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

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SUMMARY

The discussion centers on the divisibility properties within the ring of integers extended by the square root of -2, denoted as Z[sqrt(-2)]. It establishes that if sqrt(-2) divides the sum x + sqrt(-2), then it must also divide x. This is supported by the mathematical property that if an integer divides a sum and one of the summands, it must divide the other summand as well. The participants clarify this concept through examples and proofs, reinforcing the understanding of divisibility in this specific algebraic structure.

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Poirot1
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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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