Does free bimodule exist? Let $R,S$ be rings and $A$ be a set.
Does there exist a free $(R,S)$-bimodule $F(A)$ on $A$?
How do I construct it?
Is it just $\oplus_{i\in A} (R\times S)$?
 A: An $(R,S)$-bimodule is the same thing as a left $R\otimes_{\Bbb{Z}}S^\mathrm{op}$ module, so the free $(R,S)$-bimodule on a set $A$ (exists and) is
$$\bigoplus_A R\otimes_{\Bbb{Z}}S^\mathrm{op}$$
A: It exists and is simply $\bigoplus_{i\in A} R\otimes_\mathbb Z S$. The left $R$-module structure over $R\otimes_\mathbb Z S$ is defined as follows (and right $s$-module structure comes similarily):
For each $r\in R$, Define $L_r:R\times S\to R\otimes_\mathbb Z S,\quad (x,y)\mapsto(rx,y)$. It's not difficult to check that $L_r$ is bilinear, so it uniquely determines a linear map $L_{r}^{^\prime}:R\otimes_\mathbb Z S\to R\otimes_\mathbb Z S,\quad x\otimes y\to rx\otimes y$. It's also easy to show that $R\to \text{Hom}_\mathbb Z(R\otimes_\mathbb Z S,R\otimes_\mathbb Z S):\quad  r\mapsto L_{r}^{^\prime}$  is a ring homomorphism.
Finally, the universal property of tensor product implies that $R\otimes_\mathbb Z S$ is a free $(R,S)$-bimodule on a singleton and $\bigoplus_{i\in A} R\otimes_\mathbb Z S$ is a free $(R,S)$-bimodule on $A$. 
