Exhibit an upper triangular basis for a given ideal in an algebraic integer ring

Prove that \{\omega_1 = 1+2i, \omega_2 = 1+3i\} is an integral basis for K = \mathbb{Q}(i). Find a basis \{\alpha_1,\alpha_2\} (over \mathbb{Z}) for the ideal (1+i) in \mathbb{Z}[i] such that \alpha_1 = c_{1,1}\omega_1 + c_{1,2}\omega_2, \alpha_2 = c_{2,2}\omega_2, and c_{1,1},c_{2,2} > 0.

First we claim that \omega_1 and \omega_2 are linearly independent over \mathbb{Q}. Indeed, if a\omega_1 + b\omega_2 = 0, then a+b = 2a+3b = 0, and thus a = b = 0. So \{\omega_1,\omega_2\} is a basis for K (being a linearly independent set in the two dimensional \mathbb{Q}-vector space K). We claim that every integer in K is of the form a\omega_1 + b\omega_2 where a,b \in \mathbb{Z}. Indeed, we have a+bi = (3a-b)\omega_1 + (b-2a)\omega_2. So \{\omega_1,\omega_2\} is an integral basis.

Next we claim that \{2\omega_1,\omega_2\} is a basis for (1+i) over \mathbb{Z}. Note that \omega_2 = (1+i)(2+i) and 2\omega_1 = (1+i)(3+i). If \zeta = (1+i)(a+bi) \in (1+i), then evidently \zeta = (3b-a)\omega_2 + (a-2b)2\omega_1. Next, note that if 2h\omega_1 + k\omega_2 = 0, then h+2k = 3h+4k = 0, so that h = k = 0. That is, \{\alpha_1 = 2\omega_1, \alpha_2 = \omega_2\} is a basis for (1+i) over \mathbb{Z}, and certainly is upper triangular.

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