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

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).

## 1 Answer

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.