## Every unital ring homomorphism induces a bimodule

Let $R$ and $S$ be rings with 1 and let $\varphi : R \rightarrow S$ be a unital ring homomorphism. Verify that the (right) action $s \cdot r = s\varphi(r)$ makes $S$ into a unital right $R$-module, and that moreover $S$ is a $(S,R)$-bimodule.

Note that $s \cdot (r_1 + r_2) = s \varphi(r_1 + r_2)$ $= s(\varphi(r_1) + \varphi(r_2))$ $= s\varphi(r_1) + s\varphi(r_2)$ $= s \cdot r_1 + s \cdot r_2$, that $s \cdot (r_1r_2) = s\varphi(r_1r_2)$ $= s\varphi(r_1)\varphi(r_2)$ $= (s \cdot r_1)\varphi(r_2)$ $= (s \cdot r_1) \cdot r_2$, that $(s_1 + s_2) \cdot r = (s_1 + s_2) \varphi(r)$ $= s_1 \varphi(r) + s_2 \varphi(r)$ $= s_1 \cdot r + s_2 \cdot r$, and that $s \cdot 1_R = s \varphi(1_R) = s 1_S$ $= s$. So this action makes $S$ a right $R$-module.

Moreover, note that for all $s,a \in S$ and $r \in R$, we have $(s \cdot a) \cdot r = sa \cdot r$ $= (sa) \varphi(r)$ $= s(a\varphi(r))$ $= s (a \cdot r)$ $= s \cdot (a \cdot r)$, since multiplication in $S$ is associative. Thus $S$ is an $(S,R)$-bimodule.