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I was reading Aluffi's book algebra chapter 0 , in theorem 6.9 he gives us six equivalent condition to define (finite) galois extension , and i'm stuck in one implication:

If $F$ is a finite extension of $k$ then $|Aut_k(F)|=[F:k]$ implies that $F$ is normal and separable extension of $k$

The textbook told me to use this theorem:

Let $F$ be a finite and separable extension of $k$ then$|Aut_k(F)|\le [F:k]$ with equality holds iff $F$ is a normal extension of$k$.

But i can only use this theorem to prove the-other-direction implication . How can i prove the above question? Thanks.

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  • $\begingroup$ Corollary 5.20 explains this: the inequality follows from 1.7, with equality iff the minimal pol. of the simple extension factors in $\;F\;$ into different linear factors, which is the same as saying the extension's separable and, since it then it'd be the splitting field of the irreducible $\;f\;$, it also would be normal...It then has a "conversely" part. $\endgroup$ – DonAntonio Mar 14 '17 at 8:14
  • $\begingroup$ @DonAntonio but in 1.7 we talk about simple extension, it's because finite + separable implies simple so we get 5.20, in the implication i asked we haven't knew the extension is separable nor simple $\endgroup$ – Idele Mar 14 '17 at 8:32
  • $\begingroup$ @hc As fas as I see it the simple extension is needed only in one direction: the one you already did. For the other one it is not, as far as I can see. It is slightly messy as one needs to know Aluffi's book and I don't... $\endgroup$ – DonAntonio Mar 14 '17 at 9:19

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