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I need to find a generator for $V$, being $V$ the $\mathbb Z[i]$-module generated by $v_1,v_2$ where:

$(1+i)v_1 + (2-i)v_2 = 0$

$3v_1 + 5iv_2 = 0$

[EDIT] To clarify $V = \frac{\mathbb Z[i]^2}{A\mathbb Z[i]}$ where A is the traspose matrix of coefficients of the system above. Then $V$ is the $\mathbb Z[i]-$module presented by $A$

I have come to the conclusion that $V \cong \mathbb Z[i]/(11-8i)$ by calculating Smith´s form of the matrix formed from the traspose of the coefficient matrix from the system above.

Given a cyclic group, e.g. $\mathbb Z/(3)$; I know that the element $[1]$ generates de whole group, because $[1]$ has order $3$ there, and I wonder if, using this logic, $[1]$ generates the cyclic group mentioned above ($\Bbb Z[i]/(11-8i)$).

Once I know this, I know the only thing I´ve got to do to find a generator for $V$; is apply the inverse isomorphisms, which would only be a matter of calculus.

I could really use some help, thanks beforehand!

dominnpm_
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  • @DietrichBurde because we would call a group cyclic, when it is isomorph to another group with one and only one direct factor in its form according to the first theorem of structure. (At least is what I have understood) – dominnpm_ Dec 13 '21 at 19:46
  • We have $\Bbb Z[i]/(11-8i)\cong \Bbb Z_{185}$, see here. – Dietrich Burde Dec 13 '21 at 19:52

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