Let $\Omega$ be an algebraically closed field with characteristic $p>0$, a subfield $K\subset \Omega$ and $L:=K(\tau^p, \eta^p)$. If $\tau$ is transcendental over $K$ and $\eta$ is transcendental over $K(\tau)$, prove that:
$1)$ $[K(\tau, \eta):L]=p^2$;
$2)$ $[L(\zeta):L]=p$ for any $\zeta\in K(\tau,\eta)-L$;
$3)$ $K(\tau, \eta)|L$ has no primitive element.
Here's where I'm at:
$1)$ It is clear that $\eta, ..., \eta^{p-1}\in K(\tau^p, \eta)-L$ and because $\eta$ is transcendent over $K(\tau)$, we have that $1,\eta, ..., \eta^{p-1}$ are $L$-linear independent, so $[K(\tau^p, \eta):L]=p$. Analogously, $\tau, ..., \tau^{p-1}\in K(\tau, \eta)-K(\tau^p, \eta)$ and by transcendecy of $\eta$ over $K(\tau)$, we get linear independency, so $[K(\tau, \eta):L]=p^2$.
$3)$ If there existed $\alpha\in K(\tau, \eta)$ with $K(\tau, \eta)=L(\alpha)$, we would have from $2)$ that $[L(\alpha):L]=p$ (absurd, because $[K(\tau, \eta):L]=p^2$).
$2)$ this is where I'm stuck, and I don't know what the characteristic $p>0$ has to do with anything.
Any tips? Thanks!
Let $\zeta \in K(\tau,\eta) - L$. Then there exists an $f \in L(X,Y)$ such that $\zeta = f(\tau,\eta)$. We obtain: $$ \zeta^p = f(\tau,\eta)^p = f(\tau^p,\eta^p) \in L $$ by applying Frobenius. This shows $[L(\zeta):L] \leq p$. Now $L(\zeta)/L$ is a proper field extension, so $[L(\zeta):L] > 1$ and since $[L(\zeta):L]\mid p^2$ by 1) we have $[L(\zeta):L] = p$.