Computing the index $\left(\mathbb Z\left[\frac{1+\sqrt{5}}{2}\right]:\mathbb Z \left[\sqrt{5}\right]\right)$?

Question

If $\mathcal O=\mathbb Z\left[\frac{1+\sqrt{5}}{2}\right]$, I'd like to show that $\left(\mathcal O:\mathbb Z \left[\sqrt{5}\right]\right)=2$. It's to show that Dedekind's factorisation theorem doesn't apply to the prime 2. I feel like this should be some basic computation and I'm just over thinking it/forgetting some of my algebra, but what is the best way to do this? I tried to define a map from $\mathcal O$ to $\mathbb F_2$ with kernel $\mathbb Z\left[\sqrt{2}\right]$ but things weren't working out. I was also thinking of looking at these things as rank 2 $\mathbb Z$-modules, but this seemed like overkill maybe. Thanks.

Answer 1

Let $\theta=\frac{1+\sqrt{5}}{2}$. Then $\mathbb Z[\sqrt{5}]=\mathbb Z 1 + \mathbb Z 2\theta$ and $\mathcal O = \mathbb Z 1 + \mathbb Z \theta$. Therefore, $\left(\mathcal O:\mathbb Z [\sqrt{5}]\right)= 2$.

Equivalently, write $$ \pmatrix{1 \\ \sqrt 5} = \pmatrix{\hphantom{-}1 & 0 \\ -1 & 2} \pmatrix{1 \\ \frac{1+\sqrt{5}}{2}} $$ and note that the determinant of the matrix is $2$.

Answer 2

For a quadratic number field $K=\mathbf Q(\sqrt d)$, where $d\in \mathbf Z$ is square free, you know that its ring of integers $O_K$ is the ring $\mathbf Z[(1+\sqrt d)/2]$ if $d \equiv 1$ mod $4$, $\mathbf Z[\sqrt d]$ otherwise. This can be uniformly written as $O_K=\mathbf Z[(\delta +\sqrt \delta)/2]$, where $\delta$ is the discriminant of $K$. Considering only the additive structure, $O_K$ can be viewed as a $\mathbf Z$ - module, necessarily without torsion (because included in a field), hence $\mathbf Z$ - free with $\mathbf Z$-basis {${1,(\delta +\sqrt \delta)/2}$}. Given a subring $R$ of $O_K$ which is a submodule of $\mathbf Z$-rank $2$, you ask for the index $f=(O_K : R)$. Example : $K=\mathbf Q(\sqrt 5), R=\mathbf Z[\sqrt 5]$.

Let us show that {$1, f\delta$} is a $\mathbf Z$-basis of $R$. By definition of the index, $fO_K \subset R$, so $\mathbf Z +fO_K \subset R$. But $\mathbf Z +fO_K$ has $\mathbf Z$-basis {$1, f\delta$}, hence obviously $(O_K : \mathbf Z +fO_K)=f$, and then $R=\mathbf Z +fO_K$, done. We can apply this in the opposite sense, starting from a basis of $R$ to catch the index. In your example, we get $f=2$.