Z[i]-module Question

**Erdos32212** · December 8th 2008, 06:03 PM

Let $M$ be the $\mathbb{Z}[i]$ -module generated by the elements $v_1$ , $v_2$ such that $(1+i)v_1+(2-i)v_2=0$ and $3v_1+5iv_2=0$ . Find an integer $r \geq 0$ and a torsion $\mathbb{Z}[i]$ -module $T$ such that $M \cong \mathbb{Z}[i]^r \times T$ .

**NonCommAlg** · December 9th 2008, 07:15 AM

Originally Posted by Erdos32212

Let $M$ be the $\mathbb{Z}[i]$ -module generated by the elements $v_1$ , $v_2$ such that $(1+i)v_1+(2-i)v_2=0$ and $3v_1+5iv_2=0$ . Find an integer $r \geq 0$ and a torsion $\mathbb{Z}[i]$ -module $T$ such that $M \cong \mathbb{Z}[i]^r \times T$ .

$r=0$ because $M$ is a torsion $\mathbb{Z}[i]$ module. this is very easy to see: $0=3(1-i)[(1+i)v_1 + (2-i)v_2]-2(3v_1+5iv_2)=(3-19i)v_2.$ thus $v_2$ is torsion. similarly you can show that $v_1$ is

torsion too. thus $M$ is a torsion module. $\Box$

**Erdos32212** · December 9th 2008, 01:47 PM

Thanks! I have only a few questions. How do we show that $v_1$ is torsion? I know how you did it for $v_2$ , but getting things to cancel in complex analysis is not so easy [I am having trouble getting $v_2$ to cancel]. Also, did we find the torsion $\mathbb{Z}[i]$ -module such that $M \cong \mathbb{Z}[i]^r \times T$ or does $T=M$ ?.

**NonCommAlg** · December 9th 2008, 06:13 PM

Originally Posted by Erdos32212

Thanks! I have only a few questions. How do we show that $v_1$ is torsion? I know how you did it for $v_2$ , but getting things to cancel in complex analysis is not so easy [I am having trouble getting $v_2$ to cancel].

$0=(2-i)(3v_1+5iv_2) - 5i[(1+i)v_1 + (2-i)v_2]=(11-8i)v_1.$ so since every $v \in M$ is a linear combination of $v_1,v_2,$ we will have $(11-8i)(3-19i)v=0.$

Also, did we find the torsion $\mathbb{Z}[i]$ -module such that $M \cong \mathbb{Z}[i]^r \times T$ or does $T=M$ ?.

since $M$ is torsion, we can just let $r=0$ and $T=M.$

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