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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.

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  • $\begingroup$ 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, 2017 at 0:42
  • $\begingroup$ @NoahSchweber What is the connection with Morita equivalence? $\endgroup$
    – rschwieb
    Jan 23, 2017 at 22:39

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

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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.
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