In the above, I don't understand why the author needs to use another theorem to show that $f$ is separable. Theorem 3.4.5 says that in a finite field, say $F$, the Frobenius automorphism gives us $F = F^p$, entailing that every irreducible polynomial in $F[x]$ is separable. My thinking is, if $E$ exhausts all possible $p^n$ distinct roots of $f$, then $f$ just cannot have repeated roots. This is pretty much the last statement in the paragraph. So, was the author being redundant?

  • $\begingroup$ Why can $f$ not have repeated roots if all $p^n$ elements of $E$ are roots of $f$? I don't see how these two are related... $\endgroup$ – Servaes Oct 26 '15 at 15:02
  • $\begingroup$ I agree with Andy Tam: Lagrange shows that $f$ has $p^n$ distinct roots in $E$ and its degree is $p^n$. Hence it must be separable. $\endgroup$ – Hagen Knaf Oct 26 '15 at 15:22
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    $\begingroup$ Books are written by living breathing human beings, not by machines. Accordingly, we may find inefficiencies and infelicities, and even (gasp) errors in books, even highly regarded ones. $\endgroup$ – Lubin Oct 27 '15 at 14:11

You're correct. This is not really needed and you know right away that f is separable. In general, the idea you have outlined is a common and excellent way to show that a polynomial is separable.

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Beni Bogosel Nov 14 '15 at 23:30
  • $\begingroup$ How does this fail to provide an answer? They asked if they could shortcut the proof, and I said yes. $\endgroup$ – Stella Biderman Nov 17 '15 at 5:46
  • $\begingroup$ I don't remember making this review. Galois theory is not among my favorite subjects neither... Maybe your answer got mixed up with something because it is too short, like a comment. $\endgroup$ – Beni Bogosel Nov 17 '15 at 13:37
  • $\begingroup$ Ok, TY .... Just making sure. $\endgroup$ – Andy Tam Nov 19 '15 at 0:41

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