Why the Polynomial Ring Is a Simple Weyl Module

Q: Why must the monomial have maximal total degree?

Because ∂^α kills x^γ unless γ ≥ α componentwise, and any such γ other than α itself has strictly larger total degree. Maximality of |α| rules those out, leaving a single surviving term. If |α| is not maximal, higher-degree monomials survive and the result is a polynomial rather than a constant.

Q: Does the argument need K to be algebraically closed, or infinite?

No. Only α! ≠ 0 in K is used, which holds for every field of characteristic zero, including Q . Nothing about roots, closure or cardinality enters.

Q: Does simplicity fail for every prime characteristic, or only for small p?

For every prime. Given p , the polynomial x_1^p generates the proper submodule x_1^pK[X] , because ∂_1(x_1^ph) = x_1^p∂_1 h . Large p postpones the failure to higher degrees but does not prevent it.

Q: Is K[X] the only simple A_n-module?

Far from it. K[∂] is another, the twists K[X]_σ give infinitely many pairwise non-isomorphic ones, the delta module A_1/A_1x is another, and for n ≥ 2 there exist simple modules that are not even holonomic. A complete classification is known only for n = 1 , and even there it is delicate.

Q: Can I deform the action by arbitrary polynomials g_i?

No. You need ∂ g_i/∂ x_j = ∂ g_j/∂ x_i , equivalently that (g_1, … ,g_n) is the gradient of a single polynomial. Without it the left ideal ∑ _i A_n(∂_i - g_i) contains a non-zero polynomial and the quotient degenerates, possibly to zero. Taking g_i ∈ K[x_i] is the standard sufficient condition.

Q: Does simplicity of K[X] prove that A_n is a simple ring?

Not by itself. Having a faithful simple module makes A_n a primitive ring, which is weaker than simple. Simplicity of A_n needs its own commutator argument. The implication that is genuinely used runs the other way: because A_n is a simple ring, every non-zero A_n -module is faithful.

Overview

The polynomial module $K[X] = K[x_1, … ,x_n]$ carries the action of $A_n$ in which the variables multiply and the partials differentiate. This page settles the question that makes it interesting: it has no submodules apart from $0$ and itself, provided $char K = 0$ .

Equivalently, every non-zero polynomial generates the whole module. That is not obvious. A submodule must be closed under multiplication by every variable, which pushes degree up, and under every partial derivative, which pushes degree down. The claim is that starting from any single non-zero $f$ and pushing down far enough, one always arrives at a non-zero constant — after which multiplication recovers everything.

The mechanism is a one-line calculation. Pick a monomial $x^α$ of maximal total degree occurring in $f$ , with coefficient $a ≠ 0$ . Then $∂^α ⋅ f = α!,a$ , because every other monomial of $f$ is annihilated by $∂^α$ . In characteristic zero $α!,a ≠ 0$ and the argument is finished. In characteristic $p$ the factorial can vanish, and the statement is false: $x_1^pK[X]$ is a proper non-zero submodule.

Three consequences carry forward. The left ideal $∑ _i A_n∂_i$ is a maximal left ideal of $A_n$ . The endomorphism ring of $K[X]$ is exactly $K$ , so no non-scalar symmetry of the module exists. And simplicity survives twisting by an automorphism, which is how an infinite family of pairwise non-isomorphic simple P4-modules is produced from this single example.

Definition

Simple (irreducible) module

A left $R$ -module $M$ is simple, or irreducible, if $M ≠ 0$ and the only submodules of $M$ are $0$ and $M$ . Some texts abbreviate this to " $M$ has no proper submodules"; that phrasing is only correct if "proper" is read as "different from $0$ and from $M$ ", and the condition $M ≠ 0$ must be imposed separately, since the zero module vacuously has no such submodules but is never called simple.

Simplicity of the polynomial moduleCoutinho (5.1.2)

Let $K$ be a field of characteristic zero. Then $K[X]$ is a simple left $A_n(K)$ -module. It is also a torsion module, and $K[X] ≅ A_n/ ∑ _i=1^n A_n∂_i$ .

What "simple" is not

Simplicity of the module $K[X]$ and simplicity of the ring $A_n$ are different statements about different objects. Simplicity of P2 says the only two-sided ideals are $0$ and $A_n$ ; it is proved by a commutator argument and is not what is being proved here. The two interact only in one direction: because $A_n$ is a simple ring, every non-zero $A_n$ -module is faithful.

Core Concepts

Why maximal degree is the right choice

Suppose $x^α$ occurs in $f$ and we apply $∂^α$ . Any other monomial $x^γ$ of $f$ contributes only if $γ ≥ α$ componentwise, and in that case $|γ| ≥ |α|$ with equality precisely when $γ = α$ . So if $|α|$ is the largest total degree occurring in $f$ , no other monomial can contribute at all, and the result is the single constant $α! a$ . Any monomial of maximal total degree will do; when several are tied, any one of them works and different choices give different constants.

Where characteristic zero enters, and only there

The combinatorics above is characteristic-free. The one place a field hypothesis is used is the assertion $α! ≠ 0$ in $K$ . When $char K = p$ , the factorial $α!$ vanishes as soon as some $α_i ≥ p$ , and the whole argument collapses at exactly that point — not gradually, but for a specific reason that can be pinpointed.

Simple and torsion at the same time

A simple module over a ring that is not a division ring is automatically a torsion module: if some non-zero element had zero annihilator, the module would be isomorphic to the ring as a left module, forcing the ring to have no left ideals besides $0$ and itself. $A_n$ is not a division ring — the only units are the non-zero scalars, because $A_n$ is a filtered domain whose degree function is additive — so $K[X]$ being simple already implies it is torsion, with no extra work.

Simplicity is stronger than holonomicity

It is easy to conflate "as small as possible" in the sense of dimension with "as small as possible" in the sense of submodules. They are different. $K[X][1/f]$ is holonomic, of the same dimension $n$ as $K[X]$ , yet it contains $K[X]$ properly and so is not simple. Conversely there exist simple modules that are not holonomic. Neither property implies the other.

Construction and Proof

Simple modules over a non-division ringCoutinho (5.1.1)

Let $R$ be a ring with identity and $M$ a simple left $R$ -module.

For every non-zero $u ∈ M$ , $M ≅ R/ann_R(u)$ ; in particular $M$ is cyclic and $ann_R(u)$ is a maximal left ideal.
If $R$ is not a division ring, then $M$ is a torsion module.

Proof

For (1), let $φ : R \to M$ be $φ(r) = ru$ . It is $R$ -linear, and its image is a submodule containing $u ≠ 0$ , hence equal to $M$ by simplicity. Its kernel is $ann_R(u)$ , so $M ≅ R/ann_R(u)$ ; the quotient being simple is precisely maximality of the left ideal.

For (2), suppose $ann_R(u) = 0$ for some $u ≠ 0$ . By (1), $M ≅ R$ as left $R$ -modules, so the lattice of left ideals of $R$ matches the lattice of submodules of $M$ , which is trivial. A ring with identity whose only left ideals are $0$ and $R$ is a division ring. Contradiction; hence every non-zero element has non-zero annihilator.

$K[X]$ is simpleCoutinho (5.1.2)

Let $char K = 0$ and let $N ⊂ eq K[X]$ be a non-zero $A_n$ -submodule. Then $N = K[X]$ .

Proof

Choose $0 ≠ f ∈ N$ and write $f = ∑ _γ c_γ x^γ$ . Let $d = f$ and choose $α$ with $|α| = d$ and $a := c_α ≠ 0$ .

Apply $∂^α$ . By (5.7) a monomial $x^γ$ survives only if $γ ≥ α$ componentwise; but then $|γ| ≥ |α| = d$ , and since $d$ is the top degree of $f$ we must have $|γ| = d$ , which combined with $γ ≥ α$ forces $γ = α$ . So exactly one term survives and $∂^α ⋅ f = α!,a$ .

Since $char K = 0$ , $α! ≠ 0$ in $K$ , and $a ≠ 0$ , so $α! a$ is a non-zero constant. Submodules are closed under the action, so this constant lies in $N$ , and dividing by it gives $1 ∈ N$ . Finally $N ⊃ eq A_n ⋅ 1 = K[X]$ , so $N = K[X]$ . As $K[X] ≠ 0$ , it is simple.

Torsion now follows from the Lemma, since $A_n$ is not a division ring: it is a domain whose only units are the non-zero scalars, so $x_1$ has no inverse. Alternatively, torsion is visible directly, since $∂_1^,d+1$ annihilates any $f$ of degree $d$ .

The endomorphism ring

$End_A_n(K[X]) = K$ . Indeed an $A_n$ -linear $ψ$ is determined by $ψ(1)$ , and applying $ψ$ to $∂_i ⋅ 1 = 0$ gives $∂_i ⋅ ψ(1) = 0$ for every $i$ , so $ψ(1)$ is a constant $λ$ and $ψ$ is multiplication by $λ$ . This is a sharpening of Schur's lemma in this case: the endomorphism ring is not merely a division ring but the ground field itself.

Key Equations

The whole proof rests on the action of $∂^α$ on a monomial:

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

(5.7)

Write $f = ∑ _γ c_γ x^γ$ with $c_α = a ≠ 0$ and $|γ| ≤ |α|$ for every $γ$ with $c_γ ≠ 0$ . Then (5.7) collapses the sum to a single term:

∂^α ⋅ f ;=; ∑ _γ ≥ α c_γ (γ!)/((γ-α)!),x^γ-α ;=; α!,a .

(5.8)

The generating operator is therefore explicit: to send $f$ to a prescribed target $g ∈ K[X]$ , use

D_f → g ;=; (1)/(α!,a); g, ∂^α ∈ A_n, D_f → g ⋅ f = g .

(5.9)

Here $g$ denotes the multiplication operator by $g$ ; the formula makes the statement "every non-zero $f$ generates" completely constructive.

Simplicity is equivalent to a statement about left ideals, and the resulting statement is the one usually quoted:

∑ _i=1^n A_n ∂_i is a maximal left ideal of A_n, End_A_n(K[X]) = K .

(5.10)

Variable Definitions

$K$: the ground field, of characteristic zero unless stated otherwise
$K[X]$: the polynomial ring $K[x_1, … ,x_n]$ , regarded as a left $A_n$ -module
$N$: a non-zero submodule of $K[X]$ , the object shown to be all of $K[X]$
$f$: a non-zero element of $N$ , written $f = ∑ _γ c_γ x^γ$
$α$: the multi-index of a monomial of maximal total degree occurring in $f$
$a$: the coefficient $c_α$ of that monomial, non-zero by choice
$α!$: the product $α_1! ⋅ sα_n!$ , a non-zero element of $K$ exactly when $char K = 0$ or $char K > _i α_i$
$g_i$: polynomials used to deform the derivative action, giving $A_n/ ∑ _i A_n(∂_i - g_i)$
$End_A_n(M)$: the ring of $A_n$ -linear maps from $M$ to itself

Properties and Behaviour

Maximality of the ideal of the partials

$J = ∑ _i=1^n A_n∂_i$ is a maximal left ideal of $A_n$ , since $A_n/J ≅ K[X]$ is simple. More generally $ann_A_n(f)$ is a maximal left ideal for every non-zero $f ∈ K[X]$ , and these ideals are usually distinct for different $f$ even though all the quotients are isomorphic.

The deformed modules are simple tooCoutinho, Ch. 5 §1; Exercise 5.4.2

Let $g_1, … ,g_n ∈ K[X]$ satisfy the integrability condition $∂ g_i/∂ x_j = ∂ g_j/∂ x_i$ for all $i, j$ — for instance, $g_i ∈ K[x_i]$ for each $i$ . Then $M_g := A_n/ ∑ _i A_n(∂_i - g_i)$ is a simple $A_n$ -module, isomorphic as a $K$ -vector space to $K[X]$ , with

x_j ⋅ f = x_j f, ∂_i ⋅ f = (∂ f)/(∂ x_i) + g_i f .

Proof, and why it is short

Integrability makes $x_i ↦ x_i$ , $∂_i ↦ ∂_i + g_i$ a well-defined endomorphism of $A_n$ — the only relation to check beyond $[∂_i + g_i, x_j] = δ_ij$ is $[∂_i + g_i, ∂_j + g_j] = ∂ g_j/∂ x_i - ∂ g_i/∂ x_j = 0$ — and it is an automorphism because $∂_i = (∂_i + g_i) - g_i$ lies in the image and $A_n$ is simple, so the kernel vanishes. Hence $M_g$ is the twist $K[X]_σ$ .

Simplicity is then immediate without invoking the twisting machinery. A subspace of $K[X]$ is an $M_g$ -submodule exactly when it is closed under multiplication by every $x_j$ and under $f ↦ ∂ f/∂ x_i + g_i f$ . Since multiplication by the polynomial $g_i$ is already available, the second condition is equivalent to closure under $∂/∂ x_i$ . So $M_g$ and $K[X]$ have literally the same submodules, and $K[X]$ is simple.

Simple modules are the building blocks, but not the whole story

It is not true that every $A_n$ -module is a direct sum of simple modules: $A_n$ is not a semisimple ring. The correct statement is that the well-behaved modules — the holonomic ones — have finite composition series, so they are built from simple modules by iterated extensions rather than by direct sums. The module $K[X][1/f]$ , which contains $K[X]$ as a submodule without a complement, is the standard illustration.

Examples and Special Cases

$n = 1$ , and the fastest route to a constant

For $f ∈ K[x]$ of degree $d$ with leading coefficient $a$ , $∂^d ⋅ f = d!,a$ . Nothing subtler is needed: in one variable the multi-index of maximal degree is unique. So the submodule generated by any non-zero $f$ contains $d! a$ , hence $1$ , hence everything.

The companion module $K[∂]$

$K[∂] ≅ A_n/ ∑ _i A_n x_i$ is also simple. One can repeat the argument with the roles of $x$ and $∂$ exchanged, but it is cheaper to note that $K[∂]$ is the Fourier twist of $K[X]$ and that twisting preserves simplicity. This is the standard example of getting a second theorem for free from an automorphism.

Characteristic $p$ : $x_1^pK[X]$ is a submodule

Let $char K = p > 0$ . Then $∂_1(x_1^p h) = p x_1^p-1h + x_1^p ∂_1 h = x_1^p ∂_1 h$ , and the other partials pass through $x_1^p$ untouched, so $x_1^pK[X]$ is closed under all the generators. It is a proper non-zero submodule and $K[X]$ is not simple. Iterating, $K[X] ⊃ x_1^pK[X] ⊃ x_1^2pK[X] ⊃ ⋅ s$ is an infinite strictly descending chain, so the module does not even have finite length. This mirrors the failure of simplicity of P6 in characteristic $p$ .

A larger function module that is not simple

Over $ℂ$ in one variable, the module of holomorphic functions $H(U)$ on an open $U ⊂ eq C$ contains $C[z]$ as a proper non-zero submodule, so it is not simple. It is not a torsion module either: $( (z))$ satisfies no polynomial differential equation. Enlarging the space of functions destroys both properties at once.

Simplicity is not inherited by localisations

$K[x][1/x]$ is an $A_1$ -module containing $K[x]$ properly, so it is not simple. Its composition factors are $K[x]$ and the delta module $K[x][1/x]/K[x] ≅ A_1/A_1x$ , and both of those are simple. This is the shortest example of a non-trivial composition series over $A_1$ .

Worked Example

Generating from an arbitrary polynomial, and watching it fail in characteristic 2

Step 1 - one variable, an explicit generator
Work over $ℚ$ with $n = 1$ and $f = x^3 - 2x + 5$ . The degree is $3$ and the leading coefficient is $1$ , so $α = 3$ and $a = 1$ . Differentiating three times: $∂ f = 3x^2 - 2$ , $∂^2 f = 6x$ , $∂^3 f = 6$ . That matches $α!,a = 3! ⋅ 1 = 6$ .
Step 2 - hitting a prescribed target
By (5.9) the operator that carries $f$ to a chosen $g$ is $D = 16,g,∂^3$ . Take $g = x^2 - 1$ :

$D ⋅ f = 16(x^2-1),∂^3 ⋅ (x^3 - 2x + 5) = 16(x^2-1) ⋅ 6 = x^2 - 1 = g .$

So $A_1 f ∋ g$ for every $g$ , and $A_1 f = Q[x]$ . Note that the division by $6$ is the only step that could fail over another field.
Step 3 - two variables, and why the degree must be maximal
Take $n = 2$ and $f = x_1^2 + x_1x_2$ , of degree $2$ with two monomials tied at the top. Both admissible choices work:

$∂_1^2 ⋅ f = 2 + 0 = 2 (α = (2,0)), ∂_1∂_2 ⋅ f = 0 + 1 = 1 (α = (1,1)),$

and the constants obtained, $2 = 2! ⋅ 0! ⋅ 1$ and $1 = 1! ⋅ 1! ⋅ 1$ , are exactly the values $α!,c_α$ predicted by (5.8). A non-maximal index is useless: $α = (1,0)$ gives $∂_1 ⋅ f = 2x_1 + x_2$ , not a constant, because the surviving monomials are those of degree above $|α|$ .
Step 4 - the same polynomial over a field of characteristic 2
Now let $K = F_2$ , $n = 1$ , and $f = x^2$ . Here $α = 2$ and $α! = 2 = 0$ in $K$ , so (5.8) returns $∂^2 ⋅ f = 0$ : the constant that the proof needs simply is not there. Directly, $∂(x^2) = 2x = 0$ , so the submodule generated by $f$ is

$A_1 ⋅ x^2 = F_2[x],x^2 = h ∈ F_2[x] : h ∈ x^2F_2[x] ,$

which is closed under $∂$ because $∂(x^2 h) = x^2 ∂ h$ . It is proper — it does not contain $1$ or $x$ — and non-zero, so $F_2[x]$ is not a simple $A_1(F_2)$ -module.

Result

Over $ℚ$ , $16(x^2-1)∂^3$ sends $x^3-2x+5$ to $x^2-1$ , and the same recipe reaches every target, so $x^3-2x+5$ generates $Q[x]$ . Over $F_2$ the polynomial $x^2$ generates only the proper submodule $x^2F_2[x]$ . The single point of difference is whether $α!$ is invertible in the ground field.

Applications and Industry Use

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

Every non-zero map out of $K[X]$ is injective. If $ψ : K[X] → M$ is $A_n$ -linear and non-zero, its kernel is a proper submodule, hence $0$ . This is used constantly when comparing the polynomial module with larger function modules: the inclusion $K[X] ↪ H(U)$ is forced to be the only interesting map up to scalars.
Building an infinite supply of simple modules. Since simplicity is preserved under twisting by an automorphism and $A_n$ has a large automorphism group, the single theorem here generates families such as $K[X]_σ_r$ , which are pairwise non-isomorphic.
Composition series of concrete modules. Identifying $K[X]$ and the delta module as simple is what lets one write down the composition factors of $K[x][1/x]$ and, more generally, of localisations along a hypersurface.
Solution spaces. $End_A_n(K[X]) = K$ says that the system $∂_1 f = … = ∂_n f = 0$ has a one-dimensional solution space in $K[X]$ , that is, holonomic rank $1$ . Any computation reporting otherwise is wrong.
Representation theory of $A_n$ . Classifying simple $A_n$ -modules is a hard open problem even for $n = 1$ beyond the known families; $K[X]$ and its twists are the base of every known list.

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.

Simplicity turns several potentially expensive questions into cheap ones.

Is a given non-zero $f$ a generator? Yes, always, in characteristic zero. The certificate is the operator $1α!,a∂^α$ from (5.9), read off from the support of $f$ in constant time once the top-degree monomials are known.
Is a proposed submodule proper? If it is non-zero it is everything, so the only test needed is whether it contains a non-zero element. No Gröbner basis is required.
Is $ann_A_n(f)$ maximal? Yes, automatically. Computing generators for it is a genuine syzygy computation in the Weyl algebra, but knowing that the quotient is simple often removes the need to compute at all.

The one place where implementations must be careful is the ground field. Systems that work over a finite field for efficiency, or that reduce modulo a prime to control coefficient growth, silently leave the setting in which this theorem holds: the reduction of a Weyl algebra computation modulo $p$ is not governed by the same module theory. Packages such as Dmodules in Macaulay2 and dmod.lib in Singular therefore work over $ℚ$ or a number field for D-module computations.

Failure Modes and Common Mistakes

Deforming the derivative action without the integrability condition

It is often stated that for any $g_1, … ,g_n ∈ K[X]$ the module $A_n/ ∑ _i A_n(∂_i - g_i)$ is again a copy of $K[X]$ with a modified derivative action, and is simple. That is false without a hypothesis. The left ideal contains the commutator $[∂_i - g_i, ∂_j - g_j] = ∂ g_i/∂ x_j - ∂ g_j/∂ x_i$ , which is a polynomial. If that polynomial is non-zero, the ideal meets $K[X]$ non-trivially and the quotient collapses.

Concretely, take $n = 2$ , $g_1 = x_2$ , $g_2 = 0$ . Then $[∂_1 - x_2, ∂_2] = -[x_2,∂_2] = 1$ , so the ideal contains $1$ , equals $A_2$ , and the quotient is the zero module — not a simple module at all. The correct hypothesis is $∂ g_i/∂ x_j = ∂ g_j/∂ x_i$ , satisfied automatically when each $g_i$ lies in $K[x_i]$ .

Choosing a monomial of maximal degree in one variable rather than in total degree

For $f = x_1^2 + x_1x_2^3$ the monomial with the largest exponent of $x_1$ is $x_1^2$ , but $∂_1^2 ⋅ f = 2$ only because the second monomial happens to be killed. Change the example to $f = x_1^2x_2 + x_1^2$ : taking $α = (2,0)$ gives $∂_1^2 ⋅ f = 2x_2 + 2$ , which is not a constant. Total degree, not degree in a single variable, is what makes the argument work.

Confusing simplicity of the module with simplicity of the ring

The two theorems have the same adjective and different content. Simplicity of $A_n$ concerns two-sided ideals and is proved by showing that the commutator with $x_i$ or $∂_i$ lowers degree. Simplicity of $K[X]$ concerns submodules and is proved by the differentiation argument here. Neither proof is a special case of the other, and citing one for the other is a common slip in write-ups.

Assuming that simple implies holonomic, or the reverse

$K[X][1/f]$ is holonomic and not simple. There are also simple P1-modules of dimension greater than P2 for $n ≥ 2$ ; Stafford constructed such examples. Simplicity is a lattice-theoretic condition and holonomicity a growth condition, and they constrain each other only weakly.

Reading "no proper submodules" literally

Under the usual reading of "proper" as "not equal to the whole module", every module has the proper submodule $0$ , so no module at all would be simple. The intended reading is "no submodules other than $0$ and $M$ ", together with $M ≠ 0$ . The source's phrasing is the standard abbreviation; the zero module is excluded by convention, not by the words.

Historical Notes

That the polynomial representation of the canonical commutation relations admits no invariant subspace is old, and in the analytic setting it is the algebraic shadow of the Stone-von Neumann theorem, which says the irreducible unitary representation of the Heisenberg relations is unique up to equivalence. The purely algebraic statement, with the differentiation argument given above, belongs to the ring theory of $A_n$ developed from the 1960s onwards, and appears in Dixmier's 1968 study of $A_1$ .

The wider question — classify all simple $A_n$ -modules — turned out to be much harder than this example suggests. Block obtained a classification of the simple $A_1$ -modules around 1980 in terms of irreducible elements of a localisation of $A_1$ , and Stafford showed in the 1980s that $A_n$ for $n ≥ 2$ has simple modules that are not holonomic, so the simple objects are not confined to the well-behaved part of the theory. The construction on the twisted-modules page is the elementary end of this story.

Comparison

Which of the standard modules are simple, and why not when they are not.
Module	Simple?	Reason
$K[X]$ , $char K = 0$	yes	$∂^α$ extracts the non-zero constant $α! a$
$K[X]$ , $char K = p$	no	$x_1^pK[X]$ is a submodule; $α!$ can vanish
$K[∂] ≅ A_n/ ∑ _i A_nx_i$	yes	Fourier twist of $K[X]$ ; twisting preserves simplicity
$K[X]_σ$ , $σ ∈ Aut(A_n)$	yes	twisting preserves the submodule lattice
$K[x][1/x]$	no	contains $K[x]$ ; composition factors $K[x]$ and $A_1/A_1x$
$A_n$ as a module over itself	no	$A_n∂_1$ is a proper non-zero left ideal
$H(U)$ over $A_1(C)$	no	contains $C[z]$ ; also not a torsion module

Key Takeaways

Key points

Over a field of characteristic zero, $K[X]$ is a simple left $A_n$ -module: it has no submodules but $0$ and itself.
Proof: for $0 ≠ f$ , differentiate by a multi-index $α$ of a monomial of maximal total degree; all other terms die and $∂^α f = α!,c_α ≠ 0$ .
The construction is effective: $1α!,c_α,g,∂^α$ sends $f$ to any prescribed target $g$ .
Consequences: $∑ _i A_n∂_i$ is a maximal left ideal, $K[X]$ is a torsion module, and $End_A_n(K[X]) = K$ .
Characteristic $p$ is genuinely excluded: $x_1^pK[X]$ is a proper non-zero submodule and $K[X]$ has infinite length.
The deformed modules $A_n/ ∑ _i A_n(∂_i - g_i)$ are simple too, but only when the $g_i$ satisfy $∂ g_i/∂ x_j = ∂ g_j/∂ x_i$ ; otherwise the quotient can be zero.
Simplicity and holonomicity are independent conditions: neither implies the other.

FAQs

Why must the monomial have maximal total degree?

Because $∂^α$ kills $x^γ$ unless $γ ≥ α$ componentwise, and any such $γ$ other than $α$ itself has strictly larger total degree. Maximality of $|α|$ rules those out, leaving a single surviving term. If $|α|$ is not maximal, higher-degree monomials survive and the result is a polynomial rather than a constant.

Does the argument need

K

to be algebraically closed, or infinite?

No. Only $α! ≠ 0$ in $K$ is used, which holds for every field of characteristic zero, including $ℚ$ . Nothing about roots, closure or cardinality enters.

Does simplicity fail for every prime characteristic, or only for small

p

For every prime. Given $p$ , the polynomial $x_1^p$ generates the proper submodule $x_1^pK[X]$ , because $∂_1(x_1^ph) = x_1^p∂_1 h$ . Large $p$ postpones the failure to higher degrees but does not prevent it.

K[X]

the only simple

A_n

-module?

Far from it. $K[∂]$ is another, the twists $K[X]_σ$ give infinitely many pairwise non-isomorphic ones, the delta module $A_1/A_1x$ is another, and for $n ≥ 2$ there exist simple modules that are not even holonomic. A complete classification is known only for $n = 1$ , and even there it is delicate.

How does simplicity relate to the module being torsion?

One implies the other here. A simple module over a ring that is not a division ring is a torsion module, by the Lemma; $A_n$ is a domain whose units are just the non-zero scalars, so it is not a division ring. The converse fails: $K[x][1/x]$ is torsion but not simple.

Why is

End_A_n(K[X])

equal to

K

and not just some division ring?

Schur's lemma alone gives a division ring. Here one can do better because the module is cyclic with a known annihilator: an endomorphism is determined by the image of $1$ , and that image must be killed by every $∂_i$ , hence is a constant. So the endomorphism ring is exactly the scalars.

Can I deform the action by arbitrary polynomials

g_i

No. You need $∂ g_i/∂ x_j = ∂ g_j/∂ x_i$ , equivalently that $(g_1, … ,g_n)$ is the gradient of a single polynomial. Without it the left ideal $∑ _i A_n(∂_i - g_i)$ contains a non-zero polynomial and the quotient degenerates, possibly to zero. Taking $g_i ∈ K[x_i]$ is the standard sufficient condition.

Does simplicity of

K[X]

prove that

A_n

is a simple ring?

Not by itself. Having a faithful simple module makes $A_n$ a primitive ring, which is weaker than simple. Simplicity of $A_n$ needs its own commutator argument. The implication that is genuinely used runs the other way: because $A_n$ is a simple ring, every non-zero $A_n$ -module is faithful.

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); Exercise 5.4.2 for the deformed modules.
J. Dixmier, Sur les algèbres de Weyl, Bulletin de la Société Mathématique de France 96 (1968), 209-242 - the first systematic study of $A_1$ and its modules.
R. E. Block, The irreducible representations of the Lie algebra $sl(2)$ and of the Weyl algebra, Advances in Mathematics 39 (1981), 69-110 - classification of the simple $A_1$ -modules.
J. T. Stafford, Non-holonomic modules over Weyl algebras and enveloping algebras, Inventiones Mathematicae 79 (1985), 619-638 - simple modules that are not holonomic.
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. 6, for simple modules, primitive rings and the units of $A_n$ .
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1, for the module theory of $A_n$ in the filtered setting.
P. M. Cohn, Algebra, Volume 1, second edition, Wiley, 1982 - Ch. 10, for simple modules, Schur's lemma and the isomorphism theorems.
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization - notation for partial derivatives and set operations.

AI Suggested Questions

Write down the operator sending $x_1^3x_2 - x_2^2$ to $1$ and verify the computation term by term.
Show that $ann_A_1(x)$ contains $x∂ - 1$ and $∂^2$ , and explain why it must be a maximal left ideal.
Prove that the only units of $A_n$ are the non-zero scalars, using the additivity of the degree function.
Exhibit the full submodule lattice of $F_p[x]$ over $A_1(F_p)$ for $p = 3$ as far as degree $9$ .
Verify directly that $[∂_1 - x_2, ∂_2] = 1$ and conclude that $A_2/(A_2(∂_1 - x_2) + A_2∂_2)$ is the zero module.
Compute the composition factors of $K[x][1/x]$ over $A_1$ and identify each with a module named on this page.
Show that $K[∂]$ is simple by repeating the differentiation argument with the roles of $x$ and $∂$ exchanged.