Question

Let$\mathit{I}\mathbf{\ne }\mathit{R}$ be an ideal in a commutative ring$\mathit{R}$ with identity. Prove that$\raisebox{1ex}{$\mathbf{R}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{I}$}\right.$ is.an integral domain if and only if whenever$\mathit{a}\mathit{b}\mathbf{\in }\mathit{I}$ either$\mathit{a}\mathbf{\in }\mathit{I}$ or$\mathit{b}\mathbf{\in }\mathit{I}$

Short answer

It is proved that $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$is an integral domain if and only if whenever$ab\in I$ either$a\in I$ or $b\in I$.

Step-by-step solution

Integral domain:

Integral domain is defined as a non-trivial ring$R$ with identity if it is commutative and contains no divisor of zero, that means if$a,b\in R$ and$ab=0$ then, either$a=0$ or $\mathit{b}\mathbf{=}\mathbf{0}$.

Proof

Let$\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$ be an integral domain, then for all$\left(a+I\right),\left(b+I\right)\in \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$ if $\left(a+I\right)\left(b+I\right)=I$

Then, either$\left(a+I\right)=I$ or, $\left(b+I\right)=I$, and we know that$\left(x+I\right)=I$ if and only if $x\in I$.

So, this is equivalent to say that whenever$ab\in I$ either or $b\in I$.

Conversely, let$ab\in I$ implies either$a\in I$ or $b\in I$.

Now, if $a,b\in I$then we must have,$\left(a+I\right)$,$\left(b+I\right)\in \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$,

It is given that $ab\in I$, so we can write, $\left(a+I\right)\left(b+I\right)=ab+I=I$,implies, either$\left(a+I\right)=I$ or, $\left(b+I\right)=I$,

Which proves that,$\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$I$}\right.$,is an integral domain.