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The whole question looks like-

Let, $x^p-x-1$ be a polynomial over a field $F$ of characteristic $p\ne 0$ and $\alpha$ be a root of it. Prove that $F(\alpha)$ is separable extension over $F$.


I have tried a bit of the problem which goes as follows-
Note that, if $F$ is finite i.e. $F\simeq\Bbb{Z}_p$ then the result is obvious, since any irreducible polynomial over a finite field cannot have multiple root.
Again, $f’(x)=-1\ne0$, hence $f(x)$ has all roots simple.
Now, let $p(x)\in F[x]$ is the minimal polynomial of $\alpha$ over $F$. Then $p(x)|f(x)\implies$ any root of $p(x)$ is a root of $f(x)\implies p(x)$ cannot have multiple roots.
So, I get $\alpha$ is separable over $F$. From this I cannot get any idea how to show $F(\alpha)$ is separable over $F$.
Can anybody complete this proof? Thanks for assistance in advance.

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  • $\begingroup$ AN extension generated by separable elements is a separable extension. $\endgroup$ – Lord Shark the Unknown Dec 5 '18 at 12:54
  • $\begingroup$ And how do you deal with the case $F = \mathbb{F}_p(t)$ ? To get the intuition for separability you probably need the concept of fixed field. Distinct roots implies we'll have enough automorphisms (of the splitting field) for $F$ being the fixed field of the Galois group. $\endgroup$ – reuns Dec 5 '18 at 12:58
  • $\begingroup$ Lord Shark the Unknown, yes intuitively looks like. But I can't write the proof properly. $\endgroup$ – Biswarup Saha Dec 5 '18 at 13:07

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