Question

Let $\mathbf{f}\mathbf{:}\mathbf{R}\mathbf{\to }\mathbf{S}$be a homomorphism of rings with kernel $\mathbf{K}$. Let$\mathbf{I}$ be an ideal in$\mathbf{R}$ such that role="math" localid="1655793443658" $\mathbf{I}\mathbf{\subseteq }\mathbf{K}$. Show thatrole="math" localid="1655793422563" $\overline{\mathbf{f}}\mathbf{:}\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{I}$}\right.\mathbf{\to }\mathbf{S}$ given byrole="math" localid="1655793400465" $\overline{\mathbf{f}}\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{I}\mathbf{)}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$ is a well defined homomorphism.

Short answer

It has been proved that$\overline{\mathrm{f}}:\raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{I}$}\right.\to \mathrm{S}$ given by $\overline{\mathrm{f}}\left(\mathrm{r}+\mathrm{I}\right)=\mathrm{f}\left(\mathrm{r}\right)$is a well defined homomorphism.

Step-by-step solution

Homomorphism of Rings

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

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

Kernel of Homomorphism 

Let $\mathbf{f}\mathbf{:}\mathbf{R}\mathbf{\to }{\mathbf{R}}^{\mathbf{\text{'}}}$ be a homomorphism. Then, the kernel$\mathbf{K}$ of homomorphism$\mathbf{f}$ is an ideal in ring$\mathbf{R}$

Conclusion

For being a homomorphism, first a mapping must be well-defined

Let $\mathrm{a},\mathrm{b}\in \mathrm{R}$in such a way that$\left(\mathrm{a}+\mathrm{I}\right)=\left(\mathrm{b}+\mathrm{I}\right)$ in$\raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{I}$}\right.$ , which means that$\mathrm{b}-\mathrm{a}\in \mathrm{I}\subseteq \mathrm{K}$ .

Consider in particular$\mathrm{f}\left(\mathrm{b}-\mathrm{a}\right)=0$ then,

$\begin{array}{rcl}\overline{\mathrm{f}}\left(\mathrm{a}+\mathrm{I}\right)& =& \left(\mathrm{a}\right)\\ & =& \mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}-\mathrm{a}\right)\\ & =& \mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)-\mathrm{f}\left(\mathrm{a}\right)\\ & =& \mathrm{f}\left(\mathrm{b}\right)\\ & =& \overline{\mathrm{f}}\left(\mathrm{b}+\mathrm{I}\right)\end{array}$

So,$\overline{\mathrm{f}}$ is well-defined.

Again,

$\begin{array}{rcl}\overline{\mathrm{f}}\left(\mathrm{a}+\mathrm{b}+\mathrm{I}\right)& =& \mathrm{f}\left(\mathrm{a}+\mathrm{b}\right)\\ & =& \mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)\\ & =& \overline{\mathrm{f}}\left(\mathrm{a}+\mathrm{I}\right)+\overline{\mathrm{f}}\left(\mathrm{b}+\mathrm{I}\right)\end{array}$

And,

$\begin{array}{rcl}\overline{\mathrm{f}}\left(\mathrm{ab}+\mathrm{I}\right)& =& \mathrm{f}\left(\mathrm{ab}\right)\\ & =& \mathrm{f}\left(\mathrm{a}\right)\mathrm{f}\left(\mathrm{b}\right)\\ & =& \overline{\mathrm{f}}\left(\mathrm{a}+\mathrm{I}\right)\overline{\mathrm{f}}\left(\mathrm{b}+\mathrm{I}\right)\end{array}$

Which shows that$\mathrm{f}:\raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{I}$}\right.\to \mathrm{S}$ is a well-defined homomorphism.