The Noetherian Property in Short Exact Sequences

Q: Is it enough to know that M/N is Noetherian?

No. Take N = M non-Noetherian, so that M/N = 0 . Both outer terms must be controlled; the quotient alone sees nothing inside N .

Q: What does the constructive proof buy over the chain proof?

Explicit generators and a bound on how many are needed: p + q , where p generators suffice inside N and q in M/N . This is what an implementation actually computes, and it is how bounds such as "submodules of A_n^k need at most 2k generators" are derived from Stafford's two-generator theorem.

Q: Is the class of finitely generated modules closed under extensions?

Yes - generators of N together with lifts of generators of M/N generate M . What fails for finitely generated modules is closure under submodules, and that single failure is why the Noetherian condition exists.

Q: How is this used to prove that finitely generated modules over a Noetherian ring are Noetherian?

A module generated by k elements is a quotient of R^k . Finite direct sums of Noetherian modules are Noetherian, so R^k is; quotients of Noetherian modules are Noetherian, so the module is. Both steps are corollaries of the proposition on this page.

Overview

Knowing that one module is Noetherian is of limited use. What makes the property practical is that it propagates: along submodules, along quotients, and - crucially - backwards along extensions. This page proves the propagation rule and draws out the consequences.

The statement is short. If $N$ is a submodule of $M$ , then $M$ is Noetherian if and only if $N$ and $M/N$ are both Noetherian. Two of the three implications are routine: a submodule of a Noetherian module has fewer submodules to worry about, and a quotient has fewer still, once one knows that submodules of $M/N$ come from submodules of $M$ . The content is the converse. That $N$ and $M/N$ being Noetherian forces $M$ to be Noetherian is what lets the property be built up rather than merely inherited.

The converse turns on a small lemma about comparable submodules: if $P_1 ⊂ eq P_2$ agree both modulo $N$ and in their intersection with $N$ , then they are equal. That lemma converts "two chains stabilise" into "one chain stabilises", which is the whole proof.

Everything downstream uses this. Finite direct sums of Noetherian modules are Noetherian, so $R^k$ is Noetherian whenever the ring $R$ is; that is the step from a Noetherian ring to all its finitely generated modules, and it is what makes every finitely generated module over the Weyl algebra Noetherian once P2 itself is known to be.

Definition

Recall the terms. A left $R$ -module is Noetherian when every submodule is finitely generated, equivalently when ascending chains of submodules stabilise. A short exact sequence of left $R$ -modules is a diagram

0 → M' ⟶ ι M ⟶ π M'' → 0

(8.5)

with $ι$ injective, $π$ surjective and $imι = π$ .

Up to isomorphism this is the same data as a module $M$ together with a submodule $N = ι(M') ≅ M'$ , with $M'' ≅ M/N$ . The two languages are used interchangeably below.

Extension

Given modules $M^{'}$ and $M^{''}$ , an extension of $M^{''}$ by $M^{'}$ is a module $M$ fitting into a short exact sequence (8.5). The extension splits if $M ≅ M' ⊕ M''$ compatibly with $ι$ and $π$ . A property is closed under extensions if $M^{'}$ and $M^{''}$ having it forces $M$ to have it, split or not.

Construction and Proof

The comparison lemma

Comparable submodules agreeing modulo $N$ and inside $N$ are equalCoutinho (8.1.2)

Let $M$ be a left $R$ -module and let $N, P_1, P_2$ be submodules with $P_1 ⊂ eq P_2$ . If

N + P_1 = N + P_2 and N ∩ P_1 = N ∩ P_2 ,

then $P_1 = P_2$ .

Proof

Only $P_2 ⊂ eq P_1$ needs proof. Let $u ∈ P_2$ . Then $u ∈ N + P_2 = N + P_1$ , so $u = y + w$ with $y ∈ N$ and $w ∈ P_1$ . Now $w ∈ P_1 ⊂ eq P_2$ , so

y = u - w ∈ P_2 ∩ N = P_1 ∩ N ⊂ eq P_1 .

Hence $u = y + w$ is a sum of two elements of $P_1$ , so $u ∈ P_1$ . $□$

Comparability is not decoration

Without the hypothesis $P_1 ⊂ eq P_2$ the lemma is false. In $M = K^2$ over $K$ , take $N = K(1,1)$ , $P_1 = K(1,0)$ and $P_2 = K(0,1)$ . Then $N + P_1 = N + P_2 = K^2$ and $N ∩ P_1 = N ∩ P_2 = 0$ , yet $P_1 ≠ P_2$ . Knowing a submodule's image and its intersection pins it down only among submodules already known to be nested.

The main proposition

Noetherianity in a short exact sequenceCoutinho (8.1.3)

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

$M$ is Noetherian if and only if both $N$ and $M/N$ are Noetherian.
If $M = N + N'$ for submodules $N, N^{'}$ that are both Noetherian, then $M$ is Noetherian.

Proof of (1), the easy direction

Suppose $M$ is Noetherian. Every submodule of $N$ is a submodule of $M$ , hence finitely generated, so $N$ is Noetherian. A submodule of $M/N$ has the form $L/N$ for a unique submodule $L$ with $N ⊂ eq L ⊂ eq M$ , by the correspondence theorem. Since $L$ is finitely generated, the images of its generators generate $L/N$ . So $M/N$ is Noetherian.

Proof of (1), the converse

Assume $N$ and $M/N$ are Noetherian and let $L_1 ⊂ eq L_2 ⊂ eq …$ be an ascending chain of submodules of $M$ . Two derived chains are available.

L_1 ∩ N ⊂ eq L_2 ∩ N ⊂ eq … ⊂ eq N, (N + L_1)/(N) ⊂ eq (N + L_2)/(N) ⊂ eq … ⊂ eq (M)/(N) .

The first is a chain in $N$ , so it stabilises at some index $s$ ; the second is a chain in $M/N$ , so it stabilises at some index $r$ . Put $t = s, r$ . For every $i ≥ t$ we then have

N + L_t = N + L_i and N ∩ L_t = N ∩ L_i ,

with $L_t ⊂ eq L_i$ . The comparison lemma applies with $P_1 = L_t$ , $P_2 = L_i$ and gives $L_t = L_i$ . So the original chain stabilises, and $M$ satisfies the ascending chain condition, hence is Noetherian. $□$

Proof of (2)

By the second isomorphism theorem, $M/N = (N + N')/N ≅ N'/(N ∩ N')$ , a quotient of the Noetherian module $N^{'}$ , hence Noetherian by part (1). Now $N$ and $M/N$ are both Noetherian, so $M$ is Noetherian, again by part (1). $□$

A second proof of the converse, without chains

Direct construction of generators

It is worth seeing the converse proved constructively, because it is the version that is actually implemented. Let $L ⊂ eq M$ be a submodule. Then $L ∩ N$ is a submodule of $N$ , so it is generated by finitely many elements $v_1, … , v_p$ . And

(L)/(L ∩ N) ≅ (L + N)/(N) ⊂ eq (M)/(N)

is a submodule of the Noetherian module $M/N$ , so it is generated by finitely many classes $w_1, … , w_q$ ; choose representatives $w_1, … , w_q ∈ L$ .

Claim: $L = Rv_1 + … + Rv_p + Rw_1 + … + Rw_q$ . Given $u ∈ L$ , its class in $L/(L ∩ N)$ is $∑ _j b_j w_j$ for some $b_j ∈ R$ , so $u - ∑ _j b_j w_j$ lies in $L ∩ N$ and is therefore $∑ _i a_i v_i$ for some $a_i ∈ R$ . Hence $u = ∑ _i a_i v_i + ∑ _j b_j w_j$ , as claimed, and $L$ is generated by $p + q$ elements. $□$

What the second proof adds

The chain proof shows only that generators exist. The generator proof exhibits them, and it records a bound: if every submodule of $N$ needs at most $p$ generators and every submodule of $M/N$ at most $q$ , then every submodule of $M$ needs at most $p + q$ . Applied to $R^k$ this gives the familiar bound of $k ⋅ p$ generators for submodules of a free module of rank $k$ over a ring whose left ideals need at most $p$ generators - for $A_n$ , where $p = 2$ by Stafford's theorem, submodules of $A_n^k$ need at most $2k$ generators.

Variable Definitions

$R$: an associative ring with identity; modules are left $R$ -modules
$M$: the middle module of the short exact sequence
$N$: the distinguished submodule of $M$ , identified with $M^{'}$
$M/N$: the quotient module, identified with $M^{''}$
$ι$ , $π$: the injection $M^{'} ↪ M$ and the surjection $M ↠ M^{''}$
$L_i$: the terms of an ascending chain of submodules of $M$
$P_1, P_2$: comparable submodules in the comparison lemma, with $P_1 ⊂ eq P_2$
$v_i$ , $w_j$: generators of $L ∩ N$ and lifts of generators of $L/(L ∩ N)$
$A_n$: the $n$ -th Weyl algebra over a field $K$

Properties and Behaviour

Finite direct sums

If $M_1, … , M_k$ are Noetherian left $R$ -modules then $M_1 ⊕ … ⊕ M_k$ is Noetherian. Induct on $k$ : the sequence $0 → M_1 → M_1 ⊕ … ⊕ M_k → M_2 ⊕ … ⊕ M_k → 0$ is exact, and both ends are Noetherian by hypothesis and induction.

Free modules of finite rank

If $R$ is left Noetherian - that is, Noetherian as a left module over itself - then $R^k$ is a Noetherian left $R$ -module for every finite $k$ . This is the corollary that carries the whole theory: it is the reason a finitely generated module over a left Noetherian ring is Noetherian, since such a module is a quotient of some $R^k$ .

The Noetherian modules form a Serre subcategory

Within the category of left $R$ -modules, the full subcategory of Noetherian modules is closed under submodules, quotients and extensions. Classes with these three closure properties are called Serre subcategories, and they are exactly the classes one can quotient by to form a well-behaved quotient category. The same three properties hold for finite-length modules and for Artinian modules; they fail for finitely generated modules, which are not closed under submodules.

Sums and intersections of Noetherian submodules

If $N$ and $N^{'}$ are Noetherian submodules of any module $M$ , then $N + N'$ is Noetherian by part (2) of the proposition, and $N ∩ N'$ is Noetherian as a submodule of $N$ . The set of Noetherian submodules of $M$ is therefore closed under finite sums and intersections; its union over all such submodules is the largest Noetherian submodule only when the family is directed and finite, which it need not be.

Additivity of length

If $M$ has finite length then so do $N$ and $M/N$ , and $ℓ(M) = ℓ(N) + ℓ(M/N)$ . Finite length is the conjunction of the Noetherian and Artinian conditions, and this additivity is the quantitative shadow of the qualitative statement proved above. It is the model for the additivity of multiplicity in exact sequences used later in the collection.

Examples and Special Cases

Building modules over the Weyl algebra

$A_n$ is Noetherian on both sides, so $A_n^k$ is a Noetherian left $A_n$ -module and every quotient $A_n^k/N$ - that is, every finitely generated $A_n$ -module - is Noetherian. Every module named in this collection that is presented by a finite matrix of operators is therefore Noetherian without further argument.

Extensions that do not split

The proposition says nothing about splitting, and this matters. The sequence $0 → K[x] → K[x, x^-1] → Q → 0$ of $A_1$ -modules, with $Q$ isomorphic to the delta module $A_1/A_1x$ , does not split, and yet the middle term is Noetherian because both ends are. This is worked out in detail below.

Two Noetherian pieces, non-Noetherian union

Finiteness in the corollary on direct sums cannot be relaxed. Each $K[x_1, … , x_N]$ is Noetherian; their increasing union $K[x_1, x_2, … ]$ is not. An infinite ascending chain of Noetherian submodules is exactly the shape of a counterexample.

Finite generation of $N$ is not enough

One might hope that $N$ finitely generated together with $M/N$ Noetherian would force $M$ Noetherian. It does not. Take $M = N = K[x_1, x_2, … ]$ as a module over itself: $N$ is cyclic, $M/N = 0$ is Noetherian, and $M$ is not. The hypothesis on $N$ has to be the full Noetherian condition, not finite generation.

Worked Example

$K[x, x^-1]$ as a Noetherian $A_1$ -module of length two

Step 1 - the module and its generator
Let $M = K[x, x^-1]$ , the ring of Laurent polynomials, with $A_1$ acting by multiplication by $x$ and by $∂ = d/dx$ . It is generated over $A_1$ by the single element $x^-1$ : applying $∂$ repeatedly gives $∂^k x^-1 = (-1)^k k!, x^-k-1$ , so all negative powers are reached, and multiplying by $x^k+1$ gives $x^k$ for $k ≥ 0$ . So $M$ is cyclic over $A_1$ , and $N = K[x]$ is an $A_1$ -submodule.
Step 2 - identify the quotient
Let $Q = M/N$ , spanned over $K$ by the classes of $x^-1, x^-2, x^-3, …$ . Write $v$ for the class of $x^-1$ . Then $x,v = 0$ in $Q$ , since $x ⋅ x^-1 = 1 ∈ N$ . So there is a surjection of $A_1$ -modules

$A_1/A_1x ⟶ Q, 1 ⟼ v,$

surjective because $∂^k v = (-1)^k k!,[x^-k-1]$ exhausts a spanning set of $Q$ . Since $A_1/A_1x$ is a simple $A_1$ -module and $Q ≠ 0$ , the surjection is an isomorphism: $Q$ is the delta module.
Step 3 - apply the proposition
$N = K[x]$ is a simple P1-module, hence Noetherian; $Q ≅ A_1/A_1x$ is simple, hence Noetherian. By the proposition, $M$ is Noetherian. Moreover $0 ⊂ N ⊂ M$ is a chain with simple quotients, so $ℓ(M) = 2$ .
Step 4 - list every submodule
Let $L ⊂ eq M$ be a submodule. If $L ⊂ eq N$ then $L = 0$ or $L = N$ by simplicity of $N$ . Otherwise the image of $L$ in $Q$ is non-zero, hence all of $Q$ by simplicity, so $L + N = M$ . If in addition $L ∩ N = 0$ then $L ≅ Q$ , so $M$ would contain a non-zero element killed by $x$ - impossible, since $M$ is a domain under multiplication. Hence $L ∩ N = N$ , that is $N ⊂ eq L$ , and then $L = L + N = M$ . The complete list of submodules is $0$ , $K[x]$ , $K[x,x^-1]$ .
Step 5 - the extension does not split
Step 4 shows there is no submodule isomorphic to $Q$ , so $M ≅ K[x] ⊕ Q$ as an extension: the sequence $0 → K[x] → K[x,x^-1] → Q → 0$ is non-split. The proposition never asked for a splitting, which is precisely why it is useful; a closure result that required split extensions would be nearly vacuous.
Step 6 - change the ring and the answer changes
Regard the same set $K[x,x^-1]$ as a module over $K[x]$ instead. The submodules $x^-kK[x]$ form a strictly ascending chain

$K[x] ⊂ neq x^-1K[x] ⊂ neq x^-2K[x] ⊂ neq …$

that never stabilises, since $x^-k-1 ∉ x^-kK[x]$ . So $K[x,x^-1]$ is not a Noetherian $K[x]$ -module. These sets are not stable under $∂$ , which is why they do not obstruct the $A_1$ -statement: the larger ring has fewer submodules to control.

Result

$K[x,x^-1]$ is a cyclic, Noetherian $A_1$ -module of length $2$ , with exactly three submodules $0 ⊂ K[x] ⊂ K[x,x^-1]$ , arising as a non-split extension of the delta module $A_1/A_1x$ by $K[x]$ . Over $K[x]$ the same set is not Noetherian. Noetherianity was obtained purely from the two simple outer terms, with no information about how they are glued.

Applications and Industry Use

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

From ring to modules. $R$ left Noetherian $\Rightarrow$ $R^k$ Noetherian $\Rightarrow$ every finitely generated left $R$ -module Noetherian. Both arrows are instances of this page's proposition, and the chain is completed on the Noetherian rings page.
Filtered-to-graded transfer. Lifting the property from the associated graded ring needs that submodules inherit an induced filtration and that the graded pieces assemble into an exact sequence; the closure results here are what make the induced objects usable.
Dimension theory. Dimension and multiplicity behave additively in short exact sequences, and the proofs mirror the argument given here: control the sub, control the quotient, deduce the middle.
Finite length of holonomic modules. That holonomic modules have finite length is proved by combining additivity of multiplicity with the two-out-of-three property; without extension-closure the descending chain argument would not close.
Software. Module operations in computer algebra systems - kernels, images, sums, intersections, quotients - all produce finitely generated objects only because the ambient ring is Noetherian and the closure properties above hold.

Limits of Validity

The two-out-of-three shape is asymmetric. From $M$ Noetherian one gets both outer terms; from the outer terms one gets $M$ . But from $M/N$ Noetherian and $M$ arbitrary one gets nothing about $N$ , and from $N$ Noetherian alone one gets nothing about $M$ .
Finite generation cannot replace Noetherianity in the hypothesis. $M = N = K[x_1,x_2, … ]$ over itself has $N$ cyclic, $M/N = 0$ , and $M$ not Noetherian.
Infinite direct sums and infinite unions escape. Only finitely many Noetherian pieces can be assembled. A direct limit of Noetherian modules is generally not Noetherian, even when all the transition maps are injective.
Splitting is a separate question. The proposition gives no information about whether an extension splits, and the worked example shows a non-split one. Conversely, splitting adds no strength to the conclusion.
The statement is about a fixed ring. Restricting scalars along a ring map $S \to R$ can destroy the property, as the last step of the worked example shows: Noetherian over $A_1$ , not Noetherian over $K[x]$ .

Failure Modes and Common Mistakes

Dropping comparability in the lemma

The lemma requires $P_1 ⊂ eq P_2$ . Two submodules can have the same image in $M/N$ and the same intersection with $N$ without being equal - $K(1,0)$ and $K(0,1)$ inside $K^2$ with $N = K(1,1)$ . The lemma is applied only to consecutive terms of a chain, where comparability is free; applying it elsewhere is a genuine error, and one that is easy to make because the conclusion looks like a general uniqueness statement.

Stabilising the two derived chains at different indices

The chain $L_i ∩ N$ stabilises at some $s$ and the chain $(N + L_i)/N$ at some $r$ , and there is no reason for $s = r$ . Both conditions are needed simultaneously, so one must pass to $t = s, r$ before invoking the lemma. Arguments that use a single index quietly assume the two stabilisation points coincide.

Assuming a Noetherian module is a finite direct sum of simple ones

Noetherian modules of finite length are built from simple modules by extensions, not by direct sums. The worked example produces a length-two module that is not a direct sum of its composition factors. Semisimplicity is a much stronger condition, and over $A_n$ it usually fails: $A_1/A_1x^2$ , for instance, is a non-split self-extension of the delta module.

Using the result on an infinite filtration

A module exhausted by an increasing family of Noetherian submodules $M = ⋃ _i M_i$ is not Noetherian in general; the proposition applies to a single submodule at a time and induction only reaches finitely far. This is the exact point at which the argument for $K[x_1, … , x_N]$ fails to extend to $K[x_1, x_2, … ]$ .

Forgetting that the correspondence theorem is doing work

The easy direction uses that every submodule of $M/N$ is $L/N$ for a submodule $L ⊃ eq N$ of $M$ . That is a theorem, not a definition, and it is what makes "quotients have fewer submodules" precise. Without it there is no reason a quotient of a Noetherian module should inherit anything.

Historical Notes

The closure properties are already present, in the language of ideals and modules over commutative rings, in Emmy Noether's 1921 treatment of the ascending chain condition; the module-theoretic formulation with short exact sequences is a product of the homological algebra developed in the 1940s and 1950s by Eilenberg, Mac Lane and Cartan.

Serre's 1953 paper on classes of abelian groups isolated exactly the three closure conditions - subobjects, quotients, extensions - as what is needed to form a quotient category, and the classes satisfying them now bear his name. That the Noetherian modules form such a class is the abstract content of this page.

In the non-commutative setting the same statements hold verbatim, and they are the standard tools in the theory of Noetherian rings developed from the 1950s onwards. For rings of differential operators the closure properties were used from the very beginning, notably in Bernstein's 1971 work, where the finiteness of the theory rests on being able to move freely between a module, a submodule and a quotient without losing control.

Comparison

The table records which finiteness conditions survive which operations. "Extension" means: the outer terms of a short exact sequence have the property, therefore the middle does.

Closure properties of finiteness conditions on modules.
Property	Submodules	Quotients	Extensions	Finite direct sums	Arbitrary direct sums
Finitely generated	no	yes	yes	yes	no
Noetherian	yes	yes	yes	yes	no
Artinian	yes	yes	yes	yes	no
Finite length	yes	yes	yes	yes	no
Simple	no	no	no	no	no
Free	no	no	no	yes	yes

The single "no" in the Noetherian row is what forces every statement in this chapter to carry a finiteness hypothesis somewhere. The "no" in the submodule column of the first row is what the Noetherian condition was invented to repair.

Key Takeaways

Key points

For $N ⊂ eq M$ : $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian.
The converse direction is the useful one; it says the class is closed under extensions, split or not.
The proof stabilises two derived chains at a common index and applies a comparison lemma valid for nested submodules only.
A constructive second proof exhibits generators of a submodule $L$ as generators of $L ∩ N$ together with lifts of generators of $L/(L ∩ N)$ , giving a bound on generator counts.
Consequences: finite sums and finite direct sums of Noetherian modules are Noetherian, and $R^k$ is Noetherian for $R$ left Noetherian.
Finiteness is essential: infinite direct sums and increasing unions of Noetherian modules need not be Noetherian.
$K[x,x^-1]$ is a length-two Noetherian $A_1$ -module arising as a non-split extension, and is not Noetherian over $K[x]$ .

FAQs

Is it enough to know that

M/N

is Noetherian?

No. Take $N = M$ non-Noetherian, so that $M/N = 0$ . Both outer terms must be controlled; the quotient alone sees nothing inside $N$ .

Does the extension have to split?

No, and that is the point. The worked example gives a non-split sequence $0 → K[x] → K[x,x^-1] → A_1/A_1x → 0$ whose middle term is Noetherian all the same. Closure under extensions is strictly stronger than closure under direct sums.

Why can't the argument be iterated to infinity?

Because each application handles one submodule, and induction on the number of applications only reaches finite stages. $K[x_1,x_2, … ]$ is the union of the Noetherian rings $K[x_1, … ,x_N]$ and is itself not Noetherian, so no limiting version of the statement can be true.

Does the same statement hold for Artinian modules?

Yes, with descending chains in place of ascending ones; the proof is the same word for word after reversing inclusions, and the comparison lemma is unchanged. Consequently finite length, which is the conjunction of the two, also satisfies two-out-of-three.

What does the constructive proof buy over the chain proof?

Explicit generators and a bound on how many are needed: $p + q$ , where $p$ generators suffice inside $N$ and $q$ in $M/N$ . This is what an implementation actually computes, and it is how bounds such as "submodules of $A_n^k$ need at most $2k$ generators" are derived from Stafford's two-generator theorem.

Is the class of finitely generated modules closed under extensions?

Yes - generators of $N$ together with lifts of generators of $M/N$ generate $M$ . What fails for finitely generated modules is closure under submodules, and that single failure is why the Noetherian condition exists.

How is this used to prove that finitely generated modules over a Noetherian ring are Noetherian?

A module generated by $k$ elements is a quotient of $R^k$ . Finite direct sums of Noetherian modules are Noetherian, so $R^k$ is; quotients of Noetherian modules are Noetherian, so the module is. Both steps are corollaries of the proposition on this page.

Does the result depend on the ring being commutative?

No. Nothing in the lemma or the proposition uses commutativity, and both statements hold for left modules over any ring with identity, and symmetrically for right modules. This is exactly why the machinery applies to the Weyl algebra.

References

S. C. Coutinho, A Primer of Algebraic D-modules, London Mathematical Society Student Texts 33, Cambridge University Press, 1995 - Ch. 8 §1, Lemma (8.1.2) and Proposition (8.1.3).
E. Noether, Idealtheorie in Ringbereichen, Mathematische Annalen 83 (1921), 24-66.
M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, 1969 - Ch. 6, Proposition 6.3 and its corollaries.
J.-P. Serre, Groupes d'homotopie et classes de groupes abéliens, Annals of Mathematics 58 (1953), 258-294 - the origin of the notion of a Serre class.
J. C. McConnell and J. C. Robson, Noncommutative Noetherian Rings, Graduate Studies in Mathematics 30, revised edition, American Mathematical Society, 2001 - Ch. 1 and Ch. 3.
J.-E. Björk, Rings of Differential Operators, North-Holland Mathematical Library 21, North-Holland, 1979 - Ch. 1, for the filtered version of these arguments.
J. T. Stafford, Module structure of Weyl algebras, Journal of the London Mathematical Society (2) 18 (1978), 429-442.
ISO 80000-2:2019, Quantities and units - Part 2: Mathematics, International Organization for Standardization.

AI Suggested Questions

Prove the Artinian analogue of the proposition by reversing all inclusions, and check that the comparison lemma is unchanged.
Show that $A_1/A_1x^2$ is a non-split self-extension of the delta module and compute its submodule lattice.
Give an example of an increasing union of Noetherian submodules whose union is not Noetherian, other than the polynomial ring in infinitely many variables.
Derive the bound of $2k$ generators for submodules of $A_n^k$ from the constructive proof and Stafford's theorem.
Show that the class of modules of finite length is a Serre subcategory and that length is additive in short exact sequences.
Find all $A_1$ -submodules of $K[x,x^-1, (x-1)^-1]$ and compute its length.
Explain why restricting scalars from $A_1$ to $K[x]$ can destroy the Noetherian property, in terms of the number of submodules.