Characterize the irreducibles in ZZ[sqrt(2)]

Characterize the irreducible elements in \mathbb{Z}[\sqrt{2}].

Suppose \zeta \in \mathbb{Z}[\sqrt{2}] is irreducible. By our proof of Theorem 8.5, it follows that (\zeta) \supseteq (p) for some unique positive rational prime p. That is, p = \zeta\eta for some \eta \in \mathbb{Z}[\sqrt{2}]. Now p^2 = N(\zeta)N(\eta), so that either N(\zeta) = \pm p or N(\zeta) = \pm p^2. (Recall that N(\zeta) is a rational integer and not 1.) Let \zeta = a+b\sqrt{2}.

Suppose N(\zeta) = \pm p. Then \pm p = a^2 - 2b^2, so that a^2 - 2b^2 is a rational prime.

Suppose N(\zeta) = \pm p^2. Now N(\eta) = \pm 1, so that (by Theorem 7.3) \eta is a unit. (In Theorem 7.9 we saw that the units in \mathbb{Z}[\sqrt{2}] have the form \pm(1+\sqrt{2})^k where k \in \mathbb{Z}.) Thus \zeta is an associate of p in \mathbb{Z}[\sqrt{2}]. We claim that in this case, there does not exist a solution (h,k) of the equation h^2 - 2k^2 = \pm p. If so, then \pm p = (h+k\sqrt{2})(h-k\sqrt{2}). But since \pm p is irreducible, either h+k\sqrt{2} or h-k\sqrt{2} is a unit- so p = \pm 1, a contradiction.

In summary, if \zeta = a+b\sqrt{2} is irreducible in \mathbb{Z}[\sqrt{2}], then either (1) N(\zeta) = \pm p is a rational prime or (2) N(\zeta) = \pm p^2 is the square of a rational prime and the equations h^2-2k^2 = \pm p have no solutions (h,k) in \mathbb{Z}.

Conversely, if N(\zeta) = p is prime, then \zeta is irreducible. If N(\zeta = \pm p^2 where p is a rational prime and if h^2-2k^2 = \pm p has no solution, then no element of \mathbb{Z}[\sqrt{2}] has norm \pm p, so that \zeta is irreducible.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

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

You are commenting using your 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: