Counterexample to primitive element theorem Let $k$ be a field of characteristic $p$ and let $\alpha,\beta$ be algebraically independent over $k$. Prove the following:
(a) $k(\alpha,\beta)$ has degree $p^2$ over $k(\alpha^p,\beta^p)$.
(b) There exists  infinitely many extensions between $k(\alpha,\beta)$ and $k(\alpha^p,\beta^p)$.
Remark: This problem appeared couple of times in MSE but anyway let me ask the following questions because they are quite subtle and I did not find these moment in previous topics. So please do not duplicate this topic!
Proof:
(a) Consider an extension $k(\alpha,\beta^p)$ of $k(\alpha^p,\beta^p)$. We'll show that $[k(\alpha,\beta^p):k(\alpha^p,\beta^p)]=p$. Let's take polynomial $p(X)=X^p-\alpha^p\in k(\alpha^p,\beta^p)[X]$ then we see that $p(X)=(X-\alpha)^p$ and $\alpha$ its root. So let $g(X)=\text{Irr}(\alpha,k(\alpha^p,\beta^p),X)$ - irreducible polynomial of $\alpha$ over $k(\alpha^p,\beta^p)$. 
Then $g(X)\mid (X-\alpha)^p$ so $g(X)=(X-\alpha)^i$ with $1\leq i \leq p$.
Suppose that $1\leq i <p$ and we know that $g(X)=(X-\alpha)^i\in k(\alpha^p,\beta^p)[X]$  then constant term $\alpha^i\in k(\alpha^p,\beta^p)$ but $\text{gcd}(i,p)=1$ then $\exists \lambda,\mu$ such that $\lambda i+\mu p=1$ and after easy manipulations we get that $\alpha \in k(\alpha^p,\beta^p)$. 
Thus $\alpha=P(\alpha^p,\beta^p)$ for some $P(x,y)\in k(x,y)$, i.e. $P(x,y)$ could be rational function.
Therefore, $(\alpha,\beta)$ is a root of equation $P(x^p,y^p)-x=0$ where $P(x,y)=\dfrac{A(x,y)}{B(x,y)}$. Finally, $(\alpha,\beta)$ is a root of equation $A(x^p,y^p)-xB(x^p,y^p)=0$. But I guess this is contradiction since $\alpha,\beta$ are algebraically independent and $A(x^p,y^p)-xB(x^p,y^p)$ is a non-zero polynomial (can anyone explain it? I was not able to understand it. Why this polynomial is non-zero?). 
This contradiction shows that $g(X)=(X-\alpha)^p$ and eventually $[k(\alpha,\beta^p):k(\alpha^p,\beta^p)]=p$.
The same reasoning I guess can be applied to $k(\alpha,\beta)$ over $k(\alpha,\beta^p)$ and this degree is also $p$. Using tower's theorem it gives us that the total degree is $p^2$.
(b) Let's show that there are infinitely many intermediate fields between $k(\alpha,\beta)$ and $k(\alpha^p,\beta^p)$. 
Let $L=k(\alpha^p,\beta^p)$ and $E=L(\alpha,\beta)=k(\alpha,\beta)$. For any $\lambda \in L$ consider intermediate field $L(\alpha+\lambda\beta)$.
There is one key moment: if $L$ is infinite then we can show that for any $\lambda\neq \mu$ we have $L(\alpha+\lambda\beta)\neq L(\alpha+\mu\beta)$. It's not so difficult. Indeed, if they are equal then we can show that $L(\alpha,\beta)=L(\alpha+\lambda\beta)$ and $p^2=[L(\alpha+\lambda\beta):L]\leq p$ which is contradiction.
Question: But what if $L$ is finite field? Note that in our case $\alpha, \beta$ are not indeterminates. They are just some fixed algebraically independent element over $k$.
I have thought on this many hours but did not come up with something good.
 A: Consider what algebraic independence means; because $\alpha$ and $\beta$ are algebraically independent over $k$:


*

*It follows that $\alpha$ is transcendental over $k(\beta)$ and $\beta$ is transcendental over $k(\alpha)$.

*In particular $\alpha$ is transcendental over $k(\beta^p)$ and $\beta$ is transcendental over $k(\alpha^p)$.

*In particular $\alpha$ and $\beta$ are both transcendental over $k$.


It follows immediately that $L:=k(\alpha^p,\beta^p)$ is an infinite field, solving your second question.
For your first question, I haven't (yet) been able to untangle everything after your sentence

...and after easy manipulations we get that $\alpha \in k(\alpha^p,\beta^p)$.

But because $\alpha$ is transcendental over $k(\beta^p)$, it follows that $\alpha\notin L=k(\alpha^p,\beta^p)$, and you are done.
Also, the algebra is slightly simpler when instead of considering the constant term of $(X-\alpha)^i$, you consider the coefficient of $X^{i-1}$, which is $-i\alpha$. If $p\nmid i$ then $-i$ is a unit in $k$, so if $(X-\alpha)^i\in L[X]$ then also $-i\alpha\in L$ and hence $\alpha\in L$, leading to a contradiction in the same way as above. This shows that $g(X)=(X-\alpha)^p$.
