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

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

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

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

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

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

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.