Reduced Rings as Subdirect Products of Domains

Q: Where exactly does the proof of (12.6) use minimality?

Only at the last step. The enlarged set S' is an m-system avoiding zero regardless of minimality, so Zorn produces a prime p' disjoint from S' and hence contained in p. Minimality is what forces p' = p and therefore S' = S. Without it one only learns that some smaller prime is completely prime.

Executive Summary

The semiprime decomposition $(12.5)$ produces prime factors, and prime rings can be far from commutative-looking — $M_{2} (k)$ is prime and full of zero divisors. Strengthening the hypothesis from semiprime to reduced upgrades the factors all the way to domains, which is the best one can ask of a decomposition into rings with no zero divisors.

The upgrade is not formal. It rests on $(12.6)$ : in a reduced ring the minimal prime ideals are completely prime, so the quotients by them are domains rather than merely prime rings. Lam's proof of that lemma is the most technical argument in §12 and turns on an m-system built from powers.

$a^{2} = 0 \Rightarrow a = 0$ Reduced

DomainsThe factors

Minimal primesThe family of ideals

(12.6)The enabling lemma

Overview

Two strengths of primeness coexist in noncommutative ring theory. An ideal $𝔭$ is prime when $𝔄 𝔅 \subseteq 𝔭$ forces $𝔄 \subseteq 𝔭$ or $𝔅 \subseteq 𝔭$ for ideals; it is completely prime when the same implication holds for elements, that is, when $R / 𝔭$ is a domain. Completely prime is strictly stronger: the zero ideal of $M_{2} (k)$ is prime but not completely prime.

completely prime ⟹ prime, prime not ⟹ completely prime

(12.6)

In commutative rings the two notions coincide, which is why the distinction has no commutative shadow.

For a general ring the minimal primes carry no element-level information, so the decomposition of $(12.5)$ stops at prime factors. The content of $(12.6)$ is that under the reduced hypothesis the two notions of primeness do coincide at the bottom of the prime spectrum, and $(12.7)$ then reads off the decomposition.

This page sits alongside Semiprime and Semiprimitive Rings as Subdirect Products, which handles the weaker hypotheses, and Subdirectly Irreducible Rings, which supplies the hypothesis-free fallback.

Learning Objectives

Distinguish prime from completely prime and give a separating example.
Prove that $x y = 0 \Leftrightarrow y x = 0$ in a reduced ring.
Prove the power-product lemma behind the m-system $S'$ .
Prove $(12.6)$ : minimal primes of a reduced ring are completely prime.
Prove $(12.7)$ in both directions and identify the family of ideals used.
Characterise the ideals $𝔄$ with $R / 𝔄$ reduced.

Definitions

Definition§12Completely prime ideal

An ideal $𝔭$ of a ring $R$ is completely prime if $R / 𝔭$ is a domain; equivalently $𝔭 \neq R$ and $a b \in 𝔭$ implies $a \in 𝔭$ or $b \in 𝔭$ . Lam also uses strongly prime for this notion. Every completely prime ideal is prime.

Reduced: $a^{2} = 0 \Rightarrow a = 0$ . Equivalently $R$ has no nonzero nilpotent element, and then no nonzero nil or nilpotent one-sided ideal either.
m-system: $\emptyset \neq S \subseteq R$ such that for $a, b \in S$ there exists $r \in R$ with $a r b \in S$ . By $(10.4)$ , $𝔭$ is prime iff $R ∖ 𝔭$ is an m-system.
${Nil}_{*} R$: The lower nilradical: the intersection of all prime ideals, equivalently of all minimal primes. It is a nil ideal, so it vanishes in a reduced ring.
Minimal prime: A prime ideal minimal under inclusion. These exist below every prime by Zorn's Lemma, since a descending chain of primes intersects in a prime.
Domain: $R \neq 0$ and $a b = 0 \Rightarrow a = 0$ or $b = 0$ . Equivalently $R$ is prime and reduced.

Domains here are not assumed commutative; a division ring and the free algebra k⟨x,y⟩ are both domains.

Core Concepts

Reduced rings have symmetric zero products

The first consequence of the hypothesis is a symmetry that fails in general rings and is used at every step below: in a reduced ring, $x y = 0$ if and only if $y x = 0$ . Indeed if $x y = 0$ then $(y x)^{2} = y (x y) x = 0$ , and reducedness kills $y x$ .

x y = 0

(y x)^{2} = y (x y) x = 0

y x = 0

The failure in general is easy to see: in $k ⟨ x, y ⟩ / (x y)$ the element $y x$ satisfies $(y x)^{2} = y (x y) x = 0$ but $y x \neq 0$ . That ring is simply not reduced, and with reducedness the symmetry evaporates.

Why the minimal primes and not all primes

Minimality is essential in $(12.6)$ . Take $R = k ⟨ x, y ⟩$ , a domain and hence reduced, and any surjection $R ↠ M_{2} (k)$ sending $x, y$ to two matrices generating $M_{2} (k)$ as a $k$ -algebra. Its kernel $𝔓$ is a prime ideal, because $M_{2} (k)$ is a prime ring, and it is not completely prime, because $M_{2} (k)$ is not a domain. Of course $𝔓 ⊋ 0$ and $0$ is prime, so $𝔓$ is not minimal. The Zorn argument in the proof works precisely because $𝔭$ cannot be shrunk.

The enlarged m-system

Given a minimal prime $𝔭$ , its complement $S = R ∖ 𝔭$ is an m-system. The proof enlarges it to

S' = {a_{1}^{n_{1}} a_{2}^{n_{2}} \dots a_{r}^{n_{r}} : r \geq 1, n_{i} \geq 1, a_{1} a_{2} \dots a_{r} \in S} \supseteq S

(12.6)

Words in which each letter of an element of $S$ may be repeated. The point is that $S'$ is still an m-system and still misses $0$ .

Minimality of $𝔭$ then forces $S' = S$ , and $S' ∋ a^{2}$ whenever $a \in S$ is exactly the statement that $R / 𝔭$ is reduced. A reduced prime ring is a domain, and the lemma is proved.

Key Results

Lemma(12.6)aPower products in a reduced ring

Let $R$ be a reduced ring and $a_{1}, \dots, a_{r} \in R$ with $a_{1} a_{2} \dots a_{r} \neq 0$ . Then $a_{1}^{n_{1}} a_{2}^{n_{2}} \dots a_{r}^{n_{r}} \neq 0$ for all integers $n_{1}, \dots, n_{r} \geq 1$ .

Proof

First, the single-letter step. Put $x = a_{1}$ and $y = a_{2} \dots a_{r}$ , so $x y \neq 0$ ; we claim $x^{2} y \neq 0$ . Suppose $x^{2} y = 0$ . By the symmetry of zero products in a reduced ring, $y x^{2} = 0$ , i.e. $(y x) x = 0$ , hence $x (y x) = 0$ by the same symmetry, i.e. $x y x = 0$ . Then $(x y)^{2} = (x y x) y = 0$ , so $x y = 0$ — a contradiction. Iterating, $a_{1}^{n_{1}} a_{2} \dots a_{r} \neq 0$ for every $n_{1} \geq 1$ .

Now rotate. Applying the symmetry with $x = a_{1}^{n_{1}}$ and $y = a_{2} \dots a_{r}$ gives $a_{2} \dots a_{r} a_{1}^{n_{1}} \neq 0$ , and the single-letter step applied to the first factor of this new word gives $a_{2}^{n_{2}} a_{3} \dots a_{r} a_{1}^{n_{1}} \neq 0$ . Repeating $r$ times and rotating back yields $a_{1}^{n_{1}} a_{2}^{n_{2}} \dots a_{r}^{n_{r}} \neq 0$ .

Lemma(12.6)Minimal primes of a reduced ring

Let $R$ be a reduced ring and let $𝔭$ be a minimal prime ideal of $R$ . Then $𝔭$ is completely prime, i.e. $R / 𝔭$ is a domain.

Proof

Let $S = R ∖ 𝔭$ , an m-system by $(10.4)$ , and let $S'$ be the set of products $a_{1}^{n_{1}} \dots a_{r}^{n_{r}}$ with $n_{i} \geq 1$ and $a_{1} \dots a_{r} \in S$ . Then $S \subseteq S'$ .

** $S'$ is an m-system.** Take $u = a_{1}^{n_{1}} \dots a_{r}^{n_{r}}$ and $v = b_{1}^{m_{1}} \dots b_{s}^{m_{s}}$ in $S'$ , witnessed by $a = a_{1} \dots a_{r} \in S$ and $b = b_{1} \dots b_{s} \in S$ . Since $S$ is an m-system there is $x \in R$ with $a x b \in S$ . The word $a_{1}, \dots, a_{r}, x, b_{1}, \dots, b_{s}$ has plain product $a x b \in S$ , so raising its letters to the exponents $n_{1}, \dots, n_{r}, 1, m_{1}, \dots, m_{s}$ gives $u x v \in S'$ .

** $0 \notin S'$ .** If $a_{1}^{n_{1}} \dots a_{r}^{n_{r}} \in S'$ then $a_{1} \dots a_{r} \in S$ , and $0 \in 𝔭$ gives $a_{1} \dots a_{r} \neq 0$ ; the power-product lemma then makes $a_{1}^{n_{1}} \dots a_{r}^{n_{r}} \neq 0$ .

Minimality closes the argument. The ideal $(0)$ is disjoint from $S'$ , so by Zorn's Lemma there is an ideal $𝔭'$ maximal with respect to $𝔭' \cap S' = \emptyset$ , and $𝔭'$ is prime by $(10.5)$ . Since $S \subseteq S'$ we get $𝔭' \subseteq R ∖ S = 𝔭$ , so minimality of $𝔭$ forces $𝔭' = 𝔭$ . Hence $𝔭 \cap S' = \emptyset$ , i.e. $S' \subseteq S$ and therefore $S' = S$ .

In particular $a \notin 𝔭$ implies $a^{2} \in S' = S$ , i.e. $a^{2} \notin 𝔭$ . So $R / 𝔭$ is reduced. Being also prime, it is a domain: if $\bar{a} \bar{b} = 0$ then $(\bar{b} \bar{r} \bar{a})^{2} = \bar{b} \bar{r} (\bar{a} \bar{b}) \bar{r} \bar{a} = 0$ for all $\bar{r}$ , so $\bar{b} \bar{R} \bar{a} = 0$ and primeness gives $\bar{a} = 0$ or $\bar{b} = 0$ .

Theorem(12.7)Reduced rings as subdirect products of domains

A nonzero ring $R$ is reduced iff $R$ can be represented as a subdirect product of domains. When $R$ is reduced one may take the factors to be $R / 𝔭_{i}$ with ${𝔭_{i}}$ the set of minimal prime ideals of $R$ .

Proof

**( $\Leftarrow$ ).** Suppose $ε : R ↪ \prod_{i} R_{i}$ is subdirect with each $R_{i}$ a domain. If $a \in R$ satisfies $a^{2} = 0$ , then each coordinate $ϕ_{i} (a)$ squares to zero in a domain, hence vanishes; so $ε (a) = 0$ and $a = 0$ . Thus $R$ is reduced.

**( $\Rightarrow$ ).** Let $R$ be reduced. The lower nilradical ${Nil}_{*} R$ is a nil ideal, so it is zero. Every prime ideal contains a minimal prime, so the minimal primes ${𝔭_{i}}$ satisfy $⋂_{i} 𝔭_{i} = {Nil}_{*} R = 0$ . The quotient maps therefore give a subdirect representation $R ↪ \prod_{i} R / 𝔭_{i}$ , and each $R / 𝔭_{i}$ is a domain by $(12.6)$ .

Corollary§12Reduced ideals

For a proper ideal $𝔄 ⊊ R$ , the quotient $R / 𝔄$ is reduced iff $𝔄$ is an intersection of completely prime ideals. This is the exact analogue of: $𝔄$ is semiprime iff $𝔄$ is an intersection of prime ideals.

Proof

If $R / 𝔄$ is reduced, apply $(12.7)$ to it: the minimal primes of $R / 𝔄$ are completely prime and meet in zero, and their preimages in $R$ are completely prime ideals meeting in $𝔄$ . Conversely, if $𝔄 = ⋂_{i} 𝔮_{i}$ with each $𝔮_{i}$ completely prime, then $R / 𝔄$ embeds in $\prod_{i} R / 𝔮_{i}$ , a product of domains, hence is reduced.

CorollaryEx. 12.7Idempotents in a reduced ring are central

Let $R$ be a reduced ring and $e = e^{2} \in R$ . Then $e \in Z (R)$ .

Proof

Represent $R$ subdirectly in a product of domains $R_{i}$ by $(12.7)$ . Each $ϕ_{i} (e)$ is an idempotent of the domain $R_{i}$ : from $ϕ_{i} (e) (ϕ_{i} (e) - 1) = 0$ we get $ϕ_{i} (e) \in {0, 1}$ , which is central in $R_{i}$ . Hence $ϕ_{i} (e r - r e) = 0$ for every $r \in R$ and every $i$ , and injectivity gives $e r = r e$ .

CorollaryEx. 12.5Strongly regular rings

Call $R$ strongly regular if for every $a \in R$ there exists $x \in R$ with $a^{2} x = a$ . Then every nonzero strongly regular ring is a subdirect product of division rings.

Proof

Strong regularity is inherited by quotients, and it forces $R$ to be reduced: if $a^{2} = 0$ then $a = a^{2} x = 0$ . So $(12.7)$ presents $R$ subdirectly in a product of domains $R / 𝔭_{i}$ , each of which is again strongly regular. In a strongly regular domain, $a \neq 0$ and $a (a x - 1) = 0$ give $a x = 1$ , and then $a (x a - 1) = 0$ gives $x a = 1$ ; so every nonzero element is invertible and each factor is a division ring.

Proof Techniques and Method

The technique behind (12.6), isolated for reuse.

Turn the ideal into an m-systemWork with

S = R ∖ 𝔭

rather than with

𝔭

. Primeness becomes a closure property of

S

, which is easier to enlarge.

Enlarge the m-systemAdd the products you want to be nonzero — here, powers of the letters. Check the m-system axiom on the enlarged set.

Show $0$ stays outThis is where the ring-theoretic hypothesis is spent: the power-product lemma is the only place reducedness is used quantitatively.

Zorn back downAn ideal maximal with respect to missing

S'

is prime and sits inside

𝔭

; minimality collapses it onto

𝔭

, forcing

S' = S

Read the conclusion elementwise

S

closed under squaring is exactly

R / 𝔭

reduced; reduced plus prime is a domain.

Note what is not used: no chain conditions, no finiteness, no commutativity, and no structure theory of prime rings. The proof is elementary and self-contained, which is why the result holds in complete generality.

Worked Example

A commutative computation: $k [x, y] / (x y)$

Let $R = k [x, y] / (x y)$ for a field $k$ . The ideal $(x y)$ is generated by a squarefree polynomial, hence radical, so $R$ is reduced. Its minimal primes are the images of $(x)$ and $(y)$ , since $(x y) = (x) \cap (y)$ in $k [x, y]$ and both are prime.

R ↪ k [y] \times k [x], f \mapsto (f (0, y), f (x, 0))

(E.1)

Both factors are domains. The kernel is $(x) \cap (y) = (x y)$ , which is zero in $R$ , so the map is injective and each coordinate is onto.

The image is ${(g, h) \in k [y] \times k [x] : g (0) = h (0)}$ , a $k$ -subspace of codimension $1$ in the product: the two coordinate axes glued at the origin. This is the algebraic content of the geometric picture, and it shows again that a subdirect product is generally a proper subring of the product.

Where the reduced hypothesis is spent

** $(12.6)$ fails without reduced.** In $R = M_{2} (k)$ the zero ideal is a minimal prime, but $R / 0 = M_{2} (k)$ has $E_{12}^{2} = 0$ and is not a domain. So minimal prime alone gives nothing at the element level.
** $(12.7)$ fails without reduced.** $ℤ / 4$ has a nonzero nilpotent, so no embedding into a product of domains can exist: the image of $2$ would be a nonzero nilpotent in some domain.
Symmetry of zero products fails without reduced. In $S = k ⟨ x, y ⟩ / (x y)$ we have $x y = 0$ while $y x \neq 0$ ; consistently, $(y x)^{2} = 0$ , so $S$ is not reduced.

A noncommutative reduced ring

Let $D$ be a noncommutative division ring and $R = {(a, b) \in D [t] \times D [t] : a (0) = b (0)}$ , two polynomial rings glued at $t = 0$ . As a subring of a product of domains, $R$ is reduced; it is not itself a domain, since $(t, 0) (0, t) = (0, 0)$ with both factors nonzero. The kernels of the two coordinate projections are completely prime — each quotient is $D [t]$ — and they meet in zero, so the defining embedding $R ↪ D [t] \times D [t]$ is already the subdirect representation by domains that $(12.7)$ promises.

Comparison and Classification

Primeness at the ideal level and at the element level
Notion	Condition on $𝔭$	Quotient $R / 𝔭$	Commutative case
Prime	$𝔄 𝔅 \subseteq 𝔭 \Rightarrow 𝔄 \subseteq 𝔭$ or $𝔅 \subseteq 𝔭$	prime ring	prime ideal
Completely prime	$a b \in 𝔭 \Rightarrow a \in 𝔭$ or $b \in 𝔭$	domain	prime ideal — the same thing
Semiprime	$𝔄^{2} \subseteq 𝔭 \Rightarrow 𝔄 \subseteq 𝔭$	semiprime ring	radical ideal
Reduced (as an ideal)	$a^{2} \in 𝔭 \Rightarrow a \in 𝔭$	reduced ring	radical ideal — the same thing

What each hypothesis buys in §12
Hypothesis	Ideals used	Factors	Reference
semiprime	prime ideals	prime rings	(12.5)(a)
semiprimitive	left primitive ideals	left primitive rings	(12.5)(b)
reduced	minimal primes, now completely prime	domains	(12.6)–(12.7)
strongly regular	minimal primes	division rings	Ex. 12.5

Each row strengthens the row above it: reduced implies semiprime, and strongly regular implies both reduced and semiprimitive.

Relationship Map

Division ringDomainReducedSemiprime

**Reduced $\Rightarrow$ semiprime**, strictly: $M_{2} (k)$ is prime, hence semiprime, and is not reduced.
**Reduced $not \Rightarrow$ semiprimitive**: $k [[x]]$ is a commutative domain with $rad = (x) \neq 0$ . The two strengthenings of semiprime are independent.
**Reduced $+$ subdirectly irreducible $\Rightarrow$ domain**, by Lam's Exercise 12.1 together with $(12.6)$ ; this is the atom-level shadow of $(12.7)$ .
**Commutative $+$ reduced $+$ subdirectly irreducible $\Rightarrow$ field**, which is $(12.4)$ — the strongest conclusion available in the section.

Weaker input

Semiprime

Factors are prime rings. Zero divisors survive: $M_{2} (k)$ can appear as a factor.

This page

Reduced

Factors are domains. No zero divisors anywhere, and idempotents become central.

Stronger input

Strongly regular

Factors are division rings. Equivalent to von Neumann regular plus reduced.

Applications and Industry Use

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

Ring theory

Central idempotents for free

That idempotents in a reduced ring are central is proved most cleanly by decomposing into domains and observing that idempotents in a domain are $0$ or $1$ . This underlies the theory of abelian regular rings and of Pierce sheaves.

Commutative algebra

Irreducible components

For a reduced commutative noetherian ring the minimal primes are finite in number and $(12.7)$ is the decomposition of the variety into irreducible components — the normalisation of a nodal curve is exactly the passage to the product.

Operator algebras

Reduced group algebras

Semiprimeness and reducedness of group algebras are longstanding questions; where a reduced group algebra is known, $(12.7)$ converts the statement into an embedding in a product of domains, which is what zero-divisor conjectures are really about.

Symbolic computation

Radical and component splitting

Computer algebra systems reduce a commutative ring by quotienting out the nilradical and then split along minimal primes; every subsequent computation runs component by component. Singular, Macaulay2 and Sage all expose this workflow.

The honest summary: $(12.7)$ is what licenses the phrase *we may assume $R$ is a domain*. Any property preserved by subrings and by products can be checked on domains and transported back, and that is how it is used in practice.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Lam's termcompletely prime, also written strongly prime in §12

Terminology hazardStrongly prime means something else in Handelman–Lawrence; prefer completely prime

ReducedStandard everywhere; semiprime is the ideal-level analogue, not a synonym

Nilradical

{Nil}_{*} R

(lower, Baer, prime radical); commutative sources write

Nil (R)

Minimal primes

Min (R)

Min Spec R

in commutative sources

ImplementationsSingular minAssGTZ, Macaulay2 minimalPrimes, Sage minimal_associated_primes

Failure Modes and Common Mistakes

Do not assume finitely many minimal primes. That is a noetherian phenomenon; in general the index set can be infinite.
Do not expect the embedding to be onto in any coordinate-free sense: only the individual coordinate maps are surjective.
Do not use the symmetry $x y = 0 \Leftrightarrow y x = 0$ outside reduced rings — it is equivalent to a genuine hypothesis, not a formal identity.
Do not confuse a reduced ideal (reduced quotient) with a reduced ring element; the terminology is inherited from the commutative radical-ideal vocabulary.

Best Practices

Check reducedness elementwise before invoking $(12.7)$ ; semiprimeness is not enough and is easier to verify by accident.
Prefer the minimal primes as the index family — they are the smallest family that works and in the noetherian commutative case they are finite.
When transporting a property to the domain factors, verify it is preserved by quotients; the subdirect map only guarantees coordinatewise surjectivity.
State explicitly whether prime means the ideal condition or the element condition when writing for a mixed audience.

Quick Reference

(12.7)

R \neq 0

reduced

iff

subdirect product of domains

(12.6)

R

reduced,

𝔭

minimal prime

\Rightarrow

R / 𝔭

a domain

Family usedthe minimal primes;

⋂_{i} 𝔭_{i} = {Nil}_{*} R = 0

Key symmetryreduced

\Rightarrow

(

x y = 0 iff y x = 0

)

Power lemma

a_{1} \dots a_{r} \neq 0 \Rightarrow a_{1}^{n_{1}} \dots a_{r}^{n_{r}} \neq 0

Ideal version

R / 𝔄

reduced

iff

𝔄

is an intersection of completely prime ideals

Idempotentsin a reduced ring every idempotent is central

Upgradestrongly regular

\Rightarrow

subdirect product of division rings

Results on this page
Reference	Statement	Hypotheses
(12.6)a	Power products stay nonzero	$R$ reduced
(12.6)	Minimal primes are completely prime	$R$ reduced; $𝔭$ minimal
(12.7)	Reduced $iff$ subdirect product of domains	$R \neq 0$
§12 remark	Reduced ideals are intersections of completely primes	$𝔄 ⊊ R$
Ex. 12.5	Strongly regular $\Rightarrow$ subdirect product of division rings	$R \neq 0$
Ex. 12.7	Idempotents are central	$R$ reduced

Frequently Asked Questions

Why is a prime ring not automatically a domain?

Because primeness is a condition on ideals, not elements. In $M_{2} (k)$ the product of any two nonzero ideals is nonzero — there are only the ideals $0$ and $R$ — yet $E_{12}^{2} = 0$ . The elementwise condition is completely primeness, and $(12.6)$ is the statement that reducedness forces the two to agree at minimal primes.

Where exactly does the proof of (12.6) use minimality?

Only at the last step. The enlarged set $S'$ is an m-system avoiding zero regardless of minimality, so Zorn produces a prime $𝔭'$ disjoint from $S'$ and hence contained in $𝔭$ . Minimality is what forces $𝔭' = 𝔭$ and therefore $S' = S$ . Without it one only learns that some smaller prime is completely prime.

Does (12.7) say anything about how many factors are needed?

Only that one factor per minimal prime suffices. For a reduced commutative noetherian ring that number is finite and equals the number of irreducible components of the associated variety. In general it can be infinite, and there is no canonical smaller family: the minimal primes are already the irredundant choice.

Is the converse direction of (12.7) really that easy?

Yes. A subring of a product of domains is reduced because nilpotency is checked coordinatewise and a domain has no nonzero nilpotents. All the difficulty is in producing the domains, which is $(12.6)$ .

How does this compare with the commutative statement everyone knows?

In commutative algebra one says: a ring is reduced iff the nilradical is zero iff it embeds in the product of its quotients by minimal primes, which are domains. All of that is immediate commutatively, because prime and completely prime coincide. The noncommutative content of $(12.6)$ – $(12.7)$ is precisely that the same statement survives without commutativity.

What breaks if I try to prove (12.6) by localisation?

Noncommutative rings need not admit localisation at a prime — the Ore condition may fail — so the standard commutative proof, which inverts everything outside $𝔭$ , has no analogue. The m-system argument is the substitute: it manipulates the multiplicative structure of the complement without ever forming a ring of fractions.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991 — §12, lemma (12.6) and theorem (12.7), pp. 207–209; §10 for m-systems (10.3)–(10.5).
T. Y. Lam, Exercises in Classical Ring Theory, Springer-Verlag — worked solutions to the §12 exercises on reduced and strongly regular rings.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988 — prime radicals, minimal primes and completely prime ideals.
K. R. Goodearl, Von Neumann Regular Rings, Pitman, 1979 — strongly regular rings and their decomposition into division rings.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition 1964 — radicals and subdirect decompositions.

AI Suggested Questions

Write out the m-system proof of $(12.6)$ for a concrete noncommutative reduced ring and inspect the set $S'$ .
Give an example of a reduced ring with infinitely many minimal primes.
Prove that a reduced ring satisfying the ascending chain condition on annihilators has finitely many minimal primes.
How does $(12.7)$ interact with polynomial extensions — is $R [x]$ reduced whenever $R$ is?
Show that a strongly regular ring is von Neumann regular, and deduce the equivalence with regular plus reduced.
What is the sheaf-theoretic form of $(12.7)$ , and how does it relate to the Pierce sheaf of a reduced ring?
Does every domain arise as a factor in the decomposition of some reduced ring that is not itself a domain?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Reduced rings have symmetric zero products

Why the minimal primes and not all primes

The enlarged m-system

06Key Results

07Proof Techniques and Method

08Worked Example

A commutative computation: k[x,y]/(xy)

Where the reduced hypothesis is spent

A noncommutative reduced ring

09Comparison and Classification

10Relationship Map

Semiprime

Reduced

Strongly regular

11Applications and Industry Use

Central idempotents for free

Irreducible components

Reduced group algebras

Radical and component splitting

12Standards and Notation

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

A commutative computation: $k [x, y] / (x y)$

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions