# On the intuitive idea behind decomposition into cyclic modules

Let $R$ be a ring and $M$ be a free $R$-module. If $M$ has a basis of size $n$ then we have $M \cong R^n$. I find this rather intuitive.

With that as a building block, I also find rather intuitive that given some relations that the generators $x_1, x_2, \cdots, x_k$ satisfy, then we can quotient $R^k$ by an ideal generated by the relations to get $M$.

For example, if $M$ is a $\mathbb{Z}$-module with generators $x,y$ such that $x-y = 0$ then $M \cong \mathbb{Z^2}/(\ (1,-1)\ )$. (Is this correct?)

Generally speaking, with $D$ an integral domain, one can also take the short exact sequence

$$0 \to D^n \stackrel{f}{\to} D^m \stackrel{g}{\to} M \to 0$$

and if $\ker{g}$ is generated by the relations of $M$ then we will get $M \cong D^m/\ker{g} = D^m/\text{im}{f}$ because the sequence is exact.

Suppose now I have $M$ a $D$-module and some relations its generators satisfy. Why is it that when I want to decompose $M$ into a direct sum of cyclic modules I write down a matrix whose columns are the relations and then find its Smith Normal Form? I get it that we are kind of performing some sort of "change of basis", but in what way does that matrix fit that short exact sequence, if in any way?

For example if $G$ is an abelian group with generators $x,y,z$ that satisfy the relations

$$\begin{cases}2x+4y = 0\\ x+y+3z = 0\end{cases}$$

then I write down the matrix

$$\begin{bmatrix} 2 & 1 \\ 4 & 1 \\ 0 & 3 \end{bmatrix}$$

and find its smith normal form,

$$\begin{bmatrix} 1 & 0 \\ 0 & 2 \\ 0 & 0 \end{bmatrix}$$

to get $G \cong \mathbb{Z}_{/2}\ \oplus\ \mathbb{Z}$ that I know is correct (at least the final result), but why do we get there in this way?

• Does this post answer your question? – André 3000 Dec 22 '17 at 1:43

You have to specify what you mean by "quotient $R^k$ with an ideal generated by the relations". Your example is not correct. By your construction, $M= \langle x \rangle$ is a cyclic module, it is nonzero unless $x=0$, but $\mathbb{Z}^2/((1,-1))=0$.