Question

Let S and l be as in Exercise 41 of Section 6.1. Prove that$\raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$I$}\right.\cong {\mathrm{\mathbb{Z}}}_2$ .

Short answer

It has been proved that$\raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$I$}\right.\cong {\mathrm{\mathbb{Z}}}_2$.

Step-by-step solution

Definition as per reference

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

Let l is an ideal in containing elements of S with even numerator.

It has been proved that S/I consists of exactly two distinct cosets.

Prove  S/ I is isomorphic to ℤ2

Since S is a commutative ring with identity, then so is S/I . (By Theorem 6.9)

Therefore, S/I is a commutative ring with identity and has only two cosets.

Hence, this ring of two elements is isomorphic to${\mathrm{\mathbb{Z}}}_2$ , which implies $\raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$I$}\right.\cong {\mathrm{\mathbb{Z}}}_2$.