Background on the $p$-adic valuation The following is described in Exercises 5.30 & 5.31 in Atiyah & Macdonald's Introduction to Commutative Algebra text:
Let $A$ be a valuation ring of a field $K$. The group $U$ of units of $A$ is a subgroup of the multiplicative group $K^{\ast}$ of $K$. Let $\Gamma = K^{\ast}/U$. If $\xi, \eta \in \Gamma$ are represented by $x,y \in K^{\ast}$, define $\xi \geq \eta$ to mean $xy^{-1} \in A$. This defines a total ordering on $\Gamma$ which is compatible with the group structure (i.e., $\xi \geq \eta \Rightarrow \xi \omega \geq \eta \omega$ for all $\omega \in \Gamma$). In other words, $\Gamma$ is a totally ordered abelian group. It is called the value group of $A$. The canonical homomorphism $v:K^{\ast} \rightarrow \Gamma$ satisfies $v(x+y) \geq$ min$(v(x),v(y))$ and $v(xy) = v(x) + v(y)$ for all $x,y \in K^{\ast}$, and is known as the valuation of $K$ with values in $\Gamma$.
I'd like to apply this result to a particular case to show that the resulting valuation coincides with the $p$-adic valuation. Below is what I've gathered so far towards this, but I have a few questions.
Let $(p)$ be the principal ideal generated by a prime number $p$. Then $\mathbb{Z}_{(p)}$, the integers localized at the prime ideal $(p) = p\mathbb{Z}$, is a valuation ring of $\mathbb{Q}$. The group of units of $\mathbb{Z}_{(p)}$ is given by $U = \{\frac{a}{b}: a,b \in \mathbb{Z}, p \nmid a, p \nmid b\}$. $U$ is a subgroup of the multiplicative group of units $\mathbb{Q}^{\ast}$ of $\mathbb{Q}$, where $\mathbb{Q}^{\ast}$ consists of all non-zero rational numbers. Forming the quotient group $\Gamma = \mathbb{Q}^{\ast}/U$, then, we have $\Gamma \cong \mathbb{Z}$, which is a totally ordered abelian group via the relation $c \leq d$ for $c,d \in \mathbb{Z}$ if and only if $d-c \in \mathbb{N}$. Every non-zero rational number $x$ (that is, every $x \in \mathbb{Q}^{\ast})$ can be written uniquely as $x = p^n \frac{e}{f}$ for integers $e$ and $f$ indivisible by $p$ and an integer $n$ (that is, $x = p^n \cdot u$ for some $u \in U$). The resulting valuation $v_p:\mathbb{Q}^{\ast} \rightarrow \mathbb{Z}$, then, is the correspondence $x \mapsto n$ (where we also define $v_p(0) = \infty$). This is known as the p-adic valuation or p-adic order, and it satisfies the two properties given in the result above.
Here are the questions that I have about this:

*

*Why is the group $\Gamma = \mathbb{Q}^{\ast}/U$ above isomorphic to $\mathbb{Z}$? What exactly does the isomorphism look like?

*Why is it true that every non-zero rational number $x$ can be written uniquely as $x = p^n \frac{e}{f}$ for integers $e$ and $f$ indivisible by $p$ and an integer $n$?

Thank you!
 A: Let's answer the second question first. We need to be a bit more precise for the statement to be true:

Claim: any $x\in\mathbb Q^*$ can be written uniquely in the form $x=p^n\frac{e}{f}$ for $n\in\mathbb Z$ and coprime integers $e,f$ such that $p\nmid e$, $p\nmid f$ and $f\in\mathbb N$.


Proof: For existence write $x=\frac{a}{b}$ for $a,b\in\mathbb Z$. Using prime factorization in $\mathbb Z$ write $a=p^k e$ and $b=p^\ell f$ for $k,\ell\in\mathbb Z^{\ge0}$ and $e,f\in\mathbb Z$. Then $x=p^{k-\ell}\frac{e}{f}$ and if necessary you divide any common factors out of $\frac{e}{f}$ so that without loss of generality $(e,f)=1$.
For uniqueness suppose $x=p^n\frac{e}{f}=p^{n'}\frac{e'}{f'}$ for $n,n',e,e'\in\mathbb Z$ and $f,f'\in\mathbb N$ with $(e,f)=(e',f')=1$ and $p$ not dividing $e,e',f,f'$. Suppose without loss of generality $n\ge n'$; multiplying you get $p^{n-n'}ef'=e'f$, and then $p$ does not divide the right side so it cannot divide the left either, and deduce $n=n'$. Now you have $ef'=e'f$. From here you see $e\mid e'f$, but because $(e,f)=1$ you can deduce $e\mid e'$. By symmetry you can also deduce $e'\mid e$, so $e=\pm e'$, and because $f,f'$ are both positive you can deduce $e$ and $e'$ must have the same sign so $e=e'$. From here you easily get $f=f'$ and you are done.

Now for the second statement, if $x\in\mathbb Q^*$ you are going to write $x=p^nu$ for $u\in U$ and $n\in\mathbb Z$ the uniquely determined integer of our previous claim. You map $\mathbb Q^*\to\mathbb Z$ by sending $x\mapsto n$. Uniqueness of $n$ means this is well-defined, and you can check that it is a homomorphism. Finally you notice that $U$ is exactly the kernel of this map and surjectivity is clear, so you conclude with the first isomorphism theorem.
