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 $\mathit{f}\left(a\right)\mathbf{=}\left(a+I,a+J\right)$

b) Is$\mathit{f}$ surjective? [Hint: Consider the case when$\mathit{R}\mathbf{=}\mathbf{\square }\mathbf{,}\mathit{I}\mathbf{=}\left(2\right)\mathbf{,}\mathit{J}\mathbf{=}\left(4\right)$

Short answer

It is proved that $f$is surjective mapping.

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 $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$

Conclusion

We will prove it by contradiction. Let$f$ be not surjective.

We will take the case$R=\square$,$I=\left(2\right)$,$J=\left(4\right)$so we get,

$\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.\times \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$J$}\right.\cong {\square }_2\times {\square }_4$and there will be no $a\in \square$, such that,

$f\left(a\right)=\left({\left[a\right]}_2,{\left[a\right]}_4\right)=\left({\left[1\right]}_2,{\left[0\right]}_4\right)$, since${\left[a\right]}_2={\left[1\right]}_2$ implies that

${\left[a\right]}_4={\left[0\right]}_2$which in turn implies that a is even, that contradicts the assumption that$f$ is not surjective.