I am trying to solve a problem in Silverman-Tate's Rational Points on Elliptic Curves:
Exercise 2.8. Let $p$ and $R$ be as in the previous exercise, and let $\sigma \ge \nu \ge 0$ be integers. Prove that the quotient group $p^\nu R/p^\sigma R$ is a cyclic group of order $p^{\sigma-\nu}$.
In the "previous exercise", $p$ is defined to be a prime number, and $R=\{x\in \mathbb{Q}:\|x\|_p \le 1\}$, where $\|\cdot\|_p$ is the $p$-adic absolute value. Although the authors did not assume a background in commutative algebra, I am personally trying to solve it in the style of commutative algebra.
My progress so far. First of all $R=\mathbb{Z}_{(p)}$, the localization of $\mathbb{Z}$ at the prime ideal $(p)$. Therefore instead of $p^\nu R/p^\sigma R$ we write $(p^\nu)_{(p)}/(p^\sigma)_{(p)}$. We need to show then
$$ (p^\nu)_{(p)}/(p^\sigma)_{(p)} \cong ((p^\nu)/(p^\sigma))_{(p)} \text{ is isomorphic to }\mathbb{Z}/(p^{\sigma-\nu}). $$
One notice that $(p^\nu)/(p^\sigma) \cong \mathbb{Z}/(p^{\sigma-\nu})$ through the exact sequence of abelian groups
$$ 0 \to (p^\sigma) \to (p^\nu) \cong \mathbb{Z} \to \mathbb{Z}/(p^{\sigma-\nu}) \to 0 $$
It remains to show that $((p^\nu)/(p^\sigma))_{(p)} \cong (\mathbb{Z}/(p^{\sigma-\nu}))_{(p)}$ is isomorphic to $\mathbb{Z}/(p^{\sigma-\nu})$.
I think I have problem showing the isomorphism mentioned in the last sentence. Is it by showing that the localization map $\mathbb{Z}/(p^{\sigma-\nu}) \to (\mathbb{Z}/(p^{\sigma-\nu}))_{(p)}$ defined by $m \mapsto m/1$ a bijection? Injectivity is clear but I am afraid surjectivity is not very obvious.
Any help will be appreiated.
Just think about what the elements of $(\mathbf Z/(p^{\sigma - v}))_{(p)}$ actually look like. Try an example: in $(\mathbf Z/(3^{5}))_{(3)}$, you're inverting elements that are not multiples of $3$, but those are already units in $\mathbf Z/(3^5)$. So are you really mystified by surjectivity of $\mathbf Z/(3^{5}) \to (\mathbf Z/(3^{5}))_{(3)}$?