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.)
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!
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.
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.
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)
@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
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?
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.
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!
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 sqrt(6)=sqrt(2)sqrt(3)=sqrt(10/5)sqrt(15/5)=sqrt(10)sqrt(15)/5 is an element of Q[sqrt(10),sqrt(15)] and so on. Thank you very much for your quick and appropriate response.
Amazing video, I've used it to do my solve my own question of f(x)=(x^2-2)(x^2-3)(x^2-5)(x^2-7) over Q. But it's so long, the amount combinations I need is crazy. Am I correct in thinking that I'd get 40 4th order groups? Also slightly confused as to how to write them all out using your format of i,j,k,l
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
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?
Your welcome! I'm more of a group theorist. Check out Keith Conrad's website - he has a series of excellent expository articles on fields and number theory.
@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
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.)
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!
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.
Your welcome! If you have any questions, I'm happy to help. Galois Theory makes more sense when one works through many examples.
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.
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)
This material is advanced undergraduate or first-year graduate. Lots of cool ideas coming together here.
@MathDoctorBob, thank you! It was a nice video, you clearly left out the easy details for viewers to follow.
@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
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?
Thanks! C2 and Z/2 both refer to the 2-element group. It depends on the book you are using.
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.
I get x^8-40x^6+352x^4-960x^2+576. No fancy tricks; just grinding it out.
very good video, do you have more videos about galois theory?
@Nasseef278 You're welcome! - Bob
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!
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 sqrt(6)=sqrt(2)sqrt(3)=sqrt(10/5)sqrt(15/5)=sqrt(10)sqrt(15)/5 is an element of Q[sqrt(10),sqrt(15)] and so on.
Thank you very much for your quick and appropriate response.
Thank you much Dr Bob.
Amazing video, I've used it to do my solve my own question of f(x)=(x^2-2)(x^2-3)(x^2-5)(x^2-7) over Q.
But it's so long, the amount combinations I need is crazy. Am I correct in thinking that I'd get 40 4th order groups? Also slightly confused as to how to write them all out using your format of i,j,k,l
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
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?
thank you so much
You're welcome!
Your welcome! I'm more of a group theorist. Check out Keith Conrad's website - he has a series of excellent expository articles on fields and number theory.