Question

Let R be a ring with identity. Show that the map $\mathbf{f}\mathbf{:}\mathbf{}\mathbf{Z}\mathbf{\to }\mathbf{R}$given by $\mathbf{f}\mathbf{\left(}\mathbf{k}\mathbf{\right)}\mathbf{=}\mathbf{k}\mathbf{·}{\mathbf{1}}_{\mathbf{R}}$is a homomorphism.

Short answer

It can be proved that f is a homomorphism.

Step-by-step solution

Proving homomorphism

In order to prove homomorphism

Let $a,b\in \mathrm{\mathbb{Z}}$

$\begin{array}{rcl}f\left(a+b\right)& =& \left(a+b\right)·1_R\\ & =& a·1_R+b·1_R\\ & =& f\left(a\right)+f\left(b\right)\end{array}$

Proving second condition

$\begin{array}{rcl}f\left(a·b\right)& =& \left(a·b\right)·1_R\\ & =& \left(a·1_R\right)·\left(b·1_R\right)\\ & =& f\left(a\right)·f\left(b\right)\end{array}$

Conclusion

Hence, f is a homomorphism.