Weird Definition about Presentations of Modules

Question

Definition 1.70

Suppose $(a_{ij})=A\in M_{m\times n}(R).$ Let $L$ be a free module of rank $m$ with basis $\{x_1,..,x_m\}.$ Let $$ y_j=\sum^{m}_{i=1}a_{ij}x_i \text{ for } j=1,...,n$$ and let $K=Ry_1+Ry_2+...+Ry_n$. Let $$ M=L/K=\{x+K:x\in L\}=\{\bar{x}:x\in L\}$$ Notice that $\{x_1,...,x_m\}$ generates $L.$ Thus, $\{\bar{x}_1,...,\bar{x}_m\}$ generates M. Notice that $y_j\in K$ for $j=1,...,n.$ Therefore, $\bar{y}_j=0$ for all $j.$ So $$ \sum^{m}_{i=1}a_{ij}\bar{x}_i=0$$ for $j=1,...,n.$

In summary. $\{\bar{x}_1,...,\bar{x}_m\}$ is a generating set for $M$ and these generators satisfy the relations $$ \sum^{m}_{i=1}a_{ij}\bar{x}_i=0 \quad j=1,..,n$$ A is called the relations matrix for $M$. $M$ is also called the module determined by generators and relations with relations matrix $A.$

Note. We could have used a different free module $L'$ with different basis $\{x'_1,...,x'_m\}$. The resulting $M'$ is isomorphic to $M.$

So the above is a definition in my class' lecture notes which I believe, when looking elsewhere on the internet, corresponds to the definition of a module presentation? I have a few questions about it that I'd very much appreciate help understanding:

(1) What exactly is the idea behind the above construction? Why create a module this way?

(2) Is $M$ a free module here? It says $\{\bar{x}_1,...,\bar{x}_m\}$ is a generating set for $M.$ But that doesn't imply $M$ is free, right? (i.e. those elements may not be independent)

(3) Why is the note at the end true? Why is $M'$ isomorphic to $M$?

Answer 1

(1) The idea is that every finitely generated module $M$ is the quotient of a free $R$-module $L$, say $R^m$, (one maps each vector in a basis of $L$ onto a generator of $M$). The kernel $K$ of this morphism (the so-called module of relations between the generators) may also be finitely generated, e.g. if $R$ is a noetherian ring). In this case $K$ is itself a quotient of another finitely generated free module $L_1=R^m$. So we obtain an exact sequence: $$L_i\longrightarrow L\longrightarrow M\longrightarrow 0.$$ This is this exact sequence your definition describes. Such a module is said to be finitely presented. Finitely presented modules play a prominent role in commutative algebra and algebraic geometry.

(2) No, $M$ is not necessarily free – actually, if it is free, it is a direct summand of $L$.

(3) The isomorphism in question results from a result known as the snake lemma, qui asserts results betwen the modules involved in a commutative diagram of exact sequences.

Answer 2

(1) The idea is that every finitely generated module $M$ is the quotient of a free $R$-module $L$, say $R^m$, (one maps each vector in a basis of $L$ onto a generator of $M$). The kernel of this morphism (the so-called module of relations between the generators) may also be finitely generated, e.g. if $R$ is

Answer 3

(1) The idea is to present a module in the form of a quotient of a free module by some relations. This kind of representations work things out quite well.

An example from group theory would be the dihedral group: a group generated by two elements $x$ and $y$ (which is the free group $\langle x,y\rangle$) with the relations $x^2=1, yx=xy^{n-1},y^{n}=1$.

Unfortunately no example of $R$-modules comes to my mind.

(2) You're right. $M$ is not free. And the dependence of its elements is determined by the relations matrix.

(3) You can take an isomorphism between $M$ and $M'$ induced from $x_i\rightarrow x'_i$ (by passing through the quotient).