Question

Suppose $\mathit{I}$and $\mathit{J}$are ideals in a ring $\mathit{R}$and let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{I}$}\right.\mathbf{\times }\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{J}$}\right.$be the function defined by $\mathit{f}\left(a\right)\mathbf{=}\left(a+I,a+J\right)$

$\mathbf{}\left(a\right)\mathbf{}\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathbf{}\mathit{f}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{a}\mathbf{}\mathit{h}\mathit{o}\mathit{m}\mathit{o}\mathit{m}\mathit{o}\mathit{r}\mathit{p}\mathit{h}\mathit{i}\mathit{s}\mathit{m}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathit{s}\mathbf{.}$

Short answer

It is proved that$f$ is a homomorphism of rings.

Step-by-step solution

Homomorphism of rings

Let $\mathit{R}$and $\mathit{R}\mathbf{\text{'}}$be two rings. A mapping$\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{R}\mathbf{\text{'}}$ is called an homomorphism of rings if,

• $\mathit{f}\left(a+b\right)\mathbf{=}\mathit{f}\left(a\right)\mathbf{+}\mathit{f}\left(b\right)$
• $\mathit{f}\left(ab\right)\mathbf{=}\mathit{f}\left(a\right)\mathit{f}\left(b\right)$, for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$

Proof

Since $I$and $J$are ideals in a ring $R$therefore, for every $a,b\in R$

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

And,

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

Clearly,$f$ is a homomorphism of rings.