Let $E/F$ be a (not necessarily finite) algebraic extension, where $F$ is finite. Now, it is known that $E/F$ is a normal extension. On the other hand, $E/F_p$ is algebraic and, since $F_p$ is perfect, $E/F_p$ is separable, which means that $E/F$ is separable (if we have $L/M/N$ and $L/N$ is separable, then $L/M$ and $M/N$ are separable). So $E/F$ is Galois. Is this right or am I doing something wrong?
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2Cf. this question – J. W. Tanner Jun 10 '24 at 01:05
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This shows that every finite extension of $F$ is cyclic. And, therefore since every algebraic extension is the union of its intermediate extensions, every algebraic extension is Galois. – Yahya Jun 10 '24 at 01:28
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Yes, that is correct. Every algebraic extension of a finite field is Galois. Exactly because every finite extension of a finite field is cyclic and Galois. You seem to be knowledgable about the Galois theory of infinite (but still algebraic) extensions. The listed facts imply that the finite quotient groups of that infinite Galois group are all cyclic.
Jyrki Lahtonen
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Posting this in an attempt to get this near-FAQ out of the unanswered queue. – Jyrki Lahtonen Jul 09 '24 at 04:31