Describe the integral bases of a given quadratic field

For each of the fields E_1 = \mathbb{Q}(-5 + \sqrt{-28}) and E_2 = \mathbb{Q}(1 - \sqrt{75/9}), do the following:

  1. Describe the integers.
  2. Give an integral basis.
  3. Describe all the other integral bases.
  4. Compute the discriminant of the field.

Note that E_1 = \mathbb{Q}(\sqrt{-7}), since -5 + \sqrt{-28} = -5 + 2\sqrt{-7} and \sqrt{-7} = (5 + 9-5 + \sqrt{-28})/2. Since -7 \cong 1 mod 4, the integers in E_1 have the form a + b(1+\sqrt{-7})/2 where a, b \in \mathbb{Z}, \{1, \omega\} is an integral basis (where \omega = (1+\sqrt{-7})/2), and the discriminant is -7.

Suppose \{\alpha,\beta\} is another integral basis; then we have A[1\ \omega]^\mathsf{T} = [\alpha\ \beta]^\mathsf{T}, where A is an invertible matrix over \mathbb{Z}. That is, \mathsf{det}(A) = \pm 1. Letting A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right], we have \alpha = a+b\omega and \beta = c+d\omega where ad-bc = \pm 1.

Similarly, E_2 = \mathbb{Q}(\sqrt{3}) since 1 - \sqrt{75/9} = 1 - (5/3)\sqrt{3} and \sqrt{3} = (-1+(1-\sqrt{75/9}))/(-5/3). Since 3 \not\equiv 1 mod 4, the integers in E_2 have the form a+b\sqrt{3} with a,b \in \mathbb{Z}, \{1, \sqrt{3}\} is an integral basis, and the discriminant is 12. If \{\alpha,\beta\} is another integral basis, then there exists an invertible matrix A such that A[1\ \sqrt{3}]^\mathsf{T} = [\alpha\ \beta]^\mathsf{T}. As before, \alpha = a+b\sqrt{3} and \beta = c+d\sqrt{3} where ad-bc = \pm 1.

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