Write $(\mathbb{Z}\times\mathbb{Z})/\mathbb{Z}(20,30)$ as a direct product of cyclic groups.

How does one use the Fundamental Theorem of Finitely Generated Modules over PID to show this? I've started based on something I read elsewhere, but I'm not sure it's right or where to go from here.

Pick generators $\overline{(1,0)}$ and $\overline{(0,1)}$. We have \begin{align} &20\overline{(1,0)} + 30\overline{(0,1)} = \overline{(0,0)}\\ &20[\overline{(1,0)}+\overline{(0,1)}]+10\overline{(0,1)} = \overline{(0,0)} \\ &20\overline{(1,1)} + 10\overline{(0,1)} = \overline{(0,0)} \end{align} Let $\overline{(1,1)}$ be a new generator. Can continue to write \begin{align} &10(2\overline{(1,1)} + \overline{(0,1)}) = \overline{(0,0)} \\ &10\overline{(2,3}) = \overline{(0,0)} \end{align} Use $\overline{(2,3)}$ as a second generator and create a homomorphism $\mathbb{Z}\times\mathbb{Z}\to(\mathbb{Z}\times\mathbb{Z})/\mathbb{Z}(20,30)$ with $(a,b) \mapsto a\overline{(1,1)} + b\overline{(2,3)}$... what result am I looking to get?

• You've essentially done it. You have found generators $u:=(1,1), v:=(2,3)$ of $\mathbb{Z}\times\mathbb{Z}$ such that $\mathbb{Z}(2,3)$ is exactly $\mathbb{Z} (10v)$. So the quotient is isomorphic to $\mathbb{Z}\times\mathbb{Z}_{10}$. – ancientmathematician Mar 23 '18 at 7:54

Notice that $Tor((\mathbb{Z}\times\mathbb{Z})/\mathbb{Z}(20,30)) = (\overline{(2,3)})$ the submodule generated by $\overline{(2,3)}$ which is isomorphic to $\Bbb{Z}/10\Bbb{Z}$. Since $(\mathbb{Z}\times\mathbb{Z})/\mathbb{Z}(20,30)$ is not all torsion, there is a $\Bbb{Z}$ component in the decomposition. Also, you can verify that the rank of the module is $1$. So $(\mathbb{Z}\times\mathbb{Z})/\mathbb{Z}(20,30) \cong \Bbb{Z} \times \Bbb{Z}_{10}$