Number Theory Z[sqrt(-2)]

[Poirot]

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]

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]

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]

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.

 

[Poirot]

Of course. Thanks 'Prove it'.

 

[The Chaz]

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?

 

[Poirot]

[HR][/HR]

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]

[HR][/HR]

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!