Question

Question: Let $l$ be an ideal in a ring R. Prove that every element in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$ is a solution of localid="1649767868958" $x^2=x$if and only if for every $\mathit{a}\mathbf{}\mathbf{\in }\mathbf{}\mathit{R}\mathbf{,}\mathbf{.}{\mathit{a}}^{\mathbf{2}}\mathbf{-}\mathit{a}\mathbf{\in }\mathit{l}\mathbf{.}$

Short answer

Answer:It is proved that every element in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$ is a solution of $x^2=x$if and only if for every , $a\in R,.a^2-a\in l.$

Step-by-step solution

Quotient Ring:

$$\begin{gathered} \text{Let}R\text{bearingand}l\text{beitsideal.Then,thesetofallcosetsofforms} \\ \text{anadditivecommutativegroup}\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.\text{inwhichadditionisdefinedby} \\ (a+l)+(b+l)=a+b+l\text{,fora,b,∈Randmultiplicationisdefinedby}(a+l)(b+l)=ab+l \end{gathered}$$

Step 2:

Assume that every element in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$is a solution of $x^2=x$. Let, $a+l\in \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$,

Then,

$$\begin{gathered} {(a+l)}^2=a+l \\ \Rightarrow a^2+l=a+l \end{gathered}$$

Which proves that, $a^2-a\in l$, since we know, $a+l=b+l\Rightarrow a-b\in l.$.

Step 3:

Conversely, suppose that, for every element, $a\in R$we have,$a^2-a\in l$ .

Now, let$a+l\in \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$,

Then,

$$\begin{gathered} {(a+l)}^2=a^2+l=a+las{a}^2-a\in l. \\ \Rightarrow {(a+l)}^2=a+l \end{gathered}$$

, which shows that $a+l$ is a solution of $x^2-x$.