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Suppose $K $ is a field and $\overline K $ an algebraic closure. Let $f $ be a $K $-automorphism of $\overline K$, let $L$ be the subfield of $\overline K $ fixed by $f $. In this post : (link), they want to prove that every finite extension of $L $ is cyclic (and hence separable).

Isn't this a counterexample ?

Let $p$ be an odd prime, $K= \mathbb F_p (t^{2p}) $. Then the polynomial $X^{2p}-t^{2p} $ is irreducible over $K $ with exactly two roots $\pm t$ in $\overline K $. So there is a $K $-automorphism $f $ of $\mathbb F_p (t) $ that sends $t $ to $-t $. This automorphism can be extended to an automorphism of $\overline K $.

Now, $L $ (the fixed field defined above) does not contain $t $ (since $f (t)\neq t $). So the extension $L(t)/L $ is finite, but not separable as $t $ is not separable.

Is this reasoning correct ? Thank you.

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That is not a counterexample.

We have $$ f(t^2)=f(t)^2=(-t)^2=t^2, $$ so $t^2\in L$.

And $t$ is separable over $\Bbb{F}_p(t^2)$ because the zeros of $X^2-t^2=(X-t)(X+t)$ are simple. Hence $t$ is also separable over $L$.

It seems to me that the confusion was to think of separability being an intrinsic property of an element when, in fact, it says something about the relation between an element and a field. While $t$ is not separable over the smaller field $K$, it is separable over the bigger field $L$.


It is not difficult to show that in your setting actually $L=\Bbb{F}_p(t^2)$.

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