Ring As a Module Over Itself

Theorem

With the scalar multiplication map :R×RR given by (r,r)rr the ring R is an R-module.

The proof is somewhat trivial, but given for completeness.

Proof

Let r,s,tR, then clearly r.(s.t)=rst=(rs).t from associativity of multiplication in R. Furthermore r.(s+t)=r(s+t)=rs+rt=r.s+r.t and (r+s).t=(r+s)t=rt+st=r.t+s.t from the distributive law. Furthermore, it is clear from the definition of the identity in a ring that 1.s=1s=s.