Valuations, Isomorphism, Local ring

Question

Let $x \in \mathbb{Q}, \, x \neq 0$, such that $x=p^r \frac{a}{b}, \quad a,b, r \in \mathbb{Z}, \quad p \nmid a, \quad p\nmid b$. Let $v_p(x):=r$ and $v_p(0):= \infty$. Also, $$\mathcal O_p= \left\{ x \in \mathbb{Q} : v_p(x) \geq 0\right\}= \left\{\frac{a}{b} : p \nmid b\right\} \quad \& \quad m_p= \left\{ x \in \mathbb{Q} : v_p(x) > 0\right\}=\left\{\frac{a}{b} : p \nmid b \, \wedge p \mid a \right\} $$ Show that $O_p / m_p \cong \mathbb{Z}/ p \mathbb{Z}$.

I have this:

$$\phi: \mathbb{Z} \rightarrow{} O_p \rightarrow O_p/m_p \;,\;\; s. t.\;\;a \mapsto \frac{a}{b} \mapsto \frac{a}{b} + m_p$$

I know that, $Ker(\phi)=p \mathbb{Z}$, but my problem is prove that $\phi $ is onto.

Any help is appreciated,

Cheers!

Answer 1

Your map, at least how I am seeing it, isn't well-defined. What is the $b$?

You should try mapping the other way:

Hint: Define

$$\phi:\mathcal{O}_p\to\mathbb{Z}/p\mathbb{Z}:\frac{a}{b}\mapsto ab^{-1}+p\mathbb{Z}$$

This is well-defined, since $b$ is invertible in $\mathbb{Z}/p\mathbb{Z}$.

Answer 2

Let $[x]\in{\cal O}/{\frak m}$ be an arbitrary equivalence class with representative $x=a/b\in{\cal O}\subset\bf Q$.

Then $p\nmid b\implies v(a)=v(a/b)\implies [a]=[a/b]=[x]$ with $a\in\bf Z$.