Uniqueness

peccavi_2006 · **Posted:** Sat, 10 Mar 2012 16:14:15 UTC

Suppose $x \in \mathbb{Q}$ . Then, apparently, x can be uniquely written in the form
$$ \frac{m}{n} \cdot p^{\nu}$ ,
where the integers m and n are coprime to p, n>0 and $\frac{m}{n}$ is in lowest terms ( $\nu$ is then the order of x).

What I was wondering, though, was what about something like $$ \frac{4}{9}$ ?
Because I could write that:
1. $$ \frac{1}{9} \cdot 2^{2}$ , so the order is 2
or
2. $$ \frac{4}{1} \cdot 3^{-2}$ , so the order is -2. But then I can't work out which part of the conditions this example is violating, because this isn't very unique.

Thank you!

Voxx · **Posted:** Sat, 10 Mar 2012 16:25:56 UTC

if you consider any rational number a/b:

a/b = (1/b)*a^1
a/b = a*(b^(-1))

so i think you must have an aditional condition so you can have a unique form to write them down... (for instance the order is also a non-negative number...)

outermeasure · **Posted:** Sat, 10 Mar 2012 16:37:31 UTC

peccavi_2006 wrote:

Suppose $x \in \mathbb{Q}$ . Then, apparently, x can be uniquely written in the form
$$ \frac{m}{n} \cdot p^{\nu}$ ,
where the integers m and n are coprime to p, n>0 and $\frac{m}{n}$ is in lowest terms ( $\nu$ is then the order of x).

What I was wondering, though, was what about something like $$ \frac{4}{9}$ ?
Because I could write that:
1. $$ \frac{1}{9} \cdot 2^{2}$ , so the order is 2
or
2. $$ \frac{4}{1} \cdot 3^{-2}$ , so the order is -2. But then I can't work out which part of the conditions this example is violating, because this isn't very unique.

Thank you!

What you have is the p-adic valuation $\mathop{\mathrm{ord}}_p(x)=\nu$ (or more generally, the order function $\mathop{\mathrm{ord}}_{\mathfrak{p}}\colon F^\times\to\mathbb{Z}$ at an irreducible prime $\mathfrak{p}$ of a factorial ring R whose field of fractions is F). The dependence on the prime p is implicit (and suppressed).

outermeasure · **Posted:** Sat, 10 Mar 2012 16:38:10 UTC

Moved from Algebra to A&NT.

peccavi_2006 · **Posted:** Sat, 10 Mar 2012 16:45:19 UTC

Oh!
So then $ord_{2}(x) = 2$ , whereas $ord_{3}(x) = -2$ ? I then assume that $ord_{p}(x) = 0 \forall p \neq 2, 3$ ?

outermeasure · **Posted:** Sat, 10 Mar 2012 16:47:14 UTC

peccavi_2006 wrote:

Oh!
So then $ord_{2}(x) = 2$ , whereas $ord_{3}(x) = -2$ ? I then assume that $ord_{p}(x) = 0 \forall p \neq 2, 3$ ?

Yes.

peccavi_2006 · **Posted:** Sat, 10 Mar 2012 16:49:07 UTC

Awesome - thanks outermeasure

Shadow · **Posted:** Sat, 10 Mar 2012 20:12:11 UTC

Voxx wrote:

if you consider any rational number a/b:

a/b = (1/b)*a^1
a/b = a*(b^(-1))

so i think you must have an aditional condition so you can have a unique form to write them down... (for instance the order is also a non-negative number...)

No, it depends on the choice of prime is all.

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Uniqueness

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