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I know that if $q=p$ (where $p$ is prime) then Gal($\mathbb{F}_{p^k}/\mathbb{F}_p)$ is cyclic of order $k$.

I heard that in general (for $q=p^m$) the galois group is cyclic of the order of the extension (i.e. that :Gal($\mathbb{F}_{q^k}/\mathbb{F}_q)=C_{[\mathbb{F}_{q^k}:\mathbb{F}_q]}$).

How can I prove this claim ?

3 Answers 3

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Note that the Galois group you are after is a subgroup of Gal$(\mathbb{F}_{q^k}/\mathbb{F}_{p})$ (if $q = p^m$). This is true for any tower $L/M/K$ where Galois groups are defined; Gal$(L/M)$ is always a subgroup of Gal$(L/K)$.

Now the Galois group Gal$(\mathbb{F}_{q^k}/\mathbb{F}_p)$ is generated by the Frobenius automorphism $Frob_p : x\mapsto x^p$. Which powers of this are in Gal$(\mathbb{F}_{q^k}/\mathbb{F}_q)$, i.e. which powers fix the field $\mathbb{F}_q$?

Well these are the powers of $Frob_p^m$. (check this)

So the Galois group you want is cyclic, generated by $Frob_p^m$. The claim about the degree follows easily.

EDIT: You do not need to worry about the extension being Galois (although if you know this then the claim about the size of the Galois group is trivial, by definition of Galois. This is the approach Benjamin Lim is using and is equally valid just slightly more theoretical).

Explicitly, we know that Gal$(\mathbb{F}_{q^k}/\mathbb{F}_p)$ is cyclic of order $mk$, generated by $Frob_p$. Thus the subgroup generated by $Frob_p^m$ has order $k$.

This agrees with:

$[\mathbb{F}_{q^k} : \mathbb{F}_q] = \frac{[\mathbb{F}_{q^k} : \mathbb{F}_p]}{[\mathbb{F}_{q} : \mathbb{F}_p]} = \frac{mk}{m} = k$.

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    Why $Aut(L/M)=Gal(L/M)$ in general ?2012-06-07
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    @Belgi Well $L/F$ is separable so it is clear that $L/M$ is separable. Furthermore $L/F$ being normal implies that $L/M$ is normal. Hence $L/M$ is Galois.2012-06-07
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    I don't know what Aut$(L/M)$ means, do you mean Aut$(L)$. If so then this is the group of all automorphisms of $L$. However the Galois group Gal$(L/M)$ is something more...it is the elements of this automorphism group that fix $M$.2012-06-07
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    Why is the extension being Galois an issue? I know you can use the fundamental theorem of Galois theory but my approach here does not need it.2012-06-07
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    @fretty The fact that every subgroup of a finite cyclic group is cyclic proves the OP's claim immediately.2012-06-07
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    Clearly everything in Gal$(L/M)$ fixes $K$ too (since $K \subseteq M$). Hence Gal$(L/M)\subseteq$ Gal$(L/K)$. Hence it is a subgroup.2012-06-07
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    Yes but it was also asked to prove something about the size of this group too.2012-06-07
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    I was just doing it in an explicit way rather than going into theory.2012-06-07
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    I am trying to prove your claim about the powers of Frob that fixes $\mathbb{F}_q$ and I am having problems with it...2012-06-07
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    Ok so everything in $\mathbb{F}_q$ satisfies $x^q = x$. This is the same as $Frob_p^m(x) = x$. Thus the powers of $Frob_p^m$ fix $\mathbb{F}_q$. Now note that no smaller power $Frob_p^i$ can fix $\mathbb{F}_q$, since otherwise everything in $\mathbb{F}_q$ would have to satisfy $x^{p^i} = x$, which cannot happen since the degree of this polynomial is too small.2012-06-07
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    @fretty Since the OP does not know the fundamental theorem I have removed that from my post.2012-06-07
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    Well it probably doesn't hurt anything to mention it. The OP will probably learn about it soon and so then will see the result in a simpler way.2012-06-07
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The proof I'm thinking of works for all finite fields simultaneously. The Frobenius map $x \mapsto x^q$ is an element of the Galois group; moreover, its fixed field is $\mathbb{F}_q$, so by the fundamental theorem it generates the Galois group. The Frobenius map cannot have order less than $k$ because $x^{q^r} - x$ has $q^r$ roots, so the Galois group is cyclic of order $k$.

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Every subgroup of a finite cyclic group is cyclic. Can you see why your claim follows from this?

You want to know why if $q = p^m$, we have that $\textrm{Gal}(\Bbb{F}_{q^n}/\Bbb{F}_q)$ is cyclic. Well you know that $\textrm{Gal}(\Bbb{F}_{q^n}/\Bbb{F}_q) \subset \textrm{Gal}(\Bbb{F}_{q^n}/\Bbb{F}_p) = \textrm{Gal}(\Bbb{F}_{p^{nm}}/\Bbb{F}_p) $ that is cyclic, so by what I said above it follows that $\textrm{Gal}(\Bbb{F}_{q^n}/\Bbb{F}_q)$ is cyclic.

Ok let us deal with the order of extensions now. You know that

$$nm = [\Bbb{F}_{p^{nm}}:\Bbb{F}_p] = [\Bbb{F}_{p^{nm}}:\Bbb{F}_{p^m}][\Bbb{F}_{p^{m}}:\Bbb{F}_p].$$

You also know that $[\Bbb{F}_{p^m} : \Bbb{F}_p] = m$ yes? So therefore it follows that

$$ [\Bbb{F}_{p^{nm}}:\Bbb{F}_{p^m}] = \frac{\Bbb{F}_{p^{nm}}:\Bbb{F}_p]}{[\Bbb{F}_{p^{m}}:\Bbb{F}_p]} = \frac{nm}{m} = n.$$

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    No...this only tells me it it cyclic...what is the relation between the size of the group and the dimension of the extension ? (by the way are both sub extensions galois ? )2012-06-07
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    @Belgi Every finite extension of a finite field is Galois. Why?2012-06-07
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    Since every finite field is perfect hence if the extension is algebraic and f.g it is the splitting field of $(x-a_1)...(x-a_n)$ where $a_i$ are the generators of the extension2012-06-07
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    @Belgi No I don't think that is correct. Hint: Every finite field is the splitting field of some polynomial.2012-06-07
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    What is wrong with what I wrote ? (and you probbly mean the polynomial $x^{q^n}-x$, right ?)2012-06-07
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    @Belgi Yes that is the polynomial we want. So you know that every finite extension of a finite field is a Galois extension. Now is it clear to you now that your claim follows immediately from what I said in my answer?2012-06-07
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    @Belgi By the way what you said does not make sense because 1) The field being perfect only guarantees separability and 2) From what you said you don't know that every finite extension of a finite field is a normal extension.2012-06-07
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    I still don't understand...so both are the splitting field of the polynomials we wrote...what does that say about the order of the subgroup ?2012-06-07
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    @Belgi Do you agree that $\textrm{Gal}(\Bbb{F}_{q^n}/\Bbb{F}_{q})$ is contained in the cyclic group that I stated above? Hence do you agree that a finite subgroup of a cyclic group is cyclic? By the way the notation in your question is very confusing, can you amend it so that I can answer your question about the order of the extension? For the moment do you see why every finite extension of a finite field is cylic?2012-06-07
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    What is the problem in the notation ? I agree with everything you wrote but I don't understand the claim about the order of the group...2012-06-07
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    I have not claimed anything about the order. Please see my edit above.2012-06-07
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    But that is the question in the post, I agree it is cyclic but we also know the order of $Gal(\mathbb{F}_{q^n}/\mathbb{F}_{q})$ is $n$ - I don't understand why...2012-06-07
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    @Belgi Please see my edit.2012-06-07
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    What in my notation is unclear ?2012-06-07
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    @Belgi You have $p's$ and $q's$ at one go in one line. I suggest you edit that.2012-06-07
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    I fixed the typo, thanks. and I am sorry, but I don't know the fundamental theorem of Galois theory (yet...)2012-06-07
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    @Belgi You should learn that ***immediately***.2012-06-07
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    I will, it will be in the next two-three lectures or so ;)2012-06-07
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    @Belgi You don't even need any fundamental theorem to see the degree of the extension being $n$. It is right in front of your face there!! :D :D :D2012-06-07
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    then why ? I really missing something here2012-06-07
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    @Belgi See the edit.2012-06-07