# Discrete Valuation

1. Dec 21, 2016

### Bashyboy

1. The problem statement, all variables and given/known data
Let $p$ and let $\nu : \mathbb{Q}^\times \rightarrow \mathbb{Z}$ be defined by $\nu (\frac{a}{b}) = \alpha$, where $\frac{a}{b} = p^\alpha \frac{c}{d}$, where $p$ divides neither $c$ nor $d$. Prove that the corresponding valuation ring $R := \{x \in \mathbb{Q}^\times ~|~ \nu(x) \ge 0 \} \cup \{0\}$ is the ring of all rational numbers whose denominators are relatively prime to $p$. Describe the units of this valuation ring.

2. Relevant equations

Let $K$ be some field. A function $\nu : K^\times \rightarrow \mathbb{Z}$ is a discrete valuation if

$\nu(ab) = \nu(a) + \nu(b)$

$\nu$ is surjective

$\nu(x+y) \ge \min \{\nu(x), \nu(y)\}$ if $x+y \neq 0$.

3. The attempt at a solution

I find this problem rather annoyingly ill-posed. For one, it seems that the author of this problem presupposes that every rational number $\frac{a}{b}$ as $p^\alpha \frac{c}{d}$ for some integers $\alpha$, $b$, and $c$ satisfying the above conditions, without offering a proof of this or asking to prove this as a part of the problem. Moreover, we aren't even asked to show that this is a well-defined discrete valuation. So, I would like to prove some of these things and then discuss the difficulty I am having with this problem.

First, let me prove every rational number $\frac{a}{b}$ can be written in the previously mentioned form. If neither $a$ nor $b$ have the prime $p$ appearing in their factorization, then $\frac{a}{b} = p^0 \frac{a}{b}$. Now, if both $a$ and $b$ have $p$ appearing in their factorization in the form $p^\alpha$ and $p^\beta$, respectively, then $\frac{a}{b} = \frac{p^\alpha c}{p^\beta d} = p^{\alpha - \beta} \frac{c}{d}$.

Now I will prove that $\nu$ is well-defined. Suppose that $p^\alpha \frac{a}{b} = p^\beta \frac{c}{d}$ but $\alpha - \beta \neq 0$. Without loss of generality, suppose $\alpha - \beta > 0$. Then $p^{\alpha - \beta}ad = bc$ implies $p^{\alpha - \beta} |cd$, and because $p|p^{\alpha-\beta}$, we have $p|cd$. But this implies either $p|c$ or $p|d$, both of which are contradictions. Hence, $\alpha = \beta$. This gives us

$\nu(p^{\alpha} \frac{a}{b}) = \alpha = \beta = \nu (p^\beta \frac{c}{d})$,

thereby establishing that $\nu$ is well-defined.

How does this sound so far? The following is where difficulty begins to emerge.

Now I am trying to show that it is a discrete valuation. Showing that $\nu$ satisfies $\nu(ab) = \nu(a) + \nu(b)$ is rather easy. For surjectivity, if $n$ is some integer, then $\nu(p^n \frac{p+1}{p-1}) = n$, where $p$ clearly divides neither the numerator nor denominator. I am having a little trouble showing $\nu$ possesses the last property: Consider $p^\alpha \frac{a}{b}$ and $p^\beta \frac{c}{d}$, and suppose that $\alpha < \beta$. Then

$\nu(p^\alpha \frac{a}{b} + p^\beta \frac{c}{d}) = \nu( p^\beta \frac{p^{\alpha - \beta} ad + cd}{bd}) = \beta \ge \min \{\alpha, \beta \} = \alpha$

However, it seems possible that $p$ could divide the numerator of $\frac{p^{\alpha - \beta} ad + cd}{bd}$...

2. Dec 21, 2016

### andrewkirk

I don't follow this step. Why does that last statement follow?

In any case, I don't think it's necessary since part of our supposition should also be that $p$ does not divide any of $a,b,c,d$ in this case. So by uniqueness of prime factorisations we get a contradiction by prime-factoring both sides of the equation $p^{\alpha - \beta}ad = bc$ and observing that the factorisation of the left-hand side has $p$ as a factor while that of the right-hand side does not.

By the way, it would be better in this paragraph to use letters other than $a,b$ because they have already been used in the definition of $\nu$ for something different, where they can be divisible by $p$, yet here we require them to not be divisible by $p$. I would be inclined to instead write: suppose $\frac ab=p^\beta\frac cd=p^\beta \frac{c'}{d'}$ where $p$ does not divide any of $c,d,c',d'$ and $\alpha,\beta$ are non-negative integers with $\alpha\geq \beta$. We can then use the prime factorisations of $p^{\alpha-\beta}cd'=c'd$ to prove that $\alpha=\beta$, thereby pacifying the constructivist logicians that are uncomfortable with proof by contradiction (not that that matters, but I think it's good to be constructivist if one can do so without undue inconvenience).

3. Dec 21, 2016

### andrewkirk

There are two errors in here. Firstly the numerator of that fraction should be $p^{\alpha - \beta} ad + cb$, not $p^{\alpha - \beta} ad + cd$. Secondly, given your assumptions, $p^{\alpha - \beta}$ will not be an integer. It's best to be consistent with notation and assumptions or else one confuses oneself. Earlier you assumed $\alpha>\beta$ but now you have assumed $\beta>\alpha$, yet you seem to still be using the earlier assumption some of the time.

If you re-write this fraction carefully, paying attention to the assumptions, you should be able to write the sum of the two rationals as $p$ raised to the lower of the two indices, multiplied by a fraction whose denominator is not divisible by $p$, and whose numerator may or may not be.
That is not a problem. If it does it just makes the value of $\nu$ applied to the sum bigger, so the inequality will still be satisfied.