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Let $K/F$ be a field extension and $\alpha \in K$ is algebraic over the field $F.$ Now suppose $\alpha$ is separable over $F.$ Then how can I show that $F(\alpha)/F$ is a separable extension, i.e., an arbitrary element of $F(\alpha)$ is a root of a separable polynomial over $F$ ?

I don't want to use embeddings and it is also clear when $F$ is a perfect field. I also know that if $\operatorname{char}(F)=p>0$ then $F(\alpha)=F(\alpha ^p).$ Using this only I want to get the result. Any help will be appreciated, many thanks.

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    $\begingroup$ I don’t know of any slick way of doing this without bringing embeddings in. Maybe somebody with a better viewpoint will do it for you, in which case I will slip away in embarrassment and delete this comment. $\endgroup$ – Lubin Jan 21 '18 at 5:46
  • $\begingroup$ Sir is it not possible to prove for any $\beta \in F(\alpha) ,$ $F(\beta)=F(\beta ^p)$ ? $\endgroup$ – user371231 Jan 21 '18 at 5:55
  • $\begingroup$ In characteristic $p$, that is surely true in a separable extension. In $F(\alpha)$, when $\alpha$ is a separable element, that will require a proof. And of course, one will have to prove that your criterion implies that the whole extension is separable. $\endgroup$ – Lubin Jan 22 '18 at 0:22
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Three methods that do not directly refer to counting field embeddings are in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/separable2.pdf. The first two ways involve tensor products of fields and the third way uses derivations.

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  • $\begingroup$ Personally, I think that those other methods are more elaborate and advanced than the method using embeddings. $\endgroup$ – Lubin Jan 22 '18 at 0:24
  • $\begingroup$ @Lubin indeed, those other methods require more background, but for some generalizations of the meaning of "separable" they can be a more useful approach. Personally I never liked arguments involving counting field embeddings over and over, and these other viewpoints have a certain elegance that I think the field embedding approach lacks, although I would not consider using these fancier methods in a first course on field theory. $\endgroup$ – KCd Jan 23 '18 at 0:41
  • $\begingroup$ I think I agree that the embeddings method is ugly, @KCd. And a pain in the neck to implement. $\endgroup$ – Lubin Jan 23 '18 at 2:45

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