Ok, boring question here (I guess, at least). Let $R=\mathbb Z[\sqrt{-5}]$. Let $M=\langle 2,1+\sqrt{-5} \rangle$ the $R$-module generated by $2$ and $1+\sqrt{-5}$. I am asked to show that $M \times M = \langle (2,1+\sqrt{-5}), (1-\sqrt{-5},2) \rangle$, i.e. that $M \times M$ is generated by the two elements $(2,1+\sqrt{-5})$ and $(1-\sqrt{-5},2)$, and that $M \times M$ is free.
I was able to prove that $(2,1+\sqrt{-5})$ and $(1-\sqrt{-5},2)$ are linearly independent, but I can't find any way to show that they generate the whole $M \times M$ module.
I tried to proceed as follows: for any $a,b,c,d,e,f,g,h \in \mathbb Z$, I must find $\alpha, \beta, \gamma, \delta \in \mathbb Z$ such that $$\big( 2(a+b\sqrt{-5})+(1-\sqrt{-5})(c+d\sqrt{-5}), 2(e+f\sqrt{-5})+(1-\sqrt{-5})(g+h\sqrt{-5}) \big)$$ $$=$$ $$\big( 2(\alpha+\beta\sqrt{-5})+(1-\sqrt{-5})(\gamma+\delta\sqrt{-5}), (1+\sqrt{-5})(\alpha+\beta\sqrt{-5})+2(\gamma+\delta\sqrt{-5}) \big).$$
After some computations, this lead me to $$\big( 2a+c+5d+(2b+d-c)\sqrt{-5}, 2e+g+5h+(2f+h-g)\sqrt{-5} \big)$$ $$=$$ $$\big( 2\alpha + \gamma + 5\delta+(2\beta+\delta-\gamma)\sqrt{-5}, 2\gamma + \alpha - 5\beta + (2\delta+\beta+\alpha)\sqrt{-5}) \big),$$
which means that I have to solve the following system: $$ \begin{cases} 2\alpha + \gamma + 5\delta= 2a+c+5d\\ 2\beta+\delta-\gamma= 2b+d-c\\ 2\gamma + \alpha - 5\beta= 2e+g+5h\\ 2\delta+\beta+\alpha= 2f+h-g\\ \end{cases} $$
and I don't know how to go on.
Any help is much appreciated! Thanks!
Let $I = \langle(2,1+\sqrt{-5}),(1-\sqrt{-5},2)\rangle$.
As you've probably seen, $2 \times 2 - (1+\sqrt{-5})(1-\sqrt{-5}) = -2 \neq 0$ so $I$ is a free $R$-module of rank $2$, and we even have an $R$-basis of $I$.
More importantly this calculation also shows that $(0,2)$ and $(2,0)$ are in $I$ (just do the combination that cancels each component in the obvious way).
Then it's immediate to see that $(0,1+\sqrt{-5})$ and $(1-\sqrt{-5},0)$ are also in $I$, and finally that $M \times M \subset I$.
The other inclusion is obvious : since $2,1+\sqrt{-5},1-\sqrt{-5} \in M$, $I \subset M \times M$.
Hence $M \times M$ is the free $R$-module of rank $2$ generated by $(2,1+\sqrt{-5})$ and $(1-\sqrt{-5},2)$