# What is the p-adic valuation of a number?

There seem to be two conflicting definitions for p-adic valuation in the literature.

Firstly, for any non-zero integer n, we have $\nu=\nu_p(n)$ is the greatest non-negative integer such that $p^\nu$ divides $n$. Secondly, we have $|n|_p$ which is defined as $1/p^\nu$. [These definitions can be extended to the rationals.]

$\nu$ is defined as the p-adic valuation in Khrennikov, Nilson, P-adic deterministic and random dynamics (for example) and $|\cdot|_p$ is defined as the p-adic valuation in Khrennikov, P-adic and group valued probabilities, in Harmonic, wavelet and p-adic analysis (for example).

Question: Is there a preferred definition for p-adic valuation?

• One of these is sometimes called the multiplicative valuation and the other the additive valuation. Nov 7, 2010 at 9:23
• Using "valuation" to name an absolute value is bad practice. No algebra reference book does that.
– BS.
Nov 7, 2010 at 10:08
• As usual, Lang has other ideas: in Algebra an absolute value is a valuation if it satisifes $|x + y| \leqq \max(|x|, |y|)$. Nov 7, 2010 at 18:31
• Generally "global variable names" with ultra-specific senses are not a good idea. Better to rely upon context. Nomenclature changes over time, too. Dec 28, 2013 at 23:45
• @BS. that "bad practice" can be found in Russian math books. See my reply below. (I know it's weird to give this comment 11 years late. Oh well.) Sep 19, 2021 at 1:42

I will explain what's going on. We call $$\lvert x\rvert_p$$ the $$p$$-adic absolute value of $$x$$ and $$v_p(x)$$ the $$p$$-adic valuation of $$x$$. The distinction that is made by the two terms "absolute value" and "valuation" is completely standard… in English. However, Khrennikov is originally from Russia and in Russian there is one term for both concepts (нормирование = normirovanie, with stress on the second syllable -- I am not making that up, but see comments below this answer about stress on derived words in Russian, like verbs becoming nouns or nouns becoming adjectives). There is a term "absolute value" in Russian, but it is not an abstract concept; it refers only to the usual absolute value on the real or complex numbers (and quaternions?). This is perhaps why Khrennikov is using the term "valuation" incorrectly to refer to an absolute value function.

(I'm giving a course in Moscow this semester and I found this point frustrating when I was preparing my initial lectures. In different books I found the same word used for an absolute value and for a valuation and couldn't find the term that exclusively means absolute value. Eventually I determined there isn't one; you just know by context what meaning is intended. Native speakers are welcome to correct me here.)

UPDATE (3 years later): I learned from a student in St. Petersburg that the mathematicians there use separate terms for an absolute value $$\lvert\cdot\rvert$$ on a field and its corresponding valuation $$v$$: they call $$\lvert x\rvert$$ the norm (норма) of $$x$$ and $$v(x)$$ the exponent (показатель) of $$x$$.

UPDATE (11 years later): Consistent with Laurent's answer, the Russian Wikipedia page for absolute value (find the English one and then click on Русский) refers to an absolute value as нормирование, a word I mentioned in the first paragraph above, and a valuation as экспоненциальное нормиорование, where the word in front of нормирование is eksponentsialnoe = exponential.

• The distinction that is made by the two terms "absolute value" and "valuation" is completelty standard... in English. It is completely standard in French as well. Nov 7, 2010 at 12:25
• "нормирование, with stress on the second syllable" Cool, I would have guessed the stress to be on the fourth, i.e. нормировáние! (It seems that both are accepted, but I prefer second now). Nov 6, 2022 at 18:38
• @Carl-FredrikNybergBrodda it's amazing that you just made that comment! One of the edits I just made to my answer was to remove a remark "I am not making this up" when I said where the stress is, since I thought nobody would care. Now I will put back an updated version. I too was surprised where the stress is. When Russian words change grammatical form, the stress often doesn't change, which is not like English (and maybe Swedish). More examples: to sum = сумми́ровать and summation = сумми́рование, not суммировáние; vector = ве́ктор and vectorial = ве́кторный, not вeкто́рный. Nov 6, 2022 at 18:56
• @KConrad Even cooler! Thanks for keeping the comment, and for the additional examples, I did not know this. Swedish sometimes keeps the stress in places English does not (e.g. "hístory" and "histórical" in English are "história" and "histórisk", respectively, in Swedish), but in other places, we don't keep it... unfortunately, as a rule, Swedish has too many exceptions to any given rule (including that one). Nov 6, 2022 at 19:06

I would say it's incorrect to call $|\cdot|_p$ a valuation; it is an absolute value or better yet a norm, but certainly not a valuation. Note also that for any $0 < \alpha < 1$, the formula $|n|=\alpha^{\nu_p(n)}$ defines a norm and it's customary but by no means obligatory to take $\alpha = 1/p$.

See http://en.wikipedia.org/wiki/Valuation_%28algebra%29 for the definition of a valuation and the remark that "Some authors use the term exponential valuation rather than "valuation". In this case the term "valuation" means "absolute value"."

• To add to the confusion, valuations are also called "order functions" by some authors. Let's hope that the usage advocated by Laurent (and Bourbaki and Fontaine) will prevail. Nov 7, 2010 at 10:22

The conflict is just that some people use the words valuation and absolute value interchangeably. The term "p-adic valuation", used correctly, refers to $\nu$, though perhaps in some areas of math the prevailing choice is the other way around.

After looking several sources, I think we should mention always $$v_p(.)$$ as $$p$$-adic "additive valuation", because we have $$v(xy)=v(x)+v(y)$$ while the $$p$$-adic norm or absolute value $$|.|_p$$ can be termed as $$p$$-adic "multiplicative valuation" because $$|xy|_p=|x|_p|y|_p.$$