Let $M$ be the only maximal ideal of the local ring $\Bbb{Z}_{(p)}$, show that $\Bbb{Z}_{(p)}/M$ is isomorphic to $\Bbb{Z}/(p)$.

Question

Question: Let $M$ be the only maximal ideal of the local ring $\Bbb{Z}_{(p)}$, show that $\Bbb{Z}_{(p)}/M$ is isomorphic to $\Bbb{Z}/(p)$.

I have proved that $M=\{\dfrac a b\in\Bbb{Z}_{(p)}:p|a\}$ is only maximal ideal of $\Bbb{Z}_{(p)}$. So $\Bbb{Z}_{(p)}/M$ is a field. I saw somewhere that, this field is isomorphic to $\Bbb{Z}/(p)$, but the proof is not trivial. Can any one give an outline of the proof of this? Thank you.

Answer 1

Let $S:=\mathbb{Z}\setminus (p)$. Thus, $\mathbb{Z}_{(p)}=S^{-1}\mathbb{Z}$ and $M=S^{-1}(p)$. Hence, $\mathbb{Z}_{(p)}/M= S^{-1}\mathbb{Z}/S^{-1}(p)\cong S^{-1}(\mathbb{Z}/(p))=S^{-1}\mathbb{Z}_{(p)}$. Now since the elements of $S$ are unite in the ring $\mathbb{Z}_{(p)}$ we have $S^{-1}\mathbb{Z}_{(p)}=\mathbb{Z}_{(p)}$, and so $\mathbb{Z}_{(p)}/M\cong \mathbb{Z}_{(p)}$.