Is there a subring of $M_2(\mathbb{R})$ ring-isomorphic to $\mathbb{H}?$

Question

In the following post Quaternions are not ring-isomorphic to 2x2 real matrices it is asked if $M_2(\mathbb{R})$ is ring-isomorphic to $\mathbb{H}$, having different negative answers.

Is it possible to use the facts presented in the mentioned post to answer negatively my question? One approach I had was to show that any non-trivial subring of $M_2(\mathbb{R})$ has some zero divisors, but this is clearly not the case (take the subring generated by $I$ and $-I$ where $I$ is the $2 \times 2$ identity matrix).

Answer 1

Isomorphic:

• As real algebras, no because both are 4 dimensional and $M_2(\Bbb{R})$ has nilpotent elements while $\Bbb{H}$ is a division ring.

• As rings, look at rational quaternions $\Bbb{Q}+i\Bbb{Q}+j\Bbb{Q}+ij\Bbb{Q}$, the ring isomorphism sends $\Bbb{Q}+i\Bbb{Q}$ to $\Bbb{Q}\pmatrix{1 & 0 \\ 0 & 1}+\Bbb{Q}P\pmatrix{0 & 1 \\-1 & 0}P^{-1}$ and it reduces to ask if some $J \in M_2(\Bbb{R})$ satisfies $JP\pmatrix{0 & 1 \\-1 & 0}P^{-1}=-P\pmatrix{0 & 1 \\-1 & 0}P^{-1}J$ which we know there isn't because otherwise we would have $M_2(\Bbb{R})\cong \Bbb{H}$ as real algebras.