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In number theory, where $p$ is a prime number and $n$ is an integer not equal to zero, what is the definition of the function $\operatorname{ord}_{p}(n)$ in the context of $p$-adic valuations?

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If $p$ is prime, $\text{ord}_p(n)$ is a notation for the $p$-adic valuation, which is the exponent of $p$ in the prime decomposition of $n$. – Joel Cohen Mar 4 '12 at 17:29
From "In number theory, for a given prime number $p$, the $p$-adic order or $p$-adic additive valuation of a number $n$ is the highest exponent $v$ such that $p^v$ divides $n$." – user2468 Mar 4 '12 at 17:36
Thanks J.D and Joel Cohen. – VVV Mar 4 '12 at 17:47
You can also denote it by $v_p(n)$ or $\upsilon_p(n)$ (as has been said on Wikipedia). It may be the better notation, because $\text{ord}_p(n)$ can also denote the order of $n$ mod $p$, i.e. the least positive integer $k$ such that $n^k\equiv 1\pmod{p}$. – user236182 Dec 15 '15 at 14:21
up vote 5 down vote accepted

The expression $\operatorname{ord}_p(n)$ denotes the largest integer $v$ such that $p^v\mid n$.

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Thank you Zev Chonoles. – VVV Mar 4 '12 at 17:47

$\rm ord_p n\:$ is the order of $\rm\:n,\:$ when $\rm\:n\:$ is viewed as a "power series" in radix $\rm\:p\:$ representation

$$\rm ord_p n\: =\: k\ \iff\ n\ =\: a_k p^k + a_{k+1} p^{k+1}+\cdots,\ \ a_k\ne 0,\ \ a_i \in \{0,1,\ldots,p-1\}$$

Therefore, said equivalently, $\rm\:ord_p n\:$ is exponent of the prime $\rm\:p\:$ in the unique factorization of $\rm\:n.\:$ The analogy between power series and adic expansions will become clearer when you study valuation theory.

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