A fact about irreducibles in ZZ[i]

Let \pi = a+bi be an irreducible Gaussian integer with a,b \neq 0. Show that if \pi is a factor of its conjugate \overline{\pi} = a-bi, then \pi is an associate of 1+i.

Let \tau = c+di and suppose \pi\tau = \overline{\pi}. This equality yields the two equations ac-bd=a and ad+bc=-b, which can be rearranged as a(c-1) = bd and ad = -b(c+1). Now ad^2 = -bd(c+1) = -a(c+1)(c-1), so that $latex d^2 + c^2 = 1. Hence \tau \in \{ 1,-1,i,-i \}.

If \tau = 1, then a+bi = a-bi, so that b = 0, a contradiction. Similarly, if \tau = -1 then -a-bi = a-bi and we have a = 0. If \tau = i, then -b+ai = a-bi, so that a = -b. If \tau = -i, then b-ai = a-bi, and we have a = b. In either case, |a| = |b|. If |a| > 1, then \pi = a+bi has a nontrivial factorization (namely a(1+i)) and so is not irreducible. Thus |a| = 1, and so \pi \in \{1+i, 1-i, -1+i, -1-i\}. These are precisely the associates of 1+i.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: