# Quaternion ring

1. Mar 11, 2008

### Gtay

1. The problem statement, all variables and given/known data
Show that the quaternion division ring H has infinitely many u satisfying u$$^{2}$$=-1

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

3. The attempt at a solution
Let u = a.1 +bi+cj+dk then u$$^{2}$$=a$$^{2}$$-b$$^{2}$$-c$$^{2}$$-d$$^{2}$$+2a(bi+cj+dk)+2bicj+2bidk+2cjdk and this = -1 provided a=0 and -b$$^{2}$$-c$$^{2}$$-d$$^{2}$$=1 and 2bicj+2bidk+2cjdk= 0 but I do not see how 2bicj+2bidk+2cjdk= 0.

2. Mar 11, 2008

### Hurkyl

Staff Emeritus

(1) You do know what the product $i \cdot j$ is, right?

(2) -b²-c²-d²=1 can never be satisfied.

3. Mar 12, 2008

### Gtay

(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 -b²-c²-d²=-1

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

4. Mar 12, 2008

### K.J.Healey

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?

5. Mar 12, 2008

### K.J.Healey

I guess the question is for your group what is [bi,cj] = ? or even {bi,cj} = ?

6. Mar 12, 2008

### Gtay

Yes!!! Thank you very much. I see it now. Poor me.

7. Mar 12, 2008

### Gtay

I do not know what [bi, cj] or {bi, cj} stand for.

8. Mar 12, 2008

### Hurkyl

Staff Emeritus
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 -b²)

9. Mar 12, 2008

### K.J.Healey

commutator and anticommutator:
[A,B] = AB-BA
{A,B} = AB+BA