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Quaternion ring

by Gtay
Tags: quaternion, ring
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Gtay
#1
Mar11-08, 10:45 PM
P: 14
1. The problem statement, all variables and given/known data
Show that the quaternion division ring H has infinitely many u satisfying u[tex]^{2}[/tex]=-1


2. Relevant equations
Elements of H is of the form a.1 +bi+cj+dk where a, b, c, d in [tex]\textsl{R}[/tex] ( reals) and i[tex]^{2}[/tex]= j[tex]^{2}[/tex]= k[tex]^{2}[/tex]=ijk = -1.


3. The attempt at a solution
Let u = a.1 +bi+cj+dk then u[tex]^{2}[/tex]=a[tex]^{2}[/tex]-b[tex]^{2}[/tex]-c[tex]^{2}[/tex]-d[tex]^{2}[/tex]+2a(bi+cj+dk)+2bicj+2bidk+2cjdk and this = -1 provided a=0 and -b[tex]^{2}[/tex]-c[tex]^{2}[/tex]-d[tex]^{2}[/tex]=1 and 2bicj+2bidk+2cjdk= 0 but I do not see how 2bicj+2bidk+2cjdk= 0.
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Hurkyl
#2
Mar11-08, 10:50 PM
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Two comments:

(1) You do know what the product [itex]i \cdot j[/itex] is, right?

(2) -bē-cē-dē=1 can never be satisfied.
Gtay
#3
Mar12-08, 12:31 AM
P: 14
Thanks for the comments
(1) The product of i.j =k but this ring is not a commutative ring and so I do not have 2bicj = 2bcij=2bck.
ij=k, jk=i, ki=j, ji=-k, kj=-i, ik=-j
(2) My bad. I meant -bē-cē-dē=-1

But still, I do not see why 2bicj+2bidk+2cjdk= 0. Is there anything that I am missing here?

K.J.Healey
#4
Mar12-08, 01:02 AM
P: 641
Quaternion ring

if its not a commutative ring then how do you have 2bicj?
Shouldnt that term then be :
i j (b*c) + j i (c*b)
If ij=-ji, and c*b=b*c then these cancel, right?
K.J.Healey
#5
Mar12-08, 01:03 AM
P: 641
I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?
Gtay
#6
Mar12-08, 01:08 AM
P: 14
Yes!!! Thank you very much. I see it now. Poor me.
Gtay
#7
Mar12-08, 01:09 AM
P: 14
Quote Quote by K.J.Healey View Post
I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?
I do not know what [bi, cj] or {bi, cj} stand for.
Hurkyl
#8
Mar12-08, 09:29 AM
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Quote Quote by Gtay View Post
(1) The product of i.j =k but this ring is not a commutative ring and so I do not have 2bicj = 2bcij=2bck.
But fortunately, the real numbers are in the center of the ring -- a fact that is usually explicitly given by describing the quaternions as an algebra over the reals. However (unless I made an error), you can actually derive this fact from the identities given.

(I note that you already assumed this fact when you simplified bibi to -bē)
K.J.Healey
#9
Mar12-08, 12:22 PM
P: 641
Quote Quote by Gtay View Post
I do not know what [bi, cj] or {bi, cj} stand for.
commutator and anticommutator:
[A,B] = AB-BA
{A,B} = AB+BA


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