Question

If $\mathit{R}$is a ring with identity, and $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$ is a homomorphism from $\mathit{R}$ to a ring $\mathit{S}$, prove that $\mathit{f}\left(1_R\right)$is an idempotent in $\mathit{S}$ . [Idempotent were defined in Exercise 3 of Section 3.2.]

Short answer

Hence it is proved that $f\left(1_R\right)$is an idempotent in$S$ .

Step-by-step solution

Property of Rings

If any ring $R$has elements such that, $a,b\in R$, then, addition and multiplication of the function of its elements is respectively given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Homomorphism

We have a homomorphism of rings as$f:R\to S$:

In this case, we get:

$$\begin{gathered} f\left(1_R\right)=f\left(1_R{1}_R\right) \\ =f\left(1_R\right)f\left(1_R\right) \\ =f{\left(1_R\right)}^2 \end{gathered}$$

Hence proved, $f\left(1_R\right)$ is an idempotent in $S$.