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I've already studied Galois Theory and I'm trying to explain to a friend why the Galois group of a finite Galois field extension $L/K$ has precisely $[L\colon K]$ elements.

However, the proof I know of this is quite general and requires some definitions first.

For the sake of simplicity, I'm avoiding Galois-theoreic words, such as separability, and telling him only about extensions of finite fields.

So, in the particular case of an extension of finite fields $\mathbb{F}_{p^{n}}/\mathbb{F}_{p}$, is there a simple argument, without having to adress other definitions, for showing that the Galois group has order $n$?

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    $\begingroup$ It's at least easy to show that there are at least $n$ elements by just exhibiting the powers of the Frobenius. $\endgroup$ – Captain Lama Apr 13 '16 at 12:47
  • $\begingroup$ Actually the question arised when I was proving that it is generated by the Frobenius. So, I need exactly the argument for showing that there is no one else there. $\endgroup$ – Shoutre Apr 13 '16 at 12:51
  • $\begingroup$ Lang does it simply by stating that it was already proven that a degree $n$ extension has Galois group of order $n$ $\endgroup$ – Shoutre Apr 13 '16 at 12:52
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Consider the series $\mathbb F_{p^n},\mathbb F_{p^{n-1}},...,\mathbb F_{p^2},\mathbb F_p$ of fields.

You have to show that there is no intermediate field between $F_{p^{i+1}}$ and $F_{p^i}$ for $i=1,...,n-1$.

This follows from $[F_{p^{i+1}}:F_{p^i}]=p$ for $i=1,...,n-1$

Therefore the field extension has exactly $n$ fields, corresponding to the elements of the galois group, which therefore must have exactly $n$ elements.

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  • $\begingroup$ Thanks for your answer. I'm not assuming Galois main theorem, though. Wish I could find a really elementary argument for this particular case. $\endgroup$ – Shoutre Apr 13 '16 at 13:11
  • $\begingroup$ Then, please specify "elementary" $\endgroup$ – Peter Apr 13 '16 at 13:39
  • $\begingroup$ More specifically, not assuming Galois theory. I'm just assuming basic facts about finite fields (e.g. we can view then as vector spaces over $\mathbb{F}_{p}$) and the definition of an automorphism. I want to show that there can't exist an automorphism of $\mathbb{F}_{q}$ that is not a power of the Frobenius. $\endgroup$ – Shoutre Apr 13 '16 at 13:43
  • $\begingroup$ Knowing that the size of the Galois group is $n$, where $q=p^{n}$, would give this. $\endgroup$ – Shoutre Apr 13 '16 at 13:44
  • $\begingroup$ Then assume, the order of the automorphism is divisisble by a prime $q\ne p$, and use the characteristic of the field ($p$) to show that this is impossible. $\endgroup$ – Peter Apr 13 '16 at 13:51

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