The Polynomial Ring as a Module over the Weyl Algebra

Q: Do I need to check that the action is well defined?

No, provided the Weyl algebra is defined as a subring of End_K(K[X]) , which is the definition used in this collection. The action is then the inclusion itself. If instead you define A_n by generators and relations, you must check that multiplication and differentiation satisfy those relations - which is the content of [∂_i, x_j] = δ_ij as operators, verified by the product rule.

Q: Why is ∑ _i A_n∂_i a maximal left ideal?

Because the quotient by it is K[X] , which is a simple module in characteristic zero. Left ideals containing J correspond to submodules of A_n/J , and a simple module has only the two trivial submodules.

Q: Is K[X] isomorphic to K[∂] as an A_n-module?

No. Their annihilator ideals are ∑ _i A_n∂_i and ∑ _i A_n x_i , and these are different maximal left ideals; an isomorphism of cyclic modules would force the annihilators of corresponding generators to coincide. They are, however, related by an automorphism of A_n : K[∂] is the Fourier twist of K[X] .

Q: Does the presentation change in characteristic p?

No. The proof uses only that x^α∂^β is a K -basis of A_n , which holds in any characteristic. What fails in characteristic p is simplicity, and hence the maximality of the ideal: x_1^pK[X] is a proper non-zero submodule.

Q: What is the analogue for A_n/ ∑ _i A_n(∂_i - g_i)?

As a K -vector space it is again K[X] , by the same splitting argument, but the derivative acts by f ↦ ∂ f/∂ x_i + g_i f . When the g_i satisfy the integrability condition ∂ g_i/∂ x_j = ∂ g_j/∂ x_i these modules are exactly the twists of K[X] by automorphisms fixing the x_i , and deciding when two of them are isomorphic produces infinitely many pairwise non-isomorphic simple modules.

Overview

The Weyl algebra $A_n = A_n(K)$ is not defined abstractly and then represented; it is defined as a ring of $K$ -linear operators on the polynomial ring $K[X] = K[x_1, … ,x_n]$ , namely the subring of $End_K(K[X])$ generated by the multiplication operators $x_1, … ,x_n$ and the partial derivatives $∂_1, … ,∂_n$ . The polynomial ring is therefore a left $A_n$ -module before anything is proved: it is the module the algebra was built to act on.

That makes $K[X]$ the prototype for the whole theory, and it is worth extracting exactly what kind of module it is. Three facts settle almost everything. It is cyclic, generated by the constant polynomial $1$ . Its presentation is $K[X] ≅ A_n / ∑ _i=1^n A_n ∂_i$ , because the annihilator of $1$ is precisely the left ideal generated by the partials. And it is simple in characteristic zero — every non-zero polynomial generates the whole module — which is proved on the simplicity page.

Two features are worth flagging at once because they are easy to get backwards. First, $K[X]$ is a torsion module: every polynomial is killed by some non-zero operator (take enough derivatives). Second, it is nevertheless faithful: no non-zero operator kills all of $K[X]$ , since $A_n$ sits inside $End_K(K[X])$ by definition. Torsion is a statement about elements one at a time; faithfulness is a statement about the module as a whole, and here the two point in opposite directions.

Almost every other module in this collection is built from this one. The module $K[∂] ≅ A_n/ ∑ _i A_n x_i$ is its Fourier transform; the twisted family $A_n/ ∑ _i A_n(∂_i - g_i)$ is what "multiply by $∈ t g$ " looks like algebraically; and the module of holomorphic functions is the analytic enlargement in which $K[X]$ sits as a very small submodule.

Definition

Throughout, $K$ is a field, taken of characteristic zero wherever simplicity is used, and $K[X]$ abbreviates $K[x_1, … ,x_n]$ .

The polynomial moduleCoutinho, Ch. 5 §1

The polynomial module is $K[X]$ equipped with the left $A_n$ -action determined on generators by

x_i ⋅ f = x_i f, ∂_i ⋅ f = (∂ f)/(∂ x_i) (1 ≤ i ≤ n, f ∈ K[X]).

This is well defined without any verification: the operators on the right are exactly the generators of $A_n$ inside $End_K(K[X])$ , so the action is the inclusion $A_n ↪ End_K(K[X])$ read as a module structure.

Written out on a general operator in canonical form, the action is

( ∑ _α,β c_αβ, x^α ∂^β ) ⋅ f ;=; ∑ _α,β c_αβ, x^α ∂^|β| f∂ x^β,

(5.1)

which is why the canonical form with all $x$ 's to the left of all $∂$ 's is the convenient one: it displays the operator in the form in which it is applied.

Torsion element, torsion module

Let $R$ be a ring and $M$ a left $R$ -module. An element $u ∈ M$ is a torsion element if its annihilator $ann_R(u) = a ∈ R : au = 0$ is a non-zero left ideal. If every element of $M$ is a torsion element, $M$ is a torsion module. Note that $u = 0$ is always a torsion element, with annihilator $R$ .

A caution about the word "module"

$K[X]$ appears in two different roles. As a subring of $A_n$ it is the commutative subalgebra generated by the multiplication operators. As a module it is the object defined above. The two are different structures on the same underlying set, and the module is not the ring acting on itself: $A_n$ is free of rank one over itself, while $K[X]$ is not even projective (see Properties).

Core Concepts

Why a quotient and not a subobject

A cyclic module is always a quotient of the ring by a left ideal, and reading $K[X]$ that way is what converts questions about polynomials into questions about ideals of $A_n$ . The generator is $1$ , and the relations satisfied by $1$ are exactly $∂_i ⋅ 1 = 0$ . Saying "the polynomial module is $A_n$ modulo the left ideal of the partials" is the algebraic form of the sentence "a polynomial is determined by its derivatives at the origin, and the constant function is the one all of whose derivatives vanish".

Differentiation lowers degree, and that is the whole story

Two elementary facts about $∂^β$ acting on monomials drive everything on this page and the next. Applying $∂^β$ to $x^α$ gives zero unless $β ≤ α$ componentwise, and gives the constant $α!$ when $β = α$ . Enough differentiation therefore destroys any given polynomial (torsion) but can also extract a non-zero constant from it (simplicity). Which of the two happens depends only on how many derivatives you take and in which directions.

Torsion versus faithful

It is tempting to read "torsion module" as "the module is annihilated by something". It is not. Over $A_n$ the annihilators $ann(f)$ are all non-zero, but their intersection over all $f ∈ K[X]$ is $0$ : that intersection is a two-sided ideal, and by simplicity of P4 the only two-sided ideals are $0$ and $A_n$ , while $1 ⋅ 1 = 1 ≠ 0$ .

The size of the module

$K[X]$ is infinite dimensional over $K$ , and it has to be: $A_n$ has no non-zero finite-dimensional representations in characteristic zero, because the trace of $[∂_i, x_i] = 1$ would have to be both $0$ and $dim$ . So $K[X]$ is, in a precise sense, as small as a faithful $A_n$ -module can be: with respect to the Bernstein filtration it has dimension $d(K[X]) = n$ and multiplicity $e(K[X]) = 1$ , the minimum allowed by Bernstein's inequality.

Construction and Proof

There is really only one thing to prove, and it is the presentation (5.4). The proof rests entirely on the canonical basis theorem: the monomials $x^α ∂^β$ form a $K$ -basis of $A_n$ .

Presentation of the polynomial moduleCoutinho (5.1.2)

$K[X]$ is a cyclic left $A_n$ -module generated by $1$ , the annihilator of $1$ is the left ideal $J = ∑ _i=1^n A_n ∂_i$ , and consequently $K[X] ≅ A_n/J$ as left $A_n$ -modules.

Proof

Consider the $A_n$ -linear map $φ : A_n → K[X]$ , $φ(P) = P ⋅ 1$ . It is $A_n$ -linear because the action is associative: $φ(QP) = (QP) ⋅ 1 = Q ⋅ φ(P)$ .

Surjectivity. For $f ∈ K[X]$ , the multiplication operator $f ∈ A_n$ satisfies $φ(f) = f ⋅ 1 = f$ . So $φ$ is onto and $K[X] = A_n ⋅ 1$ is cyclic.

$J ⊂ eq φ$ . Each generator satisfies $∂_i ⋅ 1 = 0$ , so for any $Q_i ∈ A_n$ we get $( ∑ _i Q_i ∂_i) ⋅ 1 = ∑ _i Q_i ⋅ 0 = 0$ .

$φ ⊂ eq J$ . Write $P$ in canonical form and group the terms by the $∂$ -exponent:

P ;=; ∑ _β f_β(x), ∂^β ;=; f_0(x) ;+; ∑ _β ≠ 0 f_β(x), ∂^β .

If $β ≠ 0$ then $β_i ≥ 1$ for some $i$ , so $∂^β = ∂^β - e_i∂_i$ and the term $f_β ∂^β$ lies in $A_n ∂_i ⊂ eq J$ . Hence $P = f_0 + Q$ with $Q ∈ J$ , and applying $φ$ gives $φ(P) = f_0$ . If $P ∈ φ$ then $f_0 = 0$ and $P = Q ∈ J$ .

So $φ = J$ , and the first isomorphism theorem for modules gives $A_n/J ≅ K[X]$ .

The decomposition $A_n = K[X] ⊕ J$

The proof shows more than the isomorphism. Every element of $A_n$ splits uniquely as a polynomial plus an element of $J$ : uniqueness holds because a polynomial lying in $J$ is annihilated by $φ$ and equals its own image. So $A_n = K[X] ⊕ J$ as $K$ -vector spaces, and the isomorphism $A_n/J → K[X]$ is just "take the $∂$ -free part of the canonical form". This makes membership in $J$ decidable by inspection, with no Gröbner basis computation required.

$J$ is a maximal left ideal

Because $K[X]$ is simple in characteristic zero, $A_n/J$ has no proper non-zero submodules, so $J$ is a maximal left ideal of $A_n$ . It is emphatically not a two-sided ideal: the two-sided ideal generated by $∂_1, … ,∂_n$ contains $[∂_1, x_1] = 1$ and is therefore all of $A_n$ .

Key Equations

With multi-index notation $α = (α_1, … ,α_n) ∈ N^n$ , $x^α = x_1^α_1 ⋅ s x_n^α_n$ and $α! = α_1! ⋅ sα_n!$ , the action on monomials is

∂^β ⋅ x^α = cases α!(α-β)!, x^α-β & if β ≤ α componentwise,[2pt] 0 & otherwise. cases

(5.2)

The two extreme cases of (5.2) are the ones used repeatedly:

∂^α ⋅ x^α = α! ∈ K^ × (char K = 0), ∂^β ⋅ x^α = 0 whenever |β| > |α|.

(5.3)

The annihilator of the generator and the resulting presentation are

ann_A_n(1) = ∑ _i=1^n A_n ∂_i, K[X] ; ≅ ; A_n / ∑ _i=1^n A_n ∂_i .

(5.4)

A torsion certificate for an arbitrary non-zero polynomial, immediate from (5.3):

∂_i^,N ⋅ f = 0 for every i and every N > f, so ann_A_n(f) ≠ 0.

(5.5)

Finally, the companion module obtained by dividing by the $x_i$ instead of the $∂_i$ :

A_n / ∑ _i=1^n A_n x_i ; ≅ ; K[∂] = K[∂_1, … ,∂_n], x_i ⋅ ∂^β = -β_i, ∂^β - e_i.

(5.6)

Here $e_i$ is the $i$ -th standard basis vector and the term is read as zero when $β_i = 0$ . On $K[∂]$ the roles are exchanged: the partials multiply and the variables differentiate, up to sign.

Variable Definitions

$K$: the ground field; characteristic zero wherever simplicity or $α! ≠ 0$ is invoked
$K[X]$: shorthand for the polynomial ring $K[x_1, … ,x_n]$
$A_n$: the $n$ -th Weyl algebra over $K$ , a subring of $End_K(K[X])$
$x_i$: the operator "multiply by $x_i$ ", and also the $i$ -th variable of $K[X]$
$∂_i$: the operator $∂/∂ x_i$ on $K[X]$
$α, β$: multi-indices in $N^n$ , with $|α| = α_1 + … + α_n$ and $α! = α_1! ⋅ sα_n!$
$x^α ∂^β$: the canonical basis monomials of $A_n$ , all variables written to the left of all partials
$J$: the left ideal $∑ _i=1^n A_n ∂_i$ , equal to $ann_A_n(1)$
$ann_R(u)$: the left annihilator $a ∈ R : au = 0$ of an element $u$ of a left $R$ -module
$g_i$: polynomials used to twist the action, giving the modules $A_n/ ∑ _i A_n(∂_i - g_i)$

Properties and Behaviour

Basic properties of $K[X]$ as an $A_n$ -moduleCoutinho (5.1.1), (5.1.2)

It is cyclic, generated by $1$ , with $ann_A_n(1) = ∑ _i A_n∂_i$ .
It is faithful: $ann_A_n(K[X]) = 0$ .
It is a torsion module: every $f$ satisfies $∂_1^N ⋅ f = 0$ for $N > f$ .
In characteristic zero it is simple, so every non-zero $f$ generates it.
It is infinite dimensional over $K$ , and as an $A_n$ -module it is neither free nor projective; over the subring $K[X] ⊂ A_n$ it is of course free of rank one.

Why $K[X]$ is not projective over $A_n$

Take $n = 1$ and the presentation $0 → A_1∂ → A_1 → K[x] → 0$ . If $K[x]$ were projective the sequence would split, exhibiting $K[x]$ as a direct summand of the free module $A_1$ . But $A_1$ is a domain, so a free module over it has no non-zero torsion elements, while every element of $K[x]$ is torsion. Hence no splitting exists. The same argument works for every $n$ .

The general torsion lemma behind (2) and (3)Coutinho (5.1.1)

Let $R$ be a ring and $M$ a simple left $R$ -module. Then (i) $M ≅ R/ann_R(u)$ for every non-zero $u ∈ M$ , and (ii) if $R$ is not a division ring, $M$ is a torsion module.

For (i), the map $r \mapsto r u$ is $R$ -linear with image a non-zero submodule, hence all of $M$ ; its kernel is $ann_R(u)$ . For (ii), if some non-zero $u$ had $ann_R(u) = 0$ then $M ≅ R$ as left modules, so $R$ would have no left ideals other than $0$ and $R$ , which for a ring with identity forces $R$ to be a division ring.

Dimension and multiplicity

Filtering $K[X]$ by the good filtration $Γ_m = B_m ⋅ 1$ induced by the generator gives $Γ_m = f : f ≤ m$ , of dimension $m+nn$ . Hence P4 $= n$ and P6 $= 1$ : the polynomial module is holonomic, and of the smallest possible multiplicity.

Examples and Special Cases

$n = 1$ : the module $K[x]$

$K[x] ≅ A_1/A_1∂$ . The generator is $1$ ; the element $x^k$ generates too, since $∂^k x^k = k!$ . A $K$ -basis is $1, x, x^2, …$ , on which $∂$ shifts down with a coefficient and $x$ shifts up. This is the algebraic skeleton of the raising and lowering operator picture.

The companion module $K[∂]$

Dividing by the other half of the generators gives $A_n/ ∑ _i A_n x_i$ , which as a $K$ -vector space has basis the classes of $∂^β$ , so it may be identified with the polynomial ring $K[∂_1, … ,∂_n]$ . Here the $∂_i$ act by multiplication. To see how $x_i$ acts, commute it past $∂^β$ using $[x_i, ∂_i^,k] = -k,∂_i^,k-1$ : modulo $∑ _j A_n x_j$ the term $∂^β x_i$ dies and one is left with $x_i ⋅ ∂^β = -β_i ∂^β - e_i$ . So $x_i$ acts as $-∂/∂ ξ_i$ in the symbol variables. This module is the Fourier transform of $K[X]$ .

Twisting the derivative actionCoutinho, Ch. 5 §1 and §2

Fix $g_1, … ,g_n ∈ K[X]$ and let $J_g = ∑ _i A_n(∂_i - g_i)$ . Every element of $A_n$ can still be written as $f + Q$ with $f ∈ K[X]$ and $Q ∈ J_g$ , so $A_n/J_g ≅ K[X]$ as $K$ -vector spaces, by the same " $∂$ -free part" recipe. The $x_i$ -action agrees with the usual one, but the derivative action does not:

∂_i ⋅ (f + J_g) ;=; ( (∂ f)/(∂ x_i) + g_i f ) + J_g .

Analytically this is differentiation conjugated by $! ( ∈ t g )$ , which is exactly why the $g_i$ cannot be arbitrary if one wants an automorphism of $A_n$ behind the change: the integrability condition $∂ g_j/∂ x_i = ∂ g_i / ∂ x_j$ is needed. Taking each $g_i ∈ K[x_i]$ is the easy way to satisfy it.

A module that is cyclic but not simple

For non-constant $f ∈ K[X]$ the localisation $K[X][1/f]$ is again a left $A_n$ -module, by the quotient rule, and it is finitely generated. It is not simple: it contains $K[X]$ as a proper non-zero submodule. It is nevertheless still holonomic, which shows how much weaker holonomicity is than simplicity.

Characteristic $p$ breaks simplicity, not the module

The action and the presentation (5.4) hold over any field, since the canonical basis theorem does. Simplicity does not. Over $K$ of characteristic $p$ , the subspace $x_1^p K[X]$ is an $A_n$ -submodule, because $∂_1(x_1^p h) = p x_1^p-1 h + x_1^p ∂_1 h = x_1^p ∂_1 h$ . It is proper and non-zero, so $K[X]$ is not simple. This is the module-level shadow of the failure of simplicity of P6 itself in characteristic $p$ .

Worked Example

Everything at once, for $n = 2$ and $f = x_1^2 x_2 + 3x_1$

Step 1 - the derivative table
Work in $K[x_1,x_2]$ over $ℚ$ , with $f = x_1^2x_2 + 3x_1$ of total degree $3$ . Differentiating directly:

Operator $D$ $D ⋅ f$
$∂_1$ $2x_1x_2 + 3$
$∂_2$ $x_1^2$
$∂_1^2$ $2x_2$
$∂_1∂_2$ $2x_1$
$∂_2^2$ $0$
$∂_1^3$ $0$
$∂_1^2∂_2$ $2$

Each entry is a one-line check. For instance $∂_1^2 ∂_2 f$ : first $∂_2 f = x_1^2$ (the term $3x_1$ has no $x_2$ ), then $∂_1^2 x_1^2 = 2$ .
Step 2 - $f$ is a torsion element
The table already exhibits two non-zero operators killing $f$ , namely $∂_1^3$ and $∂_2^2$ , in line with (5.5). A less obvious one comes from comparing two entries: $x_2 ⋅ (∂_1∂_2 f) = 2x_1x_2$ and $x_1 ⋅ (∂_1^2 f) = 2x_1x_2$ , so

$(x_2∂_1∂_2 - x_1∂_1^2) ⋅ f = 2x_1x_2 - 2x_1x_2 = 0 .$

So $ann_A_2(f)$ is a large left ideal, not just the obvious high-order derivatives.
Step 3 - $f$ generates the whole module
The monomial of maximal degree in $f$ is $x_1^2x_2$ , with $α = (2,1)$ and coefficient $1$ . By (5.3), $∂^α x^α = α! = 2! ⋅ 1! = 2$ , while $∂^α$ kills every other monomial of $f$ (here $3x_1$ , which $∂_2$ already destroys). So $∂_1^2∂_2 ⋅ f = 2$ , confirming the last row of the table, and

$D ⋅ f = 1 with D = 12,∂_1^2∂_2 ∈ A_2 .$

Since $1$ generates $K[x_1,x_2]$ , we get $A_2 f ∋ 1$ and therefore $A_2 f = K[x_1,x_2]$ . Note where characteristic zero entered: we divided by $α! = 2$ .
Step 4 - reading off a maximal left ideal
Because $f$ generates, $K[x_1,x_2] ≅ A_2/ann_A_2(f)$ and that annihilator is maximal. One element of it is forced by Step 3: from $D f = 1$ we get $f,(Df) = f$ , hence

$( fD - 1 ) ⋅ f = f ⋅ 1 - f = 0, fD - 1 = 12(x_1^2x_2 + 3x_1)∂_1^2∂_2 - 1 .$

This operator is non-zero (its canonical form has the basis monomial $12 x_1^2x_2∂_1^2∂_2$ with non-zero coefficient) and it has order $3$ , so it is genuinely different from the operators found in Step 2.
Step 5 - the presentation in action
Take $P = x_1∂_2^2 + x_2∂_1 + x_1x_2 ∈ A_2$ . Its $∂$ -free part is $f_0 = x_1x_2$ , and indeed $P ⋅ 1 = 0 + 0 + x_1x_2$ . Splitting as in the construction, $P = x_1x_2 + Q$ with $Q = (x_1∂_2)∂_2 + x_2∂_1 ∈ J = A_2∂_1 + A_2∂_2$ . So the class of $P$ in $A_2/J$ corresponds to the polynomial $x_1x_2$ , exactly as (5.4) predicts, and no computation harder than reading the canonical form was needed.

Operator $D$	$D ⋅ f$
$∂_1$	$2x_1x_2 + 3$
$∂_2$	$x_1^2$
$∂_1^2$	$2x_2$
$∂_1∂_2$	$2x_1$
$∂_2^2$	$0$
$∂_1^3$	$0$
$∂_1^2∂_2$	$2$

Result

For $f = x_1^2x_2 + 3x_1$ : $12∂_1^2∂_2 ⋅ f = 1$ , so $A_2 f = K[x_1,x_2]$ and $f$ generates; simultaneously $∂_1^3, ∂_2^2, x_2∂_1∂_2 - x_1∂_1^2$ and $12 f ∂_1^2∂_2 - 1$ all annihilate $f$ , so $f$ is a torsion element. A single polynomial is thus both a generator of the whole module and a torsion element - the combination that characterises a simple torsion module over a non-division ring.

Applications and Industry Use

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

The polynomial module is the coefficient object against which most concrete D-module computations are run.

Polynomial solutions of a system. If $P_1, … ,P_r ∈ A_n$ and $M = A_n/ ∑ _j A_n P_j$ is the module of the system, then $Hom_A_n(M, K[X])$ is canonically the space of polynomial solutions $f : P_j f = 0 ∀ j$ , because a homomorphism is determined by the image of the generator and the relations become the equations. This is the identity that turns solving into a Hom computation; see solutions as homomorphisms.
Holonomic rank and D-finiteness. Replacing $K[X]$ by larger solution modules - rational functions, formal power series, holomorphic functions - keeps the same formalism and changes only how many solutions are found. $K[X]$ is the smallest useful choice and detects exactly the polynomial solutions.
Symbolic summation and integration. The closure properties used by Zeilberger's algorithm are proved for modules built from $K[X]$ by localisation and twisting, so the properties of this module propagate directly into the algorithms.
Quantum mechanics. The action here is the polynomial-algebraic form of the Schrödinger representation, with position acting by multiplication and momentum by differentiation; see the quantum origins page. The absence of finite-dimensional representations is the algebraic content of the fact that position and momentum cannot both be bounded operators.

Design Considerations

For a mathematical object, design considerations are the modelling choices: which ring, which filtration, which category to work in.

Which side the module acts on

$K[X]$ is naturally a left module because operators compose on the left of the function they act on. The presentation (5.4) uses left ideals throughout, and $∑ _i A_n∂_i$ is not the same set as $∑ _i ∂_i A_n$ . When a construction later needs a right module - for instance in direct images - the correct move is to apply the transposition anti-automorphism or a side-changing functor, not to reinterpret the same ideal on the other side.

Which presentation to carry

The same module has many presentations, and they are not equally useful. $A_n/ ∑ _i A_n∂_i$ makes the constant $1$ the generator and makes membership in the ideal decidable by inspection. Taking a different generator, say $x^γ$ , produces a much more complicated annihilator that encodes the same module. As a rule, present a cyclic module by a generator whose annihilator you can describe, not by whichever generator appears first.

Polynomials or symbols

$K[X]$ and $K[∂]$ are exchanged by the Fourier automorphism but are not isomorphic as $A_n$ -modules: in one the generator is killed by all the $∂_i$ , in the other by all the $x_i$ . Choosing between them is choosing which half of the generators you want to act by multiplication, and constant-coefficient problems are usually easier on $K[∂]$ .

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 come up constantly, and only one of them is expensive.

Apply an operator to a polynomial. Put the operator in canonical form and use (5.1) term by term. Cost is linear in the number of terms of the operator times the cost of differentiating a polynomial.
Decide membership in $J = ∑ _i A_n∂_i$ . Free: put the operator in canonical form and check whether the $∂$ -free part is zero. No Gröbner basis is needed, because $x^α∂^β : β ≠ 0$ is already a basis of $J$ .
Compute $ann_A_n(f)$ for a given polynomial $f$ . This is the expensive one. It is a syzygy computation in the Weyl algebra and needs a non-commutative Gröbner basis with respect to a term order refining the chosen filtration. The answer is always a maximal left ideal in characteristic zero, but writing generators for it is not cheap.

Every general-purpose D-module package represents this module the same way, as the cyclic module presented by the ideal of the partials, and reports $d = n$ , $e = 1$ , holonomic rank $1$ . Implementations include the Dmodules package in Macaulay2, dmod.lib in Singular, and the ore_algebra package in SageMath. A useful correctness check on any of them: the reported holonomic rank of $A_n/ ∑ _i A_n∂_i$ must be $1$ , since the only solutions of $∂_i f = 0$ for all $i$ are the constants.

Failure Modes and Common Mistakes

Writing "the ideal generated by $∂_1, … ,∂_n$ " without saying which kind

The left ideal $∑ _i A_n∂_i$ is proper and maximal. The two-sided ideal generated by the same elements is all of $A_n$ , since it contains $∂_1 x_1 - x_1 ∂_1 = 1$ . So the phrase " $A_n$ modulo the ideal generated by the partials" is either the polynomial module or the zero module depending on a word that is often left out. Always write the left ideal explicitly.

Treating $K[X]$ as a subring of $A_n$ when a module is meant

As a subring, $K[X] ⊂ A_n$ is closed under multiplication and contains no information about differentiation. As a module, $K[X]$ is a quotient of $A_n$ on which the partials act non-trivially. Statements such as " $K[X]$ is a free $A_n$ -module of rank one" arise from conflating them and are false: the generator $1$ has non-zero annihilator, and $K[X]$ is not even projective.

Assuming the annihilator of a polynomial is generated by high-order derivatives

For $f$ of degree $d$ the operators $∂^β$ with $|β| > d$ certainly annihilate $f$ , but the left ideal $I$ they generate is not the annihilator. $I$ kills every polynomial of degree at most $d$ at once, so it sits inside the two distinct maximal left ideals $ann(1)$ and $ann(f)$ - distinct because $∂_1$ lies in the first and not the second when $f$ has a linear term. A left ideal contained in two different maximal left ideals is not itself maximal. A genuine annihilator always contains operators mixing $x$ 's and $∂$ 's, as Steps 2 and 4 of the worked example show.

Applying an operator without first putting it in canonical form

$∂_1 x_1$ and $x_1∂_1$ are different operators: on $f = 1$ the first gives $1$ and the second gives $0$ . Formula (5.1) is valid only once all variables are to the left of all partials. Reordering must be done with $[∂_i, x_j] = δ_ij$ , never by treating the symbols as commuting.

Historical Notes

The action of multiplication and differentiation on polynomials is far older than D-module theory; it is what Heisenberg's commutation relation $[∂, x] = 1$ describes, and Hermann Weyl's 1928 analysis of the canonical commutation relations is why the algebra carries his name. The observation that no finite-dimensional space can carry such an action - the trace argument - is essentially contemporary with the relation itself, and it is the first indication that the polynomial module, or something at least as large, is unavoidable.

The algebraic reading of the polynomial ring as a cyclic module with a specified annihilator belongs to the ring-theoretic study of $A_n$ that began in earnest with Dixmier's 1968 paper on $A_1$ . Classification results for $A_1$ -modules were obtained by Block around 1980; the general problem of classifying simple $A_n$ -modules remains open, which is one reason the explicit families built from $K[X]$ by twisting retain their interest. In the D-module tradition proper, the polynomial module is the structure sheaf of affine space read algebraically, and Bernstein's 1971-72 work made it the base case from which the localisations $K[X][1/f]$ and the modules $K[X][1/f]f^s$ are built.

Comparison

The standard first modules over $A_n$ (over a field of characteristic zero; the last row is over $A_1(C)$ ).
Module	Presentation	Cyclic	Simple	Torsion	$d(M)$
$K[X]$	$A_n/ ∑ _i A_n∂_i$	yes	yes	yes	$n$
$K[∂]$	$A_n/ ∑ _i A_n x_i$	yes	yes	yes	$n$
$K[X]_σ$ , $σ$ an automorphism	$A_n/σ^-1( ∑ _i A_n∂_i)$	yes	yes	yes	$n$
$A_n$ itself	$A_n/0$	yes	no	no	$2n$
$K[X][1/f]$ , $f ≠ 0$ non-constant	cyclic, holonomic	yes	no	yes	$n$
$H(U)$ , holomorphic functions	none finite	no	no	no	not defined

Key Takeaways

Key points

$K[X]$ is a left $A_n$ -module by construction: $A_n$ was defined as a subring of $End_K(K[X])$ , with $x_i$ multiplying and $∂_i$ differentiating.
It is cyclic on the generator $1$ , whose annihilator is the left ideal $∑ _i=1^n A_n∂_i$ , giving $K[X] ≅ A_n/ ∑ _i A_n∂_i$ .
The proof is one application of the canonical basis theorem: split an operator into its $∂$ -free part plus a member of the ideal.
Every polynomial is a torsion element, yet the module is faithful - torsion is a condition on elements, not on the module.
In characteristic zero the module is simple, and the ideal $∑ _i A_n∂_i$ is therefore a maximal left ideal; in characteristic $p$ simplicity fails, though the presentation survives.
Dividing by the $x_i$ instead gives $K[∂]$ , the Fourier transform of $K[X]$ , on which the partials multiply and the variables differentiate with a sign.
$K[X]$ is holonomic with $d = n$ and $e = 1$ , the smallest invariants a non-zero module over $A_n$ can have.

FAQs

Do I need to check that the action is well defined?

No, provided the Weyl algebra is defined as a subring of $End_K(K[X])$ , which is the definition used in this collection. The action is then the inclusion itself. If instead you define $A_n$ by generators and relations, you must check that multiplication and differentiation satisfy those relations - which is the content of $[∂_i, x_j] = δ_ij$ as operators, verified by the product rule.

How can

K[X]

be a torsion module and faithful at the same time?

Torsion means each element separately has a non-zero annihilator; faithfulness means the intersection of all those annihilators is zero. For $f$ of degree $d$ , $∂_1^d+1$ kills $f$ but kills nothing of higher degree, so the annihilators shrink to nothing as $f$ ranges over the module. Over a commutative domain this behaviour cannot occur for a finitely generated module, which is where the intuition misleads.

K[X]

finitely generated as an

A_n

-module?

Yes, by a single element - it is cyclic. That is compatible with being infinite dimensional over $K$ , because $A_n$ is itself infinite dimensional over $K$ .

Why is

∑ _i A_n∂_i

a maximal left ideal?

Because the quotient by it is $K[X]$ , which is a simple module in characteristic zero. Left ideals containing $J$ correspond to submodules of $A_n/J$ , and a simple module has only the two trivial submodules.

What replaces

K[X]

when I want more solutions than the polynomial ones?

Formal power series, convergent power series, rational functions $K[X][1/f]$ , or - over $ℂ$ in one variable - the holomorphic functions $H(U)$ . Each is an $A_n$ -module by the same formulas, and the solution space of a system grows as the module does. The polynomial module is the smallest of them.

K[X]

isomorphic to

K[∂]

as an

A_n

-module?

No. Their annihilator ideals are $∑ _i A_n∂_i$ and $∑ _i A_n x_i$ , and these are different maximal left ideals; an isomorphism of cyclic modules would force the annihilators of corresponding generators to coincide. They are, however, related by an automorphism of $A_n$ : $K[∂]$ is the Fourier twist of $K[X]$ .

Does the presentation change in characteristic

p

No. The proof uses only that $x^α∂^β$ is a $K$ -basis of $A_n$ , which holds in any characteristic. What fails in characteristic $p$ is simplicity, and hence the maximality of the ideal: $x_1^pK[X]$ is a proper non-zero submodule.

What is the analogue for

A_n/ ∑ _i A_n(∂_i - g_i)

As a $K$ -vector space it is again $K[X]$ , by the same splitting argument, but the derivative acts by $f ↦ ∂ f/∂ x_i + g_i f$ . When the $g_i$ satisfy the integrability condition $∂ g_i/∂ x_j = ∂ g_j/∂ x_i$ these modules are exactly the twists of $K[X]$ by automorphisms fixing the $x_i$ , and deciding when two of them are isomorphic produces infinitely many pairwise non-isomorphic simple modules.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 5 §1, Lemma (5.1.1) and Proposition (5.1.2).
S. C. Coutinho, A Primer of Algebraic D-modules, Cambridge University Press, 1995 - Ch. 1, for the construction of $A_n$ inside $End_K(K[X])$ and the canonical basis.
P. M. Cohn, Algebra, Volume 1, second edition, Wiley, 1982 - Ch. 10, for the module-theoretic background assumed here (cyclic modules, annihilators, the isomorphism theorems).
J. Dixmier, Sur les algèbres de Weyl, Bulletin de la Société Mathématique de France 96 (1968), 209-242 - the systematic study of $A_1$ and its modules.
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1, for the polynomial module in the filtered setting.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, revised edition, Graduate Studies in Mathematics 30, American Mathematical Society, 2001 - Ch. 1 and Ch. 8, for annihilators and the ring-theoretic context.
R. Hotta, K. Takeuchi and T. Tanisaki, D-modules, Perverse Sheaves, and Representation Theory, Progress in Mathematics 236, Birkhäuser, 2008 - Ch. 1, for the polynomial module as the structure sheaf of affine space.
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization - notation for $∂$ , $ℕ$ and operator names.

AI Suggested Questions

Show directly that $x^α ∂^β : β ≠ 0$ is a $K$ -basis of the left ideal $∑ _i A_n∂_i$ .
Compute a generating set for $ann_A_1(x^2)$ and check that the quotient is isomorphic to $K[x]$ .
Verify the identity $x_i ⋅ ∂^β = -β_i∂^β-e_i$ in $A_n/ ∑ _j A_n x_j$ for $n = 1$ and $β = 3$ .
Prove that $K[X]$ is not a projective $A_n$ -module for general $n$ , extending the argument given for $n = 1$ .
Show that in characteristic $p$ the submodules $x_1^kpK[X]$ form a strictly decreasing chain, so $K[X]$ is not even of finite length.
Compute $Hom_A_1(A_1/A_1(x∂ - 2), K[x])$ and interpret the answer as a space of polynomial solutions.
Give a good filtration on $K[X]$ different from $Γ_m = B_m ⋅ 1$ and check that the dimension and multiplicity are unchanged.

The Polynomial Ring as a Module over the Weyl Algebra

Overview

Definition

Core Concepts

Why a quotient and not a subobject

Differentiation lowers degree, and that is the whole story

Torsion versus faithful

The size of the module

Construction and Proof

Key Equations

Variable Definitions

Properties and Behaviour

Why K[X] is not projective over An

Examples and Special Cases

Worked Example

Applications and Industry Use

Design Considerations

Which side the module acts on

Which presentation to carry

Polynomials or symbols

Computational Notes

Failure Modes and Common Mistakes

Historical Notes

Comparison

Key Takeaways

Key points

FAQs

Related Engineering Topics

References

AI Suggested Questions

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