Put $F := \mathbb{F}_p(t) = \left\{\frac{f(t)}{g(t)} : f(t), g(t) \in \mathbb{F}_p[t], g(t) \neq 0\right\}$.
Now in one book the author considers the so-called Frobenius homomorphism $\mathcal{F}:F \to F$ defined by $\mathcal{F}(x) = x^p$. Then he proceeds to show that $t$ is not separable over $\text{Im}(\mathcal{F}) =: K$.
However in another set of notes, essentially the same example is given, but the author starts by "considering $\alpha$ s.t. $\alpha^p = t$" and then he shows $\alpha$ is not separable over $F$.
Question: In the book, it is clear that we have the extension $F/K$. However, in the set of notes, in what extension of $F$ does $\alpha$ belong to ? Naively $\alpha$ is a $p$-th root of $t$, so I thought the extension would have to be some field of functions...?
When $a$ is not a $p$-th power in $F$, we can always define $F[\sqrt[p]{a}] = F[X]/(X^p-a)$. We can also specify that we are working within an algebraic closure.
In characteristic $p$, $p$-th roots are unique, so this is even somewhat canonical.