I am trying to study some module theory using the book "Algebra" by Michael Artin (2nd Edition, to be precise), and I can't really fathom what is written in Section 14.5.
Left multiplication by an $m \times n$ matrix defines a homomorphism of $R$-modules $A: R^n \rightarrow R^m$. Its image consists of all linear combinations of the columns of $A$ with coefficients in the ring, and we may denote the image by $AR^n$. We say that the quotient module $V=R^m/AR^n$ is presented by the matrix $A$. More generally, we call any isomorphism $\sigma: R^m/AR^n \rightarrow V$ a presentation of a module $V$, and we say that the matrix $A$ is a presentation matrix for $V$ if there is such an isomorphism.
For example, the cyclic group $C_5$ of order $5$ is presented as a $\mathbb{Z}-$module by the $1\times 1$ integer matrix $[5]$, because $C_5$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$.
The above two paragraphs seem to be clear.
We use the canonical map $\pi: R^m \rightarrow V=R^m/AR^n$ to interpret the quotient module $V=R^m/AR^n$, as follows:
Proposition 14.5.1
(a) $V$ is generated by a set of elements B $=(v_1, \cdots, v_m)$, the images of the standard basis elements of $R^m$.
(b) If $Y=(y_1, \cdots, y_m)^t$ is a column vector in $R^m$, the element B$Y=v_1y_1+\cdots + v_my_m$ of $V$ is zero if and only if $Y$ is a linear combination of the columns of $A$, with coefficients in $R$ - if and only if there exists a column vector $X$ with entries in $R$ such that $Y=AX$.
Proof. The images of the standard basis generate $V$ because the map $\pi$ is surjective. Its kernel is the submodule $AR^n$. This submodule consists precisely of the linear combinations of the columns of $A$.
First, could you expound on the proof please? The proposition is obvious when applied to the above example with $R^m=R^n=\mathbb{Z}$, $V=C_5$, and $A=[5]$; but it doesn't help me to carry out more detailed proof.
If a module $V$ is generated by a set B $=(v_1, \cdots, v_m)$, we call an element $Y$ of $R^m$ such that B$Y=0$ a relation vector, or simply a relation among the generators. We may also refer to the equation $v_1y_1+\cdots+v_my_m=0$ as a relation, meaning that the left side yields $0$ when it is evaluated in $V$. A set $S$ of relations is a complete set if every relation is a linear combination of $S$ with coefficients in the ring.
Example 14.5.2 The $\mathbb{Z}-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete set of relations
$$ 3v_1+2v_2+v_3=0\\ 8v_1+4v_2+2v_3=0\\ 7v_1+6v_2+2v_3=0\\ 9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$ A=\begin{bmatrix} 3 & 8 & 7 & 9 \\ 2 & 4 & 6 & 6 \\ 1 & 2 & 2 & 1 \\ \end{bmatrix}. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2, v_3)A=(0, 0, 0, 0)$.
Here comes the second question. How did he find out immediately what matrix presents the group $V$? Artin gives an algorithm for determining the presentation matrix below this example, but since this example precedes it, the reader should be able to determine this matrix without the algorithm. So, in this case $R=\mathbb{Z}$, but what are $m$ and $n$? How do I find this matrix $A$ neatly?