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I'm familiar with the notation $\mathrm{ord}_a(x)$, when $x$ is an integer. However, I'm reading a book where this notation is used with $x$ rational. I'm not sure how to interpret this?

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If it refers to the multiplicative order, reduce x mod a. –  Qiaochu Yuan Nov 16 '10 at 19:48
    
The situation is x is rational, 2x is an integer and p is an odd prime. The book says that it is clear that ord_p(x)=>0. Do you know what this means? –  Jason Smith Nov 16 '10 at 19:58
    
In this situation, $x$ can be written as $\frac{m}{2}$. Then, for any odd prime $p$, only the numerator can contribute to the order, i.e. if $m=p^rm'$ where $m'$ is relatively prime to $p$, then, $ord_p(x)=r\geq 0$ –  Timothy Wagner Nov 16 '10 at 20:01

2 Answers 2

up vote 2 down vote accepted

The notation is defined and explained in some detail here: see pages 11-13.

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Thats awesome thanks! –  Jason Smith Nov 16 '10 at 21:53
    
+1. Could you please explain why, as far as I understand, there are two different notations for, what seems to me, the same concept? I mean the $p-$ adic valuation of the rational number $p/q$: $\nu_p(p/q).$ –  Américo Tavares Nov 16 '10 at 21:59
    
@Américo: yes, these are two different notations for the same concept. It is reasonable to write it as $v_p$ because it is is a discrete **v**aluation. It is also traditional to refer to the exponent of $p$ in the factorization of $\frac{a}{b}$ as the "order at $p$ of $\frac{a}{b}$'', which makes the $\operatorname{ord}_p$ notation reasonable. (I myself use both.) For better or worse, having multiple notations for the same thing is pretty common in mathematics... –  Pete L. Clark Nov 16 '10 at 22:22
    
Thanks! It confirms my understanding. I have used $\nu$ instead of $v$ because I see that notation many times. –  Américo Tavares Nov 16 '10 at 22:33

For any rational number $x$ and any prime integer $p$, we can write $x=p^r\frac{m}{n}$ uniquely (where $r$ is an integer thanks to Yuval for clarifying that), where $m,n$ are integers relatively prime to $p$. Then, we write $ord_p(x)=r$

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Thanks, that answers the question. –  Jason Smith Nov 16 '10 at 21:09
    
Just to clarify: $r$ is an integer that could be negative, e.g. $\mathrm{ord}_p(1/p) = -1$. –  Yuval Filmus Nov 16 '10 at 21:20

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