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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, I think there is an error in your computations. E.g. the determinant of the original matrix is $-11 + 8i$, which is not the same as $3 - 19i$ (even up to a unit). Regards,2012-07-04
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    Ok, I'm going to make my calculations again. Regardless, can you tell me if the idea is correct?2012-07-04
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    Dear Gustavo, Yes, the idea is correct. Regards,2012-07-04
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    I made my calculations again, I noticed that in one particular step I was multiplying a row by a factor which was not an unit (in $\mathbb{Z}[i]$); now the result is the matrix $$\left[\begin{array}{cc}1&0\\0&-11+8i\end{array}\right]$$ just as you pointed through the determinant. Thus the result is that V is isomorphic to $\frac{\mathbb{Z}[1]}{<-11+8i>}$. Thanks!2012-07-04
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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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