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Given the extension in the title is finite, show that there exists an $n\geq 0 $ and $y \in L$ such that $y^{p^n} = t$ and $\mathbb F_p(t)(y) \leq L$ is a separable extension.

I have currently the result that if $x$ is inseparable over $K =\mathbb F_p(t)$ then $K(x)$ contains a $p$th root of $t$.

Here I am asked to show that essentially $L $ contains a $p^n$th root of $t$ for some non-negative $n$. I suspect that maybe I can manage to use some kind of inductive argument but I can't quite see how to set it up?

If $K(t) = K\leq L$ is separable to begin with, then we know that we can take $n=0$ since $t\in L$.

If not, then $\exists x \in L$ such that $x$ is inseparable over $K$.

then $\exists x_1 \in K(x)$ such that $x_1^p = t$.

If $K(x_1) \leq L$ is separable then we are done. Otherwise,...

At this point I want to be able to say "rinse and repeat" with $x_1$ in place of $t$, so that if we find a $y \in L $ such that $y^p = x_1$ then $y^{p^2} = t$. Continue until we find a $y$ such that $K(y) \leq L$ is separable.

However I don't know if doing such a thing is legitimate, or even if it is a process that is guaranteed to terminate.

How might I be able to polish off this argument?

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