Compute the norm and discriminant of an ideal in a given algebraic integer ring

Let K = \mathbb{Q}(i) and \mathcal{O} = \mathbb{Z}[i]. Let A = (2) and B = (2+i) be ideals in \mathcal{O}. For I \in \{A,B\}, find a basis for I and compute the discriminant and norm of I. Compute \mathcal{O}/I.


We claim that \{2,2i\} is a basis for A over \mathbb{Z}. Indeed it is clear that (2,2i)_\mathbb{Z} = (2), and if 2a+2bi = 0, then a = b = 0. Then the discriminant of A is \mathsf{det}^2 \left[ \begin{array}{cc} 2 & 2i \\ 2 & -2i \end{array} \right] = -64 and (using Theorem 9.10 in TAN) N(A) = -4.

Let a + bi \in \mathcal{O} be arbitrary, and say a \equiv a_0 and b \equiv b_0 mod 2, where a_0,b_0 \in \{0,1\}. Certainly a+bi \equiv a_0+b_0i mod A, so that \mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{i}, \overline{1+i}\}. Since N(A) = 4, we know that these cosets are distinct.

We claim now that \{2+i,1-2i\} is a basis for B over \mathbb{Z}. Indeed, note that 1-2i = -i(2+i) \in B, and if \zeta = (2+i)(a+bi) \in B, then zeta = a(2+i) - b(1-2i). Moreover, if a(2+i) + b(1-2i) = 0, then 2a+b = a-2b = 0, so that 5b = 0, and thus b = a = 0. Then the discriminant of B is \mathsf{det}^2 \left[ \begin{array}{cc} 2+i & 1-2i \\ 2-i & 1+2i \end{array} \right] = -100 and the norm of B is N(B) = 5.

Note that N(2+i) = 5 \in B. If a+bi \in \mathcal{O} is arbitrary, then a+bi = a-2b + b(2-i) \equiv a-2b mod B. If a-2b \equiv k mod 5, with k \in \{0,1,2,3,4\}, then a+bi \equiv k mod B. Thus \mathcal{O}/B = \{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}. Since N(B) = 5, these cosets are distinct.

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