Question

Let T and I be as in Exercise 42 of Section 6.1. Prove that $\mathit{T}\mathbf{/}\mathit{I}\cong {\mathrm{\mathbb{Z}}}_p$.

Short answer

It has been proved that $T/I\cong {\mathrm{\mathbb{Z}}}_p$.

Step-by-step solution

Definition as per reference

pis a prime number and T is a ring of rational numbers (in lowest terms) whose denominators are not divisible by p .

I is an ideal in T containing elements of T whose numerator is divisible by p.

It has been proved that T/I consists of exactly p distinct cosets.

Suppose a map φ : T→Zp

Define a map $\phi :T\to Z_p$such that $\phi \left(r/s\right)={\left[r\right]}_p{\left[s\right]}_p^{-1}$.

It is a well defined map.

Prove that φ is a homomorphism

Let $r/s,r\text{'}/s\text{'}\in T$then $\phi \left(r/s+r\text{'}/s\text{'}\right)=\phi \left(rs\text{'}+sr\text{'}/ss\text{'}\right)$

Now,

$\begin{array}{rcl}\phi \left(rs\text{'}+sr\text{'}/ss\text{'}\right)& =& {\left[rs\text{'}+sr\text{'}\right]}_p{\left[ss\text{'}\right]}_p^{-1}\\ & =& \left[rs\text{'}\right]{\left[ss\text{'}\right]}_p^{-1}+\left[sr\text{'}\right]{\left[ss\text{'}\right]}_p^{-1}\\ & =& \phi \left(r/s\right)+\phi \left(r\text{'}/s\text{'}\right)\end{array}$

Similarly, $\phi \left(\left(r/s\right)·\left(r\text{'}/s\text{'}\right)\right)=\phi \left(r/s\right)·\phi \left(r\text{'}/s\text{'}\right)$

Therefore, $\phi$is a homomorphism.

Prove that φ is surjective

It is a surjective map since $\phi \left(r\right)={\left[r\right]}_p$for any $r\in \mathrm{\mathbb{Z}}$.

Also, $Ker\left(\phi \right)=I$.

Conclusion

Hence, By First Theorem of Isomorphism we can conclude that $T/I\cong {\mathrm{\mathbb{Z}}}_p$