How a multiplication operation is defined in $4$-tuple

Question

Let $R$ be a commutative ring. Let $S$ be a collection of pairs $(a,b)$ of elements of $R$. Define addition and multiplication as follows - $$ (a,b)+(a',b') = (a+a',b+b') $$ and $$(a, b) · (a', b') = (a · a' +b · b' · r, a · b' +a' · b) $$ where $r \in R$

Likewise, now I have to define a multiplication rule for 4-tuple $$(0, 1, 0, 0)^2 = r ; (0, 0, 1, 0)^2 = s ; (0, 1, 0, 0)·(0, 0, 1, 0) = (0, 0, 0, 1) = −(0, 0, 1, 0)·(0, 1, 0, 0)$$ for some $r,s \in R$. (possibly different $r,s$ than for 2-tuple case)

I don't understand how this is defined. How did we got $(0, 1, 0, 0)·(0, 0, 1, 0) = (0, 0, 0, 1)$?

Answer 1

As far as I can see, you mean to talk about the generalized quaternions as seen in this post. You don't deduce any of those relations: you need them all to be given so that you can multiply.

To find the general form of the product of two $4$-tuples, you just enforce linearity of multiplication on those generators with those relations.

So, $$(\alpha_1, \alpha_2,\alpha_3,\alpha_4)(\beta_1, \beta_2,\beta_3,\beta_4) \\ =(\alpha_1(1,0,0,0)+\alpha_2(0,1,0,0)+\alpha_3(0,0,1,0)+\alpha_4(0,0,0,1))(\beta_1(1,0,0,0)+\beta_2(0,1,0,0)+\beta_3(0,0,1,0)+\beta_4(0,0,0,1)) \\ =\alpha_1\beta_1 (1,0,0,0) +\alpha_1\beta_2(0,1,0,0)+\alpha_1\beta_3(0,0,1,0)+\alpha_1\beta_4(0,0,0,1) \\ + \alpha_2\beta_1(0,1,0,0)+r\alpha_2\beta_2(1,0,0,0)+\alpha_2\beta_3(0,0,0,1)+r\alpha_2\beta_4(0,0,1,0) \\ +\alpha_3\beta_1 (0,0,1,0)-\alpha_3\beta_2(0,0,0,1)+s\alpha_3\beta_3(1,0,0,0)-s\alpha_3\beta_4(0,1,0,0) \\ + \alpha_4\beta_1 (0,0,0,1) -r\alpha_4\beta_2(0,0,1,0)+s\alpha_4\beta_3(0,1,0,0)-rs\alpha_4\beta_4(1,0,0,0) \\ =(\alpha_1\beta_1+r\alpha_2\beta_2+s\alpha_3\beta_3-rs\alpha_4\beta_4)(1, 0, 0, 0) \\ +(\alpha_1\beta_2+\alpha_2\beta_1-s\alpha_3\beta_4+s\alpha_4\beta_3)(0, 1, 0, 0) \\ +(\alpha_1\beta_3+r\alpha_2\beta_4+\alpha_3\beta_1-r\alpha_4\beta_2)(0, 0, 1, 0) \\ +(\alpha_1\beta_4+\alpha_2\beta_3-\alpha_3\beta_2+\alpha_4\beta_1)(0, 0, 0, 1) $$

So there in the last step, you have the generali mutliplication rule. I proofread as best I could a couple times... hopefully if any typos are here someone will help. (At least in the normal case when $r=s=-1$ the equations hold correctly, I think.)

Answer 2

What's missing in the description is that the ring you make the set of $4$-tuple into should be an $R$-algebra and that elements are $$ (a,b,c,d)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) $$ or you wouldn't be able to do anything with the given relations.

If you write $u=(1,0,0,0)$, $i=(0,1,0,0)$, $j=(0,0,1,0)$, $k=(0,0,0,1)$, another implicit assumption should be that $u$ is the ring identity and that elements $a\in R$ are identified with $au=(a,0,0,0)$. The given relations now read $$ i^2=r,\quad j^2=s,\quad ij=k=-ji $$ The ring properties now help in deducing the other “elementary” products: $$ ik=i(ij)=i^2j=rj,\quad ik=i(-ji)=-(ij)i=-ki,\\ jk=j(ij)=(ji)j=-kj,\quad jk=j(-ji)=-j^2i=-si $$ Moreover, being the ring an $R$-algebra means elements of $R$ (or $4$-tuples of the form $(a,0,0,0)$) commute with every element.

Now, considering $(a,b,c,d)=a+bi+cj+dk$ and $(a',b',c',d')=a'+b'i+c'j+d'k$ you can write the product $$ (a,b,c,d)(a',b',c',d')=(a+bi+cj+dk)(a'+b'i+c'j+d'k) $$ using distributivity, commutativity of “scalars” and the above elementary products. Just tedious work.