Ring As a Module Over Itself

Theorem

With the scalar multiplication map \(\cdot : R \times R \to R\) given by \((r, r') \mapsto rr'\) the ring \(R\) is an \(R\)-module.

The proof is somewhat trivial, but given for completeness.

Proof

Let \(r, s, t \in R\), 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 \(1.s = 1s = s\).