## Scalar multiplication by a fixed central element is an R-module endomorphism

Let $R$ be a ring with 1, and fix an element $z \in Z(R)$. Let $M$ be a left $R$-module. Prove that the mapping $m \mapsto z \cdot m$ is an $R$-module endomorphism of $M$. Prove further that if $R$ is commutative, the mapping $R \rightarrow \mathsf{End}_R(M)$ given by $r \mapsto r \cdot \mathsf{id}$ is a ring homomorphism.

Let $\varphi_z(m) = z \cdot m$, and let $m,n \in M$ and $r \in R$. Then $\varphi_z(m + r \cdot n) = z \cdot (m + r \cdot n)$ $= z \cdot m + z \cdot (r \cdot n)$ $= z \cdot m + zr \cdot n$ $= z \cdot m + rz \cdot n$ $= z \cdot m + r \cdot (z \cdot n)$ $= \varphi_z(m) + r \cdot \varphi_z(n)$. Thus $\varphi_z$ is an $R$-module endomorphism of $M$.

Now suppose $R$ is commutative and define $\Phi : R \rightarrow \mathsf{End}_R(M)$ by $\Phi(z) = \varphi_z$. Let $z,w \in R$ and let $m \in M$. Note that $\Phi(z + w)(m) = \varphi_{z+w}(m)$ $= (z + w) \cdot m$ $= z \cdot m + w \cdot m$ $= \varphi_z(m) + \varphi_w(m)$ $= \Phi(z)(m) + \Phi(w)(m)$ $= (\Phi(z) + \Phi(w))(m)$. Thus $\Phi(z+w) = \Phi(z) + \Phi(w)$. Similarly, $\Phi(zw)(m) = \varphi_{zw}(m)$ $= zw \cdot m$ $= z \cdot (w \cdot m)$ $= \varphi_z(\varphi_w(m))$ $= \Phi(z)(\Phi(w)(m))$ $= (\Phi(z) \circ \Phi(w))(m)$, so that $\Phi(zw) = \Phi(z) \circ \Phi(w)$. Thus $\Phi$ is a ring homomorphism. Note also that $\Phi(1)(m) = \varphi_1(m)$ $= 1 \cdot m$ $= m$ $= \mathsf{id}(m)$, so that $\Phi(1) = \mathsf{id}$. Thus $\Phi$ is in fact a unital ring homomorphism.