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If I were to visualize the ideal $(2, 3+3i)$ of $Z[i]$ on the complex plane, I would find a gcd of $2$ and $3+3i$ (for example $1+i$) and the ideal $(2, 3+3i)$ is identical to $(1+i)Z[i]$, which forms a lattice generated by $(1+i)$ and $(1+i)i$. But $6$ and $2+2\sqrt{-5}$ has no gcd in $Z[\sqrt{-5}]$, and I'm not sure $(6, 2+2\sqrt{-5})$ forms a lattice on the complex plane. Is any (especially non-principal) ideal a lattice? If so, how can I effectively find out the basis of the lattice?

Thanks in advance

user26857
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sacch
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  • Surely it is a lattice, because it is a free abelian group of rank two. As an abelian group your example ideal is generated by $6,6\sqrt{-5}$,$2+2\sqrt{-5}$ and $(2+2\sqrt{-5})\sqrt{-5}=-10+2\sqrt{-5}$. Can you make progress and find a basis for the free abelian group? Some of the proofs for the fact that a submodule of a f.g. free module over a PID actually can be read as giving an algorithm for finding a basis for that submodule. That may be overkill here, but it's not hard. Here, of course $\Bbb{Z}$ takes the role of that PID, and $\Bbb{Z}[\sqrt{-5}]$ the role of the f.g. free module. – Jyrki Lahtonen Sep 27 '15 at 10:41
  • Hint: The proof I'm referring to examines the set of first coordinates, observes that they form an ideal $I$ of the PID, locates an element of the submodule whose first coordinate is a generator of $I$, and then proceeds to study the smaller submodule consisting of elements with a vanishing first coordinate. Rinse. Repeat (here rank is two, so...). – Jyrki Lahtonen Sep 27 '15 at 10:45

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