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

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.

• How do you define this index? – Bernard Feb 21 '19 at 19:42
• @Bernard I think the OP wants to find the index of $\mathbb Z[\sqrt 5 ]$ as a subgroup of $\mathcal O$, i.e. the OP want to find the ideal norm. – Kenny Wong Feb 21 '19 at 19:45
• Not sure if this is the first appearance. Take a look anyway. – Jyrki Lahtonen Feb 24 '19 at 5:05
• This feels a bit better. And this seems to be specifically about rings of integers in a number field. – Jyrki Lahtonen Feb 24 '19 at 5:14
• @JyrkiLahtonen on this material and my one-time mse blog, I just ordered this new book by an author I like (for ternary forms; the book will mostly relate binary forms and quadratic number fields) bookstore.ams.org/dol-52?_zs=JL6BH1&_zl=Llbu4 OR bookstore.ams.org/dol-52 author J. L. Lehman, book Quadratic Number Theory, described as excellent for a second number theory class. – Will Jagy Feb 25 '19 at 17:55

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 $$2$$.
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 $$f=2$$.