## 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.