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).

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