Generating element of local ring in a purely inseparable extension Let $F$ be an algebraic function field (in one variable) over the finite field $\mathbb{F}_{p}$. Let $F'$ be a purely inseparable extension of $F$ of degree $p$. In particular, $F'$ is a simple extension of $F$.
Let $\mathcal{P}$ be a place of $F$. Then there is only one place $\mathcal{P}'$ of $F'$ lying above $\mathcal{P}$, and $\mathcal{P}$ is totally ramified in $F'$ with ramification index $p$. 
So the integral closure of the local ring $\mathcal{O}_{\mathcal{P}} \subseteq F$ in $F'$ is simply $\mathcal{O}_{\mathcal{P}'}$. My question is: does there exist a generating element $\beta \in F'$ of $F'$ over $F$ that also works as a generating element of $\mathcal{O}_{\mathcal{P}'}$ as an  $\mathcal{O}_{\mathcal{P}}$-module?  
If $F'$ were separable over $F$, then given a basis of $F'$ over $F$ we can use the trace functional to construct a dual basis and sandwich $\mathcal{O}_{\mathcal{P}'}$ between the modules over $\mathcal{O}_{\mathcal{P}}$ generated by the basis and that generated by the dual basis, and then use the fact that $\mathcal{O}_{\mathcal{P}}$ is a principal ideal domain to establish equality.
But what do we do if $F'$ is inseparable over $F$?
 A: This is not the central question you ask, but I made a claim that at least one highly respected mathematician has doubted, namely that if $F$ is a function field in one variable over a finite field, then $F^{1/p}$ is the only inseparable extension of $F$ of degree $p$. Let’s see why.
To save myself typing, I’ll write $\Bbb F_q=k$, $x^{1/p}=\xi$, $y^{1/p}=\eta$. Let’s suppose that $y$ is of degree $m$ over $k(x)$, so that $k(x,y)$ is separable of degree $m$ over $k(x)$. Then $\bigl[k(\xi,\eta):k(\xi)\bigr]=\bigl[k(x,y):k(x)\bigr]=m$. Now since $y\notin k(\xi)$, we will also have $\bigl[k(\xi,y):k(\xi)\bigr]=m$. (It may be that there are problems if $m$ is composite, but in Georges’ example, it’s certain that $y$ is of degree $p$ over $k(\xi)$.) But we now see that $k(\xi)\subset k(\xi,y)\subset k(\xi,\eta)$, and both bigger fields are of the the same degree $m$ over the smallest field. Thus the two bigger fields are equal. That is, in your notation, I’ve shown that $y^{1/p}\in F(x^{1/p})$. Similarly, $x^{1/p}\in F(y^{1/p})$.
But to put the icing on the cake, I’ll exhibit, in Georges’ proposed counterexample, $\xi\in F(\eta)=k(x,y^{1/p})$ and $\eta\in F(\xi)=k(x^{1/p},y)$.
Indeed
\begin{align}
\xi^p+\xi\eta^{p-1}+\eta^p&=0&\xi^p+\xi\eta^p/\eta+\eta^p&=0\\
\xi\eta^{p-1}&=-y-x&\xi y/\eta&=-x-y\\
\xi&=-\frac{x+y}{\eta^{p-1}}&\eta&=-\frac{\xi y}{x+y}
\end{align}
But now Georges, if $K$ is a field finitely generated
over a perfect field $k$, with transcendence degree $d$, is it not true that the field degree $[K:K^p]=p^d$ ? This is something that I have always firmly believed, though I’m sure without ever having proved it in full detail. Your proposed counterexample made me doubt perhaps fifty years’ unsupported mathematical belief.
