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Number Theory Z[sqrt(-2)]

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Poirot

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Feb 15, 2012
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Why does it follow that if x is an integer such that sqrt(-2)|x+sqrt(-2) in Z[sqrt(-2)], then
sqrt(-2)|x?
 

Prove It

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Jan 26, 2012
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Why does it follow that if x is an integer such that sqrt(-2)|x+sqrt(-2) in Z[sqrt(-2)], then
sqrt(-2)|x?
Wouldn't it be because in order for ANY number to evenly divide into a sum, it must divide into each term in the sum evenly?
 

Sudharaka

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Feb 5, 2012
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Why does it follow that if x is an integer such that sqrt(-2)|x+sqrt(-2) in Z[sqrt(-2)], then
sqrt(-2)|x?
Hi Poirot, :)

Can you please elaborate more about your question. Do you mean,

\[\frac{x}{\sqrt{-2}}+\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}\]

then prove that, \(\sqrt{-2}|x\) ?
 

Prove It

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Jan 26, 2012
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Hi Poirot, :)

Can you please elaborate more about your question. Do you mean,

\[\frac{x}{\sqrt{-2}}+\sqrt{-2}\in \mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} ; a\in \mathbb Z,b\in \mathbb Z \}\]

then prove that, \(\sqrt{-2}|x\) ?
I believe the OP is asking that if \(\displaystyle \displaystyle \sqrt{-2}\) evenly divides \(\displaystyle \displaystyle x + \sqrt{-2}\), then to show that \(\displaystyle \displaystyle \sqrt{-2}\) evenly divides x.
 
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Poirot

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Feb 15, 2012
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Of course. Thanks 'Prove it'.
 

The Chaz

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Jan 26, 2012
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Wouldn't it be because in order for ANY number to evenly divide into a sum, it must divide into each term in the sum evenly?
I don't follow. We have $5|2+3$, yet five divides neither of the summands.
Are you referencing a certain property of this ring?
 
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Poirot

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Feb 15, 2012
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I don't follow. We have $5|2+3$, yet five divides neither of the summands.
Are you referencing a certain property of this ring?
To be more precise, if x|y+z and x|y, then x|z.
Pf. xm=y+z and xn=y for some m,n. Thus z=x(m-n) so that x|z
 

The Chaz

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Jan 26, 2012
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To be more precise, if x|y+z and x|y, then x|z.
Pf. xm=y+z and xn=y for some m,n. Thus z=x(m-n) so that x|z
Agreed!