Let $M = \langle \alpha \rangle$ be a cyclic $\mathbb{Z}$-module of order 12.
My goal is to list the pure submodules of $M$. A submodule $N$ is pure in $M$ if for $y \in N$ and $a \in \mathbb{Z}$ and there exists $x\in M$ such that $ax=y$, then actually $\exists z\in N$ such that $az=y$.
I am really rusty on my algebra. My thoughts are this.
$\mathbb{Z}$-modules are in correspondence with abelian groups. Since $M$ is cyclic of order 12, then I will stick to what I know and look at $\mathbb{Z}_{12}$. The non-trivial subgroups of this are $\mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_4, \mathbb{Z}_6$.
I read somewhere that pure submodules are summands of the module $M$. (That is, if $N$ is pure then $\exists L$ disjoint from $N$ such that $M = N+L$). And, the converse is true. If $N$ is a summand, then it is pure. The proof is somewhere on this site, but I am having trouble finding it.
From my knowledge of group theory, the only way to write $\mathbb{Z}_{12}$ as a direct product $\mathbb{Z}_{12} = \mathbb{Z}_4 \times \mathbb{Z}_3$. I suppose this means that the only way to write $M$ as a direct sum of its submodules is in the same way. Thus, the only pure submodules are those corresponding to $\mathbb{Z}_4$ and $\mathbb{Z}_3$. I guess to be pedantic, $0$ and $M$ are also pure.
Are these thoughts correct?
Best Answer
An easy way to check is to note that $\mathbb{Z}_2$ is not pure (with elements of $\mathbb{Z}_{12}$ in bold, $2\cdot\mathbf{3} = \mathbf{6}$, but for the two elements of the copy of $\mathbb{Z_2}$ we have $2\cdot\mathbf{0} = \mathbf{0} \neq \mathbf{6}$ and $2\cdot\mathbf{6} = \mathbf{0} \neq \mathbf{6}$, so there is no element of $\mathbb{Z}_2$ that doubles to give $\mathbf{6}$), and the same for $\mathbb{Z}_6$ ($2 \cdot \mathbf{3} = \mathbf{6}$, but doubling any element of the copy of $\mathbb{Z}_{6}$ in $\mathbb{Z}_12$ gives either $0$, $4$, or $8$, so there is no element of $\mathbb{Z}_6$ that doubles to give $\mathbf{6}$.