Finding the Galois group of the polynomial [imath]x^4+x^3+1[/imath] over the finite fields [imath]\mathbb{F}_2,\mathbb{F}_3,\mathbb{F}_4[/imath]

[MathNugget]

Pretty much the title.

First, I prove [imath]x^4+x^3+1[/imath] is irreducible over [imath]\mathbb{F}_2[/imath]. it has no root, it's not divisible by [imath]x^2+x+1[/imath], the only irreducible polynomial in [imath]\mathbb{F}_2[X][/imath] of degree 2.

The theorems I believe are useful here are:
1) if [imath]k \subset K[/imath] is a finite extension, then [imath]G(K / k)[/imath] is cyclic group.
2) k finite field, [imath]f \in k[X][/imath] irreducible polynomial. Then [imath]k[X]/(f)[/imath] is splitting field of f over k. The roots of f are [imath]\hat{X}=\alpha, \alpha^2, ..., \alpha^{q^{n-1}}[/imath], and [imath]q=\lvert k \rvert[/imath] and n=deg(f).

And now...I am supposed to find the automorphisms of that splitting field, I guess. They're supposed to be the elements of the Galois group...

 

[blamocur]

I used SymPy to play with this example, and realized how little I remember about field extensions and their Galois groups. Since it's been more than a week since op I am posting results for [imath]k=\mathbb F_2[/imath] and [imath]k=\mathbb F_5[/imath] (I haven't done anything for [imath]\mathbb F_4[/imath]).
First, a reminder: any automorphism of [imath]G(K/k)[/imath] must map the roots of the generating polynomial to its other roots.

Case [imath]k = \mathbb F_2[/imath]: There are 4 Roots of [imath]P(x) = x^4+x^3+1[/imath] in the extension [imath]K/\mathbb F_2[/imath]:
[math]\;\;r_0(x) = x, \;\;r_1(x) = x^2, \;\;r_2(x) = x^3+1, \;\;r_3(x) =x^3+x^2+x.[/math]They form a cyclic group of 4th order (i.e. additive [imath]\mathbb Z_4[/imath]) generated by either automorphism [imath]\sigma : x \rightarrow x^2[/imath] or [imath]\sigma : x \rightarrow x^3+x^2+x[/imath]

Case [imath]k=\mathbb F_5[/imath] and [imath]P(x) = x^4+x^3+1[/imath] the is different since [imath]P(2) =0[/imath] in [imath]\mathbb F_5[/imath], i.e. [imath]P(x)[/imath] is not irreducible over [imath]\mathbb F_5[/imath] and [imath]P(x) = Q(x)(x-2)[/imath] where [imath]Q(x) = x^3+3x^2+x+2[/imath].
[imath]Q(x)[/imath], in turn, has three roots in [imath]K[/imath]:
[math]\;\;r_0(x) = x, \;\;r_1(x) = 2x^2+x-2, \;\;r_2(x) = x^3-x+1.[/math]Obviously, the group is a cyclic group of order 3, as all finite groups of order 3 are.

 

[blamocur]

Ooops: I looked into [imath]k=\mathbb F_5[/imath] but not [imath]k=\mathbb F_3[/imath].
And for [imath]\mathbb F_5[/imath] the root of [imath]Q(x)[/imath] cannot have degrees higher than 2. Will post an update later.

 

[blamocur]

For [imath]k=\mathbb F_3[/imath] the polynomial is not irreducible either since [imath]P(1) = 0[/imath], so in there we get
[math]x^4+x^3+1 = Q(x)(x-1) = (x^3-x^2-x-1)(x-1)[/math]The roots of [imath]Q(x)[/imath] in the factor ring [imath]\mathbb F_3[X] / (X^3-X^2-X-1)[/imath] are : [imath]\;\;x[/imath], [imath]\;\;-x^2+x[/imath] and [imath]\;\;x^2+x+1[/imath].

Hope the above is correct, but by all means feel free to check this.

 

[MathNugget]

For [imath]k=\mathbb F_3[/imath] the polynomial is not irreducible either since [imath]P(1) = 0[/imath], so in there we get
[math]x^4+x^3+1 = Q(x)(x-1) = (x^3-x^2-x-1)(x-1)[/math]The roots of [imath]Q(x)[/imath] in the factor ring [imath]\mathbb F_3[X] / (X^3-X^2-X-1)[/imath] are : [imath]\;\;x[/imath], [imath]\;\;-x^2+x[/imath] and [imath]\;\;x^2+x+1[/imath].

Hope the above is correct, but by all means feel free to check this.

Well, I passed the exam . And I scored fairly high on this exercise (it's easy to solve for [imath]\mathbb{F}_p[/imath], p prime, but apparently not so easy for composite numbers . Here's (some) things I used:
1) In [imath]\mathbb{F}_p[/imath], the roots of an irreducible nth degree polynomial are [imath]\alpha, \alpha^p,...{\alpha^p}^{n-1}[/imath]. Here, p is a prime. This is proved with Frobenius automorphism, which in turn is proved with a proposition saying "if the field is finite, the Frobenius endomorphism is bijective(= an automorphism)". Here, [imath]\alpha^4+\alpha^3+1=0[/imath]. Raise to power p, in field of characteristic p, you get easily [imath]\alpha^p[/imath] is root, and the process is repeated.

2) Like you did, the task starts with reducing the polynomial to something without roots in the field.

3) We use that the Galois group has as many elements as the degree of the field extension (=the degree of the irreducible polynomial we're left with after step 2) ).

I agree with everything you said, though. And I realize I said same things over 3 weeks ago, but now I know how to use them to solve the problem, for a prime p. I also added some details (things make a bit more sense now). I forgot about this thread, sorry team for the lack of updates.

 

[blamocur]

Well, I passed the exam

Congratulations!

 

[MathNugget]

Congratulations!

Thank you . Saved the best for last though, differential geometry. Let's see if I can get ready in 48h for an exam I failed 3 times . I'll drop some bombs on the forum (some terrible exercises).