How to compute the value of the discriminant of $\mathcal{O}_K$

Question

I found a $\mathbb{Z}$-basis $\left\{1, \sqrt{2}, \sqrt{-1}, (\sqrt{2}+\sqrt{-2})/2 \right\}$ for ring of integers $\mathcal{O}_K$ of some number filed $K$ and I need to compute the discriminant of $\mathcal{O}_K$. I know that the formula is $$\det(Tr(a_ia_j), 1 \leq i, j \leq n)$$ where $a_i$ are just numbers in the basis $\left\{1, \sqrt{2}, \sqrt{-1}, (\sqrt{2}+\sqrt{-2})/2 \right\}$. I found that computing each $Tr(a_ia_j)$ is quite tedious to do it by hand and I feel like there should be an easier way to do it?

Answer 1

This is not tedious as you think, since $Tr(a+bi+c\sqrt{2}+di\sqrt{2})=4a$.

The first 3 elements of your basis are then pairwise orthogonal. Moreover, $1$ and $i$ are also orthogonal to $(\sqrt{2}+i\sqrt{2})/2$, so really you just have 5 easy computations to do...

It is then easy to compute the matrix you are interested in. We get $$\pmatrix{4 &0 &0 & 0\cr0 &8 & 0 & 4\cr0 & 0& -4& 0\cr 0 & 4 & 0&0 },$$ which has determinant $4\times (-4)\times (-16)=2^8$.

Side remark. Taking the field of fractions, I see that your number field is $K=\mathbb{Q}(i,\sqrt{2})=\mathbb{Q}(\zeta_8)$, and a integral basis and the discriminant of cyclotomic fields are well-known( so you could just use the basis $\zeta_8^k, k=0,...,3$ instead.) The result I found is consistent with this fact.