Why does a strictly descending chain of the ideals A_n exist as soon as _k V is infinite?

Infinite dimension supplies independent vectors v_1, v_2, … without end. For each n, (n+1)-transitivity produces a ring element annihilating v_1, …, v_n while moving v_n+1; that element lies in A_n but not in A_n+1. Transitivity is exactly the tool that manufactures separating elements.

Deriving Wedderburn–Artin from the Density Theorem

Q: Does this really prove Wedderburn–Artin, or does it assume it?

It proves it, for simple left artinian rings, and the argument is logically independent of the socle-decomposition proof in §3. The only inputs are Schur's Lemma, the Density Theorem and the observation that a simple ring is left primitive. The extension to general semisimple rings still needs the decomposition into finitely many simple components.

Q: Is a left primitive ring determined by the pair (V, k)?

Only in the artinian branch, where R is the full endomorphism ring M_n(k). In general R is merely some dense subring, and many non-isomorphic dense subrings of the same End(V_k) exist — the full ring, scalars plus finite rank, and Kaplansky's centre-prescribing rings all live inside the same End(V_k).

Q: Why is checking 2-transitivity enough to get density?

Because 2-transitivity forces End(_R V) to be exactly k, and once the endomorphism ring is correct the Density Theorem applies and delivers m-transitivity for all m. This is (11.20)(2), and it is the practical test: verify one condition on pairs of vectors rather than an infinite family of conditions.

Q: Can a left primitive ring be commutative?

Only if it is a field. A commutative left primitive ring has a faithful simple module R/m, and m annihilates it, so faithfulness forces m = 0. Consistently, (11.19) then gives _k V = 1 and R ≅ M_1(k) = k.

Q: Does the theorem give a canonical embedding of R into a matrix ring of infinite size?

It gives an embedding into End(V_k), which after choosing a basis of V_k becomes a ring of column-finite infinite matrices. The embedding depends on both the module V and the basis, so it is a useful coordinate system rather than a canonical form.

Executive Summary

Let $R$ be left primitive with faithful simple left module $V$ , and let $k = End (_{R} V)$ , a division ring. Faithfulness embeds $R$ in $End (V_{k})$ ; density says the embedded copy is $m$ -transitive for every $m$ . A dichotomy follows immediately: either $\dim_{k} V$ is finite, and then $R ≅ M_{n} (k)$ and $R$ is left artinian, or $\dim_{k} V$ is infinite and $R$ is not left artinian.

The finite branch is the Wedderburn–Artin theorem for simple left artinian rings, obtained without ever mentioning a chain condition in the proof. The infinite branch is a consolation prize with real content: $R$ still carries matrix rings of every size as subquotients.

$M_{n} (k)$ Artinian case

DenseGeneral case

2Transitivity needed for density

$(11.19)$ Structure theorem

Overview

The classical route to Wedderburn–Artin decomposes a semisimple ring into isotypic components and identifies each component by an endomorphism-ring computation. It needs the descending chain condition from the first line. The density route reverses the logic: prove a transitivity statement that holds for every left primitive ring, then observe that a chain condition forces the transitivity to saturate.

R left primitive ⟹ R ↪ End (V_{k}) densely, \dim_{k} V < \infty ⟺ R left artinian

(11.19)

The structure theorem and the dichotomy it produces.

Two ingredients feed in. Schur's Lemma makes $k = End (_{R} V)$ a division ring, so $V_{k}$ is an honest vector space; the Density Theorem supplies transitivity. Everything else on this page is extracted from those two facts by linear algebra.

Because a simple ring is left primitive and a left artinian left primitive ring is simple, the artinian branch is not a special case reluctantly recovered — it is exactly the classical theorem, proved more generally. The relationship between these classes is laid out in Primitive, Simple, Prime and Semiprimitive: How the Classes Relate.

Learning Objectives

State $(11.19)$ with the hypotheses on $R$ , $V$ and $k$ spelled out.
Prove the equivalence $\dim_{k} V < \infty \Leftrightarrow R$ left artinian for a left primitive $R$ .
Construct the strictly descending chain of left ideals in the infinite-dimensional case.
Identify the matrix sections $R_{n} / 𝔄_{n} ≅ M_{n} (k)$ and say why they are only subquotients.
Deduce the Wedderburn–Artin theorem for simple left artinian rings.
Use $(11.20)$ to certify density by checking $2$ -transitivity alone.

Definitions

$m$ -transitive: $S \subseteq End (V_{k})$ is $m$ -transitive if for every $n \leq m$ , every $k$ -linearly independent $v_{1}, \dots, v_{n}$ and every $w_{1}, \dots, w_{n} \in V$ there is $s \in S$ with $s (v_{i}) = w_{i}$ for all $i$ .
Dense ring of linear transformations: A subring of $End (V_{k})$ that is $m$ -transitive for every finite $m$ .
$V_{n}$: For a fixed independent sequence $v_{1}, v_{2}, \dots$ , the finite-dimensional subspace $V_{n} = \sum_{i = 1}^{n} v_{i} k$ .
$R_{n}$ and $𝔄_{n}$: $R_{n} = {r \in R : r V_{n} \subseteq V_{n}}$ , a subring of $R$ ; $𝔄_{n} = {r \in R : r V_{n} = 0}$ , an ideal of $R_{n}$ and a left ideal of $R$ .
Left primitive ring: A ring with a faithful simple left module; equivalently, a ring having a maximal left ideal containing no nonzero two-sided ideal.

All rings have an identity. $V$ is written as a right $k$ -vector space, and $R$ acts on the left; transformations are composed accordingly.

Core Concepts

From faithful simple module to a ring of matrices

Faithfulness of $V$ says $ann (V) = 0$ , so $ρ : R \to End (V_{k})$ is injective and $R$ may be identified with a ring of linear transformations of $V_{k}$ . Simplicity of $V$ says this ring is $1$ -transitive. Density upgrades $1$ -transitivity to $m$ -transitivity for every $m$ . So a left primitive ring is, up to isomorphism, a ring of matrices — possibly of infinite size and possibly a proper subring of all of them.

Why finite dimension collapses the theorem

If $\dim_{k} V = n < \infty$ , then $V_{k}$ is finitely generated over $k$ and $(11.17)$ makes $ρ$ onto. Since $End (V_{k}) ≅ M_{n} (k)$ for an $n$ -dimensional right $k$ -space, we get $R ≅ M_{n} (k)$ — and matrix rings over division rings are left and right artinian, so the chain condition arrives as a conclusion rather than a hypothesis.

Why infinite dimension destroys the chain condition

If $\dim_{k} V$ is infinite, choose independent $v_{1}, v_{2}, \dots$ and let $𝔄_{n}$ annihilate $V_{n}$ . Each $𝔄_{n}$ is a left ideal of $R$ , and $(n + 1)$ -transitivity supplies an $r$ killing $v_{1}, \dots, v_{n}$ but not $v_{n + 1}$ . That single element separates $𝔄_{n}$ from $𝔄_{n + 1}$ , and the resulting strictly descending chain rules out the descending chain condition.

𝔄_{1}

𝔄_{2}

𝔄_{3}

\dots

Key Results

Theorem(11.19)Structure theorem for left primitive rings

Let $R$ be a left primitive ring, $V$ a faithful simple left $R$ -module, and $k = End (_{R} V)$ — a division ring by Schur's Lemma. Then $R$ is isomorphic to a dense ring of linear transformations on the right $k$ -vector space $V$ . Moreover:

If $R$ is left artinian, then $n : = \dim_{k} V$ is finite and $R ≅ M_{n} (k)$ .
If $R$ is not left artinian, then $\dim_{k} V$ is infinite, and for every integer $n > 0$ there is a subring $R_{n} \subseteq R$ admitting a surjective ring homomorphism onto $M_{n} (k)$ .

Proof

Embedding. Since $V$ is faithful, $\ker ρ = ann (V) = 0$ , so $ρ : R \to E : = End (V_{k})$ is injective. Since $V$ is simple it is semisimple, so the Density Theorem $(11.16)$ applies, and by $(11.18)$ density of the action is the same as $ρ (R)$ being a dense ring of linear transformations on $V_{k}$ . This is the first assertion.

Finite-dimensional case. Suppose $\dim_{k} V = n < \infty$ . Then $V_{k}$ is finitely generated, so $ρ$ is onto by $(11.17)$ ; being also injective, $ρ : R ≅ E ≅ M_{n} (k)$ . In particular $R$ is left (and right) artinian.

Infinite-dimensional case. Suppose $\dim_{k} V$ is infinite. Fix $k$ -independent $v_{1}, v_{2}, \dots$ and set $V_{n} = \sum_{i \leq n} v_{i} k$ , together with

R_{n} = {r \in R : r V_{n} \subseteq V_{n}}, 𝔄_{n} = {r \in R : r V_{n} = 0} .

Restriction to $V_{n}$ gives a ring homomorphism $R_{n} \to End ((V_{n})_{k}) ≅ M_{n} (k)$ with kernel $𝔄_{n}$ . It is onto: given $τ \in End ((V_{n})_{k})$ , $n$ -transitivity produces $r \in R$ with $r v_{i} = τ (v_{i})$ for $i \leq n$ ; then $r V_{n} = τ (V_{n}) \subseteq V_{n}$ , so $r \in R_{n}$ and $r$ restricts to $τ$ . Hence $R_{n} / 𝔄_{n} ≅ M_{n} (k)$ , which is assertion (2).

No chain condition. Each $𝔄_{n}$ is a left ideal of $R$ and clearly $𝔄_{1} \supseteq 𝔄_{2} \supseteq \dots$ . By $(n + 1)$ -transitivity applied to the independent vectors $v_{1}, \dots, v_{n + 1}$ there is $r \in R$ with $r v_{1} = \dots = r v_{n} = 0$ and $r v_{n + 1} \neq 0$ ; thus $r \in 𝔄_{n} ∖ 𝔄_{n + 1}$ and the chain is strictly descending. So $R$ is not left artinian.

The last paragraph also proves the converse implication needed for (1): if $\dim_{k} V$ were infinite, $R$ would fail to be left artinian. So a left artinian left primitive ring has $\dim_{k} V$ finite, and the two cases are exhaustive and mutually exclusive.

Corollary—Wedderburn–Artin for simple left artinian rings

Let $R$ be a simple ring which is left artinian. Then $R ≅ M_{n} (k)$ for some integer $n \geq 1$ and some division ring $k$ , and $k ≅ End (_{R} V)$ for the unique simple left $R$ -module $V$ up to isomorphism.

Proof

Since $R \neq 0$ it has a maximal left ideal $𝔪$ , and $M = R / 𝔪$ is a simple left module. Its annihilator is a two-sided ideal not containing $1$ , hence proper, hence zero by simplicity of $R$ ; so $M$ is faithful and $R$ is left primitive. Applying $(11.19) (1)$ with the given left artinian hypothesis yields $\dim_{k} V = n < \infty$ and $R ≅ M_{n} (k)$ with $k = End (_{R} V)$ . Uniqueness of the simple module follows because $M_{n} (k)$ is semisimple with a single isotypic component.

Corollary—The semisimple case

If $R$ is a semisimple ring, then $R ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ for division rings $D_{i}$ . Density supplies each factor: the decomposition of $R$ into finitely many simple components is the input, and each component is simple left artinian, hence a matrix ring by the previous corollary.

Theorem(11.20)Two-transitivity suffices

Let $k$ be a division ring, $V \neq 0$ a right $k$ -vector space, and $R \subseteq E = End (V_{k})$ a subring. Then:

$R$ is $1$ -transitive if and only if $_{R} V$ is a simple module; in that case $R$ is a left primitive ring.
The following are equivalent: (a) $R$ is $2$ -transitive; (b) $R$ is $1$ -transitive and $End (_{R} V) = k$ ; (c) $R$ is dense in $E$ .

Proof

(1) $1$ -transitivity says $R v = V$ for every $0 \neq v \in V$ , which is simplicity of $_{R} V$ ; the action is faithful because $R$ sits inside $E$ , so $V$ is a faithful simple left $R$ -module and $R$ is left primitive.

(2) For (b) $\Rightarrow$ (c): by (1), $_{R} V$ is simple, hence semisimple, and the hypothesis identifies $k$ with $End (_{R} V)$ , so the Density Theorem $(11.16)$ together with $(11.18)$ gives density. (c) $\Rightarrow$ (a) is immediate, since density includes $2$ -transitivity. The remaining implication (a) $\Rightarrow$ (b) is the double-centraliser step: $2$ -transitivity forces every $R$ -endomorphism of $V$ to be right multiplication by a scalar in $k$ ; it is proved in Double Centralizers and Density for Bimodules.

Remark—What (2) of (11.19) does not say

The surjection is from a subring $R_{n}$ , not from $R$ . A simple non-artinian left primitive ring — the first Weyl algebra in characteristic $0$ , for instance — has no proper two-sided quotients at all, so no homomorphism from $R$ onto $M_{n} (k)$ can exist for $n$ with $M_{n} (k) not ≅ R$ .

Proof Techniques and Method

How this proof works, and which moves transfer to other arguments.

Move 1

Faithful plus simple equals concrete

Any faithful simple module turns an abstract ring into a ring of linear transformations. All later arguments are then linear algebra over a division ring.

Move 2

Transitivity manufactures elements

Need an element killing some vectors but not another? Ask $(n + 1)$ -transitivity for it. This is how the strictly descending chain is built, and how most exercises on primitive rings are solved.

Move 3

Stabiliser subring, annihilator ideal

The pair $R_{n} \supseteq 𝔄_{n}$ turns a finite-dimensional piece of $V$ into a matrix section of $R$ . Subquotients, not quotients — the distinction matters.

The strategic point is the order of quantifiers. The classical Wedderburn–Artin proof fixes the chain condition and derives structure; the density proof derives structure first and lets the chain condition decide how far the structure goes. That is why the same argument covers non-artinian rings without modification.

Worked Example

Finite-dimensional check

Let $R = M_{2} (ℚ)$ and let $V = ℚ^{2}$ be the module of column vectors, with $ℚ$ acting on the right by scalars. $V$ is simple and faithful, and $End (_{R} V) = ℚ$ because a matrix commuting with all of $M_{2} (ℚ)$ is scalar. Here $\dim_{k} V = 2 < \infty$ , so $(11.19) (1)$ predicts $R ≅ M_{2} (ℚ)$ — which it is. The theorem is consistent, and the content in this case is that no proper subring of $M_{2} (ℚ)$ can be $2$ -transitive on $ℚ^{2}$ .

Infinite-dimensional: scalars plus finite rank

Let $k$ be a field, $V = ⨁_{i = 1}^{\infty} e_{i} k$ , and $E = End (V_{k})$ . Let $F \subseteq E$ be the set of transformations of finite rank and put

R = k \cdot 1 + F \subseteq E .

(E.1)

Here $k \cdot 1$ denotes the scalar transformations $v \mapsto v a$ . Over a division ring one must restrict to $a \in Z (k)$ , since only central scalars act $k$ -linearly.

$F$ is an ideal of $E$ , so $R$ is a subring. It is dense: given $k$ -independent $v_{1}, \dots, v_{m}$ and arbitrary $w_{1}, \dots, w_{m}$ , extend to a basis and define $τ (v_{i}) = w_{i}$ with $τ$ zero on the remaining basis vectors. Then $τ$ has rank at most $m$ , so $τ \in F \subseteq R$ . Hence $R$ is $m$ -transitive for every $m$ , so $_{R} V$ is simple and faithful, $R$ is left primitive, and by $(11.20)$ we also get $End (_{R} V) = k$ .

Now run $(11.19) (2)$ concretely with $v_{i} = e_{i}$ . Writing $E_{i j}$ for the rank-one transformation $e_{j} \mapsto e_{i}$ killing the other basis vectors:

$𝔄_{n} = {r \in R : r V_{n} = 0}$ contains $E_{n + 1, n + 1}$ , which does not annihilate $e_{n + 1}$ ; so $𝔄_{n} ⊋ 𝔄_{n + 1}$ and $R$ is not left artinian.
$R_{n} = {r \in R : r V_{n} \subseteq V_{n}}$ contains every $E_{i j}$ with $i, j \leq n$ , and restriction gives $R_{n} / 𝔄_{n} ≅ M_{n} (k)$ — matrix rings of every size appear as sections.
$soc (R) = F$ : the minimal left ideals are $R e$ for rank-one idempotents $e$ , so this left primitive ring has nonzero socle and is therefore right primitive as well.

Construction(11.21)Kaplansky: prescribing the centre

Let $A$ be an integral domain with quotient field $k$ , and let $V = ⨁_{i \geq 1} e_{i} k$ . Let $R \subseteq End (V_{k})$ consist of those transformations whose matrix is a finite matrix $M$ over $k$ in the top-left corner and the scalar $a \in A$ down the rest of the diagonal. Then $R$ is dense in $End (V_{k})$ , hence left primitive, and $Z (R) = {a \cdot 1 : a \in A} ≅ A$ .

*Why the centre is $A$ :* if such an $r$ with corner block $M \in M_{m} (k)$ is central, then $M$ commutes with all of $M_{m} (k)$ , so $M = b \cdot I_{m}$ for some $b \in k$ ; repeating the argument one size up forces $b = a$ . Since a left primitive ring is prime, its centre is automatically a domain, so this construction shows every integral domain occurs.

Frameworks and Models

Left primitive rings organise by two independent binary features: whether $\dim_{k} V$ is finite, and whether the socle is nonzero.

Four kinds of left primitive ring
	$\dim_{k} V$ finite	Nonzero socle	Left artinian	Example
Simple artinian	yes	yes	yes	$M_{n} (D)$
Full endomorphism ring	no	yes	no	$End (V_{k})$ , $\dim_{k} V$ infinite
Scalars plus finite rank	no	yes	no	$k \cdot 1 + F$
Socle-free	no	no	no	Weyl algebra $A_{1}$ ; free algebra $k ⟨ x, y ⟩$

Four kinds of left primitive ring

The first row is the only one where $R$ is determined by $(V, k)$ alone. In the last row the ring may even have non-isomorphic faithful simple left modules, so the representation as a dense ring of transformations is a choice, not an invariant.

Process and Workflow

You have a left primitive ring $R$ with faithful simple $V$ and $k = End (_{R} V)$ . What can you conclude?

$\dim_{k} V = n < \infty$

R ≅ M_{n} (k)

R

is simple and both left and right artinian, all simple left

R

-modules are isomorphic to

V

, and every one-sided ideal is generated by an idempotent.

$\dim_{k} V$ infinite, socle $\neq 0$

R

is not artinian on either side, but

R

is also right primitive by the minimal-ideal theorem

(11.11)

, and the faithful simple modules are unique up to isomorphism.

$\dim_{k} V$ infinite, socle $= 0$

R

has no minimal one-sided ideals; nothing forces right primitivity, and

R

may carry several non-isomorphic faithful simple left modules. This is where one-sided counterexamples live.

Produce a faithful simple moduleUsually

R / 𝔪

for a maximal left ideal

𝔪

containing no nonzero two-sided ideal.

Compute $k = End (_{R} V)$ Schur guarantees a division ring; identifying it is the substantive step.

Test $2$ -transitivityBy

(11.20)

this alone certifies density — no need to verify

m

-transitivity for every

m

Measure $\dim_{k} V$ Finite gives

R ≅ M_{n} (k)

; infinite gives the matrix sections and rules out the descending chain condition.

Comparison and Classification

The two branches of $(11.19)$ side by side
Feature	$\dim_{k} V = n < \infty$	$\dim_{k} V$ infinite
Image of $ρ$	all of $End (V_{k})$	proper dense subring
Chain conditions	left and right artinian	neither left nor right artinian
Ring structure	$R ≅ M_{n} (k)$ , simple	need not be simple; need not be noetherian
Matrix rings inside $R$	$R$ itself	subquotients $R_{n} / 𝔄_{n} ≅ M_{n} (k)$ for all $n$
Simple modules	one isomorphism class	possibly several, even among faithful ones
Right primitivity	automatic	not automatic

Which implications hold in which class of rings
	General ring	Has a minimal left ideal	Left artinian	Commutative
left primitive $\Rightarrow$ prime	yes	yes	yes	yes
prime $\Rightarrow$ left primitive	no	yes	yes	no
left primitive $\Rightarrow$ simple	no	no	yes	yes
left primitive $\Rightarrow$ right primitive	no	yes	yes	yes

Which implications hold in which class of rings

Relationship Map

All ringsno structure assumed

Semiprimeno nonzero nilpotent ideals

Prime

𝔄 𝔅 = 0 \Rightarrow 𝔄 = 0

𝔅 = 0

Left primitivedense ring of linear transformations

(11.19)

Simpleno proper nonzero two-sided ideals

Simple left artinian

R ≅ M_{n} (k)

k

a division ring

Each containment is strict in general, and each becomes an equality once the descending chain condition on left ideals is imposed — that collapse is the content of $(11.7)$ and the reason Wedderburn–Artin looks so much stronger than it is.

Density Theorem

(11.16)

(11.17)

: f.g. over

k

gives surjectivity

(11.19)

: structure of left primitive ringsWedderburn–Artin for simple artinian rings

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Which faithful simple module? For a socle-free left primitive ring the choice is real: different $V$ give different concrete realisations of $R$ as transformations. Fix $V$ once and record it, as you would fix a basis.
Which side? Primitivity is not left-right symmetric, so a theorem proved for left primitive rings does not transfer by opposite-ring formalities unless the ring is known to be right primitive too. Rings with a minimal one-sided ideal are the safe case.
Artinian or not? Assuming left artinian is not a technical convenience here: it changes the conclusion from dense to equal. If you only need matrix behaviour on finitely many vectors, do not assume it.
Which invariant to carry? $\dim_{k} V$ is the useful numerical invariant in the artinian branch. Outside it, the family of matrix sections $R_{n} / 𝔄_{n}$ carries the information instead, and it depends on the chosen independent sequence.

Failure Modes and Common Mistakes

Do not assume the division ring $k$ is central in $R$ or commutative; it is $End (_{R} V)$ and can be any division ring.
Do not deduce right artinian from left artinian in general; here it happens to follow because the conclusion $M_{n} (k)$ is artinian on both sides, not because the hypothesis is symmetric.
Do not conclude simplicity from left primitivity. $End (V_{k})$ with $\dim_{k} V$ infinite is left primitive and has a proper nonzero ideal — the finite-rank transformations.
Do not expect the Wedderburn–Artin uniqueness statement to survive into the infinite-dimensional branch: the pair $(V, k)$ need not be unique when the socle vanishes.

Quick Reference

Hypotheses

R

left primitive,

V

faithful simple,

k = End (_{R} V)

Always

R ↪ End (V_{k})

as a dense ring of transformations

Dichotomy

R

left artinian

\Leftrightarrow

\dim_{k} V < \infty

Artinian branch

R ≅ M_{n} (k)

with

n = \dim_{k} V

Non-artinian branch

R_{n} / 𝔄_{n} ≅ M_{n} (k)

for every

n \geq 1

Chain

𝔄_{1} ⊋ 𝔄_{2} ⊋ \dots

left ideals

Density test

2

-transitive

\Leftrightarrow

dense

(11.20)

Centre

Z (R)

is a domain and can be any domain

(11.21)

Which result to quote
Goal	Quote	Hypotheses
Realise $R$ as transformations	$(11.19)$ , first assertion	$R$ left primitive, $V$ faithful simple
Get $R ≅ M_{n} (k)$	$(11.19) (1)$	additionally $R$ left artinian
Find matrix sections	$(11.19) (2)$	$\dim_{k} V$ infinite
Certify density cheaply	$(11.20) (2)$	$R \subseteq End (V_{k})$ , $2$ -transitive
Prescribe the centre	$(11.21)$	$A$ a domain with quotient field $k$

Frequently Asked Questions

Does this really prove Wedderburn–Artin, or does it assume it?

It proves it, for simple left artinian rings, and the argument is logically independent of the socle-decomposition proof in §3. The only inputs are Schur's Lemma, the Density Theorem and the observation that a simple ring is left primitive. The extension to general semisimple rings still needs the decomposition into finitely many simple components.

Why does a strictly descending chain of the ideals

𝔄_{n}

exist as soon as

\dim_{k} V

is infinite?

Infinite dimension supplies independent vectors $v_{1}, v_{2}, \dots$ without end. For each $n$ , $(n + 1)$ -transitivity produces a ring element annihilating $v_{1}, \dots, v_{n}$ while moving $v_{n + 1}$ ; that element lies in $𝔄_{n}$ but not in $𝔄_{n + 1}$ . Transitivity is exactly the tool that manufactures separating elements.

Is a left primitive ring determined by the pair

(V, k)

Only in the artinian branch, where $R$ is the full endomorphism ring $M_{n} (k)$ . In general $R$ is merely some dense subring, and many non-isomorphic dense subrings of the same $End (V_{k})$ exist — the full ring, scalars plus finite rank, and Kaplansky's centre-prescribing rings all live inside the same $End (V_{k})$ .

Why is checking

2

-transitivity enough to get density?

Because $2$ -transitivity forces $End (_{R} V)$ to be exactly $k$ , and once the endomorphism ring is correct the Density Theorem applies and delivers $m$ -transitivity for all $m$ . This is $(11.20) (2)$ , and it is the practical test: verify one condition on pairs of vectors rather than an infinite family of conditions.

Can a left primitive ring be commutative?

Only if it is a field. A commutative left primitive ring has a faithful simple module $R / 𝔪$ , and $𝔪$ annihilates it, so faithfulness forces $𝔪 = 0$ . Consistently, $(11.19)$ then gives $\dim_{k} V = 1$ and $R ≅ M_{1} (k) = k$ .

Does the theorem give a canonical embedding of

R

into a matrix ring of infinite size?

It gives an embedding into $End (V_{k})$ , which after choosing a basis of $V_{k}$ becomes a ring of column-finite infinite matrices. The embedding depends on both the module $V$ and the basis, so it is a useful coordinate system rather than a canonical form.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §11, (11.19)–(11.21) (pp. 194–196).
T. Y. Lam, A First Course in Noncommutative Rings, §3, for the classical proof of the Wedderburn–Artin Theorem (pp. 33–37).
N. Jacobson, “Structure theory of simple rings without finiteness assumptions”, Transactions of the American Mathematical Society 57 (1945), 228–245.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
I. N. Herstein, Noncommutative Rings, Carus Mathematical Monographs 15, Mathematical Association of America, 1968, Chapter 2.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Show directly that $End (V_{k})$ with $\dim_{k} V$ infinite is left primitive but not simple, and identify all its two-sided ideals.
Work out the matrix sections $R_{n} / 𝔄_{n}$ for the Weyl algebra acting on $k [y]$ .
Which dense subrings of $End (V_{k})$ are simple, and how does the socle detect that?
Give a left primitive ring with two non-isomorphic faithful simple left modules and compare the resulting division rings.
How does $(11.19)$ interact with Morita equivalence — are dense subrings of $End (V_{k})$ Morita equivalent to $k$ ?
State and prove the right-handed version of $(11.19)$ and explain why it is not a formal consequence of the left one.
Use Kaplansky's construction to realise $ℤ [t]$ as the centre of a left primitive ring.

Wedderburn–Artin via Density

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

From faithful simple module to a ring of matrices

Why finite dimension collapses the theorem

Why infinite dimension destroys the chain condition

Key Results

Proof Techniques and Method

Faithful plus simple equals concrete

Transitivity manufactures elements

Stabiliser subring, annihilator ideal

Worked Example

Finite-dimensional check

Infinite-dimensional: scalars plus finite rank

Frameworks and Models

Process and Workflow

Comparison and Classification

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

From faithful simple module to a ring of matrices

Why finite dimension collapses the theorem

Why infinite dimension destroys the chain condition

06Key Results

07Proof Techniques and Method

Faithful plus simple equals concrete

Transitivity manufactures elements

Stabiliser subring, annihilator ideal

08Worked Example

Finite-dimensional check

Infinite-dimensional: scalars plus finite rank

09Frameworks and Models

10Process and Workflow

11Comparison and Classification

12Relationship Map

13Design Considerations

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Frameworks and Models

Process and Workflow

Comparison and Classification

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions