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In the set $\mathbb{R}\times \mathbb{R}^3$ addition is defined by components. We define multiplication $*$ by $$(\lambda,\mathbf{x})*(\mu,\mathbf{y})=(\lambda\mu - \mathbf{x}\cdot \mathbf{y}, \lambda \mathbf{y}+\mu\mathbf{x}+\mathbf{x}\times \mathbf{y}).$$ You need not prove that $(\mathbb{R}\times \mathbb{R}^3,+,*)$ is a division ring. Also let $\mathbf{a}\in \mathbb{R}^3$ with $\lVert\mathbf{a}\rVert=1$.

  1. Find the multiplicative identity of the division ring $(\mathbb{R}\times \mathbb{R}^3,+,*)$.
  2. Find the inverse of $(1,\mathbf{a})$.
  3. Find all elements of the division ring $\mathbb{R}\times \mathbb{R}^3$ that commute with $(0,\mathbf{a})$, and prove that they also form a division ring, which is isomorphic to the division ring $\mathbb{C}$.

I know that posting full problems isn't desirable here, but I have absolutely no clue on how to tackle this. I would be thankful if anyone could help me solve this.

A ring $K$ is a division ring $\Leftrightarrow$ $K\setminus\{0\}$ is a group for multiplication.

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  • $\begingroup$ @MorganRodgers I have tried that, but I don't get much further. No, pretty sure I need to find the unit for multiplication $*$. $\endgroup$
    – lork251
    Jan 22 '19 at 1:15
  • $\begingroup$ I didn't get much past writing out the multiplication and equalling the components. Then I got some vector equations, and I don't know how to solve them. $\endgroup$
    – lork251
    Jan 22 '19 at 1:23
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    $\begingroup$ $(1,\mathbf{0})$? Thanks! $\endgroup$
    – lork251
    Jan 22 '19 at 1:27
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If you take an arbitrary element $(\lambda, \mathbf{x})$ and assume that $(\mu,\mathbf{y})$ is the multiplicative identity, then $$(\lambda, \mathbf{x})*(\mu,\mathbf{y}) = (\lambda\mu-\mathbf{x}\mathbf{y},\lambda\mathbf{y}+\mu\mathbf{x}+\mathbf{x}\times\mathbf{y}) = (\lambda, \mathbf{x}).$$ This gives two equations to solve, since you are assuming this is in fact a division ring the solution should be unique. You should check that the obvious solution to the first equation also works in the second.

This should help to solve 2., since now you know what $(1,\mathbf{a})*(\lambda,\mathbf{x})$ needs to equal for $(\lambda,\mathbf{x})$ to be the inverse of $(1, \mathbf{a})$.

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  • $\begingroup$ Any clues for 3.? $\endgroup$
    – lork251
    Jan 22 '19 at 1:47
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You got $1$ (it's $(1,0)$).

For $2$, $(1,a)\cdot (\mu,y)=(1\mu-a\cdot y,1y+\mu a+a×y)=(1,0)$. So $(\frac12, -\frac12a)$ does the trick.

Finally, $(0,a)\cdot (\mu,y)=(\mu,y)\cdot (0,a)\implies (-a\cdot y,\mu a+a× y)=(-y\cdot a,\mu a+y×a)$.

This implies $(\mu,y)=(\mu,ta)$.

We get $\Bbb R^2 \cong \Bbb C$.

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  • $\begingroup$ $(1,\mathbf{a})*(1,-\mathbf{a})\neq (1,\mathbf{0})$! There is an error in your first component of multiplicating $+$ should be $-$. $\endgroup$
    – lork251
    Jan 22 '19 at 2:12
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    $\begingroup$ Ok. I messed up. I'm getting $(\frac12,-\frac12a)$ with the correction. $\endgroup$
    – user403337
    Jan 22 '19 at 2:52

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