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 Post subject: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:14:15 UTC 
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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!

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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:25:56 UTC 
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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...)


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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:37:31 UTC 
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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).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:38:10 UTC 
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Moved from Algebra to A&NT.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:45:19 UTC 
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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?

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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:47:14 UTC 
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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.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 16:49:07 UTC 
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Awesome - thanks outermeasure :D

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 Post subject: Re: Uniqueness
PostPosted: Sat, 10 Mar 2012 20:12:11 UTC 
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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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