Let $F$ be a field with characteristic $p>0$ (where $p$ is a prime). Let $F^{'}=F(X,Y)$ be the field of fractions of the polynomial ring $F[X,Y]$. Let $K:=F(X^{1/p},Y^{1/p})$. Show that there is no $\beta \in K$ such that $K=F(\beta)$.
How do I prove this claim? Any help will be appreciated.
The idea is to show that the extension $K/F'$ has degree $p^2$, and any simple extension $F'(\alpha)$ has degree $p$, so $K/F'$ is not simple. To show that $K/F'$ has degree $p^2$, note that $$\left[K : F'\right] = \left[K : F\left(X^{1/p}, Y\right)\right] \cdot \left[F\left(X^{1/p}, Y\right) : F'\right]$$See if you can show $\left[K : F\left(X^{1/p}, Y\right)\right] = p$ and $\left[F\left(X^{1/p}, Y\right) : F'\right] = p$. For the first, take the minimal polynomial $f(t) = t^{p} - Y = \left(t - Y^{1/p}\right)^{p}$. You can verify this is the minimal polynomial since $Y$ is irreducible over $F'$, so by Eisenstein's criterion, $f(t)$ is irreducible. Since it is degree $p$, the extension is degree $p$ as well. The argument for the other one is similar - I'll leave that to you. Once you've done that, you need to show that for any $\alpha \in K$, the degree $[F'(\alpha) : F] \leq p$.
You can do this by constructing the minimal polynomial explicitly. Consider the polynomial $g(t) = t^{p} - \alpha^{p}$. Notice that $\alpha^{p} \in F'$, since we can write $$\alpha = a_0 + a_1X^{1/p} + a_2Y^{1/p} + a_3X^{1/p}Y^{1/p} + ... + a_{nm}X^{n/p}Y^{m/p}$$ with $a_i \in F$. Because $F'$ is a field of characteristic $p$, raising the whole expressionto the power of $p$ is equivalent to raising each term to the power of $p$ individually, so $\alpha^p = a_0^p + a_1^pX + a_2^pY + a_3^pXY + ... + a_{nm}^pX^{n}Y^{m} \in F'$. So, $g(t) = t^{p} - \alpha^{p} \in F'[t]$. Since this has $\alpha$ as a root, the minimal polynomial of $\alpha$, $p(t)$, must divide $g(t)$. Therefore, $\textrm{deg}(p(t))\leq \textrm{deg}(g(t)) = p$, so we get $[F'(\alpha) : F] \leq p$, meaning $K/F'$ is not simple, as desired.
