Question

Let$\mathbf{I}$and $\mathbf{K}$ be ideals in a ring$\mathbf{R}$ , with $\mathrm{K}\subseteq \mathrm{I}$. Prove that$\mathrm{I}/\mathrm{K}=\left\{\mathrm{a}+\mathrm{K}:\mathrm{a}\in \mathrm{I}\right\}$ is an ideal in the quotient ring.$\mathrm{R}/\mathrm{K}$

Short answer

It is proved that.$\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.=\left\{\mathrm{a}+\mathrm{K}:\mathrm{a}\in \mathrm{I}\right\}$is an ideal in the quotient ring$\raisebox{1ex}{$\mathrm{R}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

Step-by-step solution

Theorem

A non-empty subset $\mathbf{I}$ of a ring$\mathbf{R}$ is an ideal if and only if it has these properties:

1. If$\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{I}$ then,$\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{\in }\mathbf{I}$
2. If$\mathbf{r}\mathbf{\in }\mathbf{R}$ and$\mathbf{a}\mathbf{\in }\mathbf{I}$ then,$\mathbf{ra}\mathbf{\in }\mathbf{I}$ and $\mathbf{ar}\mathbf{\in }\mathbf{I}$

Quotient Ring

Let $\mathbf{R}$be a ring and$\mathbf{I}$be its ideal Then, the set of all cosets of$\mathbf{I}$forms an additive commutative group$\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{I}$}\right.$in which addition is defined by

$\left(a+I\right)\mathbf{+}\left(b+I\right)\mathbf{=}\left(a+b\right)\mathbf{+}\mathbf{I}$, for$\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{R}$and multiplication is defined by

$\left(a+I\right)\left(b+I\right)\mathbf{=}\mathbf{ab}\mathbf{+}\mathbf{I}$

Proof

Given that,$\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.=\left\{\mathrm{a}+\mathrm{K}:\mathrm{a}\in \mathrm{I}\right\}$.

Firstly, we know that, $0\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$, which shows that $\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$is non-empty.

Let, $\mathrm{a}+\mathrm{K},\mathrm{b}+\mathrm{K}\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$ be arbitrary then, we get.

$\left(\mathrm{a}+\mathrm{K}\right)-\left(\mathrm{b}+\mathrm{K}\right)=\left(\mathrm{a}-\mathrm{b}\right)+\mathrm{K}\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

Now, assume there is some $\mathrm{r}+\mathrm{K}\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$ for which,

$\left(\mathrm{a}+\mathrm{K}\right)\left(\mathrm{r}+\mathrm{K}\right)=\mathrm{ar}+\mathrm{K}\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$and

$\left(\mathrm{r}+\mathrm{K}\right)\left(\mathrm{a}+\mathrm{K}\right)=\mathrm{ra}+\mathrm{K}\in \raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$.

So, by above theorem, $\raisebox{1ex}{$\mathrm{I}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$is an ideal.