Holonomic Modules Have Finite Length

Q: Does finite length follow from finite generation alone?

No. A_n is generated by one element and has infinite descending chains of submodules. Finite generation gives the ascending chain condition over a noetherian ring; the descending one is extra, and holonomicity is what provides it.

Q: Is the inequality ℓ(M) ≤ e(M) ever an equality?

Yes, whenever every composition factor has multiplicity 1 . K[x][1/x] is the standard example: length 2 , multiplicity 2 , factors K[x] and A_1/A_1x . Equality fails as soon as some factor has multiplicity 2 or more.

Overview

Every finitely generated module over $A_n$ is noetherian, because P1 is a noetherian ring. Very few of them are artinian: $A_n$ itself is not, and neither is any non-zero left ideal. What this page proves is that holonomic modules are artinian, and that the descending chain condition comes with an explicit numerical bound.

The mechanism is short. If $M$ is holonomic and $N ⊂ eq M$ , then $N$ and $M/N$ are holonomic too, so all three have dimension $n$ and multiplicity is additive: $e(M) = e(N) + e(M/N)$ . Multiplicities of non-zero modules are positive integers. So each proper step down a chain of submodules consumes at least one unit of multiplicity, and a chain cannot be longer than $e(M)$ steps.

Being both noetherian and artinian is exactly the condition for a module to possess a composition series, and the Jordan-Hölder theorem then makes the number of factors an invariant, the length $ℓ(M)$ . The argument above gives the bound

ℓ(M) ≤ e(M),

(10.6)

and the immediate corollary is that a holonomic module of multiplicity $1$ is simple. The bound is not an equality in general: the worked example computes a case where $ℓ = e = 2$ , and then one where $ℓ = 1$ while $e = 2$ .

Finite length is the property everything downstream is built on. It gives cyclicity, it makes the category of holonomic modules behave like a category of finite-dimensional representations, and it is the reason the descending chain used to produce the Bernstein-Sato polynomial terminates.

Definition

Artinian module

Let $R$ be a ring and $M$ a left $R$ -module. $M$ is artinian if every descending chain of submodules

N_1 ⊃ eq N_2 ⊃ eq N_3 ⊃ eq ⋅ s

is eventually stationary: there is $k$ with $N_j = N_k$ for all $j ≥ k$ . This is the mirror image of the noetherian condition, which asks the same of ascending chains.

Composition series and length

A composition series of $M$ is a chain of submodules

0 = N_0 ⊂ neq N_1 ⊂ neq ⋅ s ⊂ neq N_r = M

in which every quotient $N_i/N_i-1$ is simple. The integer $r$ is the length $ℓ(M)$ . The simple quotients are the composition factors.

Basic properties of artinian modulesCoutinho (10.2.1)

Let $M$ be a left $R$ -module and $N ⊂ eq M$ a submodule.

$M$ is artinian if and only if every non-empty set of submodules of $M$ has a minimal element with respect to inclusion.
$M$ is artinian if and only if both $N$ and $M/N$ are artinian.
If $M = N + N'$ with $N$ and $N^{'}$ artinian submodules, then $M$ is artinian.

The proofs are the exact duals of the corresponding statements for noetherian modules, obtained by reversing every inclusion; Coutinho leaves them as exercises and so does this page. Statement (1) needs a weak form of choice, exactly as its noetherian counterpart does.

Core Concepts

Multiplicity as a budget

The right way to think about the proof is that $e(M)$ is a budget of size $e(M)$ , denominated in positive integers. Every time you pass from a submodule to a strictly smaller submodule inside a holonomic module, the quotient is a non-zero holonomic module and therefore costs at least $1$ . The budget is finite, so the descent must stop. Nothing about the Weyl algebra is used beyond the two facts that dimensions cannot drop below $n$ and that multiplicity is additive when dimensions agree.

Why noetherian is not enough

Noetherianity is cheap: every finitely generated module over a noetherian ring has it. It bounds nothing about how far down one can go. $A_n$ is a noetherian module over itself, and the chain

A_n x_n ⊃ neq A_n x_n^2 ⊃ neq A_n x_n^3 ⊃ neq ⋅ s

descends forever. Each inclusion is strict: if $x_n^k = P x_n^k+1$ for some $P ∈ A_n$ , then $(1 - P x_n) x_n^k = 0$ , and since P3 is a domain this forces $P x_n = 1$ , impossible because the only units of $A_n$ are the non-zero scalars. So artinianness is genuinely extra information, and holonomicity is what supplies it.

What finite length buys

A module of finite length is determined, up to composition factors and multiplicities of occurrence, by a finite list of simple modules. That converts questions about a holonomic module into questions about finitely many simple ones. It also makes induction on length a legitimate proof technique, and the proof that holonomic modules are cyclic is exactly such an induction.

Length against multiplicity

The two invariants are both additive on short exact sequences of holonomic modules, and $ℓ ≤ e$ because every simple holonomic module has $e ≥ 1$ . They agree precisely when every composition factor has multiplicity $1$ . Multiplicity is the finer invariant, but it depends on the choice of filtration convention, whereas length is purely module-theoretic.

Construction and Proof

Holonomic modules are artinianCoutinho (10.2.2)

Let $M$ be a holonomic left $A_n$ -module. Then every strictly descending chain of submodules of $M$ has at most $e(M)$ steps. In particular $M$ is artinian.

Proof

Let

M = N_0 ⊃ neq N_1 ⊃ neq ⋅ s ⊃ neq N_r

be a strictly descending chain. By closure under submodules and quotients, each $N_i$ is holonomic and each factor $N_i/N_i+1$ is holonomic and non-zero. All of these therefore have dimension exactly $n$ , so additivity of multiplicity applies at every step:

e(N_i) = e(N_i+1) + e(N_i/N_i+1), i = 0,1, … ,r-1.

Summing over $i$ and telescoping,

e(M) = ∑ _i=0^r-1 e(N_i/N_i+1) + e(N_r) ≥ ∑ _i=0^r-1 1 = r,

because each $N_i/N_i+1$ is a non-zero holonomic module and so has multiplicity at least $1$ , while $e(N_r) ≥ 0$ . Hence $r ≤ e(M)$ : no strictly descending chain in $M$ is longer than $e(M)$ , and in particular there is no infinite one.

Existence of a composition series

A module that is both artinian and noetherian has a composition series.

Proof

Build the chain upwards. Suppose $N_0 = 0 ⊂ neq N_1 ⊂ neq ⋅ s ⊂ neq N_k$ has been constructed with every $N_i/N_i-1$ simple, and $N_k ≠ M$ . The set of non-zero submodules of $M/N_k$ is non-empty, so by part (1) of the artinian criterion it has a minimal element, which is therefore simple; pull it back to a submodule $N_k+1$ with $N_k+1/N_k$ simple. The process cannot continue indefinitely, because the $N_k$ form a strictly ascending chain and $M$ is noetherian. It therefore halts, and it can only halt at $N_r = M$ .

Holonomic modules have finite length

A holonomic $A_n$ -module is noetherian (it is finitely generated over a noetherian ring) and artinian (by the theorem above), hence has a composition series. By the Jordan-Hölder theorem any two composition series of a module have the same length and the same multiset of composition factors up to isomorphism, so $ℓ(M)$ is well defined.

Length is bounded by multiplicityCoutinho (10.2.3)

For a holonomic $A_n$ -module $M$ , $ℓ(M) ≤ e(M)$ .

A composition series is in particular a strictly descending chain of length $ℓ(M)$ read from the top down, so the count in the proof of the theorem applies verbatim. Coutinho labels this a scholium rather than a theorem, because no new argument is needed: the bound falls out of the chain estimate already made.

Multiplicity one implies simpleCoutinho (10.2.4)

A holonomic $A_n$ -module with $e(M) = 1$ is simple.

Suppose $0 ≠ N ⊂ eq M$ . Both $N$ and $M/N$ are holonomic, so $1 = e(M) = e(N) + e(M/N)$ with $e(N) ≥ 1$ . Hence $e(M/N) = 0$ , which for a holonomic module means $M/N = 0$ , that is $N = M$ .

Key Equations

The three numerical facts the argument uses, in the order it uses them:

d(M) = , d(N), d(M/N) ,,

(10.7)

so a submodule or quotient of a holonomic module cannot escape dimension $n$ .

d(N) = d(M/N) ⟹ e(M) = e(N) + e(M/N),

(10.8)

additivity of multiplicity, valid whenever the two dimensions agree — which for holonomic modules is automatic.

M ≠ 0 holonomic ⟹ e(M) ∈ 1, 2, 3, … ,

(10.9)

positivity and integrality of multiplicity: it is $n!$ times the leading coefficient of a Hilbert polynomial that takes integer values.

Putting them together gives, for any strictly descending chain of length $r$ inside a holonomic $M$ ,

r ≤ ∑ _i=0^r-1 e(N_i/N_i+1) ≤ e(M), hence ℓ(M) ≤ e(M).

(10.10)

Variable Definitions

$A_n$: the $n$ -th Weyl algebra over a field $K$ of characteristic zero
$M$: a holonomic left $A_n$ -module
$N_i$: the terms of a chain of submodules of $M$
$d(M)$: the dimension of $M$ , equal to $n$ when $M$ is holonomic and non-zero
$e(M)$: the multiplicity of $M$ ; a positive integer for non-zero holonomic $M$
$ℓ(M)$: the length of $M$ , the number of factors in any composition series
$Γ_i$: the $i$ -th piece of a good filtration, used when a multiplicity has to be computed

Properties and Behaviour

Length is additive

If $0 → M' → M → M'' → 0$ is an exact sequence of holonomic $A_n$ -modules, then $ℓ(M) = ℓ(M') + ℓ(M'')$ , exactly as $e(M) = e(M') + e(M'')$ . Concatenating a composition series of $M^{'}$ with the pullback of one of $M^{''}$ produces a composition series of $M$ .

The holonomic category is artinian and noetherian

Holonomic $A_n$ -modules form a full abelian subcategory of the category of $A_n$ -modules that is closed under subobjects, quotients and extensions, and every object has finite length. Consequently every holonomic module is a finite iterated extension of simple holonomic modules, and $Hom$ between holonomic modules is computed factor by factor. See the category page.

Simple holonomic modules are the atoms

Every non-zero holonomic module has at least one simple holonomic submodule, namely a minimal element of the set of non-zero submodules. Note the converse trap: for $n ≥ 2$ not every simple $A_n$ -module is holonomic, so "simple holonomic" is a strictly smaller class than "simple". See non-holonomic irreducible modules.

The Weyl algebra is not left artinian

$A_n$ regarded as a left module over itself is noetherian but not artinian, as the chain $A_n x_n ⊃ neq A_n x_n^2 ⊃ neq ⋅ s$ shows. More generally no non-zero left ideal of $A_n$ is artinian, and no $A_n$ -module having a free submodule of positive rank is artinian. This negative fact is not a curiosity; it is the hypothesis that drives the proof that holonomic modules are cyclic.

Examples and Special Cases

Multiplicity one, length one

$K[x_1, … ,x_n]$ is holonomic with $e = 1$ , hence simple, hence of length $1$ . Its simplicity can also be proved by hand, by applying $∂^α$ to a non-zero polynomial in a putative submodule until a non-zero constant appears; see the direct proof. The multiplicity argument gives it in one line.

The Dirac delta module

$A_1/A_1 x$ has Hilbert polynomial $i+22 - i+12 = i+1$ , so $d = 1$ and $e = 1$ : it is simple. It is the algebraic incarnation of the delta distribution, the class of $1$ playing the role of $δ$ and the relation $xδ = 0$ being the defining property.

Length two, multiplicity two

$K[x][1/x]$ over $A_1$ has $e = 2$ and length $2$ , with composition factors $K[x]$ and $A_1/A_1x$ . The worked example below carries out the computation.

Length one, multiplicity two

For $λ ∈ K$ with $λ ∉ Z$ , the module $A_1/A_1(x∂ - λ)$ has $e = 2$ but is simple, so $ℓ = 1 < 2 = e$ . Concretely it is the module $K[x,x^-1],x^λ$ of formal expressions $x^λ + k$ , $k ∈ Z$ . The bound (10.6) is therefore not an equality in general.

A finitely generated module that is not artinian

$A_n$ itself, for $n ≥ 1$ . It is cyclic, finitely generated and noetherian, but the chain of left ideals $A_n x_n^k$ descends strictly forever. Being finitely generated says nothing about descending chains; only holonomicity does.

Artinian without noetherian, in general algebra

The two chain conditions are genuinely independent. The Prüfer group $Z[1/p]/Z$ is an artinian $ℤ$ -module that is not noetherian. Nothing like it occurs among finitely generated $A_n$ -modules, because noetherianity there is automatic; but the general statement "artinian implies noetherian" is false and must not be quoted.

Worked Example

The composition series of $K[x][1/x]$ over $A_1$

Step 1 - the module and its multiplicity
Let $M = K[x][1/x]$ , the $K$ -span of $x^k$ for all $k ∈ Z$ , with $x$ acting by multiplication and $∂$ by differentiation. Filter it by

$Γ_k = , f/x^k : f ∈ K[x], f ≤ 2k ,, _K Γ_k = 2k+1.$

This is the filtration of the localisation theorem with $p = x$ , $p = 1$ ; it satisfies $B_1Γ_k ⊂ eq Γ_k+1$ and exhausts $M$ . The theorem gives $d(M) = 1$ and $e(M) ≤ ( p + 1)^1 = 2$ . The count $_K Γ_k = 2k+1$ is exact, so if this filtration is good the multiplicity is $1! ⋅ 2 = 2$ ; Steps 2 to 4 confirm $e(M) = 2$ independently.
Step 2 - a submodule and its quotient
$K[x] ⊂ eq M$ is an $A_1$ -submodule, with $e(K[x]) = 1$ . Let $N = M/K[x]$ , spanned by the classes of $x^-1, x^-2, x^-3, …$ . Write $δ$ for the class of $x^-1$ . Then

$x ⋅ δ = x ⋅ x^-1 = 1 = 0 N, ∂^,j x^-1 = (-1)^j, j!; x^-1-j.$

So $δ$ generates $N$ , it is killed by $x$ , and the elements $∂^,jδ$ , $j ≥ 0$ , are non-zero scalar multiples of the classes of $x^-1-j$ and hence form a $K$ -basis of $N$ . The surjection $A_1/A_1x → N$ sending $1 ↦ δ$ therefore matches basis to basis, so $N ≅ A_1/A_1x$ .
Step 3 - both factors are simple
$e(K[x]) = 1$ and $e(A_1/A_1 x) = 1$ , the latter from the Hilbert polynomial $i+22 - i+12 = i+1$ . By the corollary above, a holonomic module of multiplicity $1$ is simple, so both $K[x]$ and $N$ are simple. Hence

$0 ⊂ neq K[x] ⊂ neq K[x][1/x]$

is a composition series, and $ℓ(M) = 2$ .
Step 4 - check the multiplicity by additivity
The sequence $0 → K[x] → M → N → 0$ has all three terms holonomic of dimension $1$ , so $e(M) = e(K[x]) + e(N) = 1 + 1 = 2$ . This agrees with the leading coefficient read off in Step 1, where $_KΓ_k = 2k+1$ gave $e(M) = 2$ . Two independent computations, same answer.

So here $ℓ(M) = e(M) = 2$ : the bound (10.6) is attained.
Step 5 - a case where the bound is strict
Take $λ ∉ Z$ and $M_λ = A_1/A_1(x∂ - λ)$ . Since $x∂ - λ$ has Bernstein degree $2$ , the Hilbert polynomial is $i+22 - i2 = 2i+1$ ; check $i = 0$ : $1 - 0 = 1$ , correct since $B_0$ meets the ideal in $0$ ; $i = 2$ : $6 - 1 = 5$ , correct since $B_2$ meets the ideal in the line spanned by $x∂ - λ$ . So $d(M_λ) = 1$ and $e(M_λ) = 2$ .

But $M_λ$ is simple. Identify it with $⨁ _k ∈ Z K, x^λ+k$ , on which $x∂$ acts diagonally with the pairwise distinct eigenvalues $λ + k$ . Any non-zero submodule is spanned by eigenvectors, hence contains some $x^λ+k$ ; and since $λ + j ≠ 0$ for every $j ∈ Z$ , repeated application of $x$ and $∂$ reaches every $x^λ+j$ . So $ℓ(M_λ) = 1 < 2 = e(M_λ)$ .

Result

$K[x][1/x]$ is holonomic with $e = 2$ and composition series $0 ⊂ neq K[x] ⊂ neq K[x][1/x]$ , whose factors are $K[x]$ and the delta module $A_1/A_1x$ , each of multiplicity $1$ ; so $ℓ = e = 2$ . By contrast $A_1/A_1(x∂ - λ)$ with $λ ∉ Z$ has $e = 2$ but $ℓ = 1$ . The inequality $ℓ(M) ≤ e(M)$ is sharp but not an equality.

Applications and Industry Use

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

Existence of the b-function. The construction of the Bernstein-Sato polynomial applies finite length to the descending chain $A_n(K(s)),p^s ⊃ eq A_n(K(s)),p ⋅ p^s ⊃ eq A_n(K(s)),p^2 ⋅ p^s ⊃ eq ⋅ s$ . Because the module is holonomic the chain stabilises, and stabilisation is precisely the functional equation.
Cyclicity. The proof that every holonomic module is cyclic is an induction on length; without finite length there is no induction to run.
Decomposition of solution spaces. Composition factors of a holonomic module correspond to the irreducible pieces of the system of equations it encodes, and the finite list of factors is what makes classification statements possible.
Algorithmics. A decomposition series computed once can be reused: operations on a holonomic module reduce to operations on finitely many simple factors, which is the structural reason algorithms on holonomic functions terminate.
Representation theory. Category $𝒪$ and its relatives are categories of finite-length modules; the D-module realisation transports finite length from the algebraic side to the geometric one.

Limits of Validity

The bound needs holonomicity, not just finite generation. For a finitely generated module of dimension $d > n$ , quotients can have smaller dimension, additivity of multiplicity fails, and no chain bound follows. $A_n$ is the standard counterexample.
Characteristic zero. The whole chain runs through Bernstein's inequality. In characteristic $p$ the inequality is false, and $A_n$ has finite-dimensional modules, so the numerology changes completely.
$e$ depends on the filtration convention. The multiplicity computed with the Bernstein filtration and the multiplicity computed with the order filtration can differ, and the bound $ℓ ≤ e$ holds for each of them separately. Quote a multiplicity together with the filtration it came from.
Finite length does not give semisimplicity. $A_1/A_1∂^2$ has length $2$ ; it is a genuine extension and the theory says nothing about whether such an extension splits. Composition factors determine a module only up to extension classes.

A note on the citation for uniqueness of length

Coutinho attributes the uniqueness of the length of a composition series to an "Artin-Schreier theorem", citing Cohn's Algebra. That is a slip. The result that any two composition series have the same length and isomorphic factors is the Jordan-Hölder theorem, usually deduced from the Schreier refinement theorem; Artin-Schreier theory is an unrelated body of results about real closed fields and about degree- $p$ extensions in characteristic $p$ . The mathematics is standard and correct; only the name is wrong.

Failure Modes and Common Mistakes

Assuming length equals multiplicity

The theorem is an inequality. $A_1/A_1(x∂ - λ)$ with $λ ∉ Z$ is simple of multiplicity $2$ . Equality holds exactly when every composition factor has multiplicity $1$ , which happens often enough among familiar examples to breed a false expectation.

Using additivity of multiplicity without checking dimensions agree

The identity $e(M) = e(N) + e(M/N)$ requires $d(N) = d(M/N)$ . Inside a holonomic module that is automatic, since every non-zero subquotient has dimension $n$ . Outside, it fails: for $0 → A_n∂_n → A_n → A_n/A_n∂_n → 0$ the dimensions are $2n$ , $2n$ , $2n-1$ , and adding multiplicities is meaningless.

Confusing the descending chain condition with finite generation

Finite generation is the ascending condition in disguise and comes free over a noetherian ring. It gives no control at all over descending chains: $A_n$ is cyclic and has infinite descending chains of left ideals. Artinianness is the extra hypothesis, and it is exactly what holonomicity supplies.

Expecting the same statement over $D(X)$ without hypotheses

Finite length of holonomic modules generalises to coherent $D_X$ -modules on a smooth variety, where the proof runs through the theory of characteristic cycles rather than through Bernstein filtration multiplicities. On a singular variety $D(X)$ may fail to be noetherian and the argument has no analogue; do not transplant the statement without checking the ring.

Reading "artinian" as "finite dimensional"

An artinian $A_n$ -module is still infinite dimensional over $K$ whenever it is non-zero, because $A_n$ has no non-zero finite-dimensional representations in characteristic zero. The descending chain condition constrains submodules, not $K$ -dimension.

Historical Notes

The descending chain condition was introduced by Emil Artin in 1927 for algebras over a field, in the work that produced the Artin-Wedderburn classification. The dual condition had been isolated by Emmy Noether in 1921. Their symmetry is superficial: over a ring, artinian implies noetherian (Hopkins-Levitzki) but not the other way round, and for modules the two conditions are independent.

The Jordan-Hölder theorem is older still, going back to Jordan's 1869 work on permutation groups and Hölder's 1889 completion of it. Schreier's refinement theorem of 1928 gives the clean modern proof, and it is Schreier's name — not Artin-Schreier's — that belongs next to this result.

The application to holonomic modules is Bernstein's, and it is the step that turned his dimension inequality into a usable theory. Once one knows the modules of minimal dimension are artinian, all the structural theorems of Chapter 10 follow with very little further work; that economy is what makes the elementary route through the Bernstein filtration worth keeping alongside the geometric one.

Comparison

Which chain conditions hold where.
Module over $A_n$	Noetherian	Artinian	Finite length	$e$	$ℓ$
$K[x_1, … ,x_n]$	yes	yes	yes	$1$	$1$
$A_1/A_1x$	yes	yes	yes	$1$	$1$
$K[x][1/x]$ over $A_1$	yes	yes	yes	$2$	$2$
$A_1/A_1(x∂-λ)$ , $λ ∉ Z$	yes	yes	yes	$2$	$1$
$A_n$ , $n ≥ 1$	yes	no	no	$1$	infinite
a non-zero left ideal of $A_n$	yes	no	no	$≥ 1$	infinite
$K(x_1, … ,x_n)$	no	no	no	undefined	undefined

Key Takeaways

Key points

Inside a holonomic module every subquotient is holonomic of dimension $n$ , so multiplicity is additive at every step of any chain.
Multiplicities of non-zero holonomic modules are positive integers, so a strictly descending chain of length $r$ forces $r ≤ e(M)$ .
Hence holonomic modules are artinian; they are automatically noetherian, so they have composition series.
By Jordan-Hölder the length $ℓ(M)$ is well defined, and $ℓ(M) ≤ e(M)$ .
A holonomic module of multiplicity $1$ is simple; $K[x_1, … ,x_n]$ and $A_1/A_1x$ are examples.
The bound can be strict: $A_1/A_1(x∂-λ)$ with $λ ∉ Z$ has $ℓ = 1$ , $e = 2$ .
$A_n$ is noetherian but not artinian, which is why holonomicity is a real restriction and why the cyclicity theorem works.

FAQs

Does finite length follow from finite generation alone?

No. $A_n$ is generated by one element and has infinite descending chains of submodules. Finite generation gives the ascending chain condition over a noetherian ring; the descending one is extra, and holonomicity is what provides it.

Is the inequality

ℓ(M) ≤ e(M)

ever an equality?

Yes, whenever every composition factor has multiplicity $1$ . $K[x][1/x]$ is the standard example: length $2$ , multiplicity $2$ , factors $K[x]$ and $A_1/A_1x$ . Equality fails as soon as some factor has multiplicity $2$ or more.

Can a holonomic module have multiplicity

0

Only the zero module. For a non-zero holonomic module the Hilbert polynomial has degree $n$ with positive leading coefficient, and $e(M) = n!$ times that coefficient is a positive integer.

Do composition factors of a holonomic module have to be holonomic?

Yes. They are subquotients of a holonomic module, and the class is closed under subquotients. Note this does not say every simple $A_n$ -module is holonomic; for $n ≥ 2$ there are simple modules of larger dimension, which simply never occur as factors of a holonomic module.

Why is Coutinho's statement called a scholium?

A scholium is a remark that follows from an argument already given rather than from a new one. The chain bound $r ≤ e(M)$ proved for artinianness already contains the length bound; nothing further is needed, so the result is recorded as an observation rather than as a theorem.

Does

A_n

being non-artinian matter, or is it just a remark?

It matters. The cyclicity theorem for holonomic modules has as a hypothesis that the ring is not left artinian; that is what forces the map $a \mapsto a u$ to have a non-zero kernel, which is the pivot of the whole proof.

How do I compute the length of a given holonomic module?

In general it is hard. The multiplicity is a Hilbert polynomial computation and gives an upper bound; deciding whether a module is simple, and finding its factors, requires more, and is implemented only in special cases. For $A_1$ -modules and for localisations there are effective methods based on the b-function.

Is the category of holonomic modules semisimple?

No. It is artinian and noetherian, but extensions between simple holonomic modules exist and are non-split. $A_1/A_1∂^2$ has two composition factors isomorphic to $K[x]$ ; whether such an extension splits is a genuine question, not an automatic consequence of finite length.

Does the theorem hold for right modules?

Yes. The transposition anti-automorphism preserves Bernstein degree and turns left modules into right modules, so dimension, multiplicity, and every statement on this page transfer unchanged.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 10 §2, results (10.2.1) to (10.2.4); Ch. 9 §3, additivity (9.3.2).
I. N. Bernstein, The analytic continuation of generalized functions with respect to a parameter, Functional Analysis and its Applications 6 (1972), 273-285.
P. M. Cohn, Algebra, Volume 2, second edition, Wiley, 1989 - the Schreier refinement and Jordan-Hölder theorems.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Graduate Studies in Mathematics 30, revised edition, American Mathematical Society, 2001 - Ch. 4 and Ch. 8, for chain conditions and Gelfand-Kirillov dimension.
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1 §5, for length and multiplicity of holonomic modules.
R. Hotta, K. Takeuchi and T. Tanisaki, D-modules, Perverse Sheaves, and Representation Theory, Progress in Mathematics 236, Birkhäuser, 2008 - Ch. 3, for finite length of holonomic $D_X$ -modules.
I. Reiten, An introduction to the representation theory of artin algebras, Bulletin of the London Mathematical Society 17 (1985), 209-233 - background on artinian rings, the reference Coutinho points to at this juncture.
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization.

AI Suggested Questions

Write out the composition series of $K[x][1/(x^2-1)]$ over $A_1$ and check that its length equals its multiplicity.
Prove parts (2) and (3) of the artinian criterion by dualising the corresponding noetherian arguments.
Show that $A_1/A_1∂^2$ has length $2$ and identify both composition factors.
Find a holonomic $A_1$ -module of multiplicity $3$ and length $2$ .
Explain why the chain $A_n x_n ⊃ neq A_n x_n^2 ⊃ neq ⋅ s$ is strictly descending, using that $A_n$ is a domain with only scalar units.
For $λ ∈ Z$ , determine the length of $A_1/A_1(x∂ - λ)$ and compare with the case $λ ∉ Z$ .
State and prove additivity of length in a short exact sequence of holonomic modules.