Question

Let $\mathit{S}$andbe as in Exercise$\mathbf{41}\mathbf{}$of Section $\mathbf{6}\mathbf{.}\mathbf{1}$. Prove thatrole="math" localid="1657340969490" $\mathit{S}\mathbf{/}\mathit{I}\mathbf{\cong }{\mathit{Z}}_{\mathbf{2}}$

Short answer

It has been proved that$\mathit{S}\mathbf{/}\mathit{I}\mathbf{\cong }{\mathit{Z}}_{\mathbf{2}}$

Step-by-step solution

Definition as per reference

Let$\mathit{S}$be the set of rational numbers (in lowest terms) with odd denominators and is a subring of $\mathit{Q}$

Let $\mathit{I}$be an ideal in containing elements of $\mathit{S}$with even numerator.

It has been proved that $\mathit{S}\mathbf{/}\mathit{I}$consists of exactly two distinct co-sets.

ProveS/Iis isomorphic to Z2

Since$\mathit{S}$is a commutative ring with identity, then so is$\mathit{S}\mathbf{/}\mathit{I}$(By Theorem $\mathbf{6}\mathbf{.}\mathbf{9}$)

Therefore,$\mathit{S}\mathbf{/}\mathit{I}$is a commutative ring with identity and has only two cosets.

Hence, this ring of two elements is isomorphic to${\mathit{Z}}_{\mathbf{2}}$,which implies$\mathit{S}\mathbf{/}\mathit{I}\mathbf{\cong }{\mathit{Z}}_{\mathbf{2}}$