The Jacobson Density Theorem

Q: Why is the theorem stated for semisimple modules when the interesting case is a simple module?

Because the proof of the simple case runs through a module that is not simple. Bourbaki's argument replaces n vectors in V by one vector in V^n, and V^n is semisimple rather than simple. Stating (11.16) for semisimple modules costs nothing and makes the proof self-contained.

Q: Does density say the image of R is all of End(V_k)?

No. It says the image is dense in the finite topology: indistinguishable from the whole ring by any finite set of test vectors. Equality holds precisely when V is finitely generated as a right k-module, which is (11.17). The first Weyl algebra acting on k_0[y] is dense but very far from surjective.

Q: What goes wrong if I use a smaller division ring in place of End(_R V)?

Everything. Shrinking k enlarges End(V_k), so the target of the density statement grows while the image of R stays put. The field C acting on itself as a 2-dimensional real vector space is the standard warning: simple as a module, but a 2-dimensional subring of M_2(R), hence not dense over R.

Q: Is there a right-handed version of the theorem?

Yes, and it is the same theorem read in R^op. A semisimple right R-module V with k = End(V_R) acting on the left makes R act densely on _k V. What is genuinely asymmetric is not density but primitivity: a ring can have a faithful simple left module and no faithful simple right module.

Q: Where exactly is semisimplicity used in the proof?

In exactly one line of (11.15): to produce a complement to the submodule W, so that the projection onto W is an R-endomorphism and therefore an element of k. Elements of End(V_k) commute with k by definition, so they preserve W = Ve. No other step uses the hypothesis.

Q: Does the theorem need R to have an identity?

The proof as given uses it once, to conclude that (v_1, …, v_n) lies in the cyclic submodule R (v_1, …, v_n). For rings without identity the statement is repaired by working with modules that are unital in the appropriate sense, but every result in this collection assumes an identity.

Executive Summary

Let $V$ be a semisimple left $R$ -module and let $k = End (_{R} V)$ act on $V$ from the right, so $V$ is an $(R, k)$ -bimodule. The Density Theorem says the image of $R$ in $E = End (V_{k})$ is large: given any $f \in E$ and any finitely many $v_{1}, \dots, v_{n} \in V$ , some single $r \in R$ satisfies $r v_{i} = f (v_{i})$ for all $i$ .

This is a finite-interpolation statement, not a surjectivity statement. The image of $R$ need not be all of $E$ — but it is dense in $E$ for the finite topology, and it is all of $E$ once $V$ is finitely generated over $k$ . That last clause is the entire content of Wedderburn–Artin for left artinian simple rings.

1945Jacobson

SemisimpleHypothesis on

V

$V^{n}$ Bourbaki's trick

$End (V_{k})$ Target ring

Overview

Wedderburn–Artin classifies semisimple rings by declaring them to be finite products of matrix rings over division rings. That statement collapses without a chain condition. The Density Theorem is the replacement: it keeps the conclusion locally — on every finite set of vectors the ring behaves like a full matrix ring — and pays for the loss of chain conditions by weakening equality to approximation.

The setting is a bimodule $_{R} V_{k}$ . Two rings act on the same abelian group from opposite sides, and each is constrained by the other. Writing $ρ : R \to End (V_{k})$ for the natural map, the theorem measures how much of the right-hand centraliser the left-hand ring can see.

\forall f \in End (V_{k}), \forall v_{1}, \dots, v_{n} \in V, \exists r \in R : r v_{i} = f (v_{i}) (1 \leq i \leq n)

(11.16)

Density: finitely many prescribed values can always be realised by a single ring element.

When $V$ is simple, Schur's Lemma makes $k$ a division ring, so $V_{k}$ is a right vector space and density becomes a statement of linear algebra: the image of $R$ is $m$ -transitive for every $m$ . That specialisation is what powers Primitive Rings and Primitive Ideals and the structure theorem discussed in Deriving Wedderburn–Artin from the Density Theorem.

Learning Objectives

Set up the bimodule $_{R} V_{k}$ with $k = End (_{R} V)$ acting on the right.
State $(11.16)$ with full hypotheses and say exactly where semisimplicity of $V$ enters.
Prove the stabilisation lemma $(11.15)$ using a projection as an element of $k$ .
Reproduce the Bourbaki proof by transporting the problem to $V^{n}$ over $M_{n} (k)$ .
Deduce $(11.17)$ : $ρ$ is onto when $V_{k}$ is finitely generated.
Interpret density as topological density in the finite topology on $End (V_{k})$ .

Definitions

Definition(11.16)Dense action on a bimodule

Let $R$ and $k$ be rings and let $V$ be an $(R, k)$ -bimodule; write $E = End (V_{k})$ for the ring of endomorphisms of $V$ as a right $k$ -module, acting on the left. Then $R$ acts densely on $V_{k}$ if for every $f \in E$ and every finite list $v_{1}, \dots, v_{n} \in V$ there exists $r \in R$ with $r v_{i} = f (v_{i})$ for $i = 1, \dots, n$ .

$k = End (_{R} V)$: The ring of $R$ -module endomorphisms of $V$ , written on the right of $V$ so that composition matches multiplication and $V$ becomes an $(R, k)$ -bimodule.
$E = End (V_{k})$: The ring of additive maps $f$ on $V$ with $f (v a) = f (v) a$ for all $a \in k$ ; it contains the image of $R$ and is the double centraliser of $R$ on $V$ .
$ρ : R \to E$: The natural map $ρ (r) (v) = r v$ . Its kernel is $ann (V)$ , so $ρ$ is injective exactly when $V$ is faithful.
Finite topology on $E$: Basic open sets are ${g \in E : g (u_{i}) = w_{i}, 1 \leq i \leq n}$ for finite families of vectors; a subring is dense in this topology precisely when it acts densely.
$m$ -transitive: For $S \subseteq End (V_{k})$ with $k$ a division ring: any $n \leq m$ linearly independent vectors can be sent to any prescribed $n$ vectors by a single element of $S$ .

Rings have identity and modules are unital. Semisimple module means a direct sum of simple submodules; simple modules are nonzero.

Core Concepts

Two centralisers, one abelian group

Fix a left $R$ -module $V$ and regard $V$ merely as an abelian group with endomorphism ring $End (V)$ . The image $ρ (R)$ sits inside it; $k = End (_{R} V)$ is precisely the centraliser of $ρ (R)$ ; and $E = End (V_{k})$ is the centraliser of $k$ . So $E$ is the double centraliser of $R$ , and always $ρ (R) \subseteq E$ .

ρ (R)

k = End (_{R} V)

E = End (V_{k})

Density says that for semisimple $V$ the inclusion $ρ (R) \subseteq E$ is as tight as it can be without being equality: no element of $E$ is distinguishable from an element of $ρ (R)$ by finitely many test vectors. The bimodule viewpoint is developed further in Double Centralizers and Density for Bimodules.

Why the topology is the right language

Give $End (V_{k})$ the finite topology, in which a map is near $f$ when it agrees with $f$ on a prescribed finite set. Two maps that agree on every finite set agree everywhere, so the topology is Hausdorff, and the definition of dense action is exactly the statement that $ρ (R)$ has closure $E$ . Density is therefore genuinely topological, not merely a figure of speech.

Semisimplicity is what makes submodules complemented

The proof needs one thing from the hypothesis on $V$ : every $R$ -submodule is a direct summand. A complement gives a projection, the projection is an $R$ -endomorphism, hence an element of $k$ , and elements of $E$ commute with $k$ by definition. That short chain is the whole mechanism — everything else in the proof is bookkeeping to reduce the general case to it.

Key Results

Lemma(11.15)Submodules are stable under the double centraliser

Let $V$ be a semisimple left $R$ -module, $k = End (_{R} V)$ and $E = End (V_{k})$ . Then every $R$ -submodule $W \subseteq V$ satisfies $f (W) \subseteq W$ for all $f \in E$ ; that is, $W$ is an $E$ -submodule. Conversely every $E$ -submodule is an $R$ -submodule, since $ρ (R) \subseteq E$ .

Proof

By semisimplicity choose an $R$ -submodule $W'$ with $V = W \oplus W'$ , and let $e : V \to V$ be the projection onto $W$ along $W'$ . Since $W$ and $W'$ are $R$ -submodules, $e$ is an $R$ -endomorphism, so $e \in k$ ; writing $k$ on the right, $W = V e$ . Now take $f \in E$ . Right $k$ -linearity gives

f (W) = f (V e) = f (V) e \subseteq V e = W .

For the converse, each $ρ (r)$ lies in $E$ because $r (v a) = (r v) a$ for $a \in k$ , by definition of $k$ as a ring of $R$ -endomorphisms.

Theorem(11.16)Density Theorem (Jacobson, Chevalley)

Let $R$ be a ring and let $V$ be a semisimple left $R$ -module. Put $k = End (_{R} V)$ , so that $V$ is an $(R, k)$ -bimodule. Then $R$ acts densely on $V_{k}$ : for every $f \in End (V_{k})$ and every finite list $v_{1}, \dots, v_{n} \in V$ there is an $r \in R$ with $r v_{i} = f (v_{i})$ for all $i$ .

Proof

Fix $f$ and $v_{1}, \dots, v_{n}$ . Work with $V^{n}$ , a finite direct sum of semisimple modules, hence semisimple. Its $R$ -endomorphism ring is a matrix ring over $k$ :

\tilde{k} : = End (_{R} V^{n}) ≅ M_{n} (k),

acting on the right of $V^{n}$ by $(w ε)_{j} = \sum_{i = 1}^{n} w_{i} ε_{i j}$ for $w = (w_{1}, \dots, w_{n})$ and $ε = (ε_{i j}) \in M_{n} (k)$ .

**Step 1: lift $f$ to $V^{n}$ .** Define $\tilde{f} (w_{1}, \dots, w_{n}) = (f (w_{1}), \dots, f (w_{n}))$ . Then $\tilde{f} \in End ((V^{n})_{\tilde{k}})$ , because for each coordinate $j$ ,

\tilde{f} (w ε)_{j} = f (\sum_{i} w_{i} ε_{i j}) = \sum_{i} f (w_{i}) ε_{i j} = (\tilde{f} (w) ε)_{j},

using only that $f$ is additive and right $k$ -linear.

Step 2: apply the Lemma. Let $W = R \cdot (v_{1}, \dots, v_{n})$ , the cyclic $R$ -submodule of $V^{n}$ generated by the given vector. By $(11.15)$ applied to the semisimple module $V^{n}$ , $W$ is stable under $End ((V^{n})_{\tilde{k}})$ , hence under $\tilde{f}$ .

Step 3: read off the conclusion. Since $R$ has an identity, $(v_{1}, \dots, v_{n}) \in W$ , so $\tilde{f} (v_{1}, \dots, v_{n}) = (f (v_{1}), \dots, f (v_{n}))$ lies in $W = R \cdot (v_{1}, \dots, v_{n})$ . Thus there exists $r \in R$ with $r v_{i} = f (v_{i})$ for every $i$ .

Corollary(11.17)Finitely generated over

k

forces surjectivity

With $R$ , $V$ , $k$ , $E$ as in $(11.16)$ : if $V$ is finitely generated as a right $k$ -module, then the natural map $ρ : R \to E = End (V_{k})$ is onto. If moreover $V$ is faithful, $ρ$ is an isomorphism, so $R ≅ End (V_{k})$ .

Proof

Let $v_{1}, \dots, v_{n}$ generate $V$ as a right $k$ -module and let $f \in E$ . By $(11.16)$ pick $r \in R$ with $r v_{i} = f (v_{i})$ for all $i$ . Any $v \in V$ can be written $v = \sum_{i} v_{i} a_{i}$ with $a_{i} \in k$ , and both $ρ (r)$ and $f$ are right $k$ -linear, so

r v = \sum_{i} (r v_{i}) a_{i} = \sum_{i} f (v_{i}) a_{i} = f (\sum_{i} v_{i} a_{i}) = f (v) .

Hence $ρ (r) = f$ . The last sentence follows since $\ker ρ = ann (V)$ .

Corollary—Simple modules of finite

k

-dimension

Let $V$ be a simple left $R$ -module and $k = End (_{R} V)$ , a division ring by Schur's Lemma. If $n : = \dim_{k} V$ is finite, then $R / ann (V) ≅ M_{n} (k)$ . If $\dim_{k} V$ is infinite, $ρ (R)$ is a proper dense subring of $End (V_{k})$ .

Proof

A simple module is semisimple, so $(11.16)$ applies. If $\dim_{k} V = n < \infty$ then $V_{k}$ is finitely generated and $(11.17)$ makes $ρ$ onto; its kernel is $ann (V)$ and $End (V_{k}) ≅ M_{n} (k)$ for an $n$ -dimensional right $k$ -space. The infinite-dimensional case is treated in the structure theorem $(11.19)$ : there $ρ (R)$ cannot be all of $End (V_{k})$ because a surjection would force $R$ to be left artinian while the transitivity of the action produces a strictly descending chain of left ideals.

Remark—Faithfulness is not a hypothesis

$(11.16)$ says nothing about $\ker ρ$ . Applied to an arbitrary simple module it describes the primitive quotient $R / ann (V)$ rather than $R$ itself, which is exactly why left primitive ideals — treated in Primitive Rings and Primitive Ideals — are the right objects to carry the theory.

Proof Techniques and Method

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

Move 1

A complement is an element of $k$

Semisimplicity converts a submodule into an idempotent $e \in End (_{R} V)$ with $W = V e$ . Anything commuting with $k$ then preserves $W$ automatically.

Move 2

Diagonal embedding kills the quantifier

A statement about $n$ vectors in $V$ becomes a statement about one vector in $V^{n}$ . The cost is that $V^{n}$ is only semisimple — which is why the theorem is stated at that generality.

Move 3

Cyclic submodule as a certificate

Membership of $\tilde{f} (v)$ in $R \cdot v$ is the existence of the required $r$ . Turning an existence claim into a submodule containment is the reusable trick.

Move 2 is due to Bourbaki and replaces Jacobson's original induction on the number of vectors. The older argument proves $m$ -transitivity by induction, and is worth knowing because it exposes what fails without semisimplicity; the Bourbaki form is shorter and coordinate-free.

A fourth habit is worth naming: always check on which side each ring acts. The proof uses $f (v a) = f (v) a$ and $r (v a) = (r v) a$ in consecutive lines, and both are definitional only if $k$ is written on the right.

Worked Example

The Weyl algebra acting on polynomials

Let $k_{0}$ be a field of characteristic $0$ and let $A_{1} = k_{0} ⟨ x, y ⟩ / (x y - y x - 1)$ be the first Weyl algebra. Let $V = k_{0} [y]$ , with $y$ acting by multiplication and $x$ acting by formal differentiation $d / d y$ . The relation holds: for $p \in V$ ,

(x y - y x) p = \frac{d}{d y} (y p) - y \frac{d p}{d y} = p + y \frac{d p}{d y} - y \frac{d p}{d y} = p .

(E.1)

So $V$ really is a left $A_{1}$ -module.

Step 1 — $V$ is simple

Let $0 \neq N \subseteq V$ be a submodule and pick $0 \neq p \in N$ of degree $d$ with leading coefficient $c$ . Then $x^{d} p = d! c \neq 0$ in characteristic $0$ , so $N$ contains a nonzero constant, hence $1$ , hence every $y^{m}$ . Thus $N = V$ .

Step 2 — the endomorphism ring is $k_{0}$

Let $ϕ \in End (_{A_{1}} V)$ and set $q = ϕ (1)$ . Commuting with multiplication by $y$ gives $ϕ (p) = p q$ for every $p$ . Commuting with $x$ gives $0 = ϕ (x \cdot 1) = x \cdot q = d q / d y$ , so $q \in k_{0}$ by characteristic $0$ . Hence $k : = End (_{A_{1}} V) = k_{0}$ and $V_{k} = k_{0} [y]$ is an infinite-dimensional $k_{0}$ -space.

Step 3 — read the Density Theorem

$(11.16)$ now says: for any $k_{0}$ -linearly independent polynomials $p_{1}, \dots, p_{n}$ and any prescribed $q_{1}, \dots, q_{n} \in k_{0} [y]$ , there is a single differential operator $T \in A_{1}$ with $T p_{i} = q_{i}$ for all $i$ . Take $p_{1} = 1$ , $p_{2} = y$ and the operators

u = 1 - y x, w = x in A_{1} .

(E.2)

Then $u \cdot 1 = 1 - y \frac{d}{d y} (1) = 1$ and $u \cdot y = y - y \cdot 1 = 0$ , while $w \cdot 1 = 0$ and $w \cdot y = 1$ . So $u, w$ behave as a dual basis for ${1, y}$ , and for prescribed targets $a, b \in k_{0} [y]$ the operator

T = a (1 - y x) + b x satisfies T \cdot 1 = a, T \cdot y = b .

(E.3)

Here $a$ and $b$ denote multiplication operators, i.e. the corresponding polynomials in $y$ inside $A_{1}$ .

Concretely with $a = y^{2}$ and $b = 0$ : $T = y^{2} - y^{3} x$ , and indeed $T \cdot 1 = y^{2}$ while $T \cdot y = y^{3} - y^{3} = 0$ .

Process and Workflow

Check semisimplicityConfirm

V

is a sum of simple submodules. If

V

is simple this is free; if

V =_{R} R

it means

R

is a semisimple ring.

Compute $k = End (_{R} V)$ Do not guess it. For simple

V

it is a division ring, but identifying which one is the real work — see the Weyl algebra above.

Decide finite or infinite $\dim_{k} V$ Finite gives the sharp conclusion

R / ann (V) ≅ M_{n} (k)

via

(11.17)

; infinite gives only density.

Use density on a chosen finite setChoose test vectors adapted to the problem, produce the interpolating

r

, and draw the conclusion before the choice is forgotten.

Step 4 is where density earns its keep in practice: most applications choose two or three vectors, obtain one ring element, and derive a contradiction or a construction from it. Lam's exercises on rings satisfying $a (a b - b a) = (a b - b a) a$ and on $1 + a^{2}$ being a unit both run exactly this way.

Comparison and Classification

What the theorem gives, by hypothesis
Module $V$	$k = End (_{R} V)$	Conclusion	Is $ρ$ onto?
Simple, $\dim_{k} V = n < \infty$	division ring	$m$ -transitive for all $m$	yes: $R / ann (V) ≅ M_{n} (k)$
Simple, $\dim_{k} V$ infinite	division ring	$m$ -transitive for all $m$	no — proper dense subring
Semisimple, $V_{k}$ finitely generated	any ring	finite interpolation	yes, by $(11.17)$
Semisimple, $V_{k}$ not finitely generated	any ring	finite interpolation	not in general
Not semisimple	any ring	no conclusion	no — see the triangular example below

Which hypothesis each conclusion actually needs
	$V$ semisimple	$V$ simple	$V_{k}$ f.g.	$V$ faithful
Submodules are $E$ -stable $(11.15)$	yes	no	no	no
Density $(11.16)$	yes	no	no	no
$k$ is a division ring	no	yes	no	no
$ρ$ onto $(11.17)$	yes	no	yes	no
$R ≅ End (V_{k})$	yes	no	yes	yes

Which hypothesis each conclusion actually needs

Relationship Map

Density sits between Schur's Lemma, which supplies the division ring, and the structure theorem for left primitive rings, which consumes it.

Schur's Lemma

k = End (_{R} V)

a division ringDensity Theorem

(11.16)

Structure of left primitive rings

(11.19)

Wedderburn–Artin for simple left artinian rings

Density Theorem $(11.16)$ — semisimple $V$ , $k = End (_{R} V)$
- specialises to
  - Burnside's theorem: $k$ algebraically closed, $\dim_{k} V < \infty$ , $R$ acting irreducibly $\Rightarrow$ $ρ (R) = {End}_{k} (V)$
  - Wedderburn–Artin for simple left artinian rings
  - Artin–Whaples theorem for simple rings with centre $k$
- requires
  - semisimplicity of $V$ (used only through complements)
  - $k$ taken as the full endomorphism ring
  - an identity in $R$ , so that $v \in R \cdot v$
- does not require
  - any chain condition on $R$
  - faithfulness of $V$
  - commutativity or finite dimension anywhere

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Representation theory

Burnside and absolute irreducibility

If $k$ is an algebraically closed field, $V$ a finite-dimensional $k$ -space and $R \subseteq {End}_{k} (V)$ a $k$ -subalgebra acting irreducibly, then $End (_{R} V) = k$ and density forces $R = {End}_{k} (V)$ . This is the standard test for absolute irreducibility of a representation.

Computational algebra

MeatAxe and module recognition

Algorithms that decide irreducibility of a matrix module over a finite field work by spinning up vectors and computing the endomorphism algebra; density is the theoretical guarantee that an irreducible module of endomorphism ring $k$ is faithfully described by the full matrix algebra over $k$ .

Operator algebras

Bicommutant analogy

Von Neumann's bicommutant theorem is the same shape: a self-adjoint algebra of operators is dense in its double commutant for a weak topology. The algebraic theorem is the purely ring-theoretic skeleton of that statement.

Symbolic computation

Differential operators

For the Weyl algebra acting on polynomials, density says any finite interpolation problem for linear maps is solvable by a differential operator — the algebraic backdrop to D-module implementations in computer algebra systems.

The honest summary: density is infrastructure inside algebra. Its downstream users are structure theorems, representation-theoretic tests and the algorithms built on them, not applications outside mathematics.

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

In the finite-dimensional setting the theorem is effective. Let $A \subseteq M_{n} (F)$ be the algebra generated by $t$ given matrices over a field $F$ , acting on $V = F^{n}$ .

Irreducibility. Spinning up a vector — repeatedly applying generators and reducing — decides whether the cyclic submodule is all of $V$ in $O (t n^{3})$ field operations, and Norton's criterion turns one such computation into a full irreducibility test.
The endomorphism ring. $End (_{A} V)$ is the solution space of the linear system $g_{i} X = X g_{i}$ over the generators: $t n^{2}$ equations in $n^{2}$ unknowns, solvable in $O (t n^{6})$ naively and much faster in practice using a spun-up basis.
Realising an interpolation. Given $v_{1}, \dots, v_{m}$ and targets, the element $r$ promised by density is found by linear algebra inside the spanning set of $A$ ; the cost is dominated by computing a basis of $A$ , which can have dimension up to $n^{2}$ .
Infinite dimension. No general algorithm exists: for an arbitrary finitely presented ring even deciding whether a given module is simple is undecidable, so density is a theoretical guarantee only.

Failure Modes and Common Mistakes

Do not read $m$ -transitivity as saying the $n$ vectors may be arbitrary: for $m$ -transitivity the source vectors must be linearly independent over $k$ . The target vectors are unrestricted.
Do not assume the interpolating $r$ is unique — it is unique modulo the left ideal annihilating the chosen vectors, which is usually large.
Do not confuse $End (_{R} V)$ with $End (V_{k})$ ; they act on opposite sides and are almost never isomorphic.
Do not apply the theorem to a module you have only shown to be indecomposable. Indecomposable is not semisimple.

Historical Notes and Lessons Learned

1905Burnside's irreducibility theoremBurnside proves that an irreducible algebra of matrices over an algebraically closed field is the full matrix algebra — the finite-dimensional shadow of density.
1929Von Neumann's bicommutant theoremIn operator algebras, a self-adjoint unital algebra of bounded operators is shown to be dense in its double commutant. The formal analogy with the algebraic statement is exact in shape if not in proof.
1945Jacobson's density theoremJacobson removes all finiteness assumptions, proving that a ring acting faithfully and irreducibly on a module acts densely relative to the endomorphism division ring, and builds the structure theory of primitive rings on it.
1954Chevalley's versionChevalley gives an independent treatment in his work on spinors, which is why the result carries both names.
1950sBourbaki's proofThe reduction to a cyclic submodule of $V^{n}$ over $M_{n} (k)$ replaces the original induction on the number of vectors and is the version reproduced in most modern texts.

The methodological lesson is the same one the Jacobson radical teaches: define and prove things through the action on modules rather than through internal features of the ring. Chain conditions then become optional extras that sharpen a conclusion instead of prerequisites that make it possible.

Quick Reference

Setting

_{R} V_{k}

with

k = End (_{R} V)

E = End (V_{k})

Hypothesis

_{R} V

semisimple — nothing else

Conclusion

\forall f \in E

\forall v_{1}, \dots, v_{n}

\exists r

r v_{i} = f (v_{i})

Topology

ρ (R)

is dense in

E

for the finite topology

Simple case

k

a division ring; density

=

m

-transitivity for all

m

Surjectivity

V_{k}

finitely generated

\Rightarrow

ρ

onto

(11.17)

Faithful case

V

faithful and

\dim_{k} V = n < \infty

\Rightarrow

R ≅ M_{n} (k)

Key lemma

(11.15)

R

-submodules of a semisimple

V

are

E

-submodules

Statement to quote, by situation
You need	Quote	Hypotheses to verify
Finite interpolation	$(11.16)$	$_{R} V$ semisimple; $k$ is the full $End (_{R} V)$
A surjection onto $End (V_{k})$	$(11.17)$	as above, plus $V_{k}$ finitely generated
$R ≅ M_{n} (k)$	$(11.17)$ + faithfulness	$V$ faithful simple, $\dim_{k} V = n < \infty$
Stability of submodules	$(11.15)$	$_{R} V$ semisimple
$k$ is a division ring	Schur's Lemma	$V$ simple

Frequently Asked Questions

Why is the theorem stated for semisimple modules when the interesting case is a simple module?

Because the proof of the simple case runs through a module that is not simple. Bourbaki's argument replaces $n$ vectors in $V$ by one vector in $V^{n}$ , and $V^{n}$ is semisimple rather than simple. Stating $(11.16)$ for semisimple modules costs nothing and makes the proof self-contained.

Does density say the image of

R

is all of

End (V_{k})

No. It says the image is dense in the finite topology: indistinguishable from the whole ring by any finite set of test vectors. Equality holds precisely when $V$ is finitely generated as a right $k$ -module, which is $(11.17)$ . The first Weyl algebra acting on $k_{0} [y]$ is dense but very far from surjective.

What goes wrong if I use a smaller division ring in place of

End (_{R} V)

Everything. Shrinking $k$ enlarges $End (V_{k})$ , so the target of the density statement grows while the image of $R$ stays put. The field $ℂ$ acting on itself as a $2$ -dimensional real vector space is the standard warning: simple as a module, but a $2$ -dimensional subring of $M_{2} (ℝ)$ , hence not dense over $ℝ$ .

Is there a right-handed version of the theorem?

Yes, and it is the same theorem read in $R^{op}$ . A semisimple right $R$ -module $V$ with $k = End (V_{R})$ acting on the left makes $R$ act densely on $_{k} V$ . What is genuinely asymmetric is not density but primitivity: a ring can have a faithful simple left module and no faithful simple right module.

Where exactly is semisimplicity used in the proof?

In exactly one line of $(11.15)$ : to produce a complement to the submodule $W$ , so that the projection onto $W$ is an $R$ -endomorphism and therefore an element of $k$ . Elements of $End (V_{k})$ commute with $k$ by definition, so they preserve $W = V e$ . No other step uses the hypothesis.

Does the theorem need

R

to have an identity?

The proof as given uses it once, to conclude that $(v_{1}, \dots, v_{n})$ lies in the cyclic submodule $R \cdot (v_{1}, \dots, v_{n})$ . For rings without identity the statement is repaired by working with modules that are unital in the appropriate sense, but every result in this collection assumes an identity.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §11, especially (11.15)–(11.17) (pp. 191–194).
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.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Write out Jacobson's original induction proof of $m$ -transitivity and compare it with the Bourbaki argument.
Show that $End (V_{k})$ is complete in the finite topology and identify the closure of an arbitrary subring.
Derive Burnside's theorem on irreducible matrix algebras from the Density Theorem, with all hypotheses.
Give an example of a simple module whose endomorphism division ring is noncommutative, and compute it.
How does the Density Theorem specialise when $R$ is a group algebra $k G$ and $V$ an irreducible representation?
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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two centralisers, one abelian group

Why the topology is the right language

Semisimplicity is what makes submodules complemented

06Key Results

07Proof Techniques and Method

A complement is an element of k

Diagonal embedding kills the quantifier

Cyclic submodule as a certificate

08Worked Example

The Weyl algebra acting on polynomials

Step 1 — V is simple

Step 2 — the endomorphism ring is k0

Step 3 — read the Density Theorem

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Burnside and absolute irreducibility

MeatAxe and module recognition

Bicommutant analogy

Differential operators

13Computational Notes

14Failure Modes and Common Mistakes

15Historical Notes and Lessons Learned

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

A complement is an element of $k$

Worked Example

Step 1 — $V$ is simple

Step 2 — the endomorphism ring is $k_{0}$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions