Consider the $\mathbb{Z}$-module $M=\mathbb{Z}^2$, $$ M_1=\{(a_1,a_2):2a_1+3a_2=0\} \text{ and } M_2=\{(a_1,a_2):a_1+a_2=0\}$$ Prove that $M$ is the internal direct sum of $M_1$ and $M_2$.
To prove the above I need that any $(n,m)\in \mathbb{Z}^2$ is expressable as the sum of an element of $M_1$ and $M_2$.
I don't want anyone to prove that for me, but I'm confused as to what form the elements of $M_1$ and $M_2$ take. I've tried a few different things, but heres what I have so far:
$a_1+a_2=0\Rightarrow a_1=-a_2$ so $M_2$ are elements in $\mathbb{Z}^2$ of the form $a_1(1,-1)$ for $a_1\in \mathbb{Z}$.
Then $2a_1+3a_2=0\Rightarrow a_1=\frac{-3a_2}{2}$ so $M_1$ are elements in $\mathbb{Z}^2$ of the form $a_2(\frac{-3}{2},1)$.
But I do not believe the above is correct since $(\frac{-3}{2},1)$ is not an element of $\mathbb{Z}^2$. (Or is it?)
By solving their system of equations it is easy enough to show that $m_1+m_2=0$ only when $0=m_1=m_2$ for $m_1\in M_1$ and $m_2\in M_2$. So I'm not concerned about that part.
Any clarification is greatly appreciated.
You're quite right about $\left(-\frac32,1\right)$ not being an element of $\Bbb Z^2,$ since $-\frac32$ isn't an integer. What you'll find, though, is that $a_2$ is even for all $(a_1,a_2)\in M_1,$ so instead, the elements of $M_1$ are the integer multiples of $(-3,2).$