# Determining the presentation matrix for a module

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 $$, 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 $A=$; 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?

I know this is an old question but it's not yet answered. I'm also just an algebra student, but I'll take a stab at this.

First, let's just make sure we're on the same page with what Artin is saying. We're given a ring $R$. Now, let's say we have a finitely generated $R$-module $M$ over this ring $R$. If there are no relations between the generators of the module, then the module is isomorphic to the free module $R^m$.

Now how do we use this to inform thinking about modules that $do$ have relations. Let's say we have a finitely generated module $M$ that can be finitely generated by a set of generators $B = (v_1, ... , v_m)$. Symbolically, the entire module can be thought of as $BR^m$, all $R-$ linear combinations of the generators. But notice something here with $BR^m$... it's that many of these elements are going to be zero. We need a way of capturing which of the elements of this are zero. To do this, we'll send all of the elements of $R^m$ which are the $coordinates$ of the zero elements of $M$ to zero. These are all the elements $AR^n$. See that the columns of the $A$ matrix are indeed the relation vectors for the module. Another way to think of this is that

$$BAR^n = 0 \text{ in the module}$$

Well that's precisely what quotienting is! If $R^m$ is the space of all the coordinate vectors by which we can compose the basis of generators, then we want to send all of the coordinate vectors that are relation vectors to zero! So we set $AR^n$ to zero. It's the $columns$ of $A$ which are the coordinate vectors i.e. relation vectors for $M$. So we get that $M \cong R^m/(AR^n)$.

After meditating on this for a little bit, Artin's rules for determining the presentation matrix should follow pretty directly.

• Why "many of these elements are going to be zero"? $B$ is a basis, and therefore the only event in which an element of $BR^m$ is zero is when we take the zero vector in $R^m$. – Cary Mar 30 '18 at 2:10
• By basis here I mean a complete set of generators. Here we are encoding the information that one of the elements of B can be generated by R-linear combinations of the other elements. I think this is accomplished by setting $$BAR^n$$ to zero. Does that make sense? – Adam Mar 31 '18 at 4:03
• If by a basis you mean a set of generators, then this answers my question in the previous comment. Also I believe you meant "relation vector" (as opposed to "relation") in a few places (if we follow Artin's terminology). Thank you for your answer, it make things more clear, but I guess I still have to think about it for some time. – Cary Mar 31 '18 at 17:09