## 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 $ad-bc = \pm 1$.