p-Adic valuation

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

I've always wanted to learn p-Adic numbers, so I'm having a read through the notes at www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf.

I'm a little confused by the p-adic ordinal $\text{ord}_p x = \text{max}\{ r:p^r | x \}$

1) They claim that $\text{ord}_p \frac{a}{b} = \text{ord}_p a - \text{ord}_p b$ . I presume this is obvious, but I just can't see it

2) I tried a test just to see how things work. $\text{ord}_5 \frac{2}{15}$

So this should be $\text{ord}_5 2 - \text{ord}_5 15$

Since 2 is prime $\text{ord}_5 2 = 0$

Since $5=3 \times 15$ and 5^2 = 25 > 15

we have that $\text{ord}_5 15=1$

Thus $\text{ord}_5 \frac{2}{15}=-1$

But 1/5 does not divide 2/15. What am I doing wrong?

outermeasure · **Posted:** Fri, 24 Jun 2011 09:40:29 UTC

qwirk wrote:

I've always wanted to learn p-Adic numbers, so I'm having a read through the notes at www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf.

I'm a little confused by the p-adic ordinal $\text{ord}_p x = \text{max}\{ r:p^r | x \}$

1) They claim that $\text{ord}_p \frac{a}{b} = \text{ord}_p a - \text{ord}_p b$ . I presume this is obvious, but I just can't see it

2) I tried a test just to see how things work. $\text{ord}_5 \frac{2}{15}$

So this should be $\text{ord}_5 2 - \text{ord}_5 15$

Since 2 is prime $\text{ord}_5 2 = 0$

Since $5=3 \times 15$ and 5^2 = 25 > 15

we have that $\text{ord}_5 15=1$

Thus $\text{ord}_5 \frac{2}{15}=-1$

But 1/5 does not divide 2/15. What am I doing wrong?

You are thinking about it in the wrong way.

2/15 divided by 1/5 gives 2/3, which is a unit in $\mathbb{Z}_{(5)}$ . (Recall: $\mathbb{Z}_{(5)}$ is the localisation of $\mathbb{Z}$ at the prime ideal (5)

, i.e. $\mathbb{Z}_{(5)}=\left\lbrace\frac{a}{b}\mid a,b\in\mathbb{Z}, 5\nmid b\right\rbrace$ )

We have the discrete valuation $\mathop{\mathrm{ord}}_p\colon\mathbb{Z}-\lbrace 0\rbrace\to\mathbb{N}$ (where obviously $\mathop{\mathrm{ord}}_p(a/b)=\mathop{\mathrm{ord}}_p(a)-\mathop{\mathrm{ord}}_p(b)$ for all nonzero a,b with $b\mid a$ ), and we extend this to a homomorphism of groups $\mathop{\mathrm{ord}}_p\colon\mathbb{Q}^\times\to\mathbb{Z}$ , so we also have $\mathop{\mathrm{ord}}_p(a/b)=\mathop{\mathrm{ord}}_p(a)-\mathop{\mathrm{ord}}_p(b)$ for all $a,b\in\mathbb{Q}^\times$ .

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Ok so when we say p^r|x

we are not talking about divisibility over Z, but rather over some other integral domain?

(On a side note you just answered a lingering question I had - I often see the term "localised at a prime", but did note know what it meant)

outermeasure · **Posted:** Fri, 24 Jun 2011 11:50:33 UTC

qwirk wrote:

Ok so when we say p^r|x

we are not talking about divisibility over Z, but rather over some other integral domain?

(On a side note you just answered a lingering question I had - I often see the term "localised at a prime", but did note know what it meant)

Sorry to confuse you further: depends on how you view things.

If you are thinking about $\mathop{\mathrm{ord}}_p\colon\mathbb{Z}-\lbrace 0\rbrace\to\mathbb{N}$ , then it is divisibility over $\mathbb{Z}$ . However, this is not going to be very useful when you start doing more fun things because, for example, we want to construct the p-adic integers $\mathbb{Z}_p$ and do things over there or other overrings.

So, how can we do this? This is probably the place to introduce discrete valuation rings (DVRs) and valuation rings. A discrete valuation ring R is a PID with a unique maximal ideal, so we can put a discrete valuation $v\colon K=\mathop{\mathrm{Frac}}(R)\to\mathbb{Z}\cup\{\infty\}$ with the properties:

$v(a)=\infty\iff a=0$
$v(a+b)\geq\min\{v(a),v(b)\}$

and $R=\{x\in\mathop{\mathrm{Frac}}(R)\mid v(x)\geq 0\}$ .

A valuation is when you relax v to take value in a totally ordered group, and a valuation ring is when you relax the PID to just being an integral domain R with valuation v on K=Frac(R) such that $R=\{x\in K\mid v(x)\geq 0\}$ .

The maximal ideal $\mathfrak{m}$ allows us to define v by $v(x)=\max\{n\in\mathbb{N}\mid x\in\mathfrak{m}^n\}$ on

, and we extend to

by $v(\frac{a}{b})=v(a)-v(b)$ (check this is well-defined!). Of course, from this we can also recover $\mathfrak{m}=R\cap\{x\in K\mid v(x)>0\}$ --- note the we need the intersection with R here, otherwise you will end up with exactly where you are confused before --- because $(\mathbb{Z},\mathop{\mathrm{ord}}_p)$ is not a valuation ring.

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

Outermeasure - thanks for the detailed answer.

It might take a while to digest all that, but I appreciate it nonetheless!

qwirk · **Joined:** Fri, 3 Sep 2010 09:29:45 UTC **Posts:** 140

I think I also like the following definition (from "p-adic Numbers" by Fernando Q. Gouvea):

The p-adic valuation of $x \in \mathbb{Q}$ is given by v_p(x)

satisfying $x = p^{v_p(x)} \cdot \frac ab$ where $p \nmid ab$

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