The Euclidean norm on ZZ[sqrt(-5)]

Let N : \mathbb{Z}[\sqrt{\text{-}5}] \rightarrow \mathbb{N} be the Euclidean norm given by N(a+b\sqrt{\text{-}5}) = a^2 + 5b^2.

  1. Show that N(\alpha\beta) = N(\alpha)N(\beta) for all \alpha,\beta.
  2. Show that N(\alpha) = 1 if and only if \alpha is a unit.
  3. Show that N(\alpha) = 0 if and only if \alpha = 0.
  4. Show that if N(\alpha) is prime as a natural number then \alpha is irreducible in \mathbb{Z}[\sqrt{\text{-}5}].
  5. Show that 3 + 2\sqrt{\text{-}5} and 2 + \sqrt{\text{-}5} are prime.

Let \alpha = a_1 + a_2\sqrt{\text{-}5} and \beta = b_1 + b_2\sqrt{\text{-}5}, and note the following.

N(\alpha\beta)  =  N((a_1+a_2\sqrt{\text{-}5})(b_1+b_2\sqrt{\text{-}5}))
 =  N((a_1b_1 - 5a_2b_2) + (a_1b_2+a_2b_1)\sqrt{\text{-}5})
 =  (a_1b_1 - 5a_2b_2)^2 + 5(a_1b_2+a_2b_1)^2
 =  a_1^2b_1^2 + 5a_1^2 b_2^2 + 5a_2^2b_2^2 + 25 a_2^2b_2^2
 =  (a_1^2 + 5a_2^2)(b_1^2 + 5b_2^2)
 =  N(\alpha)N(\beta)

Suppose \alpha is a unit; then there exists \beta such that \alpha\beta = 1. Considering norms, we have N(\alpha)N(\beta) = 1 (using the previous result). So N(\alpha) = 1. Conversely, suppose N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 1; then b = 0, and we have a = \pm 1. Thus \alpha is a unit.

Certainly N(0) = 0. Now suppose N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 0; then a = b = 0.

If N(\alpha) is prime and \alpha = \beta\gamma, then N(\beta)N(\gamma) is prime. Without loss of generality, N(\beta) = 1, so that \beta is a unit. Thus \alpha is irreducible.

Note that N(3+2\sqrt{\text{-}5}) = 3^2 + 5\cdot 2^2 = 29 is prime; thus 3+2\sqrt{\text{-}5} is irreducible.

Note that a^2 + 5b^2 = 3 has no solutions in the integers since 3 is not a perfect square. In particular, no element of \mathbb{Z}[\sqrt{\text{-}5}] has norm 3. Now N(2 + \sqrt{\text{-}5}) = 9. If 2 + \sqrt{\text{-}5} = \alpha\beta, then N(\alpha)N(\beta) = 9. since no element of this ring has norm 3, without loss of generality we have N(\alpha) = 1 so that \alpha is a unit. Thus 2 + \sqrt{\text{-}5} is irreducible.

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