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Let V be a $\mathbb{Z}[i]$-module generated by elements $v_1$ and $v_2$ with the following relations:

$ (1+i)v_1+(2-i)v_2=0 $ and $ 3v_1+5iv_2=0 $

I need to write V as a sum of cyclic modules. I tried to do this the basic way: putting the matrix $ \left[\begin{array}{cc}1+i&2-i\\3&5i \end{array} \right] $ in the Smith form $ \left[\begin{array}{cc}1&0\\0&3-19i \end{array} \right] $ Now, by the Structure Theorem, we have that $V$ is isomorphic to $\frac{\mathbb{Z}[i]}{\langle 3-19i\rangle}$, which is cyclic because $\langle 3-19i\rangle$ is an ideal of $\mathbb{Z}[i]$. Therefore nothing more needs to be done.

Is that correct?

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    Dear Gustavo, You're welcome. I am writing an answer reflecting our exchange in the comments, just so that this question can move from the *unanswered* to the *answered* category. Best wishes,2012-07-05

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Here is an answer, based on the discussion in the comments:

The method is correct, but there was a calculation mistake. The correct answer is that the module $V$ is isomorphic to $\mathbb Z[i]/\langle -11 + 8i \rangle.$