Are the quaternions Morita equivalent to the real numbers?

Question

I'm wondering whether the quaternions are Morita equivalent to the real numbers. The characterisation in terms of full idempotents seems unwieldly. I can use the category-theoretic definition, but it appears for obvious choices of functors $F : {}_{\mathbb R}M \to {}_{\mathbb H}M$ (via ring extension) and $G : {}_{\mathbb H}M \to {}_{\mathbb R}M$ (via ring restriction), there cannot be an isomorphism between $F \circ G(V)$ and $V$ for any fixed vector space $V$, as $\dim(F \circ G(V)) = 4\dim(V)$. I'm thinking the answer is no.

Answer 1

If $F$ is a field, then the only rings Morita equivalent to $F$ are the matrix rings $M_n(F)$ for positive integer $n$. In particular, since $\mathbb{H}\not\cong M_n(\mathbb{R})$ for any positive integer $n$, $\mathbb{H}$ and $\mathbb{R}$ are not Morita equivalent.

Why is this true? It's a well-known characterization that $R$ is Morita equivalent to $S$ if and only if $R\cong \mathrm{End}(P)$ for some progenerator (finitely generated projective generator) $P$ in the category of $S$-modules.

If $F$ is a field, then every $F$-module is free, hence projective. And a finitely generated free $F$-module (i.e. finite dimensional $F$-vector space) is a generator if and only if its dimension is a positive integer. So $R$ is Morita equivalent to $F$ if and only if $R\cong \mathrm{End}(F^n) = M_n(F)$ for some positive integer $n$.