I want to show that the integral closure of $\mathbb{Z[\sqrt{5}]}$ in $\text{Quot}\mathbb{Z}[\sqrt{5}]=\mathbb{Q}[\sqrt{5}]$ is $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$.
my attempt: Let $\phi:=\frac{1+\sqrt{5}}{2}.$ if $a+b\sqrt{5}\in \mathbb{Q}[\sqrt{5}]$ is integral over $\mathbb{Z}[\sqrt{5}]$, then by the tower property it is also integral over $\mathbb{Z}$, as $\mathbb{Z}[\sqrt{5}]$ is integral over $\mathbb{Z}$, so we can consider $g\in\mathbb{Z}[x]$ such that $g(a+b\sqrt{5})=0$. I also know that in this case, the conjugate $a-b\sqrt{5}$ is integral too, which implies $2a$ and $a^2-5b^2\in\mathbb{Q}$ are also integral, thus, they must be integers, as $\mathbb{Z}$ is normal. It suffices to show that $a+b\sqrt{5}=c+d\phi$, for some $c,d\in\mathbb{Z}$, and for that i tried to write $c,d$ in terms of $a$, $b$ and use the fact that $2a$ and $a^2-5b^2\in \mathbb{Z}$, to show that $c,d\in\mathbb{Z}$, but i'm stucked here. Any hints? maybe another way to go about it.
Lemma: If $R$ is a normal domain with fracion field $K$ and $L|_K$ is a finite field extension, then $\alpha \in \tilde R$, the normalization of $R$ in $L$ iff the minimal monic polynomial of $\alpha$ over $K$ must have roots in $R$.
This is fairly easy to see once you observe that the coe-fficients of $f:=$ the minimal polynomial of $\alpha$ are polynomials in the roots of $f$ and hence integral (since each root is conjugate to $\alpha$ over $K$) over $R$ as well as they belong to $K$. (you will need that $R$ is normal)
Say $a+b\sqrt 5\in \mathbb Q(\sqrt 5)$ be integral over $\mathbb Z$. If $b= 0$, $a\in \mathbb Z$ by the lemma. So assume $b\neq 0$.
The minimal polynomial is $X^2-2aX+a^2-5b^2$. So $2a\in \mathbb Z$ and $a^2-5b^2\in \mathbb Z.$ Also
Let $a=\frac{m}{2}$ and $a^2-5b^2=n \implies \frac{m^2-20b^2}{4}\in \mathbb Z$ Thus $20b^2$ is an integer. It is not very hard to see that $4b^2$ must be an integer too (Write $b$ in lowest terms and see the prime factorizations of both numerator and denominator). So $b$ can at most be a half integer.
Now $a\in \mathbb Z\implies 5b^2\in \mathbb Z\implies b^2\in \mathbb Z\implies b\in \mathbb Z$
Conversely, $b\in \mathbb Z\implies a^2\in \mathbb Z\in \implies a\in \mathbb Z $
Thus either both $a,b$ are integers or both $a,b$ are strict half-integers
In the first case $a+b \sqrt 5=a-b +2b\left(\frac{1+\sqrt 5}{2} \right) \in \mathbb Z\left [\frac{1+\sqrt 5}{2} \right ]$
In the 2nd case $a+b\sqrt 5=a-b+2b\left(\frac{1+\sqrt 5}{2} \right) \in \mathbb Z\left [\frac{1+\sqrt 5}{2} \right ] \ \because $ in this case both $a-b$ and $2b$ are integers.