Section A.2 Module Theory
You can learn more about the basic theory of (commutative) rings and \(R\)-modules in any introductory algebra book, such as [DF04].
Definition A.10. Module.
Definition A.9. Let \(R\) be a ring with \(1 \neq 0\text{.}\) A left \(R\)-module is an abelian group \((M,+)\) together with an action \(R \times M \rightarrow M\) of \(R\) on \(M\text{,}\) written as \((r, m) \mapsto r m\text{,}\) such that for all \(r, s \in R\) and \(m, n \in M\) we have the following:
\((r+s) m=r m+s m\text{,}\)
\((r s) m=r(s m)\text{,}\)
\(r(m+n)=r m+r n\text{,}\) and
\(1 m=m\text{.}\)
A right \(R\)-module is an abelian group \((M,+)\) together with an action of \(R\) on \(M\text{,}\) written as \(M \times R \rightarrow M,(m, r) \mapsto m r\text{,}\) such that for all \(r, s \in R\) and \(m, n \in M\) we have
\(m(r+s)=m r+m s\text{,}\)
\(m(r s)=(m r) s\text{,}\)
\((m+n) r=m r+n r\text{,}\) and
\(m 1=m\text{.}\)
By default, we will be studying left \(R\)-modules. To make the writing less heavy, we will sometimes say \(R\)-module rather than left \(R\)-module whenever there is no ambiguity.
The definitions of submodule, quotient of modules, and homomorphism of modules are very natural and easy to guess, but here they are.
Definition A.12. Submodule.
If \(N \subseteq M\) are \(R\)-modules with compatible structures, we say that \(N\) is a submodule of \(M\text{.}\)
A map \(M \stackrel{f}{\longrightarrow} N\) between \(R\)-modules is a homomorphism of \(R\)-modules if it is a homomorphism of abelian groups that preserves the \(R\)-action, meaning \(f(r a)=r f(a)\) for all \(r \in R\) and all \(a \in M\text{.}\) We sometimes refer to \(R\)-module homomorphisms as \(R\)-module maps, or maps of \(R\)-modules. An isomorphism of \(R\)-modules is a bijective homomorphism, which we really should think about as a relabeling of the elements in our module. If two modules \(M\) and \(N\) are isomorphic, we write \(M \cong N\text{.}\)
Given an \(R\)-module \(M\) and a submodule \(N \subseteq M\text{,}\) the quotient \(M / N\) is an \(R\)-module whose elements are the equivalence classes determined by the relation on \(M\) given by \(a \sim\) \(b \Leftrightarrow a-b \in N\text{.}\) One can check that this set naturally inherits an \(R\)-module structure from the \(R\)-module structure on \(M\text{,}\) and it comes equipped with a natural canonical map \(M \longrightarrow M / N\) induced by sending \(1\) to its equivalence class.
Example A.13.
Example A.12. The modules over a field \(k\) are precisely all the \(k\)-vector spaces. Linear transformations are precisely all the \(k\)-module maps.
While vector spaces make for a great first example, be warned that many of the basic facts we are used to from linear algebra are often a little more subtle in commutative algebra. These differences are features, not bugs.
Example A.14.
Example A.13. The \(\mathbb{Z}\)-modules are precisely all the abelian groups.
Example A.15.
Example A.14. When we think of the ring \(R\) as a module over itself, the submodules of \(R\) are precisely the ideals of \(R\text{.}\)
Theorem A.16. First Isomorphism Theorem.
Any \(R\)-module homomorphism \(M \stackrel{f}{\longrightarrow} N\) satisfies \(M / \operatorname{ker} f \cong \operatorname{im} f\text{.}\)
The first big noticeable difference between vector spaces and more general \(R\)-modules is that while every vector space has a basis, most \(R\)-modules do not.
Definition A.17. Generated Modules.
A subset \(\Gamma \subseteq M\) of an \(R\)-module \(M\) is a generating set, or a set of generators, if every element in \(M\) can be written as a finite linear combination of elements in \(M\) with coefficients in \(R\text{.}\) A basis for an \(R\)-module \(M\) is a generating set \(\Gamma\) for \(M\) such that \(\sum_{i} a_{i} \gamma_{i}=0\) implies \(a_{i}=0\) for all \(i\text{.}\) An \(R\)-module is free if it has a basis.
One of the key things that makes commutative algebra so rich and beautiful is that most modules are in fact not free. In general, every \(R\)-module has a generating set - for example, \(M\) itself. Given some generating set \(\Gamma\) for \(M\text{,}\) we can always repeat the idea above and write a presentation \(\oplus_{i \in I} R \stackrel{\pi}{\longrightarrow} M\) for \(M\text{,}\) but in general the resulting map \(\pi\) will have a nontrivial kernel. A nonzero kernel element \(\left(r_{i}\right)_{i \in I} \in \operatorname{ker} \pi\) corresponds to a relation between the generators of \(M\text{.}\)
We say that a module is finitely generated if we can find a finite generating set for \(M\text{.}\) The simplest finitely generated modules are the cyclic modules.
Example A.21.
Example A.20. An \(R\)-module is cyclic if it can be generated by one element. Equivalently, we can write \(M\) as a quotient of \(R\) by some ideal \(I\text{.}\) Indeed, given a generator \(m\) for \(M\text{,}\) the kernel of the map \(R \stackrel{\pi}{\longrightarrow} M\) induced by \(1 \mapsto m\) is some ideal \(I\text{.}\) Since we assumed that \(m\) generates \(M, \pi\) is automatically surjective, and thus induces an isomorphism \(R / I \cong M\text{.}\)
Similarly, if an \(R\)-module has \(n\) generators, we can naturally think about it as a quotient of \(R^{n}\) by the submodule of relations among those \(n\) generators.
