Find a basis for a given ideal in a quadratic integer ring

Let I = (3+i,7+i) be considered as an ideal in \mathbb{Z}[i]. Find a generator and a basis for I. Draw a diagram to visualize the elements of I on the complex plane.

Note that 7+i = (1+i)(2-i) and 3+i = (1+i)(4-3i), so that I \subseteq (1+i). Conversely, since -2i(3+i) + i(7+i) = 1+i, we have (1+i) \subseteq I. Hence I = (1+i).

We claim that B = \{1+i, -1+i\} is a basis for I over \mathbb{Z}. To see this, note that -1+i = i(1+i), so that (1+i,-1+i)_\mathbb{Z} \subseteq (1+i); certainly (1+i) \subseteq (1+i, -1+i)_\mathbb{Z}, so these sets are equal. Now suppose a(1+i) + b(-1+i) = 0. Comparing coefficients, we have a-b = 0 and a+b = 0, so that a = b = 0. Thus B is free as a generating set for I over \mathbb{Z}.

We can visualize this ideal in the complex plane as in the following diagram.

(1+i) in ZZ[i]

The ideal (1+i) in the Gaussian integers

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