Question

Question: Let be an $l$ideal in a ring $R$. Prove that every element in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$has a square root if and only if for every,$\mathit{a}\mathbf{\in }\mathit{R}$, there exists $\mathit{b}\mathbf{\in }\mathit{R}$such that $\mathit{a}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{\in }\mathit{l}$.

Short answer

Answer:It is proved that every element in$\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$ has a square root if and only if for every$a\in R$ there exists$b\in R$ such that$a-b^2\in l$.

Step-by-step solution

Quotient Ring:

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

$(a+l)(b+l)=(a+b)+l$, for$a,b\in R$ and multiplication is defined by$(a+l)(b+l)=ab+l$

Step 2:

Let us consider every element in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$has a square root. It means for every element $a+l$belonging to $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$, there exists an element, $b+l$in $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$, such that,${(b+1)}^2=b^2+l=a+l$

Now, $b^2+l$subtract from both sides of the above equation, we get

$a-b^2+l=l$

, which shows that $a-b^2\in l$

Step 3:

Conversely, if we assume that for every $a\in R$, there exists a such that,$a-b^2\in l$ .

Then, ${(b+l)}^2=b^2+l$

We know that, $(a+l)=(b+l)$implies $a-b\in l$. Hence, we can write $b^2+l=a+l$

This shows that every element$\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$l$}\right.$ in has a square root.