# quaternion algebra

• Aug 27th 2011, 05:48 PM
dwsmith
quaternion algebra
Let $\mathbb{H}$ be the quaternion algebra.

Show that $\mathbb{H}$ is a division algebra, i.e.

if xy = 0, then x = 0 or y = 0.

I understand this but I am not sure how to prove it. Is it as simple as just multiplying by x and y inverse?

$x^{-1}xy=y=0$ and $xyy^{-1}=x=0$

????
• Aug 27th 2011, 07:29 PM
FernandoRevilla
Re: quaternion algebra
Essentially you are right. Suppose $x,y\in\mathbb{H}$ and $xy=0$ . We have to prove $x=0$ or $y=0$ . If $x\neq 0$ then

$xy=0\Rightarrow x^{-1}(xy)=0\Rightarrow (x^{-1}x)y=0\Rightarrow 1y=0\Rightarrow y=0$

Similar reasoning if we suppose $y\neq 0$ .
• Aug 27th 2011, 08:13 PM
dwsmith
Re: quaternion algebra
$q\in\mathbb{H}, \ q\neq 0$

How do I go about showing that $\{q, \ qi, \ qj, \ qk\}$ is a basis of $\mathbb{H}\mbox{?}$
• Aug 27th 2011, 10:36 PM
FernandoRevilla
Re: quaternion algebra
$\lambda_1q+\lambda_2(qi)+\lambda_3(qj)+\lambda_4(q k)=0$ implies $q(\lambda_11+\lambda_2i+\lambda_3j+\lambda_4k)=0$ . Now use that $\mathbb{H}$ is a division algebra.
• Aug 29th 2011, 10:02 AM
Drexel28
Re: quaternion algebra
Quote:

Originally Posted by dwsmith
Let $\mathbb{H}$ be the quaternion algebra.

Show that $\mathbb{H}$ is a division algebra, i.e.

if xy = 0, then x = 0 or y = 0.

I understand this but I am not sure how to prove it. Is it as simple as just multiplying by x and y inverse?

$x^{-1}xy=y=0$ and $xyy^{-1}=x=0$

????

Technically you're asked to prove that $\mathbb{H}$ is an integral domain. You could just note that there exists a multipicative norm on $\mathbb{H}$ which satisfies $|x|=0$ if and only if $x=0$.