Why is the second branch of Brauer's Lemma stated as A = Re rather than just *contains an idempotent*?

Because containing an idempotent is not enough for the applications. The point is that a single idempotent e ∈ A generates all of A, which is what makes A a direct summand of _R R via R = Re ⊕ R(1-e). Minimality upgrades containment to equality for free: Re is a nonzero left ideal inside a minimal one.

Minimal Left Ideals in Semiprime Rings

Q: Does (10.24) really improve on the Jacobson-radical version?

Yes, strictly. Semiprimeness is weaker than rad R = 0, since Nil_* R ⊆ rad R can be a strict inclusion — Z_(p) is semiprime with nonzero Jacobson radical. So (10.24) concludes semisimplicity from less. Under the chain condition the two hypotheses turn out to be equivalent, but that is a consequence of the theorem, not an assumption.

Q: What replaces Brauer's Lemma for minimal right ideals?

The mirror statement, proved by the same argument in R^op: a minimal right ideal A satisfies A^2 = 0 or A = eR for an idempotent e ∈ A. Since semiprimeness is left-right symmetric, the semiprime corollary is symmetric too, and in that case the left and right socles agree.

Executive Summary

A minimal left ideal is the smallest interesting piece of a ring, and there are exactly two kinds. Brauer's Lemma says a minimal left ideal $𝔄$ either satisfies $𝔄^{2} = 0$ — it is a square-zero fragment carrying no idempotent — or it is $R e$ for an idempotent $e \in 𝔄$ , in which case it is a direct summand of the ring and a projective simple module.

Semiprimeness is exactly the hypothesis that forbids the first case, since a square-zero left ideal is a nilpotent left ideal. That single deletion is enough to rerun the classical argument for semisimplicity with semiprime in place of Jacobson semisimple, giving $(10.24)$ : semisimple $=$ semiprime $+$ DCC on principal left ideals.

2Cases in Brauer's Lemma

$𝔄 = R e$ Semiprime case

$e R e$ Is a division ring

(10.24)Semisimplicity criterion

Overview

Minimal left ideals are the simple submodules of $_{R} R$ . Their sum is the left socle, and a ring is semisimple precisely when the socle is everything. The obstruction to that is not the existence of few minimal left ideals but the existence of bad ones: minimal left ideals that are not summands cannot be assembled into a decomposition of $R$ .

𝔄 minimal left ideal ⟹ 𝔄^{2} = 0 or 𝔄 = R e with e^{2} = e \in 𝔄,

(10.22)

Brauer's Lemma. No hypothesis on $R$ beyond having an identity.

This is the semiprime counterpart of a familiar argument: in a Jacobson semisimple ring, a minimal left ideal cannot be square zero either, because a square-zero left ideal is nil and therefore inside $rad R$ . Semiprimeness is a weaker hypothesis than $rad R = 0$ and does the same job here, which is why $(10.24)$ improves on $(4.14)$ .

Learning Objectives

Prove Brauer's Lemma by producing an idempotent from a nonvanishing product $𝔄 a$ .
Deduce $(10.23)$ : minimal left ideals in a semiprime ring are generated by idempotents.
Show $R = R e \oplus R (1 - e)$ and that $e R e ≅ {End}_{R} (R e)$ is a division ring.
Prove that a semiprime ring with DCC on principal left ideals is semisimple.
Decide, for $ℤ / 12$ , $M_{2} (k)$ , $T_{2} (k)$ , $ℤ$ and ${End}_{D} (V)$ , which minimal left ideals exist and which branch applies.

Definitions

Minimal left ideal: A left ideal $𝔄 \neq 0$ with no left ideal strictly between $0$ and $𝔄$ ; equivalently, $𝔄$ is simple as a left $R$ -module.
$𝔄^{2}$: The additive group generated by all products $a a'$ with $a, a' \in 𝔄$ . For a left ideal this is again a left ideal.
$soc (_{R} R)$: The left socle: the sum of all minimal left ideals of $R$ . It is a two-sided ideal.
Principal left ideal: A left ideal of the form $R a$ for a single $a \in R$ .
Semisimple ring: A ring that is a direct sum of minimal left ideals as a left module over itself; equivalently left artinian with zero Jacobson radical.

Semisimple here means semisimple as a ring, in the Wedderburn–Artin sense; some older sources use the word for what this collection calls semiprimitive.

Core Concepts

Where the idempotent comes from

The proof of Brauer's Lemma is a compressed version of an argument used throughout module theory: minimality turns a surjection into an isomorphism, and an isomorphism of a module onto itself produces a fixed point. Concretely, if $𝔄 a \neq 0$ then $𝔄 a$ is a nonzero left ideal inside $𝔄$ , so $𝔄 a = 𝔄$ ; in particular $a = e a$ for some $e \in 𝔄$ , and $e$ acts as an identity on $a$ .

The element $e^{2} - e$ then annihilates $a$ , and the annihilator ${x \in 𝔄 : x a = 0}$ is a left ideal strictly inside $𝔄$ — strictly, because $e$ is not in it. Minimality kills the annihilator and hence kills $e^{2} - e$ .

𝔄^{2} \neq 0

𝔄 a = 𝔄

for some

a

a = e a

e^{2} = e \neq 0

𝔄 = R e

What an idempotent generator buys

Once $𝔄 = R e$ with $e^{2} = e$ , three things follow immediately. First, $R = R e \oplus R (1 - e)$ as left modules, so $𝔄$ is a direct summand and in particular a projective simple module. Second, ${End}_{R} (R e) ≅ e R e$ , and since $R e$ is simple, Schur's Lemma makes $e R e$ a division ring. Third, $𝔄$ contributes a genuine block to any decomposition of $R$ — which is what makes the socle argument of $(10.24)$ terminate.

Key Results

Lemma(10.22)Brauer's Lemma

Let $𝔄$ be a minimal left ideal in a ring $R$ . Then either $𝔄^{2} = 0$ , or $𝔄 = R e$ for some idempotent $e \in 𝔄$ .

Proof

Assume $𝔄^{2} \neq 0$ . Then $𝔄 a \neq 0$ for some $a \in 𝔄$ , and in particular $a \neq 0$ . Now $𝔄 a$ is a left ideal contained in $𝔄$ and nonzero, so minimality gives $𝔄 a = 𝔄$ . Since $a \in 𝔄$ , there exists $e \in 𝔄$ with $a = e a$ .

Consider $ℑ = {x \in 𝔄 : x a = 0}$ , a left ideal contained in $𝔄$ . It is proper: $e \notin ℑ$ because $e a = a \neq 0$ . By minimality $ℑ = 0$ .

Now $e^{2} - e \in 𝔄$ and $(e^{2} - e) a = e (e a) - e a = e a - e a = 0$ , so $e^{2} - e \in ℑ = 0$ , i.e. $e$ is idempotent, and $e \neq 0$ since $e a = a \neq 0$ . Finally $R e \subseteq 𝔄$ is a nonzero left ideal, because $e = e \cdot e \in R e$ , so minimality gives $𝔄 = R e$ .

Corollary(10.23)Minimal left ideals in a semiprime ring

If $𝔄$ is a minimal left ideal in a semiprime ring $R$ , then $𝔄 = R e$ for some idempotent $e \in 𝔄$ with $e \neq 0$ . Consequently $R = R e \oplus R (1 - e)$ , the module $𝔄$ is simple and projective, and $e R e$ is a division ring.

Proof

If $𝔄^{2} = 0$ then $𝔄$ is a nonzero nilpotent left ideal, which $(10.16)$ forbids in a semiprime ring. So Brauer's Lemma leaves only the second alternative. The decomposition $R = R e \oplus R (1 - e)$ holds for any idempotent, and ${End}_{R} (R e) ≅ e R e$ ; since $R e$ is a simple module, Schur's Lemma makes this endomorphism ring a division ring.

Theorem(10.24)Semiprime criterion for semisimplicity

For any ring $R$ the following are equivalent:

$R$ is semisimple;
$R$ is semiprime and left artinian;
$R$ is semiprime and satisfies DCC on principal left ideals.

This is the exact analogue of $(4.14)$ , with Jacobson semisimple replaced by the weaker hypothesis semiprime.

Proof

**(1) $\Rightarrow$ (2).** A semisimple ring is left artinian and has $rad R = 0$ ; since ${Nil}_{*} R \subseteq rad R$ , it is semiprime. **(2) $\Rightarrow$ (3)** is immediate, principal left ideals being left ideals.

**(3) $\Rightarrow$ (1).** *Step 1: every nonzero left ideal $𝔅$ contains a minimal left ideal.* The set ${R b : 0 \neq b \in 𝔅}$ is a nonempty family of principal left ideals, so DCC provides a minimal member $R b$ . If $0 \neq ℭ \subseteq R b$ is a left ideal, pick $0 \neq c \in ℭ$ ; then $R c \subseteq R b$ is nonzero, so $R c = R b$ by minimality, whence $ℭ \supseteq R c = R b$ and $ℭ = R b$ . So $R b$ is a minimal left ideal inside $𝔅$ .

Step 2: minimal left ideals are generated by idempotents. This is $(10.23)$ , and it is the only place semiprimeness is used.

Step 3: the socle is everything. Let $𝔖 = soc (_{R} R)$ and consider the family ${R (1 - e) : e^{2} = e, R e \subseteq 𝔖}$ , nonempty because $e = 0$ qualifies. By DCC choose $e$ with $R (1 - e)$ minimal in this family, and suppose $R (1 - e) \neq 0$ . By Step 1 it contains a minimal left ideal $𝔄$ , and by Step 2 $𝔄 = R g$ with $g^{2} = g \neq 0$ . Since $g \in R (1 - e)$ we have $g e = 0$ .

Put $u = (1 - e) g$ . Then $e u = 0$ , $u e = (1 - e) g e = 0$ and $u^{2} = (1 - e) g (1 - e) g = (1 - e) (g - g e) g = (1 - e) g \cdot g = u$ , so $u$ is an idempotent orthogonal to $e$ . Moreover $u \neq 0$ : otherwise $g = e g$ , and then $g = g^{2} = (e g) (e g) = e (g e) g = 0$ . Since $u \in 𝔄$ and $R u \neq 0$ , minimality gives $R u = 𝔄$ .

Set $f = e + u$ . Orthogonality makes $f$ idempotent with $R f = R e \oplus R u \subseteq 𝔖$ . Also $1 - f = (1 - e) - u \in R (1 - e)$ , so $R (1 - f) \subseteq R (1 - e)$ . The inclusion is strict: if $R (1 - f) = R (1 - e)$ then $R = R f \oplus R (1 - f) = R f \oplus R (1 - e)$ , yet $0 \neq 𝔄 = R u \subseteq R f \cap R (1 - e)$ , contradicting directness. This contradicts the minimality of $R (1 - e)$ .

Therefore $R (1 - e) = 0$ , so $1 - e = 0$ and $R = R e \subseteq 𝔖$ . Thus $R$ is a sum of minimal left ideals, i.e. $_{R} R$ is a semisimple module, i.e. $R$ is a semisimple ring.

Remark—Left and right socles

For a semiprime ring the left socle and the right socle coincide, so the socle is an unambiguous two-sided invariant — a standard result of socle theory, and a further dividend of $(10.23)$ . Outside the semiprime world the equality can fail, and the socle must then be qualified by side; triangular rings over a pair of unequal coefficient rings supply the standard counterexamples.

Proof Techniques and Method

The reusable moves behind the proofs above.

Move 1

Minimality turns onto into iso

A nonzero left ideal inside a minimal one equals it. Applied to $𝔄 a$ , this yields a left identity for $a$ , and left identities are the raw material for idempotents.

Move 2

Kill the error term in an annihilator

To prove $e^{2} = e$ , show $e^{2} - e$ annihilates a witness and that the annihilator is a proper subideal of a minimal ideal, hence zero. This pattern recurs whenever an idempotent must be manufactured.

Move 3

Grow an idempotent

Given orthogonal idempotents $e, u$ , the sum $e + u$ is idempotent and $R (e + u) = R e \oplus R u$ . Iterating under a chain condition on the complements $R (1 - e)$ terminates and yields a decomposition of $R$ .

Move 3 is the engine of every semisimplicity by chain condition argument. Note what it requires: not merely that minimal left ideals exist, but that each is a summand. That is precisely the content of $(10.23)$ , and it is why the hypothesis is semiprimeness rather than something weaker.

Note also that the orthogonalisation step $u = (1 - e) g$ is forced: $g$ itself need not be orthogonal to $e$ , and replacing $g$ by $(1 - e) g$ is the standard correction, valid because $g e = 0$ already holds.

Worked Example

Both branches of Brauer's Lemma inside one ring

Take $R = ℤ / 12 ℤ$ , which is commutative, so left ideals are ideals. Its minimal ideals are $(4) = {0, 4, 8}$ and $(6) = {0, 6}$ .

(6)^{2} = (36) = 0, 4 \cdot 4 = 16 = 4, so (4) = R \cdot 4 with 4^{2} = 4 .

(E.1)

$(6)$ is the square-zero branch; $(4)$ is the idempotent branch, with $e = 4$ corresponding to $(0, 1)$ under $ℤ / 12 ≅ ℤ / 4 \times ℤ / 3$ .

So $ℤ / 12$ is not semiprime — consistent with ${Nil}_{*} (ℤ / 12) = (6) \neq 0$ — and $(10.23)$ correctly refuses to apply to $(6)$ . Note also $R \cdot 4 ≅ ℤ / 3$ is simple, and $R = R \cdot 4 \oplus R \cdot 9$ where $9 = 1 - 4$ satisfies $9^{2} = 81 = 9$ .

The archetype: columns in a matrix ring

In $R = M_{2} (k)$ for a field $k$ , the first column $𝔄 = R E_{11} = {(\begin{matrix} a & 0 \\ c & 0 \end{matrix})}$ is a minimal left ideal. It is idempotent-generated with $e = E_{11}$ , its square is nonzero, and $e R e = k E_{11} ≅ k$ is a division ring — Schur's Lemma in the simplest possible instance. Here $R$ is semiprime and indeed semisimple, and $R = R E_{11} \oplus R E_{22}$ .

Failure modes

$T_{2} (k)$ : the strictly upper triangular matrices form a minimal left ideal $𝔑$ with $𝔑^{2} = 0$ . The ring is left artinian but not semiprime, so $(10.24)$ correctly denies semisimplicity.
$ℤ$ : semiprime, but $2 ℤ ⊋ 4 ℤ ⊋ 8 ℤ ⊋ \dots$ shows DCC on principal ideals fails, and indeed $ℤ$ has no minimal ideals at all — its socle is zero.
${End}_{D} (V)$ for $\dim_{D} V$ infinite: semiprime (it is even primitive) with plenty of minimal left ideals $R e$ , $e$ a rank-one idempotent, but no DCC; the socle is the ideal of finite-rank endomorphisms and is a proper ideal.
$ℤ / 6 ℤ$ : semiprime and finite, hence artinian, so $(10.24)$ gives semisimplicity — and indeed $ℤ / 6 ≅ ℤ / 2 \times ℤ / 3$ .

Frameworks and Models

Classify a ring by what its minimal left ideals do. Three regimes cover everything.

Minimal left ideals of $R$
- None at all — $soc (_{R} R) = 0$
  - $ℤ$ , $k [x]$ , $k [[x]]$
  - any domain that is not a division ring
  - no chain condition on principal left ideals
- Some, all square zero or mixed — $R$ not semiprime
  - $ℤ / 12$ — one of each kind
  - $T_{n} (k)$ for $n \geq 2$
  - the square-zero ones lie in ${Nil}_{*} R$
- All idempotent-generated — $R$ semiprime
  - $M_{n} (D)$ and every semisimple ring
  - ${End}_{D} (V)$ , socle proper
  - semisimple exactly when the socle is all of $R$

The third regime splits by size of the socle, and $(10.24)$ says the split is governed by a chain condition: semiprime plus DCC on principal left ideals forces $soc (_{R} R) = R$ .

Comparison and Classification

Minimal left ideals in standard rings
Ring	Semiprime?	A minimal left ideal	Branch of $(10.22)$
$M_{n} (D)$	yes	a column $R E_{11}$	idempotent
$ℤ / 6$	yes	$(2)$ and $(3)$	idempotent ( $3^{2} = 3$ , $4^{2} = 4$ )
$ℤ / 12$	no	$(4)$ and $(6)$	one of each
$T_{2} (k)$	no	strictly upper triangular	square zero
${End}_{D} (V)$ , $\dim V$ infinite	yes	$R e$ , $e$ of rank one	idempotent
$ℤ$	yes	none exists	not applicable
$k [[x]]$	yes	none exists	not applicable

Which hypothesis delivers which conclusion
	Minimal left ideals exist	They are summands	$R$ semisimple
Semiprime	no	yes	no
$rad R = 0$	no	yes	no
DCC on principal left ideals	yes	no	no
Left artinian	yes	no	no
Semiprime + DCC on principal left ideals	yes	yes	yes

Which hypothesis delivers which conclusion

Relationship Map

All ringsBrauer's Lemma applies: two branches

Semiprimeonly the idempotent branch survives; socle is a two-sided ideal, left socle

=

right socle

Semiprime + DCC on principal left idealssocle is all of

R

Semisimple

R ≅ M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})

The same picture with Jacobson semisimple in place of semiprime is $(4.14)$ ; since $rad R = 0$ implies ${Nil}_{*} R = 0$ but not conversely, $(10.24)$ is the stronger statement. Both collapse to Wedderburn–Artin at the bottom.

𝔄

minimal left ideal

𝔄 = R e

𝔄

projective simple

𝔄

is a summand of

_{R} R

Design Considerations

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

Which chain condition? DCC on principal left ideals is much weaker than left artinian and is all that $(10.24)$ needs. State the weaker hypothesis when you can: it applies to rings of endomorphisms and to many rings of operators where full artinian-ness fails.
Which zero-radical hypothesis? If your ring is known to have no nilpotent ideals, use semiprimeness; reaching for $rad R = 0$ imposes a condition you may not be able to verify and do not need.
Which side? Semiprimeness is side-symmetric, so the left-handed statements here have right-handed twins. The socle is genuinely side-dependent in general, but not for semiprime rings.
Idempotents or modules? Working with $R e$ keeps everything inside the ring and makes decompositions explicit; working with simple modules is cleaner for functorial arguments. Brauer's Lemma is the bridge, and it is the ring-side statement.

Failure Modes and Common Mistakes

Do not assume $e R e$ is a division ring for an arbitrary idempotent; that requires $R e$ to be a minimal left ideal.
Do not confuse the socle of the ring with ${Nil}_{*} R$ ; in $ℤ / 12$ the socle contains both $(4)$ and $(6)$ while the prime radical is $(6)$ .
Do not assume the socle is a direct summand as a two-sided ideal; it is a sum of left-module summands, which is weaker.

Best Practices

When you find a minimal left ideal, immediately test $𝔄^{2}$ ; the answer decides everything else about it.
Use $(10.23)$ to convert existence statements about simple modules into explicit idempotents, which are computable objects.
State chain conditions in the weakest form your proof uses — DCC on principal left ideals rather than left artinian — so the result applies more widely.
In examples, always exhibit the complementary idempotent $1 - e$ ; it is the quickest check that the decomposition is genuine.

Quick Reference

Brauer's Lemma

𝔄

minimal left ideal

\Rightarrow 𝔄^{2} = 0

𝔄 = R e

Semiprime casealways

𝔄 = R e

e^{2} = e \neq 0

Consequence

R = R e \oplus R (1 - e)

;

𝔄

simple projective

Schur

{End}_{R} (R e) ≅ e R e

is a division ring

Semisimplicitysemiprime

+

DCC on principal left ideals

Compare

(4.14)

rad R = 0

+

same DCC

Socle

soc (_{R} R) = soc (R_{R})

for semiprime

R

Warningminimal left ideals may not exist (

ℤ

k [[x]]

)

Proof map
Step	Hypothesis used	Output
$𝔄 a = 𝔄$	minimality of $𝔄$	left identity $e$ for $a$
Annihilator vanishes	minimality of $𝔄$	$e^{2} = e$
Exclude $𝔄^{2} = 0$	semiprimeness, via (10.16)	$𝔄 = R e$
Minimal ideal inside any $𝔅$	DCC on principal left ideals	socle is large
Grow $e$ to $e + u$	orthogonalisation $u = (1 - e) g$	$R (1 - e)$ shrinks strictly
Terminate	DCC again	$R = soc (_{R} R)$

Frequently Asked Questions

Why is the second branch of Brauer's Lemma stated as

𝔄 = R e

rather than just contains an idempotent?

Because containing an idempotent is not enough for the applications. The point is that a single idempotent $e \in 𝔄$ generates all of $𝔄$ , which is what makes $𝔄$ a direct summand of $_{R} R$ via $R = R e \oplus R (1 - e)$ . Minimality upgrades containment to equality for free: $R e$ is a nonzero left ideal inside a minimal one.

Does

(10.24)

really improve on the Jacobson-radical version?

Yes, strictly. Semiprimeness is weaker than $rad R = 0$ , since ${Nil}_{*} R \subseteq rad R$ can be a strict inclusion — $ℤ_{(p)}$ is semiprime with nonzero Jacobson radical. So $(10.24)$ concludes semisimplicity from less. Under the chain condition the two hypotheses turn out to be equivalent, but that is a consequence of the theorem, not an assumption.

What replaces Brauer's Lemma for minimal right ideals?

The mirror statement, proved by the same argument in $R^{op}$ : a minimal right ideal $𝔄$ satisfies $𝔄^{2} = 0$ or $𝔄 = e R$ for an idempotent $e \in 𝔄$ . Since semiprimeness is left-right symmetric, the semiprime corollary is symmetric too, and in that case the left and right socles agree.

Where does the socle appear elsewhere in the theory?

It is the object that measures how much of a ring is semisimple. For left primitive rings the socle is either zero or a minimal two-sided ideal isomorphic to a ring of finite-rank operators, which is the content of the structure theory in Minimal Ideals and the Socle of a Primitive Ring. The module-level notion is developed in The Socle of a Module and of a Ring.

Is DCC on principal left ideals a natural hypothesis?

It is exactly the condition needed to run the descent in Step 3, and it is genuinely weaker than left artinian. It also has an independent life: rings satisfying it are the left perfect-adjacent objects studied by Bass, and DCC on principal right ideals is one of the standard equivalent forms of left perfectness.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §10, (10.22)–(10.24).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, Chapters 3 and 4 (socles and Goldie theory).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter IV.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §9 and §13.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Give a ring with a minimal left ideal but no minimal right ideal, and explain why semiprimeness rules this out.
Prove that the socle of a semiprime ring is a two-sided ideal and equals the right socle.
How does Brauer's Lemma specialise to group algebras $k G$ with $char k$ dividing $| G |$ ?
Work out the socle of ${End}_{D} (V)$ for infinite-dimensional $V$ and check it is a minimal two-sided ideal.
Explain the relationship between $(10.24)$ and Bass's characterisation of left perfect rings by DCC on principal right ideals.
Which parts of the proof of $(10.24)$ survive for rings without identity?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Where the idempotent comes from

What an idempotent generator buys

06Key Results

07Proof Techniques and Method

Minimality turns onto into iso

Kill the error term in an annihilator

Grow an idempotent

08Worked Example

Both branches of Brauer's Lemma inside one ring

The archetype: columns in a matrix ring

Failure modes

09Frameworks and Models

10Comparison and Classification

11Relationship Map

12Design Considerations

13Failure Modes and Common Mistakes

14Best Practices

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

Comparison and Classification

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions