Smith Normal Form as the best group presentation

$\newcommand{\RR}{\mathbb{R}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\FF}{\mathbb{F}} \newcommand{\coker}[1]{\text{coker} \ #1} \newcommand{\im}[1]{\text{Im} \ #1} \newcommand{\diag}[1]{\text{diag}(#1)}$

Days ago, I asked on Math.StackExchange the question: Describe, as a direct sum of cyclic groups, the cokernel of the map: $\phi: \mathbb{Z}^{3} \longrightarrow \mathbb{Z}^{3}$ , where after some hints and understand modules over PID (see handouts), I finally got an answer. However, the answer suggests me to use Smith Normal Form (SNF), so I decided to understand: What does SNF truly represent?, so let’s start with a basic definition.

Definition: Let $R$ be a ring. Every left $R$ -module $M$ is a quotient of a free left $R$ -module $N$ such that $M$ is finitely generated if and only if $N$ can be chosen to be finitely generated.

We need to remember that the rank of any free $R$ -module is uniquely determined, later this will be important in our deductions, now let’s see what a group presentation is.

Group presentation

Definition: Let $I$ be a set and $S$ a subset of the free group $F(I)$ such that $F(I)$ is generated by $I$ . Indeed, a group presentation defines the quotient group of the free group $F(I)$ by the normal subgroup generated by $S$

Now, let’s suppose we would like to get the best presentation of a given module $M$ , this is possible thanks to the SNF, which gives the answer for a finitely generated $M[X]$ -modules, to see this, let’s see how SNF builds this approach.

Let $\phi: R^{n} \longrightarrow R^{m}$ be a $R$ -map of commutative ring $R$ , where $R^{n}$ and $R^{m}$ are free $R$ -modules. Let $Y = \{y_{1}, \dots, y_{n}\}, Z = \{z_{1}, \dots, z_{m}\}$ be the basis of $R^{n}$ and $R^{m}$ , respectively. We build a matrix ${}_{Z}[\phi]_{Y}$ , called the presentation matrix, over $R$ such that

$M = \phi(y_{i}) = \sum_{j=1}^{m} a_{ji}z_{j},$

where $M \cong \coker{\phi} = R^{m}/ \im{\phi}$ .

Lemma: Let an exact sequence of $K[X]$ -modules

$0 \longrightarrow K[X]^{m} \xrightarrow{\quad \phi \quad} K[X]^{m} \xrightarrow{\quad \varphi \quad} V^{M} \longrightarrow 0,$

where $V$ is a $m$ -dimensional vector space over a field $K$ , then its matrix presentation is $XI - M$ with respect to $E$ , where $E$ is the standard basis.

Proof: For ${}_{E}[\phi]_{E}$ we write $\phi(e_̣{i})$ in terms of $E$ , then

$\phi(e_{i}) = xe_{i} - Me_{i}$ $\hspace{4em} = xe_{i} - \sum_{j} a_{ji}e_{j}$ $\hspace{6.5em} = \sum_{j} x\delta_{ij}e_{j} - \sum_{j} a_{ji}e_{j}$ $\hspace{4.5em} = \sum_{j} (x\delta_{ij} - a_{ji})e_{j}$

where $\delta_{ij}$ is the Kronecker delta. Indeed, the presentation matrix ${}_{E}[\phi]_{E} = xI - M$ . Now, let’s recall what SNF says us.

Smith Normal Form

Every nonzero $x \times m$ matrix $\Gamma$ with entries in a Euclidean ring $R$ , which is equivalent to a matrix of the form

$\left[\begin{matrix} \sum & 0 \\ 0 & 0 \end{matrix}\right],$

where $\sum = \diag{\sigma_{1}, \dots, \sigma_{q}}$ and $\sigma_{1} \vert \dots \vert \sigma_{q}$ are nonzero, such matrix is called the SNF of $\Gamma$ .

Now that we have how the presentation matrix is involved with the bases, we can do the conntection between them.

Definition: Let $R$ be a euclidean ring and let $\Gamma$ be the $n \times m$ presentation matrix associated to an $R$ -map $\phi: R^{n} \longrightarrow R^{m}$ , relative to some choice of bases, and let $M = \coker{\phi}$ .

Theorem: If $\Gamma$ is $R$ -equivalent to a SNF $\diag{\sigma_{1}, \dots, \sigma_{q}} \oplus 0$ , then those $\sigma_{1}, \dots, \sigma_{1}$ that are not units are the invariant factors of $M$ .

Proof: Suppose that $\diag{\sigma_{1}, \dots, \sigma_{q}} \oplus 0$ is in fact the SNF of $\Gamma$ , then the bases of $y_{1}, \dots, y_{n}$ of $R^{n}$ and $z_{1}, \dots, z_{m}$ of $R^{m}$ with $\phi(y_{1}) = \sigma_{1}z_{1}, \dots, \phi(y_{q}) = \sigma_{q}z_{q}$ and $\phi(y_{j}) = 0$ for all $y_{j} > q$ , if any. We see that $R/(0) \cong R$ and $R/(u) \cong \{0\}$ if $u$ is a unit. If $\sigma_{s}$ is the first $\sigma_{i}$ that is not a unit, then

$M \cong R^{m-q} \oplus \frac{R}{\sigma_{s}} \oplus \dots \oplus \frac{R}{\sigma_{q}},$

a direct sum of cyclic modules for which $\sigma_{s} \vert \dots \vert \sigma_{q}$ . Indeed $\sigma_{s}, \dots, \sigma_{q}$ are the invariant factors of $M$ .

The proof of the theorem clears why is necessary to get the SNF for answering to the question that I posted on Math.StackExchange

PS: Most of the ideas posted here were obtained of the book Advanced Modern Algebra Third Edition, Part 1 by Joseph J. Rotman, especially the proof of the theorem.