducibility of $T^p - X^p$ over $K$ and the irreducibility of $T^p - Y^p$ over $K$: $$ [L:\mathbb{F}_p(X, Y^p)] \leq [K(X):K] = [K(Y):K] = p$$.
Edit: I used the Eisenstein criterium but you can argue even easier: $ f= T^p - X^p = (T-X)^p $ and if $ f = g \cdot h $, $ g = (T-X)^k $ and analogous for $h$. But as $X^k \notin K$ for $k \in \mathbb{N} $: $ g, h $ must be units or equal to $f$.
Of course it is not wrong to state that $ T^p - X^p $ is irreducible over $ \mathbb{F}_p[X^p,Y^p] $ according to the Eisenstein criterion since $X^p$ is prime. The irreducibility over $ K $ follows with the Gauss Lemma for polynoms.
Now to the basis:
I wrote this proof according to my memory and with help of the book "Übungen zur Algebra" by Clemens Fuchs and Gisbert Wüstholz.
The first step is to show that $ \gamma^p \in K = \mathbb{F}_p(X^p, Y^p) $ for arbitrary $ \gamma \in L = \mathbb{F}_p(X,Y) $.
Let $ \gamma = \frac{\sum a_{ij} X^i Y^j}{\sum b_{kl} X^kY^l} $ with coefficients $ a_{ij}, b_{kl} \in \mathbb{F}_p $. Since $L$ has characteristic $p$, you can state that:
$$ \gamma^p = \frac{(\sum a_{ij} X^i Y^j)^p}{(\sum b_{kl} X^kY^l)^p} = \frac{\sum a_{ij}^p (X^p)^i (Y^p)^j}{\sum b_{kl}^p (X^p)^k (Y^p)^l} \in K $$
The next step is to proof that $ L = K[X,Y] $. Obviously is $ K[X,Y] $ a subring of $ L $, so we only have to show $ L \subseteq K[X,Y] $. Every $ a \in L $ has the structure: $ a = \frac f g $ for $ f,g \in \mathbb{F}_p[X,Y]$. We can write:
$$ a = \frac f g = fg^{p-1}(\frac 1 g)^p $$
Therefore: $ a \in K[X,Y] $ because $ f,g \in \mathbb{F}_p[X,Y] \subseteq K[X,Y] $ and $ (\frac 1 g)^p \in L $.
We now show: $ P = \{ X^i Y^j \}_{0 \leq i,j \leq p-1}$ is a $K$-basis for $L$.
We have shown that every element in $ \mathbb{F}_p(X,Y) $ can be written as a linear combination of monomials $ X^i Y^j $. Since $X^p,Y^p \in K $, all these monomials are $K$-multiples of elements in $P$. We know now: $P$ is a generating system for $ L $.
Let
$$ \sum\limits_{0 \leq i,j \leq p-1} f_{ij}X^iY^j = 0 $$
for $ f_{ij} \in K $. After multiplying with a common divisor we are allowed to assume that all $ f_{ij} \in \mathbb{F}_p[X^p,Y^p] $ and we can write them as:
$$ f_{ij} = \sum\limits_{k,l \geq 0} a_{ijkl} X^{pk}Y^{pl} $$
with coefficients $ a_{ijkl} \in \mathbb{F}_p $.
All together:
$$ \sum\limits_{ 0 \leq i,j \leq p-1 \; ; \; k,l \geq 0 } a_{ijkl} X^{pk+i}Y^{pl+j} = 0 $$
After all this we can now compare the coefficients and: $ a_{ijkl} = 0 $ for all $ i,j,k,l $ which means $ f_{ij} = 0 $ for all $ 0 \leq i,j \leq p - 1 $.
That's everything we need to know to say that $ P $ is linearly independent.
And therefore: $$ [L : K] = p^2 $$.
Note: There is a much more elegant way to do this since we now know much more that just the degree of the field extension. Maybe I find the time to write it here.
