Show $\mathbb{Z}\left[i,\frac{1+\sqrt{5}}{2}\right]$ is integrally closed

Question

I have seen proofs that the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{5})$ and $\mathbb{Q}(i)$ are $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ and $\mathbb{Z}[i]$ respectively by showing that $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ and $\mathbb{Z}[i]$ are Euclidean domains, and therefore integrally closed. However, the obvious norm function of $\mathbb{Z}\left[i,\frac{1+\sqrt{5}}{2}\right]$ does not seem to be a Euclidean function. How then could I show that $\mathbb{Z}\left[i, \frac{1+\sqrt{5}}{2}\right]$ is integrally closed, or equivalently, is the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(i,\sqrt{5})$?

Answer 1

Method 1: For number fields $K$ and $L$ with relatively prime discriminants, it turns out that $\mathcal O_{KL} = \mathcal O_K\mathcal O_L$, where the right side is the subring of $KL$ generated by $\mathcal O_K$ and $\mathcal O_L$.

Apply that to the fields $K = \mathbf Q(i)$ and $L = \mathbf Q(\sqrt{5})$, which have relatively prime discriminants $-4$ and $5$: the ring of integers of $KL = \mathbf Q(i,\sqrt{5})$ is $\mathbf Z[i]\mathbf Z[(1+\sqrt{5})/2] = \mathbf Z[i,(1+\sqrt{5})/2]$.

Method 2: Set $E = \mathbf Q(i,\sqrt{5}) = \mathbf Q(i)(\sqrt{5}) = F(\sqrt{5})$, where $F = \mathbf Q(i)$. The extension $E/F$ has degree $2$ and the ring $\mathcal O_F = \mathbf Z[i]$ is a PID, so $\mathcal O_E$ is a free $\mathcal O_F$-module with rank $2$. Set $\alpha = (1+\sqrt{5})/2$, so we want to show $\mathcal O_E = \mathbf Z[i,(1+\sqrt{5})/2] = \mathcal O_F[\alpha]$.

In a number field $K$ of degree $n$, if $A$ is a subring of $\mathcal O_K$ with rank $n$ as a $\mathbf Z$-module and ${\rm disc}_\mathbf Z(A)$ is a squarefree integer, then the equation ${\rm disc}(A) = [\mathcal O_K:A]^2{\rm disc}(\mathcal O_K)$ implies $[\mathcal O_K:A] = 1$, so $\mathcal O_K = A$. A similar result is true when the number field $\mathbf Q$ is replaced by any number field whose ring of integers is a PID, as follows. in an extension of number fields $E/F$ of degree $n$ where $\mathcal O_F$ is a PID, $\mathcal O_E$ is a free $\mathcal O_F$-module with rank $n$. For each $\mathcal O_F$-submodule $M$ of $\mathcal O_E$ with rank $n$ and $\mathcal O_F$-basis $\{e_1,\ldots,e_n\}$, the number $$ \det({\rm Tr}_{E/F}(e_ie_j)) $$ in $\mathcal O_F$ is well-defined up to a unit square in $\mathcal O_F$ (that is, changing the $\mathcal O_F$-basis of $M$ changes this determinant by a unit square in $\mathcal O_F$), so the principal ideal $$ {\rm disc}_{\mathcal O_F}(M) := (\det({\rm Tr}_{E/F}(e_ie_j))) $$ in $\mathcal O_F$ is independent of the choice of basis. We call this ideal the $\mathcal O_F$-discriminant of $M$, and we have the equation of ideals $$ {\rm disc}_{\mathcal O_F}(M) = [\mathcal O_E:M]_{\mathcal O_F}^2{\rm disc}_{\mathcal O_F}(\mathcal O_E), $$ where the ideal $[\mathcal O_E:M]_{\mathcal O_F}$ is the $\mathcal O_F$-index of $M$ in $\mathcal O_E$ (see Section 5, esp. Definition 5.14 here).

Let's apply this to calculating a ring of integers. If $A$ is a subring of $\mathcal O_E$ with rank $n$ as an $\mathcal O_F$-module and ${\rm disc}_{\mathcal O_F}(A)$ is a squarefree principal ideal in $\mathcal O_F$, then the equation of principal ideals $$ {\rm disc}_{\mathcal O_F}(A) = [\mathcal O_E:A]_{\mathcal O_F}^2{\rm disc}_{\mathcal O_F}(\mathcal O_E) $$ implies $[\mathcal O_E:A]_{\mathcal O_F} = (1)$, which (from the definition of $\mathcal O_F$-index) implies $\mathcal O_E = A$.

Apply this to $F = \mathbf Q(i)$ and $E = F(\sqrt{5}) = F(\alpha)$, where $\alpha = (1+\sqrt{5})/2$. Since $\alpha$ has minimal polyomial $x^2 - x - 1$ over $\mathbf Z[i]$, $$ {\rm disc}_{\mathbf Z[i]}(\mathbf Z[i][\alpha]) = ({\rm disc}(x^2 - x - 1)) = (5) = (1+2i)(1-2i), $$ which is squarefree in $\mathbf Z[i]$. Therefore $\mathcal O_E = \mathcal O_F[\alpha] = \mathbf Z[i][(1+\sqrt{5})/2]$.