calculate $II=${$xy \mid x,y \in I$} for $I=(2,1＋\sqrt{-5}) \in \Bbb{Z}[ \sqrt{-5}]$ Let $I=(2,1＋\sqrt{-5})⊂ \Bbb{Z}[ \sqrt{-5}]$ be ideal generated by $2$ and $1＋\sqrt{-5}$.
I want to calculate $II$ as a set. Product as a ideal $I・I$ is $(2)$,  but here I want to calculate $II=${$xy \mid x,y \in I$} for $I=(2,1＋\sqrt{-5}) ⊂ \Bbb{Z}[ \sqrt{-5}]$. But I'm having trouble to do that. I would be appreciated if you could tell me process of calculation.
 A: I suspect the OP may just be starting a course in algebraic number theory, so this may not help until the end, but the answer is that $z \in \mathbb{Z}[\sqrt{-5}]$ can be written as $xy$ for $x$, $y \in I$ if and only if
(1) $z$ is divisible by $2$ and
(2) $z/2$ is divisible by some nonprincipal prime ideal of $\mathbb{Z}[\sqrt{-5}]$.

Proof: First of all, since the ideal $I^2$ is $(2)$, we know that $2$ must divide $z$. Assume this from now on.
Now, suppose that $z = xy$ for $x$, $y$ in $I$. Then the ideals $(x)$ and $(y)$ factor as $IJ$ and $IK$, so $(z) = 2 JK$ or $(z/2) = JK$. Since $IJ$ and $IK$ are principal, $J$ and $K$ are non-principal, so $(z/2)$ factors into nonprincipal ideals, and thus $(z/2)$ is divisible by a nonprincipal prime.
Conversely, suppose that $(z/2) = JK$ for $J$ non-principal. Then $K$ is also non-principal. Since the class group of $\mathbb{Z}[\sqrt{-5}]$ is cyclic of order $2$, this means that $IJ$ and $IK$ are principal; say $IJ = (x)$ and $IK = (y)$. Then $(z) = (2) (z/2) = I^2 JK = (IJ) (IK) = (x) (y)$ so $z = \pm xy$. Absorbing the sign into one of $x$ or $y$, we are done. $\square$.

The nonprincipal prime ideals of $\mathbb{Z}[\sqrt{-5}]$ are the factors of those primes $p \in \mathbb{Z}$ which are $2$, $3$ or $7 \bmod 20$. So we can restate this criterion by saying that $z=a + b \sqrt{-5}$ factors as $xy$ for $x$, $y \in I$ if and only if
(1) $a$ and $b$ are even and
(2) $(a/2)^2+5(b/2)^2$ is divisible by a prime which is $2$, $3$ or $7 \bmod 20$.
