Let $k$ be a field, $R$ a unital, associative $k$-algebra and $M$ an $R$-$R$-bimodule; it is a classical fact that $\mathrm{HH}^2(R, M)$ classifies Hochschild ($k$-algebra) extensions of $R$ by $M$. Anyway it seems strange to me not to require some compatibility on $M$ such as $cm = mc$ for every $c \in k \subset R$: for example, if we put $k = R = \mathbb{C}$ and $M = \mathbb{H}$ (quaternions) with left and right multiplications bimodule structure, how is it possible to give a $\mathbb{C}$-algebra structure on $E = \mathbb{H} \oplus \mathbb{C}$ with product $$ (x, y) (x', y') = (xy' + yx', yy') $$ (this would correspond to $0 \in \mathrm{HH}^2$)? For sure $z \cdot (x, y) = (zx, zy)$ or the same on the right does not work, since $$ \left\{ \begin{aligned} (ij, i) (k, 1) &= (k-j, i) \\ (j, 1) (ik, i) &= (-k-j, i) \end{aligned} \right. \implies i \cdot [(j, 1) (k, 1)] = {?} \,. $$
ADDENDUM: there is not such $\mathbb{C}$-algebra structure: it would be a ring homomorphism $\mathbb{C} \to Z(E)$, where the center $Z(E)$ is the subring $\mathbb{C} \oplus \mathbb{R}$ (if my calculations are correct); now since necessarily $1 \mapsto (0, 1)$, we would have $$ i \mapsto (z,x) \quad\text{s.t.}\quad (0, -1) = (z, x)^2 = (2zx, x^2) $$ and this is impossible. So, where is the problem?
