Find a basis for QQ(sqrt(D)) whose elements are conjugate

Let D \in \mathbb{Z} be squarefree. Find a basis \alpha, \overline{\alpha} for \mathbb{Q}(\sqrt{D}) where \overline{\alpha} is the conjugate of \alpha.

Let \alpha = a+b\sqrt{D}. Evidently, \alpha is a root of p(x) = x^2 - 2ax + (a^2 - Db^2); the other root is a - b\sqrt{D}. If b \neq 0, then this polynomial has no rational roots, and so is irreducible over \mathbb{Q}. In this case, p(x) is the minimal polynomial of \alpha, and so \overline{\alpha} = a - b\sqrt{D}.

Note that if b = 0, then \{\alpha,\overline{\alpha}\} cannot be a basis for \mathbb{Q}(\sqrt{D}) as it contains only one element. Thus we may assume that b \neq 0.

We claim that \{\alpha, \overline{\alpha}\} is a basis for \mathbb{Q}(\sqrt{D}) so long as a \neq 0. To that end, suppose first that \{\alpha,\overline{\alpha}\} is a basis; if a = 0, then we have \alpha = b\sqrt{d} and \overline{\alpha} = -b\sqrt{D}. Now 1 = xb\sqrt{D} - yb\sqrt{D}, with x,y \in \mathbb{Q}. So \sqrt{D} = 1/b(x-y), a contradiction since \sqrt{D} is not rational. Thus a \neq 0.

Conversely, suppose a \neq 0. To show that \{\alpha,\overline{\alpha}\} is a basis, it suffices to show that it generates \mathbb{Q}(\sqrt{D}), as we know that this field has degree 2 over \mathbb{Q}. To achieve this, we will show that 1 and i are linear combinations of \alpha and \overline{\alpha}. Indeed, evidently we have 1 = \frac{1}{2a}\alpha + \frac{1}{2a}\overline{\alpha} and i = \frac{1}{2b}\alpha - \frac{1}{2b} \overline{\alpha}, as desired.

In particular, \{1+\sqrt{D}, 1-\sqrt{D}\} is a basis for \mathbb{Q}(\sqrt{D}) over \mathbb{Q}.

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