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My book's definition of an $R$-algebra is as follows:

Let $R$ be a commutative ring. Then a homomorphism $\varphi : R \to Z(S)$, where $S$ is a ring, is said to be an $R$-algebra.

My question is why $R$ must be necessarily commutative. The center condition is sufficient for the preservation of ring structure: given $\alpha : R \to Z(S)$, $r_1, r_2 \in R$, $s_1, s_2 \in S$, $$(r_1 s_1)(r_2 s_2) = r_1 (\alpha(s_1) r_2) \alpha(s_2) = (r_1 r_2) (s_1 s_2).$$ The only justification I can think of is that this allows the left and right modules to coincide, so that we need not make an arbitrary choice. But the situation is no different for modules, and I cannot see what warrants this change here. Perhaps some practical reason?

n1lp0tence
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    You might enjoy learning about quaternionic modules. – Lee Mosher Sep 22 '24 at 18:27
  • I think in many situations of interest, there is readily an algebra structure over a commutative ring. For example, in the $k$-algebra of $n\times n$ matrices over $k$ $M_n(k)$, $k$ embeds into the center of $M_n(k)$ via $\lambda \mapsto \lambda I_n$. – walkar Sep 22 '24 at 18:29

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