Question

Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$ be a homomorphism of rings with kernel. Let$\mathit{I}$ be an ideal in$\mathit{R}$ such that . $\mathit{I}\mathbf{\subseteq }\mathit{K}$Show that $\overline{\mathbf{f}}\mathbf{:}\mathit{R}\mathbf{/}\mathit{I}\mathbf{\to }\mathit{S}$given by $\overline{\mathbf{f}}\mathbf{(}\mathit{r}\mathbf{+}\mathit{I}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{r}\mathbf{\right)}$is a well defined homomorphism.

Short answer

It has been proved that $\overline{\mathbf{f}}\mathbf{:}\mathit{R}\mathbf{/}\mathit{I}\mathbf{\to }\mathit{S}$given by $\mathbf{¯}\mathit{f}\mathbf{(}\mathit{r}\mathbf{+}\mathit{I}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{r}\mathbf{\right)}$is a well defined homomorphism.

Step-by-step solution

Homomorphism of Rings

Let $\mathit{R}$and $\mathit{R}\mathbf{\text{'}}$be two rings. A mapping role="math" localid="1657298074980" $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }{\mathit{R}}^{\mathbf{\text{'}}}$is called an homomorphism of rings if,

role="math" localid="1657298976460" $\mathit{f}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}$

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

Kernel of Homomorphism

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

Conclusion

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

Let $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$in such a way that $\mathbf{(}\mathit{a}\mathbf{+}\mathit{I}\mathbf{)}\mathbf{=}\mathbf{(}\mathit{b}\mathbf{+}\mathit{I}\mathbf{)}$in $\mathit{R}\mathbf{/}\mathit{I}$, which means that

$\mathit{b}\mathbf{-}\mathit{a}\mathbf{\in }\mathit{I}\underline{\mathbf{\subset }}\mathit{K}$.

Consider in particular $\mathit{f}\mathbf{(}\mathit{b}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{=}\mathbf{0}$then,

localid="1657300903361" $\overline{\mathbf{f}}\mathbf{(}\mathit{a}\mathbf{+}\mathit{l}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}$

$\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{(}\mathit{b}\mathbf{-}\mathit{a}\mathbf{)}$

localid="1657301096604" $$\begin{gathered} \mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{-}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)} \\ \mathbf{=}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)} \\ \mathbf{=}\overline{\mathbf{f}\mathbf{(}}\mathit{b}\mathbf{+}\mathbf{1}\mathbf{)} \end{gathered}$$

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

Again,

$$\begin{gathered} \overline{\mathbf{f}}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{l}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)} \\ \mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)} \\ \mathbf{=}\overline{\mathbf{f}}\mathbf{(}\mathit{a}\mathbf{+}\mathit{l}\mathbf{)}\mathbf{+}\overline{\mathbf{f}}\mathbf{(}\mathit{b}\mathbf{+}\mathit{l}\mathbf{)} \end{gathered}$$

And,

$\overline{\mathbf{f}}\mathbf{(}\mathit{a}\mathit{b}\mathbf{+}\mathit{l}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathit{b}\mathbf{\right)}$

$\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}$

$\mathbf{=}\overline{\mathbf{f}}\mathbf{(}\mathit{a}\mathbf{+}\mathit{l}\mathbf{)}\overline{\mathbf{f}}\mathbf{(}\mathit{b}\mathbf{+}\mathit{l}\mathbf{)}$

Which shows that$\overline{\mathbf{f}}\mathbf{:}\mathit{R}\mathbf{/}\mathit{I}\mathbf{\to }\mathit{S}$is a well-defined homomorphism.