advanced29 min studyLesson 521 in Mathematics

Modules, submodules and exact sequences

Replace vector-space intuition with ring-linear structure and record generators, relations and missing image information precisely.

On this page

What you’ll understand

  • Identify submodules, annihilators and quotient modules over a stated ring.
  • Compute a kernel and cokernel as different pieces of a linear map.
  • Explain why a short exact sequence need not split over a ring.

Vector spaces allow division by every nonzero scalar. Modules retain addition and scalar multiplication while replacing the field by a ring, so that familiar basis arguments can fail. Integers acting on an abelian group already produce a rich class of examples: finite cyclic groups have torsion, whereas a free integer module does not. Throughout this module, rings are commutative with identity, ring maps preserve identity, and modules are unital. A finite module means finitely generated, not necessarily a set with finitely many elements. These conventions will be essential when finiteness enters later theorems.

Generators do not automatically form a basis

An R-module is an abelian group with a compatible R-action. A submodule is closed under addition, negatives and every scalar action, so an ideal is a submodule of R viewed over itself. A list generates M when finite R-linear combinations cover M. A basis also requires unique coefficients, which need not be possible: the class of one generates Z/6Z but has the relation six times that class equals zero. The annihilator of an element records all scalars killing it; the annihilator of M kills every element simultaneously.

A presentation R^a→R^b→M→0 records relations among b chosen generators. The first map need not be injective because relations can themselves have relations. Over a field, finite-dimensional modules are free and dimension controls much of the story. Over a general ring, torsion and nonfree ideals survive. For example, an element killed by a nonzero integer cannot be part of an integer-module basis. Keeping the scalar ring explicit prevents an abelian group isomorphism from being mistaken for an isomorphism respecting a different ring action.

Ann⁡R(m)={r∈R:rm=0},Ann⁡R(M)=⋂m∈MAnn⁡R(m),Ra⟶Rb⟶M⟶0\operatorname{Ann}_R(m)=\{r\in R:rm=0\},\qquad\operatorname{Ann}_R(M)=\bigcap_{m\in M}\operatorname{Ann}_R(m),\qquad R^a\longrightarrow R^b\longrightarrow M\longrightarrow0
The presentation is exact at Rᵇ and M: its relation image is the kernel of the generator map. No injectivity of the first arrow is implied.↔ Scroll wide equations horizontally.

Exactness compares an image with the next kernel

A sequence is exact at its middle module when the incoming image equals the outgoing kernel. A short exact sequence 0→N→M→Q→0 says that N embeds as the full kernel and every quotient class in Q is reached. It identifies Q with M/N, but does not choose representatives linearly. For a map f:M→N, the kernel is a submodule of the source, while the cokernel is N/im f and measures failure of surjectivity. Their roles differ even when both are represented by familiar cyclic groups in examples.

To split a short exact sequence, one needs a linear section Q→M of the quotient map, equivalently a compatible direct-sum decomposition M≅N⊕Q. Such a section exists for vector spaces but not for all modules. The sequence with integer multiplication by six followed by reduction modulo six is exact but not split: any homomorphism from Z/6Z into Z has zero image because Z has no nonzero six-torsion. A set-theoretic choice of residue representatives is not additive and cannot supply the missing module section.

0⟶Z⟶×6Z⟶Z/6Z⟶0,coker⁡f=N/im⁡f0\longrightarrow\mathbb Z\overset{\times6}{\longrightarrow}\mathbb Z\longrightarrow\mathbb Z/6\mathbb Z\longrightarrow0,\qquad\operatorname{coker}f=N/\operatorname{im}f
The integer sequence is exact but nonsplit. A cokernel is formed in the target, while a kernel lives in the source.↔ Scroll wide equations horizontally.

Relations connect algebra to later local tests

If R is Noetherian, a submodule of a finite module is finite, so a finite module admits a presentation with finitely many relations. Without that hypothesis, finite generation of M does not guarantee finite generation of its relation module. This is one reason the next lesson emphasizes chain conditions instead of merely counting variables. The distinction between finite generation and finite presentation will later matter for flatness and local freeness. A surjection from a finite free module supplies generators, while controlling its kernel supplies a finite presentation.

Exact sequences make these issues transportable. Localization will preserve them, allowing a global relation problem to be studied near a prime. Tensor products always preserve their right-hand quotient behavior but can destroy injectivity unless flatness is present. Completion will preserve exactness for finite modules over Noetherian rings under the later stated hypotheses. Learning to identify the specific image and kernel at each arrow is therefore more useful than memorizing the shape of a diagram: every later operation must be checked against the exactness information it is supposed to preserve.

An integer map has both relations and a missing quotient

For f:Z²→Z defined by f(a,b)=4a+6b, compute its image, kernel and cokernel.

  1. Every image is even, and f(−1,1)=2, so the image is exactly 2Z.

  2. The equation 4a+6b=0 becomes 2a=−3b. Coprimality gives a=3t,b=−2t for an integer t, so the kernel is generated by (3,−2).

  3. The cokernel is Z/2Z. The relation generator describes which source pairs map to zero, whereas the cokernel remembers that odd target integers are never reached.

The result

There is an exact sequence 0→Z→Z²→Z→Z/2Z→0, with first arrow t↦(3t,−2t) and middle arrow f. The source relation and target obstruction are distinct.

Explore the model

Interactive lab

Interactive lab · modular

A generator can carry a scalar relation

Pause and Reset, then set modulus m=12 and additive step=4. Predict the orbit 0,4,8,0, so the generated cyclic submodule has three elements and relation 3·4=0. Step to check it. For a fresh comparison Reset and set m=12,step=3; predict four elements and relation 4·3=0. This additive residue orbit illustrates a Z-module; it does not enumerate all submodules of an arbitrary ring module.

A generator can carry a scalar relationAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA generating element can have nontrivial scalar relations, unlike a vector in a field basis.

Check your understanding

For an R-linear map f:M→N, which object measures failure to reach the target?

Try it yourself

Over R=k[x], analyze 0→R/(x)→R/(x²)→R/(x)→0, where the first map sends the class of a to the class of ax and the second reduces modulo x. Is it split as an R-module sequence?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Verify image equals kernel, then compare multiplication by x on the middle module with multiplication on a proposed direct sum.

Compare with the solution

The first map is well defined and injective: ax belongs to (x²) exactly when a belongs to (x). Its image is the submodule generated by the class of x, exactly the reduction map’s kernel, and reduction is surjective. The sequence is not split. On R/(x)⊕R/(x), multiplication by x is zero everywhere, whereas in R/(x²) it sends the class of one to the nonzero class of x. An R-linear isomorphism cannot change that scalar action.

Take it with you

  • A finite module is finitely generated and can still have torsion or lack a basis.
  • Kernels encode source relations; cokernels encode missing target information.
  • Exactness does not automatically provide a linear splitting.

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