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Galois Groups of Polynomials over Finite Fields

Let P(X) = X^4 + X^3 + 1, find the Galois group of P over each of the fields F_2, F_3 and F_4.

Every Galois group of a field extension of a finite field is cyclic, this much is understood. So let's say a is a root of (an irreducible factor of) P, and adjoin it to each of our fields.

But then how do we know that, for example, P splits in F_2(a)? We don't (how can we say when it does) and so we have to adjoin further roots b, c and d, so the degree of our extension could be as high as 24, and so the splitting field of P is C_24 at most.

How do we narrow it down?
Reply 1
Avatar for nuodai
nuodai
17
Consider the case where you work over F2\mathbb{F}_2. A cunning observation is that P(Y+1)=Y4+Y3+Y2+Y+1P(Y+1) = Y^4 + Y^3 + Y^2 + Y + 1... is this familiar?

As for F3,F4\mathbb{F}_3, \mathbb{F}_4... I haven't worked these out yet. It might just be a case of brute force algebra. I'll get back to you when I've had time to think about it.
Reply 2
Avatar for Slumpy
Slumpy
16
In F_3, 1 is a root, unless I'm much mistaken.
Reply 3
Avatar for around
around
OP
13
I spotted that (after a while), so the F_3 case appears more tractable now.

F_4 is still a bitch though.
Reply 4
Avatar for Slumpy
Slumpy
16
Original post by around
I spotted that (after a while), so the F_3 case appears more tractable now.

F_4 is still a bitch though.


F_4 is a field extension of F_2. I think this should be helpful.
Edit-I'm not quite so sure now it is, but I'll have a think..
(Ok, I think it is helpful. As an extension, K/F_4 is an extension of degree at most what K/F_2 was, where K is the splitting field of the poly)
(edited 12 years ago)
Reply 5
Avatar for around
around
OP
13
Still stuck on the F3 case.

Our poly factorises into a cubic and a linear term, but how do we know that when we adjoin a root of the cubic to F_3 the cubic factor splits into a linear and a quadratic term? Or do we know this?
Original post by around
Still stuck on the F3 case.

Our poly factorises into a cubic and a linear term, but how do we know that when we adjoin a root of the cubic to F_3 the cubic factor splits into a linear and a quadratic term? Or do we know this?


Hmm, does X^4 + X^3 + 1 split over F3[X](f)\displaystyle \frac{\mathbb{F}_3[X]}{(f)}, where f=X3+2X2+2X+2f = X^3 + 2X^2 + 2X + 2?
Reply 7
Avatar for around
around
OP
13
X^4 + X^3 + 1 over F_3[X]/(f) is just 0, so yes?
Then P splits in F27\mathbb{F}_{27}, I think, and we have that Gal(FpnFp)≅Cn\displaystyle \text{Gal} \left( \frac{\mathbb{F}_{p^n}}{\mathbb{F}_p} \right) \cong C_n.

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