My attempt: let $f$ ($w$+$x$i+$y$j+$z$k) $=$ ($w$+$x$i,$y$+$z$i)

then I tried proving it is a homomorphism. $f$ is a homomorphism under addition but fails to be a homomorphism under multiplication.

Can anyone give any hint? Or any property I need to show they don't share.

Any help would be appreciated.

  • $\begingroup$ What multiplication did you define on $\mathbb{C} \times \mathbb{C}$? $\endgroup$ – Harnak Feb 3 at 10:14
  • $\begingroup$ @Harnak I'm not sure. I used $ (a, b)(c, d) = (ac, bd)$ for $a,b,c,d$ in $C$ $\endgroup$ – Flashhh Feb 3 at 10:18
  • $\begingroup$ You are trying to prove that they are isomorphic with respect to which structure? $\endgroup$ – José Carlos Santos Feb 3 at 10:21
  • $\begingroup$ @JoséCarlosSantos They are both rings, I have edited it. $\endgroup$ – Flashhh Feb 3 at 10:22
  • $\begingroup$ Then you did you use the group-isomorphism tag? $\endgroup$ – José Carlos Santos Feb 3 at 10:24

They are not isomorphic because $\mathbb{C}\times\mathbb{C}$ is commutative, whereas $\mathbb H$ is not.

  • $\begingroup$ How did you define multiplication on $\mathbb{C} \times \mathbb{C}$ ? Sorry I'm new to this topic. $\endgroup$ – Flashhh Feb 3 at 10:30
  • $\begingroup$ I used the one that you mentioned in the comments: $(a,b)\times(c,d)=(ac,bd)$. $\endgroup$ – José Carlos Santos Feb 3 at 10:32
  • $\begingroup$ Thank you so much! $\endgroup$ – Flashhh Feb 3 at 10:33

To have a ring isomorphism you need to give $\mathbb{C}^2$ the following ring structure: $$(a,b)+(c,d) = (a+c,b+d)$$ $$(a,b)(c,d) = (ac - \bar{d}b, da + b \bar{c})$$ in which case the isomorphism is given by: $$x + yi + zj + wk \mapsto (x + yi, z + wi)$$ It doesn't work with the product you mentioned in the comments above.

For more information, you can check on Cayley-Dickson construction.

  • $\begingroup$ Thank you. In this question, do we always need to give a new ring structure or use the multiplication I mentioned above by default ? $\endgroup$ – Flashhh Feb 3 at 10:43
  • $\begingroup$ That depends on what you have to do. There is no standard product in $\mathbb{C}^2$. If you want them to be ring-isomorphic, however, you have to define the product I mentioned. $\endgroup$ – Harnak Feb 3 at 10:46

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