If possible, exhibit an integral basis for a quadratic field whose elements are conjugate

If possible, exhibit and integral basis for \mathbb{Q}(\sqrt{D}) whose elements are conjugate for D \in \{ -3,-1,5\}.


Let D be a squarefree integer, and let \omega = \sqrt{D} if D \not\equiv 1 mod 4 and (1+\sqrt{D})/2 if D \equiv 1 mod 4. Recall that the integers in \mathbb{Q}(\sqrt{D}) have the form a+b\omega where a,b \in \mathbb{Q}. Let \overline{\omega} = -\sqrt{D} if D \not\equiv 1 mod 4 and (1-\sqrt{D})/2 if D \equiv 1 mod 4. Note that \omega\overline{\omega} = -D if D \not\equiv 1 mod 4 and (1-D)/4 if D \equiv 1 mod 4 (so that \omega\overline{\omega} \in \mathbb{Z}), that \omega + \overline{\omega} = 0 if D \not\equiv 1 mod 4 and 1 if D \equiv 1 mod 4, and that \omega - \overline{\omega} = 2\omega if D \not\equiv 1 mod 4 and 2\omega - 1 if D \equiv 1 mod 4.

Note that p(x) = (x - (a+b\omega))(x - (a+b \overline{\omega})) = x^2 - 2ax + (a^2 + ab(\omega+\overline{\omega}) + \omega\overline{\omega}b^2 \in \mathbb{Z}[x]. In particular, if b \neq 0 then p(x) is the minimal polynomial of a+b\omega, so that the conjugates of a+b\omega are itself and a+b\overline{\omega}.

Recall that the discriminant of \mathbb{Q}(\sqrt{D}) is 4D if D \not\equiv 1 mod 4 and D if D \equiv 1 mod 4, and that any integral basis will have this discriminant.

Suppose now that B = \{a+b\omega, a+b\overline{\omega}\} is an integral basis of \mathbb{Q}(\sqrt{D}), with b \neq 0 and a \neq 0. (Certainly if b = 0 then B cannot be a basis, much less an integral basis; if a = 0, then (for example) we cannot generate 1.) Note that a,b \in \mathbb{Z}. Then we must have

\mathsf{det} \left( \left[ \begin{array}{cc} a+b\omega & a + b\overline{\omega} \\ a + b\overline{\omega} & a+b\omega \end{array} \right] \right)^2 = \left\{ \begin{array}{crl} 4D & \mathrm{if} & D \not\equiv 1 \mod 4 \\ D & \mathrm{if} & D \equiv 1 \mod 4. \end{array} \right.

Note that (\omega - \overline{\omega})^2 = 4D if D \not\equiv 1 mod 4 and D if D \equiv 1 mod 4. Evidently, this determinant is (\omega - \overline{\omega})^2 b^2 (2a + b(\omega+\overline{\omega}))^2, and so we have (2a + b(\omega + \overline{\omega}))^2 = 1.

If D \not\equiv 1 mod 4, then \omega+\overline{\omega} = 0, and we have 4a^2 = 1. This equation has no solutions in \mathbb{Z}, and thus \mathbb{Q}(\sqrt{D}) does not have an integral basis whose elements are conjugate.

If D \equiv 1 mod 4, then \omega + \overline{\omega} = 1, and we have (2a+b)^2 = 1, so that 2a+b = \pm 1. This gives a necessary condition on a and b if we wish B to be an integral basis.

Recall that \{1,\omega\} is an integral basis of \mathbb{Q}(\sqrt{D}). If \{\alpha, \beta\} is another integral basis, then there is a (unique) invertible matrix P such that P[1\ \omega]^\mathsf{T} = [\alpha\ \beta]^\mathsf{T}. Say P = \left[ \begin{array}{cc} x & y \\ z & w \end{array} \right], \alpha = a+b\omega, and \beta = a+b\overline{\omega}; comparing coefficients, and recalling that \overline{\omega} = 1 - \omega (since D \equiv 1 mod 4), we have P = \left[ \begin{array}{cc} a & b \\ a+b & -b \end{array} \right]. Computing the determinant, we have -b(2a+b) = \pm 1. Combined with our necessary condition, b = \pm 1. If b = 1, then a = -1, and if b = -1, then a = 1. Thus our only candidates for B are B_1 = \{ -1+\omega, -1+\overline{\omega}\} and B_2 = \{ 1-\omega, 1-\overline{\omega}\}. Or, in terms of \omega, B_1 = \{-1+\omega,\omega\} and B_2 = \{1-\omega,\omega\}. It is easy to see that both B_1 and B_2 are integral bases.

To summarize, if D \equiv 1 mod 4, then \pm B is an integral basis whose elements are conjugate, where B = \{1-\omega,\omega\}. If D \not\equiv 1 mod 4, then no such integral basis exists.

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