MHB Why does sqrt(-2) divide x in Z[sqrt(-2)] if it divides x+sqrt(-2)?

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If sqrt(-2) divides x + sqrt(-2) in Z[sqrt(-2)], it implies that sqrt(-2) must also divide x. This is based on the principle that for a number to evenly divide a sum, it must divide each term in that sum. However, there are counterexamples in other contexts, such as integers, where a divisor of a sum does not necessarily divide each summand. The discussion emphasizes the need to consider properties specific to the ring Z[sqrt(-2)]. Ultimately, the conclusion is that the divisibility condition holds true within this mathematical framework.
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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?
 
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Poirot said:
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?
 
Poirot said:
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, \(\sqrt{-2}|x\) ?
 
Sudharaka said:
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 [math]\displaystyle \sqrt{-2}[/math] evenly divides [math]\displaystyle x + \sqrt{-2}[/math], then to show that [math]\displaystyle \sqrt{-2}[/math] evenly divides x.
 
Of course. Thanks 'Prove it'.
 
Prove It said:
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?
 
The Chaz said:
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
 
Poirot said:

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!
 

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