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Let $\displaystyle M$ be the $\displaystyle \mathbb{Z}[i]$-module generated by the elements $\displaystyle v_1$, $\displaystyle v_2$ such that $\displaystyle (1+i)v_1+(2-i)v_2=0$ and $\displaystyle 3v_1+5iv_2=0$. Find an integer $\displaystyle r \geq 0$ and a torsion $\displaystyle \mathbb{Z}[i]$-module $\displaystyle T$ such that $\displaystyle M \cong \mathbb{Z}[i]^r \times T$.
$\displaystyle r=0$ because $\displaystyle M$ is a torsion $\displaystyle \mathbb{Z}[i]$ module. this is very easy to see: $\displaystyle 0=3(1-i)[(1+i)v_1 + (2-i)v_2]-2(3v_1+5iv_2)=(3-19i)v_2.$ thus $\displaystyle v_2$ is torsion. similarly you can show that $\displaystyle v_1$ isQuote:
Originally Posted by Erdos32212
torsion too. thus $\displaystyle M$ is a torsion module. $\displaystyle \Box$
Thanks! I have only a few questions. How do we show that $\displaystyle v_1$ is torsion? I know how you did it for $\displaystyle v_2$, but getting things to cancel in complex analysis is not so easy [I am having trouble getting $\displaystyle v_2$ to cancel]. Also, did we find the torsion $\displaystyle \mathbb{Z}[i]$-module such that $\displaystyle M \cong \mathbb{Z}[i]^r \times T$ or does $\displaystyle T=M$?.
$\displaystyle 0=(2-i)(3v_1+5iv_2) - 5i[(1+i)v_1 + (2-i)v_2]=(11-8i)v_1.$ so since every $\displaystyle v \in M$ is a linear combination of $\displaystyle v_1,v_2,$ we will have $\displaystyle (11-8i)(3-19i)v=0.$Quote:
Originally Posted by Erdos32212
since $\displaystyle M$ is torsion, we can just let $\displaystyle r=0$ and $\displaystyle T=M.$Quote:
Also, did we find the torsion $\displaystyle \mathbb{Z}[i]$-module such that $\displaystyle M \cong \mathbb{Z}[i]^r \times T$ or does $\displaystyle T=M$?.
