FIT4.3.1. Galois Group of Order 8

Field Theory: Let K be Q[sqrt(2), sqrt(3), sqrt(5)], the splitting field of f(x) = (x^2-2)(x^2-3)(x^2-5) over Q. Find the Galois group of K over Q, find all subgroups of the Galois group, and find all subfields of K over Q.

[MathDoctorBob]

If you are just counting, convert to linear algebra over Z/2. So G = Z/2 x Z/2 x Z/2 x Z/2, which is a vector space over Z/2 under addition. Subgroups of order 4 are same as 2-dimensional subspaces, and we count these using orbits.

Quick count: SL(4, Z/2) acts transitively on the set of planes in (Z/2)^4. The stabilizer of Span (1, 0, 0, 0), (0, 1, 0, 0) is (2, 2)-block upper triangular. So we have 15.14.12.8/6.6.16 = 35 subgroups. (Happy to explain this if needed - not enough space left.)

[MathDoctorBob]

A couple of things to note: first you want to get the degree of the extension from an irreducible polynomial; the minimal polynomial for sqrt(2) + sqrt(3) + sqrt(5) has degree 8. Also degree extensions multiply so as you adjoin sqrt(2), then sqrt(3) then sqrt(5), degree is 2.2.2 = 8. Finally we can see that all obvious autos have order 2. If order 6, we would need one of order 3 by Cauchy's Theorem. Hope that helps!

[MathDoctorBob]

This material is advanced undergraduate or first-year graduate. Lots of cool ideas coming together here.

[NorceCodine]

So here's a question - if you multiply out the (x^2-2)(x^2-3)(x^2-5), its a degree-6 polynomial, and it clearly contains sqrt(2), sqrt(3), sqrt(5) as roots. Then why is the Galois-group not of order 6?

[MathDoctorBob]

The number of C2s is the number of elements of order 2; so 7.
The number of C2xC2s is the number of 2-planes in C2xC2xC2 = 7.6.4/6.4 = 7.

You need to account for the subfields that use sqrt(6), sqrt(15), sqrt(10), sqrt(30)

[MathDoctorBob]

I get x^8-40x^6+352x^4-960x^2+576. No fancy tricks; just grinding it out.

[MathDoctorBob]

@candamohcine You're welcome, and thanks again for the support. I'll try to get a few more of these up when I can. - Bob

[senkakutenanmon]

I noticed my mistake via your response.
For example Q[sqrt(6)] is also subfield of K, and Q[sqrt(6), sqrt(10)]=Q[sqrt(6), sqrt(15)]=Q[sqrt(10), sqrt(15)].
Because is an element of Q[sqrt(10), sqrt(15)] and so on.
Thank you very much for your quick and appropriate response.

[senkakutenanmon]

Q[sqrt(2), sqr(3), sqrt(5)]=Q[sqrt(2)][sqrt(3)][sqrt(5)],
and Gal(Q[sqrt(2)]/Q) and C2(cyclic group of order 2) are isomorphic.
so G and C2*C2*C2(a direct product of three C2s) are isomorphic.

From Lagrange's Theorem, the order of subgroup of G must be 1, 2, 4, 8.
order 1:{e}/order 2:three C2s/order 4:three C2*C2s/order 8:G

From Galois Correspondence Theorem, the subfields of K are
Q, Q[sqrt(2)], Q[sqrt(3)], Q[sqrt(5)], Q[sqrt(2), sqrt(3)], Q[sqrt(2), sqrt(5)], Q[sqrt(3), sqrt(5)], K

[MathDoctorBob]

In general, those are probably beyond the playlist. I can definitely help with with your example.

First note that x^8-1=(x^4-1)(x^4+1) and 1, 2, 3, 4 are the roots of x^4-1. Furthermore x^4+1 = (x^2+2)(x^2+3) over Z/5. These factors are irreducible over Z/5.

Consider F=Z/5[x]/(x^2+2)=Z/5[a] with a^2=3. Then x^2+3 is reducible over F since 3a is a root (3a)^2+3=9a^2+3=30=0. That is, F is the splitting field of x^8-1 over Z/5, with 25 elements. The Galois group is Z/2, generated by
x->x^5.

[MathDoctorBob]

Thanks! C2 and Z/2 both refer to the 2-element group. It depends on the book you are using.

[Wntero]

@MathDoctorBob, thank you! It was a nice video, you clearly left out the easy details for viewers to follow.

[MathDoctorBob]

Thanks! I'm working on a field theory (FIT) playlist right now. The most recent videos do splitting fields, factoring, and cyclotomic polynomials. I'm about 4-5 lectures from field automorphisms as a group.

[MathDoctorBob]

@Nasseef278 You're welcome! - Bob

[antiguo]

very good video, do you have more videos about galois theory?

[Iceiam]

This video is awesome! I wish you did more Galois!!
Btw you said the the galois group is Z/2 x Z/2 x z/2, but my maths teacher did the exact same question and found it to be C2 x c2 x c2
Whats up with that?
Thanks for any help!

[MathDoctorBob]

Your welcome! If you have any questions, I'm happy to help. Galois Theory makes more sense when one works through many examples.

[MathDoctorBob]

@tedtdu This problem was requested by a student. I do most of the upper level videos by request. If you have a specific problem or question, I'm happy to help. - Bob

[MathDoctorBob]

Primitive Element Theorem. There are a few ways to find this (and I have a few videos in the FIT playlist for minimal polynomials if you want to see more). The Galois group gives a way to find the other roots. we just change the signs on sqrt(2), sqrt(3), and sqrt(5). So this polynomial has roots pm sqrt(2) pm sqrt(3) pm sqrt(5).

More in a bit. I'll take a crack at it.

[tedtdu]

Eh, big Bob, I know it is great tutorial but I just do not understand what you are talk about....Any easier tutorial for Galois fields? Or do yo have written statement of the tutorial?