# What is Gal($\mathbb{F}_{q^k}/\mathbb{F}_q)$?

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 ?

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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 ? – Belgi Jun 7 '12 at 14:39
@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. – user38268 Jun 7 '12 at 14:41
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 $M$. – fretty Jun 7 '12 at 14:42
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. – fretty Jun 7 '12 at 14:42
Yes but it was also asked to prove something about the size of this group too. – fretty Jun 7 '12 at 14:49

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 ? ) – Belgi Jun 7 '12 at 14:30
@Belgi Every finite extension of a finite field is Galois. Why? – user38268 Jun 7 '12 at 14:32
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 extension – Belgi Jun 7 '12 at 14:33
@Belgi No I don't think that is correct. Hint: Every finite field is the splitting field of some polynomial. – user38268 Jun 7 '12 at 14:35
thank you forr your help! :-) – Belgi Jun 7 '12 at 15:23