A basis of algebraic integers need not be an integral basis

Exhibit a basis for an algebraic extension \mathbb{Q}(\theta) which consists of integers but which is not integral.

Suppose D is a squarefree integer congruent to 1 mod 4 and consider \mathbb{Q}(\sqrt{D}). Certainly \{1, \sqrt{D}\} is a basis of this field extension of \mathbb{Q}. Moreover, both 1 and \sqrt{D} are algebraic integers (trivially and via x^2 - D, respectively).

However, note that \frac{1}{2} + \frac{1}{2} \sqrt{D} is an algebraic integer, since it is a root of p(x) = x^2 - x + \frac{1-D}{4}. It is not, however, a \mathbb{Z}-linear combination of \{1,\sqrt{D}\}. So this basis is not integral.

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