Bernstein's Inequality

Q: Why is the lower bound n rather than 0?

Because A_n is simple and large. Simplicity forces the action on a non-zero module to be faithful, and _K B_m grows like m^2n . Fitting a space of that size into maps from Γ_m to Γ_2m requires _K Γ_m to grow at least like m^n .

Q: Is the bound sharp for every n?

Yes. K[x_1, … ,x_n] has dimension exactly n , and A_n itself has dimension exactly 2n . Every intermediate value also occurs, so no part of the interval [n,2n] is excluded.

Q: What goes wrong in characteristic p?

The p -th powers x_i^p and ∂_i^p become central, so A_n is a finite module over a polynomial ring in 2n variables and is no longer simple. It then has modules of dimension 0 - for instance quotients by maximal ideals of the centre - and the inequality fails outright.

Q: Does the inequality hold for right modules too?

Yes. The transposition anti-automorphism of A_n converts left modules into right modules while preserving Bernstein degree, so the dimension theory and the inequality transfer verbatim.

Q: Is there an analogue for rings of differential operators on other varieties?

For a smooth affine variety of dimension n in characteristic zero, yes, and the standard proof is geometric, via involutivity. For singular varieties the ring of differential operators can fail to be Noetherian or finitely generated, and no such statement should be assumed without checking the specific ring.

Overview

The dimension $d(M)$ of a finitely generated module over the Weyl algebra $A_n$ is defined as the degree of its Hilbert polynomial with respect to a good filtration. Nothing in that definition obviously restricts the value $d(M)$ can take: a priori it could be anything from $0$ to $2n$ , the dimension of $A_n$ itself.

Bernstein's inequality says that the bottom half of that range is empty. For every non-zero finitely generated $A_n$ -module, $d(M) ≥ n$ . Combined with the elementary upper bound $d(M) ≤ 2n$ , the dimension of a module over $A_n$ is confined to the interval $[n, 2n]$ , and both endpoints are attained.

The consequence that makes the inequality famous is what happens at the lower end. Modules of the minimal dimension $n$ turn out to be extremely well behaved: they have finite length, they are cyclic, and they are closed under most of the operations of the theory. They are called holonomic, and they carry almost all of the applications in this collection, from the Bernstein-Sato polynomial to automatic proof of combinatorial identities. Without the inequality, "minimal dimension" would not be a meaningful class.

The proof is short and entirely elementary. It rests on one observation: an operator of Bernstein degree at most $m$ that kills the $m$ -th piece of a good filtration must be a scalar, and a scalar that kills a non-zero space is zero. Dimension counting then does the rest. The argument is carried out in full on the proof page.

Definition

Fix a field $K$ of characteristic zero and write $A_n = A_n(K)$ for the $n$ -th Weyl algebra. The statement uses the Bernstein filtration $B_m_m ≥ 0$ of $A_n$ , in which $B_m$ is the $K$ -span of the monomials $x^α ∂^β$ with $|α| + |β| ≤ m$ .

Dimension and multiplicity

Let $M$ be a finitely generated $A_n$ -module with a good filtration $Γ = Γ_m$ . For $m ≫ 0$ the function $m ↦ _K Γ_m$ agrees with a polynomial $χ_Γ(m)$ , the Hilbert polynomial of the filtration. The dimension $d(M)$ is the degree of $χ_Γ$ , and the multiplicity $e(M)$ is $d(M)!$ times its leading coefficient. Neither depends on the choice of good filtration.

Bernstein's inequalityCoutinho (9.4.2); Bernstein 1971

Let $K$ have characteristic zero and let $M$ be a non-zero finitely generated module over $A_n(K)$ . Then

n ≤ d(M) ≤ 2n.

(9.1)

In particular $M$ cannot have dimension smaller than $n$ , however few generators it has.

Note

The upper bound is the easy half. It holds because $M$ is a quotient of a free module $A_n^r$ , and dimension does not increase along surjections, while $d(A_n) = 2n$ . The name "Bernstein's inequality" refers to the lower bound $d(M) ≥ n$ .

Core Concepts

Two competing pressures decide the dimension of a module, and the inequality is the statement that one of them always wins.

Pressure downwards: relations shrink the filtration

Every relation imposed on a module removes elements from the filtration pieces, lowering $_K Γ_m$ and so lowering the degree of the Hilbert polynomial. Impose enough relations and one might expect the growth to become very slow indeed - constant, say, giving $d(M) = 0$ .

Pressure upwards: the algebra itself is large

But the module has to accommodate the whole of $A_n$ acting on it, and $A_n$ is large: $_K B_m$ grows like $m^2n$ . It is also simple, so no non-zero operator can act as zero on a non-zero module. The action therefore embeds a space of dimension roughly $m^2n$ into a space of maps between two filtration pieces, whose dimension is roughly $χ(m)^2$ . Comparing the two growth rates forces $χ$ to grow at least like $m^n$ .

That is the entire content of the proof: the Weyl algebra is too big to act faithfully on anything that grows too slowly. Simplicity is what forbids the action from being unfaithful, and simplicity is exactly where characteristic zero enters.

Why the bound is $n$ and not something else

The exponent $2n$ in $_K B_m$ is split evenly between the source and target of the map $Γ_m → Γ_2m$ , and each contributes a factor growing like $m^d(M)$ . The comparison reads $m^2n ≤ sssim m^2 d(M)$ , which gives $d(M) ≥ n$ with no room to spare. The evenness of that split is the reason the answer is exactly half of $2n$ .

Construction and Proof

The full argument is given on the proof page. What follows is the skeleton, so that the shape of the reasoning is visible here.

Faithfulness at every filtration levelCoutinho (9.4.1)

Let $M ≠ 0$ be finitely generated over $A_n$ with a good filtration $Γ$ such that $Γ_0 ≠ 0$ . If $D ∈ B_m$ satisfies $D,Γ_m = 0$ , then $D = 0$ .

Proof of the lemma (sketch)

Induct on $m$ . For $m = 0$ , $B_0 = K$ and $Γ_0 ≠ 0$ , so a scalar killing $Γ_0$ is zero.

For $m > 0$ , suppose $D ∈ B_m$ kills $Γ_m$ . The commutators $[D, x_i]$ and $[D, ∂_i]$ lie in $B_m-1$ , because commuting with a generator drops Bernstein degree by one. For $u ∈ Γ_m-1$ we have $x_i u ∈ Γ_m$ and $∂_i u ∈ Γ_m$ , so

[D, x_i]u = D(x_i u) - x_i (Du) = 0 - 0 = 0,

and identically for $[D, ∂_i]$ . By the inductive hypothesis $[D, x_i] = [D, ∂_i] = 0$ for every $i$ , so $D$ commutes with all the generators and therefore lies in the centre of $A_n$ , which is $K$ . A scalar killing $Γ_m ⊃ eq Γ_0 ≠ 0$ is zero.

Proof of the inequality from the lemma

The lemma says the map $θ_m$ of (9.4) is injective, so $_K B_m ≤ _K Γ_m ⋅ _K Γ_2m$ , which is (9.5). For $m ≫ 0$ the right-hand side equals $χ(m)χ(2m)$ , a polynomial in $m$ of degree $2 d(M)$ ; the left-hand side is a polynomial of degree $2n$ . A polynomial of degree $2n$ bounded above by one of degree $2d(M)$ for all large $m$ forces $2n ≤ 2 d(M)$ .

For the upper bound, choose generators $u_1, … ,u_r$ of $M$ and filter by $Γ_m = B_m u_1 + … + B_m u_r$ . Then $_K Γ_m ≤ r _K B_m$ , a polynomial of degree $2n$ , so $d(M) ≤ 2n$ .

Remark

The condition $Γ_0 ≠ 0$ costs nothing: replacing $Γ$ by the shifted filtration $Γ_m+s$ for a suitable $s$ is again a good filtration with the same Hilbert polynomial up to a shift of argument, hence the same degree and leading coefficient.

Key Equations

The dimension of the $m$ -th piece of the Bernstein filtration is the number of monomials $x^α ∂^β$ of total degree at most $m$ in $2n$ variables:

_K B_m = 2n+m2n = m^2n(2n)! + O(m^2n-1).

(9.2)

For a good filtration $Γ$ on $M$ the Hilbert polynomial has the shape

χ_Γ(m) = (e(M))/(d(M)!) m^d(M) + (lower order terms),

(9.3)

and the injectivity lemma that drives the proof states that

θ_m : B_m ⟶ Hom_K(Γ_m, Γ_2m), θ_m(D)(u) = Du,

(9.4)

is injective for every $m ≥ 0$ , whenever $M ≠ 0$ and $Γ$ is a good filtration with $Γ_0 ≠ 0$ .

Comparing dimensions across (9.4) gives the inequality that is then read asymptotically:

2n+m2n ≤ _K Γ_m ⋅ _K Γ_2m = χ(m),χ(2m) (m ≫ 0).

(9.5)

The left side has degree $2n$ in $m$ ; the right side has degree $2d(M)$ . Hence $2n ≤ 2 d(M)$ .

Variable Definitions

$K$: the ground field, of characteristic zero
$A_n$: the $n$ -th Weyl algebra over $K$ , generated by $x_1, … ,x_n,∂_1, … ,∂_n$
$B_m$: the $m$ -th piece of the Bernstein filtration of $A_n$
$M$: a non-zero finitely generated left $A_n$ -module
$Γ_m$: the $m$ -th piece of a good filtration of $M$
$χ_Γ(m)$: the Hilbert polynomial: the polynomial agreeing with $_K Γ_m$ for large $m$
$d(M)$: the dimension of $M$ , that is $χ_Γ$
$e(M)$: the multiplicity of $M$ , that is $d(M)!$ times the leading coefficient of $χ_Γ$
$θ_m$: the evaluation map $B_m → Hom_K(Γ_m,Γ_2m)$ sending an operator to its action

Properties and Behaviour

Holonomic modules exist and are the extreme case

A finitely generated $A_n$ -module is called holonomic if it is zero or has $d(M) = n$ . By the inequality this is the smallest possible dimension, and $K[x_1, … ,x_n]$ shows the class is non-empty.

Holonomic modules have finite lengthCoutinho (10.2)

If $0 → M' → M → M'' → 0$ is exact with $M$ holonomic, then $M^{'}$ and $M^{''}$ are holonomic and $e(M) = e(M') + e(M'')$ . Since multiplicities are positive integers, a strictly descending chain of submodules has strictly decreasing multiplicity and so terminates. The length of $M$ is at most $e(M)$ .

No dimension gap above $n$

Every integer $d$ with $n ≤ d ≤ 2n$ occurs as the dimension of some finitely generated $A_n$ -module. Taking external products of $A_1$ -modules realises the intermediate values: if $M$ has dimension $d_1$ over $A_n_1$ and $N$ has dimension $d_2$ over $A_n_2$ , their external product has dimension $d_1 + d_2$ over $A_n_1+n_2$ .

Dimension and multiplicity of some standard $A_n$ -modules.
Module	$d(M)$	$e(M)$	Holonomic?
$A_n$	$2n$	$1$	no (for $n ≥ 1$ )
$K[x_1, … ,x_n]$	$n$	$1$	yes
$K[x][1/f]$ , $f ≠ 0$	$n$	finite	yes
$A_1/A_1(x∂ - λ)$	$1$	$2$	yes
$A_n^,r$	$2n$	$r$	no (for $n ≥ 1$ )
$0$	conventionally $- ∈ fty$	$0$	yes, by convention

Worked Example

Both endpoints are attained, for $n = 1$

Step 1 - the algebra itself, $M = A_1$
Filter $A_1$ by its own Bernstein filtration, $Γ_m = B_m$ , which is a good filtration of $A_1$ as a module over itself. A basis of $B_m$ is $x^a ∂^b : a + b ≤ m$ , so

$_K B_m = m+22 = ((m+1)(m+2))/(2) = (m^2)/(2) + (3m)/(2) + 1.$

Check the small values directly: $m = 0$ gives the single element $1$ ; $m = 1$ gives $1, x, ∂$ , so $3$ ; $m = 2$ gives $1, x, ∂, x^2, x∂, ∂^2$ , so $6$ . The formula returns $1, 3, 6$ as it should.
Step 2 - read off the invariants
The Hilbert polynomial is $χ(m) = 12m^2 + 32m + 1$ , of degree $2$ , so $d(A_1) = 2 = 2n$ . Its leading coefficient is $\frac{1}{2}$ , so $e(A_1) = 2! ⋅ 12 = 1$ . The upper endpoint of (9.1) is attained.
Step 3 - the polynomial module, $M = K[x]$
Realise $K[x]$ as $A_1 / A_1∂$ , generated by the image $\bar{1}$ of $1$ , on which $∂$ acts as $d/dx$ . Take the good filtration induced by the generator, $Γ_m = B_m ⋅ 1$ . Since $∂ ⋅ 1 = 0$ , only the monomials $x^a$ with $a ≤ m$ survive, so $Γ_m$ is the space of polynomials of degree at most $m$ and

$_K Γ_m = m + 1.$
Step 4 - read off the invariants again
Here $χ(m) = m+1$ has degree $1$ , so $d(K[x]) = 1 = n$ , and the leading coefficient is $1$ , so $e(K[x]) = 1! ⋅ 1 = 1$ . The lower endpoint of (9.1) is attained, and $K[x]$ is holonomic.
Step 5 - confirm the counting bound is consistent
For $M = K[x]$ , inequality (9.5) reads $m+22 ≤ (m+1)(2m+1)$ . At $m = 10$ this is $66 ≤ 11 ⋅ 21 = 231$ , comfortably true; both sides are quadratic in $m$ , which is precisely the borderline case $d(M) = n$ . Had $d(M)$ been $0$ , the right-hand side would have been bounded while the left grows quadratically - the contradiction the proof exploits.

Result

For $n = 1$ : $d(A_1) = 2$ with $e(A_1) = 1$ , and $d(K[x]) = 1$ with $e(K[x]) = 1$ . Both ends of $1 ≤ d(M) ≤ 2$ occur, so neither bound in Bernstein's inequality can be improved.

Applications and Industry Use

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

Bernstein's inequality is not applied directly so much as used to license the definition of holonomicity, which then does the work.

Analytic continuation of $f^s$ . Bernstein's original motivation was Gelfand's problem on meromorphic continuation of the distribution $f^s$ . The proof shows $K[x][1/f]f^s$ is holonomic over $A_n[s]$ , which produces the b-function and with it the continuation.
Automatic proof of identities. Zeilberger's algorithm and creative telescoping rest on the fact that holonomic functions form a class closed under sum, product and definite summation or integration - a closure that is proved by tracking dimensions and using finite length. See Zeilberger's method.
Representation theory. Modules arising from Lie-theoretic constructions, such as those attached to highest weight representations, are holonomic, and the finite-length conclusion is what makes character-theoretic bookkeeping possible.
Singularity theory. The roots of the b-function of a singularity are invariants of it, refining the multiplicity and connecting to the monodromy of the Milnor fibration.
Computer algebra. Dimension and multiplicity are the standard complexity measures reported by D-module packages; holonomicity is the practical decidability boundary for algorithms on differential systems.

Design Considerations

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

Two choices have to be made before the inequality can even be stated, and both affect how the theory is used afterwards.

Which filtration

The Bernstein filtration is the right choice for the elementary proof, because each $B_m$ is finite dimensional over $K$ and the counting is honest. The order filtration is the right choice for geometry, because its associated graded ring is $K[x_1, … ,x_n,ξ_1, … ,ξ_n]$ with $ξ_i$ the symbols of $∂_i$ , and the characteristic variety lives in that cotangent space. Most treatments prove the inequality with the Bernstein filtration and then transfer to the order filtration once dimension has been shown to be filtration-independent.

Which invariant to carry

Dimension alone is coarse: it takes only $n+1$ values. Multiplicity refines it and is additive in exact sequences, which is what converts "minimal dimension" into "finite length". If a construction needs to be tracked through exact sequences, carry the pair $(d, e)$ rather than $d$ ; if it needs to be tracked through geometric operations, carry the characteristic variety.

Standards and Codes

For mathematics, the relevant standards are notation, numeric and markup standards together with reference implementations.

ISO 80000-2 fixes the symbols used here: $ℕ$ , $ℚ$ , $ℂ$ for the number systems, $∂$ for partial differentiation, and upright roman type for operator names such as $dim$ and $deg$ .
ISO/IEC 40314 (MathML 3.0) is the markup standard in which every expression on this page is encoded, so that the mathematics is machine-readable and accessible to screen readers rather than an image.
There is no numbering standard for the inequality itself. The convention followed here is Coutinho's: results are cited as chapter-and-item pairs, so Bernstein's inequality is $(9.4.2)$ in that text.
Software conventions differ on one point worth pinning down: some systems report holonomic rank (the dimension of the solution space, that is the multiplicity with respect to the order filtration after localisation) rather than $e(M)$ . The two agree for many standard examples but are not the same invariant.

Material Selection

The material of a mathematical construction is its numeric substrate: the ground field, the coefficient ring and the representation used to store it.

The ground field is the load-bearing choice. Characteristic zero is required, and the usual working fields are $ℚ$ , a number field, $ℝ$ or $ℂ$ .

$ℚ$ is the field of choice for computation: exact, and Gröbner bases behave predictably. Coefficient growth during the computation, not the field, is the practical bottleneck.
$ℂ$ is the field of choice for statements connecting to analysis - solutions, monodromy, and the analytic continuation applications.
Fields of characteristic $p$ must not be used: the inequality is false there.
Neither dimension nor multiplicity changes under extension of the ground field, so it is legitimate to compute over $ℚ$ and state results over $ℂ$ .

For storage, a module is normally held as a presentation matrix over $A_n$ with entries in the canonical form $∑ c_αβ x^α ∂^β$ , which makes the Bernstein degree readable directly off the exponent vectors.

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.

Deciding holonomicity of an explicitly presented module is a Gröbner basis computation, and it is expensive.

Present the module as $A_n^r / N$ for an explicitly generated submodule $N$ .
Compute a Gröbner basis of $N$ with respect to a term order refining the Bernstein (or order) filtration, in the non-commutative setting of the Weyl algebra.
Take leading terms to obtain the associated graded module over a commutative polynomial ring in $2n$ variables.
Compute the Hilbert polynomial of that commutative graded module; its degree is $d(M)$ and $d(M)!$ times its leading coefficient is $e(M)$ .
Compare $d(M)$ with $n$ .

The cost is dominated by step 2. Gröbner basis computation in the Weyl algebra is doubly exponential in the number of variables in the worst case, and $2n$ variables appear. In practice, systems with $n$ up to about $4$ or $5$ and modest degrees are routine; larger ones need structure to be exploited. Bernstein's inequality provides a cheap sanity check on any implementation: a reported dimension below $n$ for a non-zero module is a bug, not a discovery.

Implementations include the Dmodules package in Macaulay2, dmod.lib and bfun.lib in Singular, the ore_algebra package in SageMath, and HolonomicFunctions.m in Mathematica. All of them report dimension and, where meaningful, holonomic rank.

Limits of Validity

Three hypotheses carry real weight, and the statement fails without any one of them.

Characteristic zero. The proof uses that the centre of $A_n$ is $K$ and that $A_n$ is simple. In characteristic $p$ neither holds: $x_i^p$ and $∂_i^p$ are central, $A_n$ is a finitely generated module over its centre, and modules of dimension $0$ exist. See the positive characteristic page.
Finite generation. The bound is a statement about Hilbert polynomials, which only exist for modules carrying a good filtration. An infinitely generated module such as the full field of rational functions in one variable has no good filtration and no dimension in this sense.
Non-zero. The zero module is excluded by fiat; the usual convention sets $d(0) = - ∈ fty$ so that the additivity statements in exact sequences remain uniform.

What generalises

For the ring $D(X)$ of differential operators on a smooth affine variety $X$ of dimension $n$ over a field of characteristic zero, the analogous inequality $d(M) ≥ n$ holds, and the modern proof runs through the involutivity of the characteristic variety rather than through the counting argument above. On singular varieties $D(X)$ can behave badly enough that neither the counting proof nor the geometric one applies unchanged, and the naive statement should not be assumed.

Failure Modes and Common Mistakes

Using the order filtration instead of the Bernstein filtration

The counting proof above uses the Bernstein filtration $B_m$ , in which $x$ and $∂$ both have degree $1$ , so that $_K B_m$ grows like $m^2n$ . With the order filtration, where $x$ has degree $0$ and $∂$ degree $1$ , each piece is infinite dimensional over $K$ and the Hilbert function must instead be taken over $K[x]$ . The dimension defined by the two filtrations does agree, but that is a theorem, not a definition - do not substitute one filtration for the other inside this proof.

Reading $d(M) ≥ n$ as "the characteristic variety has dimension $≥ n$ by dimension count"

The geometric statement is true, and it is the deeper reason for the inequality: the characteristic variety is involutive (Gabber's theorem), and an involutive subvariety of a $2n$ -dimensional symplectic space has dimension at least $n$ . But that is a much harder theorem than Bernstein's inequality. Quoting involutivity to prove the inequality is quoting a strictly stronger result; the elementary proof is preferred precisely because it avoids it.

Assuming a module with few generators has small dimension

$A_n$ is cyclic - it is generated by $1$ - and has the maximal dimension $2n$ . Conversely, every holonomic module is cyclic. The number of generators says nothing about the dimension; only the size of the annihilator does.

Forgetting that the Hilbert function is only eventually polynomial

$_K Γ_m$ agrees with $χ_Γ(m)$ for $m ≫ 0$ , not for all $m$ . Any argument comparing growth rates must be read asymptotically. Checking (9.5) at a single small value of $m$ proves nothing.

Historical Notes

I. N. Bernstein proved the inequality in 1971, in the course of answering a question I. M. Gelfand had raised at the 1954 International Congress of Mathematicians: does the distribution defined by $f^s$ for $ℜ(s) > 0$ continue meromorphically to the whole complex plane? Affirmative answers had been given in 1968 by Atiyah and, independently, by Bernstein and Gelfand, but both used Hironaka's resolution of singularities.

Bernstein's 1972 proof replaced resolution of singularities with the Weyl algebra and the elementary counting argument reproduced above. The change of method is the point: a deep geometric input was traded for a filtration and a dimension count. The class of modules of minimal dimension that the inequality isolates was later christened holonomic, and the theory grew around it.

The geometric explanation arrived afterwards. Sato, Kashiwara and Kawai developed the analytic side, and in 1981 Gabber proved that the characteristic variety of a coherent D-module is involutive, from which the inequality follows for smooth varieties in general. The elementary proof survives because it is short, self-contained, and gives the multiplicity bounds needed for the finite-length statement.

Comparison

The two routes to the inequality.
	Counting proof (Bernstein)	Geometric proof (via involutivity)
Filtration used	Bernstein filtration $B_m$	Order filtration $F_m$
Key input	Simplicity of $A_n$	Gabber's involutivity theorem
Length	About a page	Substantial
Generality	$A_n$ over a field of characteristic zero	Coherent modules over $D_X$ , $X$ smooth
Yields multiplicity bounds	Yes	Not directly
Explains why $n$	By dimension count	By symplectic geometry

Key Takeaways

Key points

For every non-zero finitely generated $A_n$ -module over a field of characteristic zero, $n ≤ d(M) ≤ 2n$ .
The lower bound is Bernstein's inequality; the upper bound follows because $M$ is a quotient of a free module and $d(A_n) = 2n$ .
The proof rests on one lemma: an operator in $B_m$ killing $Γ_m$ commutes with every generator, so lies in the centre $K$ , so is zero.
Both endpoints are attained: $d(A_n) = 2n$ and $d(K[x_1, … ,x_n]) = n$ .
Modules attaining the lower bound are the holonomic modules; the inequality is what makes that class well defined.
Characteristic zero is essential - the statement is false in characteristic $p$ , where $A_n$ is not simple.

FAQs

Why is the lower bound

n

rather than

0

Because $A_n$ is simple and large. Simplicity forces the action on a non-zero module to be faithful, and $_K B_m$ grows like $m^2n$ . Fitting a space of that size into maps from $Γ_m$ to $Γ_2m$ requires $_K Γ_m$ to grow at least like $m^n$ .

Does the inequality say anything about multiplicity?

Not directly - it constrains only the degree of the Hilbert polynomial. But multiplicity is what turns the dimension bound into the finite-length theorem for holonomic modules, because $e$ is a positive integer and is additive in short exact sequences.

Is the bound sharp for every

n

Yes. $K[x_1, … ,x_n]$ has dimension exactly $n$ , and $A_n$ itself has dimension exactly $2n$ . Every intermediate value also occurs, so no part of the interval $[n,2n]$ is excluded.

What goes wrong in characteristic

p

The $p$ -th powers $x_i^p$ and $∂_i^p$ become central, so $A_n$ is a finite module over a polynomial ring in $2n$ variables and is no longer simple. It then has modules of dimension $0$ - for instance quotients by maximal ideals of the centre - and the inequality fails outright.

Can I use the order filtration to prove it instead?

Not with this counting argument, because the pieces of the order filtration are infinite dimensional over $K$ . You can run a relative version over $K[x]$ , or transfer the result after proving that dimension is independent of the good filtration. The geometric route via involutivity of the characteristic variety also works but uses a much deeper theorem.

Why does the proof need

Γ_0 ≠ 0

The base case of the induction needs a non-zero space for a scalar to fail to annihilate. Shifting a good filtration by a constant is again good and changes the Hilbert polynomial only by a shift of argument, so the condition can always be arranged without affecting $d(M)$ or $e(M)$ .

Does the inequality hold for right modules too?

Yes. The transposition anti-automorphism of $A_n$ converts left modules into right modules while preserving Bernstein degree, so the dimension theory and the inequality transfer verbatim.

Is there an analogue for rings of differential operators on other varieties?

For a smooth affine variety of dimension $n$ in characteristic zero, yes, and the standard proof is geometric, via involutivity. For singular varieties the ring of differential operators can fail to be Noetherian or finitely generated, and no such statement should be assumed without checking the specific ring.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 9 §4, result (9.4.2), with the faithfulness lemma (9.4.1).
I. N. Bernstein, Modules over a ring of differential operators. Study of the fundamental solutions of equations with constant coefficients, Functional Analysis and its Applications 5 (1971), 89-101.
I. N. Bernstein, The analytic continuation of generalized functions with respect to a parameter, Functional Analysis and its Applications 6 (1972), 273-285.
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1, for the filtration-theoretic treatment.
O. Gabber, The integrability of the characteristic variety, American Journal of Mathematics 103 (1981), 445-468.
R. Hotta, K. Takeuchi and T. Tanisaki, D-modules, Perverse Sheaves, and Representation Theory, Progress in Mathematics 236, Birkhäuser, 2008 - Ch. 2, for the geometric proof.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Graduate Studies in Mathematics 30, revised edition, American Mathematical Society, 2001 - Ch. 8, for Gelfand-Kirillov dimension in this setting.
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization.
ISO/IEC 40314:2016, Information technology - Mathematical Markup Language (MathML) Version 3.0, International Organization for Standardization.

AI Suggested Questions

Work through the faithfulness lemma for $n = 1$ and $m = 2$ explicitly, showing which commutators are computed.
Compute $d(M)$ and $e(M)$ for $M = A_1/A_1(x∂ - λ)$ and check the inequality.
Show that the external product of two holonomic modules is holonomic, and that dimensions add.
Explain why $_K B_m = 2n+m2n$ by counting monomials directly.
Give an example in characteristic $p$ of a non-zero finitely generated $A_n$ -module of dimension $0$ .
Compare the multiplicity of $K[x][1/f]$ with the degree of $f$ for a few small polynomials.
State precisely what changes in the proof if the good filtration is replaced by an arbitrary exhaustive filtration.

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Fit a polynomial to the dimensions of the filtration pieces and read off degree and leading coefficient.
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Bernstein Filtration Dimension
Compute $2n+m2n$ for given $n$ and $m$ .
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