I have to find a basis for $N\subset \mathbb{Z}[i]^2$ which is generated by $\left\{(3,1),(1,3i),(3i,i+1) \right\}$ as a $\mathbb{Z}[i]$-module, which is not the same as vector spaces. How to start? I know I should use the fundamental theorem for finitely generated abelian groups, but how? Thanks everyone!
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A basis as a $\mathbb Z$-module or as a $\mathbb Z[i]$-module? Or something else? – Aaron Aug 27 '18 at 04:11
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@Aaron as $\mathbb{Z}[i]$-module, I'm going to edit. – Aug 27 '18 at 04:12
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1I would make a matrix where those generators are the rows and try to do Gaussian elimination (taking care to only do invertible row operations) until you can either eliminate some elements or see why no elements should be eliminated. – Aaron Aug 27 '18 at 04:20
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1Related. One can do this systematically using Smith normal form. – Viktor Vaughn Aug 27 '18 at 12:13
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Write $u=(3,1)$, $v=(1,3i)$ and $w=(3i,1+i)$. Then $u-3v$, $v$ and $w-3iv$ span the same $\Bbb Z[i]$-module as $u$, $v$ and $w$. Moreover, $u-3v=(0,\alpha)$ and $w-3iv=(0,\beta)$ for some $\alpha$, $\beta\in\Bbb Z[i]$, and these two elements span the same submodule as $(0,\gcd(\alpha,\beta))$ (with the gcd inside $\Bbb Z[i]$) etc.
Angina Seng
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