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.

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