Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For this question, I'm identifying the quaternions $\mathbb{H}$ as a subring of $M_2(\mathbb{C})$, so I view them as the set of matrices of form $$ \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix}. $$

I'm also viewing $\mathbb{C}$ as the subfield of scalar matrices in $M_2(\mathbb{C})$, identifying $z\in\mathbb{C}$ with the diagonal matrix with $z$ along the main diagonal.

Since $\mathbb{H}$ contains $j=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$ and $k=\begin{pmatrix} 0 & i \\ i & 0\end{pmatrix}$, I know that $$ ij+k=\begin{pmatrix} 0 & 2i\\ 0 & 0 \end{pmatrix} $$ and $$ -ij+k= \begin{pmatrix} 0 & 0\\ 2i & 0 \end{pmatrix} $$ are in the generated subring. I'm just trying to find matrices of form $\begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0\\ 0 & d \end{pmatrix}$ for $a,d\neq 0$ to conclude the generated subring is the whole ring. How can I get these remaining two pieces? Thanks.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Hint: Use linear combinations of $$ jk=\pmatrix{i&0\cr0&-i\cr}\qquad\text{and the scalar}\qquad \pmatrix{i&0\cr 0&i\cr}. $$

share|improve this answer
1  
Note that $jk$ is the quaternionic $i$ that is distinct from the scalar $i$. –  Jyrki Lahtonen Jun 13 '12 at 6:40
    
Ah, I forgot about $jk=i$. That makes it easy, thanks Jyrki! –  Noomi Holloway Jun 13 '12 at 6:45
    
Glad to help you help yourself! Welcome to Math.SE, Noomi! –  Jyrki Lahtonen Jun 13 '12 at 7:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.