Try
Integer addition allows repeated addition, but not division by every nonzero integer. In Z/6Z the nonzero element [2] satisfies 3[2]=[0]. A real vector cannot behave this way: multiplication by 1/3 would force it to be zero. Modules keep addition and scalar multiplication while allowing the scalars to form a ring rather than a field.
Throughout this lesson R is a commutative ring with identity, and modules are unital: 1m=m. A module M is an abelian group under addition, with scalar multiplication R×M→M satisfying r(m+n)=rm+rn, (r+s)m=rm+sm and (rs)m=r(sm). In particular 0m=0 and r0=0 follow from additive cancellation. A module over a field is precisely a vector space. Revisit ring conventions before using units or cancellation.
Work out 4[2], −[2] and 3[2] in Z/6Z: they are [2], [4] and [0]. This example is a Z-module, with integer scalars. It is also a module over Z/6Z itself, but the scalar ring changes what “free” means. We will always name that ring.
Predict
A submodule N⊆M is an additive subgroup closed under all R-scalars. An R-linear map f:M→P preserves addition and satisfies f(rm)=rf(m). Kernels and images are submodules: check the two closure requirements directly. A subset of R, viewed as a module over itself, is a submodule exactly when it is an ideal.
Predict whether the following familiar vector-space statements survive over Z: every generating set contains a basis; every submodule has a complementary submodule; every nonzero scalar acts injectively. The examples below will refute all three blanket claims. These failures do not invalidate module homomorphisms or quotients; they change which theorems need hypotheses.
Experiment
Define f:Z²→Z/6Z by f(a,b)=[2a+3b]. It is additive and commutes with integer multiplication. Fill in the last column before checking the supplied values.
| Input (a,b) | Integer 2a+3b | Image modulo six |
|---|
| (1,0) | 2 | [2] |
| (0,1) | 3 | [3] |
| (2,−1) | 1 | [1] |
| (3,0) | 6 | [0] |
| (0,2) | 6 | [0] |
| (4,1) | 11 | [5] |
Because [1] is attained, every residue is attained. If 2a+3b is divisible by six, reduction modulo three forces a to be a multiple of three, and reduction modulo two forces b even. Conversely those two conditions make the sum divisible by six. Thus ker f=3Z×2Z. The map identifies Z²/(3Z×2Z) with Z/6Z. Explain why six residue pairs remain: a can be reduced modulo three and b modulo two. The quotient has six elements, although both the source and the kernel are infinite.
Understand
Quotients and the first isomorphism theorem
For a submodule N define addition and scalar multiplication on cosets by (m+N)+(m′+N)=(m+m′)+N and r(m+N)=rm+N. Replacing representatives changes the resulting vector by an element of N, using additive and scalar closure, so both operations are well-defined. The module axioms follow from those of M. This resembles the ideal quotient construction, but no multiplication of module elements is required.
For a linear map f define φ:M/ker f→im f by φ(m+ker f)=f(m). Equal cosets differ by an element killed by f, so φ is well-defined. It is linear and onto its stated target. If φ(m+ker f)=0, then m∈ker f, so the coset is zero; thus it is injective. This proves the module first isomorphism theorem. It does not claim that M itself is a direct sum of kernel and image. Compare the group isomorphism theorem: the additional work is compatibility with scalars.
Integers recover abelian groups
Every abelian group has a unique unital Z-module structure: nm is repeated addition for n≥0, and negative n use additive inverses. The module axioms follow from the group laws. Conversely the underlying additive group of a Z-module is abelian, and distributivity plus 1m=m force precisely this integer action. An additive group homomorphism automatically preserves repeated addition and negatives, so is exactly a Z-linear map. This correspondence concerns abelian groups; arbitrary noncommutative groups are not modules under their group operation.
Polynomial scalars encode one operator
Let F be a field, V an F-vector space and T:V→V F-linear. Define p(t)·v=p(T)v, where p(T)=a₀I+a₁T+⋯+a_dTᵈ. Finite distributive expansion and TⁱTʲ=Tⁱ⁺ʲ prove (pq)(T)=p(T)q(T), giving a unital F[t]-module. Conversely restrict an F[t]-module to constant scalars to obtain an F-vector space, and define Tv=t·v. The module rules force every polynomial to act by evaluation at T. No finite-dimensional assumption is needed for this correspondence.
For operators T on V and U on W, an F[t]-linear map A is F-linear and satisfies AT=UA by applying it to t·v. Conversely these two conditions imply ATʲ=UʲA by induction and hence Ap(T)=p(U)A for every polynomial. Thus module maps are exactly the intertwiners of the two operators. An invariant subspace for T is precisely an F[t]-submodule. This connects invariant subspaces to polynomial algebra, without assuming that every module is a vector space over its full scalar ring.
Generators, bases and torsion
A basis over R is a family for which every element has a unique finite R-linear expansion. A module admitting a basis is free. Over an integral domain, a torsion element is one killed by some nonzero scalar; a module is torsion-free when only zero has that property. We restrict this torsion terminology here to domains, avoiding extra conventions for zero divisors.
As a Z-module, Z/6Z is generated by [1] but has no basis: every element is killed by the nonzero integer six, whereas a nonzero free Z-module is torsion-free by uniqueness of coefficients. By contrast, 2Z has the Z-basis consisting of 2: every even integer is uniquely n·2. Nevertheless it is not a direct summand of Z. A splitting would give a Z-linear retraction r:Z→2Z fixing 2. Write r(1)=2k; then r(2)=4k cannot equal 2 for an integer k. Freeness of a submodule does not guarantee a complement in its ambient module.
Apply
Use F=Q and N=((0,1),(0,0)) on Q². Predict whether one vector can generate the whole space as a Q[t]-module, even though its Q-vector-space dimension is two. Then calculate every module endomorphism A by solving AN=NA.
Check the polynomial-action example
The vector e₂ generates: e₂ and t·e₂=Ne₂=e₁ span over Q. The map Q[t]→V, p↦p(N)e₂, is onto and has kernel (t²), because p(N)e₂=a₀e₂+a₁e₁. Hence V≅Q[t]/(t²). This module is not free over Q[t]: every vector is killed by the nonzero polynomial t². For A=((a,b),(c,d)), AN=((0,a),(0,c)) and NA=((c,d),(0,0)), so c=0 and d=a. The endomorphisms are exactly ((a,b),(0,a)). Their two rational parameters do not assert a two-element Q[t]-basis of V.
For a changed case let T=diag(2,5) on Q². Find the polynomials selecting the two coordinate lines and determine the commuting maps.
Check the changed operator
(5I−T)/3=diag(1,0) and (T−2I)/3=diag(0,1), so these polynomial operators project onto the two invariant lines. A matrix commutes with T exactly when its off-diagonal entries vanish, since the eigenvalue difference is three, nonzero in Q. Thus its module endomorphisms are the diagonal matrices. The vector e₁+e₂ generates the entire Q[t]-module by the two displayed polynomials. Again it is not free over Q[t], since (t−2)(t−5) kills every vector. The field and its invertible scalars are part of this calculation.
Continue
Before transferring a vector-space argument, name the scalar ring, distinguish a generating family from a basis, and check whether an invariant complement actually exists. Quotient and isomorphism arguments remain available even when division and splitting fail.
Continue to integer presentations, where a finite list of generators and relations becomes an integer matrix. A presentation records relations; it is not automatically a basis or an exact field-style elimination problem.
