How Do Cosets in the Gaussian Integer Ring Relate to Its Ideals?

[PhysicsHelp12]

Hi so this isn't homework its in my book , i just don't get it they skipped this step


Let R=Z(i) be the ring of gaussian integers and let A=(2+i)R denote the ideal of all multiples of 2+i Describe the cosets of R/A

im just having trouble understaning this step:

"Since 2+i is in A we have i+A=-2+A"

and then it does it again "Since 5 is in A 5+A=0+A"

why is this?

thanks

 

[Office_Shredder]

The first one is, uh, wrong in its reasoning.

Note in general, if a-b is in an ideal I, then a+I = b+I. You should try proving this on your own. In particular, 5 = 5-0 is in A, so 5+A = 0+A

 

[PhysicsHelp12]

thanks Shredder:

I have no idea how to prove that I've been trying for an hour

 

[matt grime]

I don't see why office shredder called the first piece of reasoning wrong. I'll write [a] for the coset a+I. The first statement is just

1) [2+i]= [0] (certainly true)

2) [2+i] + [-2] = [0] + [-2]

3) [2+i -2] = [0 - 2]

4) =[-2]

All fine there.

What part of the second bit is troubling you? Write out what you've done.

 

[Office_Shredder]

Whoops, I thought it said i+A = 2+A. Missed the - sign there