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.

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  • Yun  On March 8, 2012 at 1:25 am

    Not a big deal, but there is a missing \ in the second line.

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