# Some questions on invariant factor decomposition of modules

Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^2$ and $N=\langle(6,4),(4,8),(4,0)\rangle\le\mathbb{Z}^2$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^2$ such that the matrix of the homomorphism w.r.t. standard bases is $A=\begin{bmatrix}6 & 4 & 4\\4 & 8 & 0\end{bmatrix}$. Then there is an equivalent matrix $D$ such that $XAY=D$ where $X=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}$, $D=\begin{bmatrix}2 & 0 & 0\\0 & 8 & 0 \end{bmatrix}$, $Y=\begin{bmatrix}1 & -2 & 2\\-1 & 2 & -1\\0 & 1 & -2\end{bmatrix}$. Since $D$ represents $\varphi$ and $N=\operatorname{im}(\varphi)$, we conclude that $M=\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$.

I am trying to understand why $N=\operatorname{im}(\varphi)$ and $\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$.

As stated in the first line of question, how can we conclude immediately that $N=\operatorname{im}(\varphi)$?
In the final step, how can we conclude from "$D$ represents $\varphi$ and $N=\operatorname{im}(\varphi)$" to "$M=\mathbb{Z}^2/\operatorname{im}(\varphi)\cong\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$" ? How exactly do we get 2 and 8? What I know is that 2 and 8 are invariant factors of $M$, but not entirely sure how do we derive that.

Thanks!

• Did you notice you did not define the hom. $\,\phi:\Bbb Z^3\to\Bbb Z^2\,$ ? Thus we cannot know why $\,\text{Im}(\phi)=N\,$ ... – DonAntonio May 21 '13 at 15:45
• What book are you reading on this subject? – DonAntonio May 21 '13 at 15:45
• @DonAntonio. Sorry I forgot to mention that. I thought the definition is not important because I have seen other example where definition is not given but the matrix can still be shown. YACP was correct, the definition is (a,b,c)->a(6,4)+b(4,8)+c(4,0). It was from a lecture note. – user71346 May 21 '13 at 23:11

$\operatorname{im}(\varphi)=\{\varphi(x_1,x_2,x_3):(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1\varphi(1,0,0)+x_2\varphi(0,1,0)+x_3\varphi(0,0,1):$ $(x_1,x_2,x_3)\in\mathbb Z^3\}=\{x_1(6,4)+x_2(4,8)+x_3(4,0):(x_1,x_2,x_3)\in\mathbb Z^3\}$ $=$ $\langle(6,4),(4,8),(4,0)\rangle=N$.
The form of the canonical diagonal matrix $D$ shows that there exists a $\mathbb Z$-basis $\{f_1,f_2\}$ of $\mathbb Z^2$ such that $\{2f_1,8f_2\}$ is a $\mathbb Z$-basis of $N$. Then $$\mathbb Z^2/N=f_1\mathbb Z\oplus f_2\mathbb Z/2f_1 \mathbb Z \oplus 8f_2\mathbb Z\simeq f_1\mathbb Z/2f_1\mathbb Z\oplus f_2\mathbb Z/8f_2\mathbb Z\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}.$$
• Thank you. I understand the last bit $f_1\mathbb Z/2f_1\mathbb Z\oplus f_2\mathbb Z/8f_2\mathbb Z\simeq\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}.$ But I am still struggling to understand why {$2f_1,8f_2$} is a basis of N. And could you kindly provide more detailed explanation of why is $\mathbb Z^2/N=f_1\mathbb Z\oplus f_2\mathbb Z/2f_1 \mathbb Z \oplus 8f_2\mathbb Z\simeq f_1\mathbb Z/2f_1\mathbb Z\oplus f_2\mathbb Z/8f_2\mathbb Z$. Thanks in advance. – user71346 May 21 '13 at 23:17
• Ah. I think I can use Structure Theorem to show $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/8\mathbb{Z}$. Am I right? but I still couldn't understand your approach. I hope to learn more than one ways to solve the problems. – user71346 May 21 '13 at 23:25