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

Show that A = \{4a+2\sqrt{2}b \ |\ a,b \in \mathbb{Z}\} is an ideal in the ring of algebraic integers in \mathbb{Q}(\sqrt{2}). Find a generator and a basis for this ideal.


First, since 2 \not\equiv 1 mod 4, the elements of A are certainly algebraic integers. (Integers in \mathbb{Q}(\sqrt{2}) have the form m+n\sqrt{2} with m,n \in \mathbb{Z}.) We claim that A = (2\sqrt{2}). To see this, note that if 4a+2\sqrt{2}b \in A, then 4a+2\sqrt{2}b = 2\sqrt{2}(b+a\sqrt{2}) \in (2\sqrt{2}). Conversely, if 2\sqrt{2}(a+b\sqrt{2}) \in (2\sqrt{2}), then 2\sqrt{2}(a+b\sqrt{2}) = 4b+2\sqrt{2}a \in A. So A = (2\sqrt{2}), and more generally, A is an ideal in the ring of integers in \mathbb{Q}(\sqrt{2}).

Now we claim that B = \{4, 2\sqrt{2}\} is a basis of A. It is certainly a generating set over \mathbb{Z}. Suppose now that 4a + 2\sqrt{2}b = 0. If b \neq 0, then \sqrt{2} = -4a/2b, so that \sqrt{2} is rational, a contradiction. So b = 0, and we have 4a = 0, so a = 0. Thus B is \mathbb{Z}-linearly independent, and hence a basis for A.

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