Is $A \otimes _R R$ isomorphic to $A$ even when $R$ does not have an identity?

Question

Let $R$ be a ring (not necessarily with an identity). For a right $R$-module $A$, I know that the tensor product $A \otimes _R R$ is isomorphic to $A$, naturally, when $R$ has an identity. Does this also hold even when $R$ does not have an identity?

Answer 1

No. If, say $R = c_{00}=\{f\colon \mathbb N\to \mathbb C\mid |\mathrm{supp}\, f|<\infty\}$ with pointwise multiplication and $A$ is any $\mathbb C$-vector space with zero $R$-module structure (meaning $a\cdot r:=0$ for every $a\in A$), then the relation of the tensor product forces $$ a\otimes r = a\otimes r\cdot s_r = a\cdot r\otimes s_r = 0. $$ Here the support projection $s_r\colon \mathbb N\to\mathbb C$ of $r$ is defined by $$ s_r(i):=\begin{cases} 1,& r(i)\neq 0,\\ 0, & r(i) = 0.\\ \end{cases} $$