I apologize if this is a duplicate, but I haven't found proper answers in the forums.

It is easy to show that the quaternions and $M_2(\mathbb R)$ are isomorphic as vector spaces over the reals, but I need to show that they are not ring-isomorphic.

I thought maybe to assume some $\varphi$ is a ring-homomorphism and get a contradiction, similar to a typical proof showing $2\mathbb Z$ is not isomorphic to $3\mathbb Z$ for example. But here I find it trickier to get a contradicting example and perhaps this isn't the right approach.

Could the isomorphism theorems be useful here? Or how can it be shown otherwise? Thanks.

  • $\begingroup$ An interesting feature is that the quaternions aren't that far off from being a matrix algebra! In particular, although $\mathbb{H}$ and $M_2(\mathbb{R})$ are not isomorphic as $\mathbb{R}$-algebras, when we turn them into $\mathbb{C}$-algebras they become isomorphic! That is, $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}\cong M_2(\mathbb{R})\otimes_\mathbb{R}\mathbb{C}$ as $\mathbb{C}$-algebras. This is related to Morita equivalence, and explains why the two rings do have a number of similar properties. $\endgroup$ – Noah Schweber Jan 22 '17 at 0:42
  • $\begingroup$ @NoahSchweber What is the connection with Morita equivalence? $\endgroup$ – rschwieb Jan 23 '17 at 22:39

Well, Hamilton's quaternions is a division ring whereas $\;M_2(\Bbb R)\;$ has lots of zero divisors...


Nothing really competes with the reason DonAntonio gave, but here are some related honorable mentions:

  1. $M_2(\mathbb R)$ has nontrivial right ideals and $\mathbb H$ does not.
  2. $M_2(\mathbb R)$ has nontrivial idempotents and $\mathbb H$ does not.
  3. A simple right $M_2(\mathbb R)$ module has dimension $2$ over $\mathbb R$ but a simple right $ \mathbb H$ module has dimension $4$ over $\mathbb R$.
  4. $M_2(\mathbb R)$ has nontrivial nilpotent elements and $\mathbb H$ does not.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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