Cyclic Weyl Modules and Their Presentations

Q: Why is a cyclic module the same thing as a quotient by a left ideal?

Because sending a ↦ au is a surjective module map R → Ru whose kernel is by definition ann(u) . Kernels of maps out of R as a left module over itself are exactly the left ideals, so the two notions carry the same data.

Q: Can the same module have both a principal and a non-principal presentation?

Yes. K[x] over A_1 is A_1/A_1∂ with generator 1 , and A_1/(A_1(x∂ - 1) + A_1∂^2) with generator x ; the second ideal is not principal. Both present the same simple module.

Q: How can a direct sum of two copies of one simple module be cyclic?

Because A_1 is simple, an operator killing the generator of one copy need not kill the other. Concretely ∂ kills 1 but not x , and that asymmetry lets (1,x) generate K[x] ⊕ K[x] . Over a commutative ring the two copies would be killed by the same maximal ideal and the trick is unavailable.

Q: Does a cyclic presentation tell me the solution space?

It tells you how to compute it: by (5.14), homomorphisms A_n/I → M are the elements of M killed by I . For A_1/A_1P with P of order k and non-vanishing leading coefficient, that space is k -dimensional over C inside the holomorphic functions on a simply connected domain.

Q: Is every finitely generated A_n-module cyclic?

No — A_n^2 is not, by the rank argument. But Stafford proved that every finitely generated module over A_n is generated by two elements, and that holonomic ones are generated by one, so the failure is as mild as it could be.

Q: What does the presentation matrix look like for a system of equations?

If the system is P_1(f) = … = P_m(f) = 0 in one unknown, the module is A_n/ ∑ _j A_nP_j , a cyclic module presented by a column of operators. Several unknowns give a genuine matrix; see systems and presentation matrices .

Overview

Almost every module named in this collection is presented the same way: choose an element, apply all the operators, and record which operators give zero. If a single element suffices to reach the whole module, the module is cyclic, and the record of relations is a left ideal $I$ of $A_n$ , with $M ≅ A_n/I$ . This is the standard way a D-module is written down, stored in a computer algebra system, and reasoned about.

Three facts make the picture worth stating carefully. First, every irreducible module is cyclic — any non-zero element generates — so simplicity is a strong form of cyclicity. Second, the ideal $I$ depends on the generator chosen, sometimes dramatically: the same module can be $A_1/A_1∂$ for one generator and a quotient by a non-principal ideal for another. Third, cyclicity is far more common over $A_n$ than intuition from commutative algebra suggests. A finite direct sum of simple modules is cyclic, and every holonomic module is cyclic — statements with no commutative analogue.

There are limits. The module of holomorphic functions is not cyclic, and neither is the free module $A_n^2$ . The obstruction in the first case is a mixture of torsion and non-torsion elements; in the second it is rank. Between those extremes, being cyclic is the normal situation.

This page collects the dictionary between cyclic modules and left ideals, proves the two basic presentations $K[X] ≅ A_n/ ∑ _i A_n∂_i$ and $K[∂] ≅ A_n/ ∑ _i A_n x_i$ , and works out in full an example that is genuinely surprising: $K[x] ⊕ K[x]$ is cyclic, with presentation $A_1/A_1∂^2$ .

Definition

Cyclic modules, annihilators, presentations

Let $R$ be a ring and $M$ a left $R$ -module. For $u ∈ M$ write $ann_R(u) = a ∈ R : au = 0$ , a left ideal. $M$ is cyclic if $M = Ru$ for some $u$ , called a generator. $M$ is finitely generated if $M = Ru_1 + … + Ru_r$ , and finitely presented if in addition the kernel of the resulting surjection $R^r → M$ is finitely generated.

For a left module the surjection $R^r → M$ sends $(a_1, … ,a_r) ↦ ∑ _i a_iu_i$ , and a map $R^s → R^r$ of left modules is right multiplication by an $s \times r$ matrix over $R$ . A presentation is therefore recorded as a matrix acting on the right.

Cyclic modules are quotients by left idealsCoutinho (5.1.1)

If $M = Ru$ then $a \mapsto a u$ is a surjective homomorphism $R \to M$ of left modules with kernel $ann_R(u)$ , so

M ≅ R/ann_R(u).

In particular, if $M$ is irreducible then $M = Ru$ for every non-zero $u ∈ M$ , so $M ≅ R/ann_R(u)$ for every such $u$ .

Irreducible modules over a non-division ring are torsionCoutinho (5.1.1)(2)

Let $R$ be a ring that is not a division ring and $M$ an irreducible left $R$ -module. Then every element of $M$ is a torsion element. Indeed if $ann_R(u) = 0$ for some $u ≠ 0$ then $M ≅ R$ by the previous lemma, so $R$ has no left ideals other than $0$ and $R$ , which for a ring with identity means $R$ is a division ring.

Which relations may be imposed

Going the other way — writing down a candidate action and checking it defines a module — is exactly the content of defining a module action from generators and relations. For $A_n$ the only relations to check are $[∂_i, x_j] = δ_ij$ and the commuting of the $x$ 's and of the $∂$ 's.

Core Concepts

A cyclic module is one unknown function and all its consequences

Read $A_n/I$ as follows: the class of $1$ is a symbol standing for an unknown function $u$ ; the elements of $I$ are the differential equations imposed on it; the module is the space of all expressions $P(u)$ , with two expressions identified when their difference is a consequence of the equations. This is precisely the D-module attached to a differential equation. Cyclicity is the statement that one unknown function suffices; a general finitely generated module needs a vector of unknowns and a matrix of equations.

The ideal is data about the generator, not about the module

Different generators give different ideals. Two left ideals $I$ and $J$ give isomorphic quotients precisely when they are similar: there is $a ∈ R$ with

J = , r ∈ R : ra ∈ I , Ra + I = R,

(5.11)

the two conditions saying that $1 ↦ a$ defines an injective and a surjective map $R/J → R/I$ . Similarity, not equality, is the right relation on presentations, and it is not easy to test.

Why cyclicity is cheap over the Weyl algebra

Over a commutative ring, $R/m ⊕ R/m$ is never cyclic: a cyclic module has a cyclic quotient by every submodule, and $(R/m)^2$ does not. Over $A_n$ this argument fails, because $A_n$ is simple: an operator killing one summand's generator need not kill the other's, and that asymmetry is enough to build a single generator for a direct sum. The mechanism is small — one element $a$ with $au = 0$ and $av ≠ 0$ — but its consequences run all the way to Stafford's theorem that every holonomic module is cyclic.

Construction and Proof

The presentation of the polynomial moduleCoutinho (5.1.2)

$K[X]$ is a cyclic $A_n$ -module generated by the constant $1$ , and $ann_A_n(1) = ∑ _i=1^n A_n∂_i$ , so $K[X] ≅ A_n/ ∑ _i A_n∂_i$ .

Proof

The element $1$ generates, since $x^α = x^α ⋅ 1$ . Let $J = ∑ _i A_n∂_i$ . Clearly $J ⊂ eq ann(1)$ , because $∂_i(1) = 0$ .

For the reverse inclusion, use the canonical form: every $P ∈ A_n$ is a $K$ -combination of monomials $x^α∂^β$ . Those with $β ≠ 0$ lie in $J$ , since $x^α ∂^β = (x^α∂^β - e_i)∂_i$ for any $i$ with $β_i > 0$ . So $P = f + Q$ with $f ∈ K[X]$ and $Q ∈ J$ . If $P ∈ ann(1)$ then $0 = P(1) = f(1) + Q(1) = f$ , whence $P = Q ∈ J$ .

Every irreducible module is cyclic and torsionCoutinho, Ch. 5, Exercise 4.3

If $M$ is an irreducible $A_n$ -module and $u ≠ 0$ , then $A_nu$ is a non-zero submodule, hence equals $M$ . Since $A_n$ is not a division ring for $n ≥ 1$ , the second lemma above makes $M$ a torsion module. Applied to $K[X]$ — which is irreducible — this says every non-zero polynomial generates the whole module, which the direct argument confirms: differentiating a polynomial $f$ enough times in each variable produces a non-zero constant.

Direct sums of simple modules are cyclicCoutinho, Ch. 5, Exercises 4.4 and 4.5

Let $R$ be a simple ring that is not a division ring, $M$ a cyclic torsion left $R$ -module and $M^{'}$ an irreducible left $R$ -module. Then $M ⊕ M'$ is cyclic. Consequently a direct sum of finitely many irreducible $A_n$ -modules is cyclic.

Proof

Let $u$ generate $M$ . Since $M$ is torsion, choose $a ≠ 0$ with $au = 0$ . Because $R$ is simple, the two-sided ideal generated by $a$ is all of $R$ , so $aM' ≠ 0$ ; choose $v ∈ M'$ with $av ≠ 0$ . Then

a,(u + v) = av ≠ 0,

so $R(u+v)$ contains the non-zero element $a v$ of the irreducible module $M^{'}$ , hence contains $Rav = M'$ , hence contains $v$ and therefore $u$ . Thus $R(u+v) ⊃ eq Ru + M' = M ⊕ M'$ .

For the consequence, induct. A finite direct sum of irreducible $A_n$ -modules is torsion: each summand is torsion by the corollary, and a finite intersection of non-zero left ideals of $A_n$ is non-zero because $A_n$ is a Noetherian domain and so satisfies the left Ore condition. Feeding the sum of the first $k$ summands in as $M$ and the next summand as $M^{'}$ gives cyclicity at each stage.

What the induction really needs

The hypothesis " $M$ torsion" is not decoration: without it there is no operator $a$ to start from. Over a general non-commutative domain the sum of two torsion elements need not be torsion — the free algebra on two generators has non-zero left ideals meeting in zero — so the Ore property of $A_n$ is doing real work in the induction step.

Key Equations

The two basic presentations, for $K[X] = K[x_1, … ,x_n]$ with its standard action and for the module $K[∂]$ in which the $∂_i$ act by multiplication:

K[X] ≅ A_n / ∑ _i=1^nA_n∂_i, K[∂] ≅ A_n / ∑ _i=1^nA_n x_i.

(5.12)

Their common generalisation, for polynomials $g_1, … ,g_n$ satisfying $∂_i g_j = ∂_j g_i$ :

A_n / ∑ _i=1^nA_n(∂_i - g_i) ; ≅ ; K[X] as K-vector spaces,

(5.13)

though not as $A_n$ -modules unless all $g_i = 0$ ; the module on the left is the twist of $K[X]$ by $∂_i ↦ ∂_i + g_i$ .

The homomorphisms out of a cyclic module, which is how presentations are used in practice:

Hom_A_n(A_n/I,, M) ; ≅ ; , m ∈ M : Im = 0 ,.

(5.14)

A general finitely generated module is a cokernel: with $Φ$ an $s \times r$ matrix over $A_n$ acting on the right,

A_n^,s ⟶ ⋅ ,Φ A_n^,r ⟶ M ⟶ 0.

(5.15)

Variable Definitions

$R$: an arbitrary ring with identity, specialised to $A_n$
$A_n$: the $n$ -th Weyl algebra over a field $K$ of characteristic zero
$K[X]$: the polynomial ring $K[x_1, … ,x_n]$ as a left $A_n$ -module
$K[∂]$: the module $A_n/ ∑ _i A_nx_i$ , on which the $∂_i$ act by multiplication
$u, v$: elements of a module, typically generators
$ann(u)$: the left ideal of operators annihilating $u$
$I, J$: left ideals of $A_n$ , presenting cyclic modules
$Φ$: a presentation matrix, acting on the right on row vectors of operators
$g_i$: polynomials defining a twist, subject to $∂_i g_j = ∂_j g_i$

Properties and Behaviour

Finitely generated implies finitely presented

$A_n$ is left Noetherian, so the kernel of any surjection $A_n^,r → M$ from a free module of finite rank is again finitely generated. Every finitely generated $A_n$ -module therefore has a presentation (5.15) by a finite matrix of operators, and the matrix is what a computer algebra system stores.

Modules that are not cyclic

$A_n^,2$ is not cyclic. Since $A_n$ is a Noetherian domain it has a division ring of fractions $D$ , and $_D(D ⊗ _A_n M)$ — the rank — is additive on short exact sequences and equals $2$ for $A_n^2$ , whereas a cyclic module has rank at most $1$ .

The module P0 of holomorphic functions is not cyclic either, for a different reason: a cyclic module is either torsion or free of rank one, and $H(U)$ is neither, since it contains both the torsion element $e^z$ and the non-torsion element $( z)$ .

Holonomic modules are cyclicStafford 1978; Coutinho, Ch. 10

Every holonomic $A_n$ -module is cyclic. This is much stronger than the direct-sum statement proved above — holonomic modules have finite length but need not be semisimple — and its proof is a genuine piece of ring theory, carried out on the dedicated page. Combined with the presentation lemma it means every holonomic module can be written as $A_n/I$ for a single left ideal $I$ .

Standard modules and their presentations.
Module	Generator	Annihilator of the generator	Cyclic?
$K[X]$	$1$	$∑ _i A_n∂_i$	yes
$K[∂]$	$1$	$∑ _i A_n x_i$	yes
$K[X]$ over $A_1$ , generator $x$	$x$	$A_1(x∂ - 1) + A_1∂^2$	yes, non-principal ideal
$K[x] ⊕ K[x]$	$(1,x)$	$A_1∂^2$	yes
$K[x][1/x]$	$1/x$	$A_1(x∂ + 1)$	yes
$A_n$	$1$	$0$	yes, not simple
$A_n^,2$	—	—	no, rank $2$
$H(U)$	—	—	no

Examples and Special Cases

A cyclic module that is not simple

$A_n$ is generated by $1$ and is therefore cyclic, but it has plenty of submodules — every left ideal. Cyclicity says nothing about simplicity in either direction.

Two non-isomorphic simple summands

Inside $H(C)$ take $M = C[x] ⊕ C[x]e^x$ , the sum of the solution modules of $u' = 0$ and $u' = u$ . The recipe gives the generator $1 + e^x$ : the operator $∂$ kills $1$ and sends $e^x$ to itself. Computing the annihilator, $P = ∑ _i f_i∂^,i$ satisfies $P(1 + e^x) = f_0 + ( ∑ _i f_i)e^x$ , which vanishes exactly when $f_0 = 0$ and $∑ _i f_i = 0$ ; that set of operators is precisely $A_1(∂^2 - ∂)$ . So

C[x] ⊕ C[x]e^x ; ≅ ; A_1/A_1(∂^2 - ∂),

the module of the equation $u'' = u'$ , whose classical solution space is spanned by $1$ and $e^x$ .

The twisted presentations

For polynomials $g_i$ with $∂_i g_j = ∂_j g_i$ , the module $A_n/ ∑ _i A_n(∂_i - g_i)$ is cyclic by construction, simple, and isomorphic to $K[X]$ as a $K$ -vector space but not as a module. These are the twists of the polynomial module; the $K$ -space isomorphism comes from writing every operator as $f + Q$ with $f ∈ K[X]$ and $Q$ in the ideal, exactly as in the proof of (5.12).

A localisation

$K[x][1/x]$ is cyclic over $A_1$ , generated by $1/x$ : applying $∂$ repeatedly gives every $x^-m$ up to a factorial, and multiplying by $x$ gives back $K[x]$ . Its annihilator is $A_1(x∂ + 1)$ . The module is not simple — $K[x]$ is a submodule — which again shows cyclicity is much weaker than irreducibility.

Worked Example

$K[x] ⊕ K[x]$ is cyclic, and its presentation is $A_1/A_1∂^2$

Step 1 - choose the generator by the recipe
Work over $A_1$ with $M = K[x] ⊕ K[x]$ . Take $u = 1$ in the first summand, and $a = ∂$ , which kills it. For the second summand pick $v$ with $∂ v ≠ 0$ ; the simplest is $v = x$ . The theorem predicts that $w = (1, x)$ generates $M$ .
Step 2 - verify that $w$ generates
$∂ ⋅ w = (0, 1)$ . Since $K[x]$ is generated by $1$ , the submodule $A_1w$ contains $(0, K[x])$ , in particular $(0,x)$ . Subtracting, $w - (0,x) = (1,0) ∈ A_1w$ , so $A_1w$ also contains $(K[x], 0)$ . Hence $A_1w = M$ and $M$ is cyclic.
Step 3 - compute the annihilator
Write $P ∈ A_1$ in canonical form $P = ∑ _i ≥ 0 f_i(x)∂^,i$ . Acting on $w$ coordinatewise,

$P ⋅ w = ( ∑ _i f_i,∂^,i(1), ∑ _i f_i,∂^,i(x) ) = (f_0, f_0x + f_1).$

This vanishes exactly when $f_0 = 0$ and then $f_1 = 0$ . So $ann(w) = ∑ _i ≥ 2f_i∂^,i = A_1∂^2$ .
Step 4 - read off the presentation
By the first lemma, $M ≅ A_1/A_1∂^2$ . A single equation, $u'' = 0$ , presents a module that is a direct sum of two copies of the simple module $K[x]$ — which is exactly the classical statement that the solutions of $u'' = 0$ form a two-dimensional space spanned by $1$ and $x$ .
Step 5 - cross-check with multiplicity
For $0 ≠ P ∈ A_1$ of Bernstein degree $d$ one has $d(A_1/A_1P) = 1$ and $e(A_1/A_1P) = d$ . Here $P = ∂^2$ has degree $2$ , so $e = 2$ . On the other side, multiplicity is additive in short exact sequences and $e(K[x]) = 1$ , so $e(K[x] ⊕ K[x]) = 2$ . The two computations agree, as they must.
Step 6 - exhibit the splitting inside the quotient
It is worth seeing the decomposition on the other side. In $A_1/A_1∂^2$ the class of $∂$ generates a submodule isomorphic to $K[x]$ , and so does the class of $1 - x∂$ , because $∂(1 - x∂) = ∂ - (x∂ + 1)∂ = -x∂^2 ∈ A_1∂^2$ . The two are distinct simple submodules, so they intersect in zero and their sum is everything.

Result

$K[x] ⊕ K[x] ≅ A_1/A_1∂^2$ , cyclic with generator $(1,x)$ and multiplicity $2$ . No commutative ring behaves this way: a module annihilated by a maximal ideal $𝔪$ is a vector space over $R/m$ , and such a module is cyclic only if it is one-dimensional. Over $A_1$ the phenomenon is routine.

Applications and Industry Use

In a mathematics topic, this section covers downstream use inside mathematics, computing and engineering rather than a manufactured product.

Reducing a system to a single equation. The classical cyclic vector theorem says that a differential system over $C(z)$ of size $k$ is equivalent to a single scalar equation of order $k$ . Cyclicity of the corresponding module is exactly that statement, and it is why the theory of scalar linear ODEs is not a special case but the general one.
Storage and interchange in computer algebra. A holonomic function is stored as an annihilating ideal plus initial conditions. Cyclicity guarantees a single-operator or single-ideal record exists, which is what makes such representations canonical enough to exchange between systems.
Linear control systems. In the behavioural approach, a linear time-varying system is a finitely presented module over an Ore algebra; controllability and flatness are properties of that module, and a cyclic presentation corresponds to a single input-output relation.
Building examples. Most of the modules used as test cases in this collection — the polynomial module, its twists, localisations, the delta module — are named by their presentations. Being fluent in passing between a generator and its ideal is the basic working skill.

Computational Notes

Read this as the manufacturing section of the template: how the object is actually built by machine, at what cost, and where the computation stops being decidable.

Three tasks recur, in increasing order of difficulty.

Compute $ann(u)$ for an explicit $u$ in an explicit module. This is a syzygy computation: express the action of a generic operator on $u$ in coordinates and solve. Over $A_n$ it is a Gröbner basis computation with respect to a term order refining the Bernstein filtration.
Decide whether $A_n/I ≅ A_n/J$ . This is similarity of left ideals, equivalent to finding $a$ satisfying (5.11). For holonomic modules there are terminating algorithms based on Hom computations; in general no procedure is known.
Decide cyclicity of a module given by a presentation matrix, and produce a generator. Stafford's theorem guarantees a generator exists in the holonomic case; making it effective is delicate, and algorithms following Hillebrand and Schmale compute one for modules over $A_n$ , with randomisation used to find a suitable element.

Practical note: the cheap invariants — dimension, multiplicity, rank — should be computed before any isomorphism test, because they refute far more candidate isomorphisms than they cost. Implementations to use are the Dmodules package in Macaulay2, dmod.lib in Singular and the ore_algebra package in SageMath.

Limits of Validity

Cyclicity is not detectable from a presentation matrix by inspection. Deciding whether a module given by a matrix is cyclic, and producing a generator, is a real computation; see the notes below.
The presentation is not unique, and equality of ideals is the wrong test. Only similarity in the sense of (5.11) characterises isomorphism, and testing it is hard even for $A_1$ .
The direct-sum theorem needs simplicity of the ring and torsion of one summand. Over a ring with non-trivial two-sided ideals, an operator killing the first generator can kill the second summand as well, and the argument collapses.
Nothing here bounds the complexity of the generator. The proofs produce an element, not a small one: the degrees of the operators involved can be much larger than those in the original presentation.
Infinitely generated modules are outside the discussion. For $H(U)$ , $C(z)$ and other unions of finitely generated modules, there is no presentation matrix at all.

Failure Modes and Common Mistakes

Treating the annihilator as an invariant of the module

The left ideal depends on the generator. Over $A_1$ the module $K[x]$ is $A_1/A_1∂$ if the generator is $1$ , but with the generator $x$ the annihilator is $A_1(x∂ - 1) + A_1∂^2$ , which is not even principal. Reporting a presentation therefore means reporting a generator too. What is invariant is the two-sided annihilator, and over the simple ring $A_n$ that is always zero — so it carries no information at all.

Reading "cyclic" as "simple", or "one equation" as "one solution"

$A_1/A_1∂^2$ is cyclic and has length $2$ ; $A_n$ is cyclic and has infinite length. A single defining equation of order $k$ generally produces a module with a $k$ -dimensional solution space, not a one-dimensional one. The number of generators of the module and the dimension of its solution space are unrelated quantities.

Assuming torsion elements always form a submodule

They do over $A_n$ , because it is a Noetherian domain and hence left Ore, so any two non-zero left ideals intersect non-trivially. Over a general non-commutative domain this fails, and with it fails the induction that makes finite direct sums of simple modules cyclic. When quoting the exercise, quote the Ore condition with it.

Mixing left and right conventions in the presentation matrix

For left modules the relations matrix acts on the right, and transposing it does not give the relations for the same module — it gives a right module, which corresponds to the left module only after applying the transposition anti-automorphism. Systems written as $Φ f = 0$ for a column vector of unknown functions correspond to the cokernel of right multiplication by $Φ$ ; getting this backwards silently replaces a module by a different one.

Historical Notes

The idea that a system of linear differential equations can be replaced by a single scalar equation is nineteenth-century in origin and was made precise as the cyclic vector theorem; Jacobson gave an algebraic proof in the 1930s for differential modules over a differential field, and the result was rediscovered repeatedly in the analytic literature. In D-module language it is a statement about cyclicity of a module over a ring of operators.

The Weyl algebra versions are due to Stafford, whose 1978 paper on the module structure of Weyl algebras proved both that every left ideal of $A_n$ is generated by two elements and that holonomic modules are cyclic. The exercises in Coutinho's Chapter 5 isolate the elementary special case — finite direct sums of irreducibles — with an argument short enough to check by hand, and that is the version proved above.

The effective side is recent by comparison. Turning Stafford's existence proofs into algorithms that output a generator, with bounds, was taken up in the 1990s and 2000s in the constructive algebra and systems theory communities, where cyclic presentations correspond to input-output representations of linear systems.

Comparison

Four finiteness conditions on a left $A_n$ -module, and how they relate.
Condition	Meaning	Implies	Typical failure
Irreducible	no proper non-zero submodules	cyclic, torsion	$A_n$ itself
Cyclic	one generator	finitely generated	$A_n^,2$ , by rank
Finitely generated	finitely many generators	finitely presented, since $A_n$ is Noetherian	$H(U)$ , $C(z)$
Holonomic	finitely generated with $d(M) = n$	cyclic and of finite length	$A_n$ , which has $d = 2n$

The surprising entry is the last: over $A_n$ the strong finiteness condition implies the weak structural one. Over a commutative Noetherian ring nothing similar holds — a module of finite length is generally not cyclic.

Key Takeaways

Key points

A cyclic module is $R u$ , and $Ru ≅ R/ann(u)$ ; presentations of cyclic modules are exactly left ideals.
The ideal depends on the generator; isomorphism of $A_n/I$ and $A_n/J$ means similarity of $I$ and $J$ , not equality.
Every irreducible module is cyclic, and over a ring that is not a division ring it is a torsion module.
$K[X] ≅ A_n/ ∑ _i A_n∂_i$ and $K[∂] ≅ A_n/ ∑ _i A_n x_i$ ; the twisted versions replace $∂_i$ by $∂_i - g_i$ .
A finite direct sum of irreducible $A_n$ -modules is cyclic, using only simplicity of $A_n$ and the Ore condition; $K[x] ⊕ K[x] ≅ A_1/A_1∂^2$ .
Stafford's theorem goes much further: every holonomic module is cyclic.
Cyclicity fails for $A_n^2$ (rank) and for $H(U)$ (a mixture of torsion and non-torsion elements).

FAQs

Why is a cyclic module the same thing as a quotient by a left ideal?

Because sending $a \mapsto a u$ is a surjective module map $R \to R u$ whose kernel is by definition $ann(u)$ . Kernels of maps out of $R$ as a left module over itself are exactly the left ideals, so the two notions carry the same data.

Can the same module have both a principal and a non-principal presentation?

Yes. $K[x]$ over $A_1$ is $A_1/A_1∂$ with generator $1$ , and $A_1/(A_1(x∂ - 1) + A_1∂^2)$ with generator $x$ ; the second ideal is not principal. Both present the same simple module.

How can a direct sum of two copies of one simple module be cyclic?

Because $A_1$ is simple, an operator killing the generator of one copy need not kill the other. Concretely $∂$ kills $1$ but not $x$ , and that asymmetry lets $(1,x)$ generate $K[x] ⊕ K[x]$ . Over a commutative ring the two copies would be killed by the same maximal ideal and the trick is unavailable.

Does a cyclic presentation tell me the solution space?

It tells you how to compute it: by (5.14), homomorphisms $A_n/I → M$ are the elements of $M$ killed by $I$ . For $A_1/A_1P$ with $P$ of order $k$ and non-vanishing leading coefficient, that space is $k$ -dimensional over $ℂ$ inside the holomorphic functions on a simply connected domain.

Is every finitely generated

A_n

-module cyclic?

No — $A_n^2$ is not, by the rank argument. But Stafford proved that every finitely generated module over $A_n$ is generated by two elements, and that holonomic ones are generated by one, so the failure is as mild as it could be.

What does the presentation matrix look like for a system of equations?

If the system is $P_1(f) = … = P_m(f) = 0$ in one unknown, the module is $A_n/ ∑ _j A_nP_j$ , a cyclic module presented by a column of operators. Several unknowns give a genuine matrix; see systems and presentation matrices.

Why insist that the generator be recorded?

Because two people can present the same module by unrelated-looking ideals. Without the generator there is no way to compare presentations except by solving the similarity problem, which is much harder than remembering which element was used.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 — Ch. 5 §1, results (5.1.1) and (5.1.2), and Exercises 4.1 to 4.5.
J. T. Stafford, Module structure of Weyl algebras, Journal of the London Mathematical Society (2) 18 (1978), 429–442 — two-generation of left ideals and cyclicity of holonomic modules.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Graduate Studies in Mathematics 30, revised edition, American Mathematical Society, 2001 — Ch. 1 and Ch. 2, for Ore conditions, rank and similarity of ideals.
P. M. Cohn, Algebra, Volume 1, 2nd edition, Wiley, 1982 — Ch. 10, the module theory Coutinho assumes.
A. Hillebrand and W. Schmale, Towards an effective version of a theorem of Stafford, Journal of Symbolic Computation 32 (2001), 699–716.
J.-E. Björk, Rings of Differential Operators, North-Holland, 1979 — Ch. 1, for filtrations, multiplicity and presentations.
N. Jacobson, Pseudo-linear transformations, Annals of Mathematics 38 (1937), 484–507 — an early algebraic form of the cyclic vector theorem.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics, International Organization for Standardization.

AI Suggested Questions

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