Modules are linear algebra over rings. They keep the language of vectors and linear maps, but bases and dimension may fail, so exact sequences become essential bookkeeping.
Modules and module homomorphisms
Definition. Let \(R\) be a ring. A left \(R\)-module \(M\) is an abelian group with scalar multiplication \(R\times M\to M\) satisfying the usual associativity, distributivity, and identity axioms. An \(R\)-linear map \(f:M\to N\) preserves addition and scalar multiplication.
Example. Vector spaces are modules over fields. Abelian groups are exactly \(\mathbb Z\)-modules. Ideals of a ring \(R\) are \(R\)-submodules of \(R\).
Submodules and quotient modules
Definition. A submodule \(N\subset M\) is an additive subgroup closed under scalar multiplication. The quotient abelian group \(M/N\) becomes an \(R\)-module by \(r(m+N)=rm+N\).
Proposition. Kernels and images of module homomorphisms are submodules, and \(M/\ker f\cong\operatorname{im}f\).
Proof. The subgroup and scalar-closure checks are direct from linearity. The isomorphism theorem is proved exactly as for groups: send \(m+\ker f\) to \(f(m)\).
Exact sequences
Definition. A sequence of modules and linear maps \[ \cdots\to M_{i-1}\xrightarrow{f_{i-1}}M_i\xrightarrow{f_i}M_{i+1}\to\cdots \] is exact at \(M_i\) if \(\operatorname{im}f_{i-1}=\ker f_i\). It is exact if it is exact at every term.
Example. For a submodule \(N\subset M\), the sequence \[ 0\to N\to M\to M/N\to0 \] is exact. It says that \(N\) is embedded in \(M\), and quotienting by its image gives \(M/N\).
Short exact sequences and splitting
Definition. A short exact sequence is an exact sequence \(0\to A\xrightarrow{i}B\xrightarrow{p}C\to0\). It splits if there exists \(s:C\to B\) with \(p\circ s=\operatorname{id}_C\), or equivalently \(r:B\to A\) with \(r\circ i=\operatorname{id}_A\).
Theorem. A short exact sequence \(0\to A\to B\to C\to0\) splits iff \(B\cong A\oplus C\) in a way compatible with the maps.
Counterexample. The sequence \[ 0\to\mathbb Z\xrightarrow{\times 2}\mathbb Z\to\mathbb Z/2\mathbb Z\to0 \] is exact as a sequence of \(\mathbb Z\)-modules but does not split, because \(\mathbb Z\) has no element of order \(2\).
Free, torsion, and finitely generated modules
Definition. An \(R\)-module is free if it has a basis, finitely generated if finitely many elements span it, and torsion over a domain if every element is killed by some nonzero scalar.
Remark. Over a field, every module is free. Over a general ring this is false: \(\mathbb Z/2\mathbb Z\) is a finitely generated \(\mathbb Z\)-module, but it is not free.
Exactness as kernels and images
A sequence \[ 0\to A\xrightarrow{f}B\xrightarrow{g}C\to0 \] is short exact precisely when \(f\) identifies \(A\) with \(\ker g\) and \(g\) identifies \(C\) with \(B/f(A)\). It splits if there is \(s:C\to B\) with \(g\circ s=\operatorname{id}_C\), equivalently a retraction \(r:B\to A\) with \(r\circ f=\operatorname{id}_A\).
Splitting lemma. For a short exact sequence, the following are equivalent: it has a section; it has a retraction; and \[ B\cong A\oplus C \] through an isomorphism compatible with the sequence.
Proof. Given a section \(s\), every \(b\in B\) decomposes as \[ b=\bigl(b-sg(b)\bigr)+sg(b), \] where the first term lies in \(\ker g=\operatorname{im}f\). The intersection of \(\operatorname{im}f\) and \(\operatorname{im}s\) is zero, giving the direct sum. Projection onto the first summand gives a retraction. Conversely, a retraction makes \(\ker r\) map isomorphically to \(C\), whose inverse gives a section.
Projective and injective modules
A module \(P\) is projective exactly when every short exact sequence ending in \(P\) splits, and exactly when \(P\) is a direct summand of a free module. Dually, an injective module makes every short exact sequence beginning with it split. These lifting and extension properties are the structural reason homological algebra studies resolutions.
Example. The sequence \[ 0\to\mathbb Z\xrightarrow{\times n}\mathbb Z \to\mathbb Z/n\mathbb Z\to0 \] does not split for \(n>1\): a section would embed a nonzero finite subgroup in \(\mathbb Z\). Thus exactness does not by itself mean direct-sum decomposition.