Inverse of the principal ideal $(\sqrt{-15})$ We work in the subring $R= \mathbb{Z} + \mathbb{Z}(\frac{1 + \sqrt{-15}}{2})$
 of $\mathbb{Q} + \mathbb{Q}(\frac{1 + \sqrt{-15}}{2})$.
The exercise asks us the give the number of principal ideals of norm 15 in $R$ and to give generators for its inverse. 
Since $\sqrt{-15} = -1 + 2 \cdot \frac{1 + \sqrt{-15}}{2}$, $\sqrt{-15} \in 
\mathbb{Z} + \mathbb{Z}(\frac{1 + \sqrt{-15}}{2})$.
Hence $(\sqrt{-15})$ is a principal ideal of $\mathbb{Z} + \mathbb{Z}(\frac{1 + \sqrt{-15}}{2})$.
We have that $(\sqrt{-15})(\frac{1}{\sqrt{-15}})=(1)$. But $\frac{1}{\sqrt{-15}} \not\in \mathbb{Z} + \mathbb{Z}(\frac{1 + \sqrt{-15}}{2})$. Does this mean that there is no inverse for $(\sqrt{-15})$ in $R$ ? I am quite confused. 
Again, any help would be appreciated. 
 A: Let $I =J= \sqrt{-15}R$ then $IJ = \{ ab, a \in I,b \in J\}$ generates an ideal : $IJ = \sqrt{-15} R\sqrt{-15} R = (-15) R R = 15 R$ (when the ideals are generated by more than one element, interpret the right multiplication by $R$ as sending  sets $\{u_1,\ldots,u_j\} $ to  $\sum_j u_jR$ not $\bigcup_j u_j R$)
For any $c\ne 0 \in R$ (we can restrict to $c \in \Bbb{Z}$)  let $\frac{J}{c} = \{ \frac{a}{c}, a\in J\}$. It is a fractional ideal. Then $I \frac{J}{15} = R$ so that $\frac{J}{15}$ is the inverse of $I$ in the group of fractional ideals.
Why is it a group ? Because $R$ is a Dedekind domain, for any prime ideal $\mathfrak{p}$ it contains a prime number $p \in \Bbb{Z}$ and $pR = \prod_j \mathfrak{p}_j$, one of the $\mathfrak{p}_i$ is $\mathfrak{p}$ so $\mathfrak{p} \frac{\prod_{j\ne i} \mathfrak{p}_j}{p} = R$.
To prove $R$ is a Dedekind domain we show the inverse is $(Frac(R):I) = \{ a \in Frac(R), aI \subset R\}$.
This doesn't hold in $S=\Bbb{Z}[2i]$ (try with $\mathfrak{p} = (2,2i)$ the only prime ideal above $(2)$ and $\mathfrak{p}^2 = (4,4i)$)
A: Hint: Let $\alpha:=\frac{1 + \sqrt{-15}}{2}$. Check that $\alpha\cdot(1-\alpha)=4$. Also recall that fractional ideals of $R$ need not be contained in $R$.
