# Show that Z[i] is a PID.

• Mar 29th 2011, 07:04 PM
JJMC89
Show that Z[i] is a PID.
Show that $\displaystyle \mathbb{Z}[i]$ is a principal ideal domain.

I have already shown that $\displaystyle \mathbb{Z}[i]$ is an integral domain. How do I show that all ideals are principal ($\displaystyle I=(r), r\in \mathbb{Z}[i]$)?
• Mar 29th 2011, 07:08 PM
tonio
Quote:

Originally Posted by JJMC89
Show that $\displaystyle \mathbb{Z}[i]$ is a principal ideal domain.

I have already shown that $\displaystyle \mathbb{Z}[i]$ is an integral domain. How do I show that all ideals are principal ($\displaystyle I=(r), r\in \mathbb{Z}[i]$)?

If you know, or can show by means of the norm, that that ring is an Euclidean one then you're done, otherwise I

can't see how to show directly that it is a PID...

Tonio
• Mar 29th 2011, 07:10 PM
JJMC89
Quote:

Originally Posted by tonio
If you know, or can show by means of the norm, that that ring is an Euclidean one then you're done, otherwise I can't see how to show directly that it is a PID...

I haven't covered Euclidean domains or norms.
• Mar 29th 2011, 07:14 PM
tonio
Quote:

Originally Posted by JJMC89
I haven't covered Euclidean domains.

Well, then you still can do something that is actually equivalent: show that the norm $\displaystyle N(a+bi):=a^2+b^2$

in $\displaystyle \mathbb{Z}[i]$ permits you to carry on "division with residue" just as the absolute value

allows us to do the same in the integers.

Once you have this proceed as with the integers to show the ring is a PID.

Tonio