## Show that a given generating set is a basis for an ideal in a quadratic integer ring

Show that $B = \{5+i, 2+3i\}$ is a basis for the ideal $(2+3i)$ in $\mathbb{Z}[i]$.

First we wish to show that $(5+i,2+3i)_\mathbb{Z} = (2+3i)$; the $(\supseteq)$ direction is clear. Now suppose $\zeta = a(5+i) + b(2+3i) = (5a+2b) + (a+3b)i$; note that $\zeta = (2+3i)(a+b-ai) \in (2+3i)$, as desired.

Now suppose we have integers $a,b \in \mathbb{Z}$ such that $(5+1)a + (2+3i)b = 0$. Comparing coefficients, we have $5a+2b = 0$ and $a+3b = 0$. Evidently this system has only the solution $a = b = 0$, so that $B$ is $\mathbb{Z}$-linearly independent. Hence $B$ is a basis for $(2+3i)$.