p-adic valuation

The title of this article has been capitalized due to technical restrictions. The correct title should be $p$ -adic valuation.

For some integer $n$ and prime $p$ , the $p$ -adic valuation of n, denoted $\nu_p(n)$ , represents the largest power of $p$ which divides $n$ . In other words, it is the value of the exponent of $p$ in the prime factorization of $n$ .

Basic Examples

$\nu_3(18)=\nu_3(2\cdot3^2)=2$ .
$\nu_5(-5)=\nu_5(-1\cdot5^1)=5$ .
$\nu_{13}(28)=\nu_{13}(2^2\cdot7\cdot13^0)=0$ .

Properties

For positive integers $x$ and $y$ , $\nu_p(xy)=\nu_p(x)+\nu_p(y).$ This property follows from the fact that $p^ap^b=p^{a+b}$ .
Furthermore, $\nu_p(x+y)\geq\min{\nu_p(x)+\nu_p(y)}.$ This follows because we can factor out $\min{\nu_p(x)+\nu_p(y)}$ copies of $p$ from the sum $x+y$ . Note that equality holds if $\nu_p(x)\neq\nu_p(y)$ , because, in this case, after factoring out $\min{\nu_p(x)+\nu_p(y)}$ copies of $p$ from the sum $x+y$ , the remaining factor cannot be congruent to $0$ modulo $p$ , because one of the terms will be congruent to $0\pmod p$ , while the other will not (because all common factors of $p$ have already been factored out).
If $n$ is a positive integer, because $n\geq p^{\nu_p(n)}$ , we deduce that $\nu_p(n)\leq \log_pn,$ because logarithms are monotone increasing for all bases greater than $1$ , which includes all primes.
Lifting the Exponent: A series of identities, among which the most prominent is: $\nu_p (x^n-y^n)=\nu_p (x-y)+\nu_p (n)$ for odd primes $p$ if $p|(x-y)$ .
Legendre's Formula: $\nu_p (n!)=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^k}\right\rfloor$ .

Extension to Rational Numbers

$\nu_p(0)$ is defined to be infinite.

Furthermore, as seen in the properties above, $\nu_p(xy)=\nu_p(x)+\nu_p(y).$ From this inspiration, we can define fractional inputs as follows: $\nu_p\left(\frac xy\right)=\nu_p(x)-\nu_p(y).$ Note that it does not matter if $\tfrac xy$ is simplified or not, because $\begin{aligned} ν_{p} (\frac{k x}{k y}) & = ν_{p} (k x) - ν_{p} (k y) \\ = (ν_{p} (k) + ν_{p} (x)) - (ν_{p} (k) + ν_{p} (y)) \\ = ν_{p} (x) - ν_{p} (y) \\ = ν_{p} (\frac{x}{y}) . \end{aligned}$

Page

Toolbox

Search

p-adic valuation

Contents

Basic Examples

Properties

Extension to Rational Numbers

See Also