m-Systems and the Characterisation of Prime Ideals

Q: Why is the definition of an m-system existential in r?

Because the corresponding condition on primes, aRb ⊆ p, is existential: it says some element of aRb escapes p. A universal version — arb ∈ S for all r — would be violated by taking r = 0 whenever 0 S, so it would have essentially no examples.

Q: Is every m-system contained in the complement of a prime?

Yes, provided 0 S, or more generally provided some ideal is disjoint from S. Apply the corollary with A = (0): there is a prime p disjoint from S, so S ⊆ R p. If 0 ∈ S no such prime exists.

Q: Does (10.5) need the ring to have an identity?

The proof as given uses (a) = RaR a, which is where the identity enters. Without an identity one replaces (a) by the ideal generated by a, namely Za + Ra + aR + RaR, and the argument goes through with more bookkeeping. Everything on this page assumes an identity.

Q: What replaces m-systems for semiprime ideals?

n-systems: c is semiprime exactly when R c is an n-system. The two notions are linked by (10.10), which shows every n-system contains an m-system through any prescribed point — the technical heart of the proof that semiprime ideals are intersections of primes.

Q: Can I always take the m-system generated by a set?

There is no canonical generated m-system, because the required r is not determined. One can close a set under *some* choice of sandwiching elements, as in the inductive construction of (10.10), but the result depends on the choices made.

Executive Summary

In commutative algebra the complement of a prime ideal is multiplicatively closed, and conversely an ideal maximal among those missing a multiplicatively closed set is prime. Both halves survive in the noncommutative world once multiplicatively closed is weakened to ** $m$ -system**: for $a, b \in S$ one asks only that $a r b \in S$ for some $r \in R$ .

That weakening is forced by the $a R b$ test for primeness, and it is exactly what is needed. The pay-off is a manufacturing process for prime ideals: choose an $m$ -system you want to avoid, apply Zorn's Lemma, and the resulting maximal ideal is prime.

$a r b \in S$ Defining condition

$R ∖ 𝔭$ Canonical example

ZornHow primes are produced

${a^{2^{i}}}$ Key non-multiplicative example

Overview

Prime Ideals in Noncommutative Rings established that $𝔭$ is prime iff $a R b \subseteq 𝔭$ forces $a \in 𝔭$ or $b \in 𝔭$ . Negate that statement: $𝔭$ is prime iff whenever $a$ and $b$ both lie outside $𝔭$ , some $a r b$ also lies outside $𝔭$ . The complement is therefore closed under a sandwiched product rather than an ordinary one.

S \neq \emptyset and \forall a, b \in S \exists r \in R : a r b \in S

(10.3)

The definition of an $m$ -system. No closure under addition, no closure under multiplication, no requirement that $1 \in S$ .

There is a cost. In commutative algebra a multiplicatively closed set can be inverted, giving $S^{- 1} R$ ; an $m$ -system generally cannot, because inverting a set in a noncommutative ring requires the Ore condition. $m$ -Systems are a tool for finding primes, not for localising at them.

Learning Objectives

State $(10.3)$ and verify it for multiplicatively closed sets and for ${a, a^{2}, a^{4}, a^{8}, \dots}$ .
Prove $(10.4)$ : $𝔭$ is prime iff $R ∖ 𝔭$ is an $m$ -system.
Prove $(10.5)$ : an ideal maximal with respect to missing an $m$ -system is prime.
Assemble the Zorn argument that produces such a maximal ideal.
Deduce that every non-nilpotent $a \in R$ lies outside some prime ideal.
Verify directly that $M_{n} (D)$ is a prime ring using matrix units.

Definitions

Definition(10.3)m-System

A nonempty subset $S \subseteq R$ is an ** $m$ -system** if for all $a, b \in S$ there exists $r \in R$ with $a r b \in S$ .

Multiplicatively closed: $a b \in S$ for all $a, b \in S$ , with $S \neq \emptyset$ . Taking $r = 1$ shows every such $S$ is an $m$ -system.
$n$ -system: For every $a \in S$ there is $r \in R$ with $a r a \in S$ . Setting $b = a$ in $(10.3)$ shows every $m$ -system is an $n$ -system; the converse fails, and repairing it is $(10.10)$ .
Disjoint from an ideal: $S \cap 𝔄 = \emptyset$ . If $0 \in S$ no ideal at all is disjoint from $S$ , since every ideal contains $0$ .
Saturation: Unlike the commutative case there is no useful saturation operation on $m$ -systems; they are used as they are found.

An m-system is a bare set: it carries no additive structure and need not contain the identity.

Core Concepts

Why not just multiplicatively closed sets

If $𝔭$ is prime and $a, b \notin 𝔭$ , nothing forces $a b \notin 𝔭$ — in $M_{2} (k)$ the ideal $(0)$ is prime yet $e_{11} e_{22} = 0$ . So complements of primes are not multiplicatively closed, and a theory built on multiplicative closure would have no examples. Inserting an unspecified $r$ makes the complement of every prime an example, by construction.

The powers of a single element

The set $S = {a, a^{2}, a^{4}, a^{8}, \dots} = {a^{2^{i}} : i \geq 0}$ is an $m$ -system in any ring: given $a^{2^{i}}$ and $a^{2^{j}}$ with $i \leq j$ , take $r = a^{2^{j} - 2^{i}}$ , so that

a^{2^{i}} \cdot a^{2^{j} - 2^{i}} \cdot a^{2^{j}} = a^{2^{j} + 2^{j}} = a^{2^{j + 1}} \in S .

(10.3a)

The exponents double, which is why the set of all powers is not needed — and indeed $S$ is not multiplicatively closed, since $a \cdot a^{2} = a^{3} \notin S$ in general.

This modest example does real work: $S$ misses $0$ precisely when $a$ is not nilpotent, and that is the hinge of the inclusion $\sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄}$ in The Radical of an Ideal as an Intersection of Primes.

Subsets of $R$ no closure assumed

$n$ -systems

\forall a \exists r : a r a \in S

— complements of semiprime ideals

$m$ -systems

\forall a, b \exists r : a r b \in S

— complements of prime ideals

Multiplicatively closed sets

a b \in S

— the commutative-style case,

r = 1

Both inclusions are strict: ${a^{2^{i}}}$ is an $m$ -system that is not multiplicatively closed, and in $M_{2} (k)$ the set ${e_{12}}$ is an $n$ -system — indeed $e_{12} e_{21} e_{12} = e_{12}$ — but so is any set containing a single element $x$ with $x R x ∋ x$ .

Key Results

Corollary(10.4)Primes are complements of m-systems

Let $R$ be a ring and $𝔭 \subseteq R$ an ideal. Then $𝔭$ is prime if and only if $R ∖ 𝔭$ is an $m$ -system.

Proof

Suppose $𝔭$ is prime. Then $𝔭 \neq R$ , so $R ∖ 𝔭 \neq \emptyset$ . Let $a, b \notin 𝔭$ . By $(10.2) (3)$ , $a R b \subseteq 𝔭$ would force $a \in 𝔭$ or $b \in 𝔭$ ; hence $a R b not \subseteq 𝔭$ and there is $r \in R$ with $a r b \notin 𝔭$ , i.e. $a r b \in R ∖ 𝔭$ .

Conversely, suppose $R ∖ 𝔭$ is an $m$ -system. Nonemptiness gives $𝔭 \neq R$ . If $a, b \notin 𝔭$ then some $a r b \notin 𝔭$ , so $a R b not \subseteq 𝔭$ . Contrapositively $a R b \subseteq 𝔭$ implies $a \in 𝔭$ or $b \in 𝔭$ , which is $(10.2) (3)$ , hence $𝔭$ is prime.

Proposition(10.5)Maximal disjoint ideals are prime

Let $S \subseteq R$ be an $m$ -system and let $𝔭$ be an ideal of $R$ that is maximal with respect to the property $𝔭 \cap S = \emptyset$ . Then $𝔭$ is a prime ideal.

Proof

First, $𝔭 \neq R$ : since $S \neq \emptyset$ , an ideal disjoint from $S$ cannot be all of $R$ . We verify $(10.2) (2)$ . Suppose $a \notin 𝔭$ and $b \notin 𝔭$ but $(a) (b) \subseteq 𝔭$ .

The ideals $𝔭 + (a)$ and $𝔭 + (b)$ strictly contain $𝔭$ , so by maximality neither is disjoint from $S$ : choose $s \in S \cap (𝔭 + (a))$ and $s' \in S \cap (𝔭 + (b))$ . As $S$ is an $m$ -system there is $r \in R$ with $s r s' \in S$ . But

s r s' \in (𝔭 + (a)) R (𝔭 + (b)) \subseteq 𝔭 + (a) R (b) \subseteq 𝔭 + (a) (b) \subseteq 𝔭,

because every term involving $𝔭$ is absorbed by the ideal $𝔭$ and $(a) R (b) \subseteq (a) (b)$ . So $s r s' \in 𝔭 \cap S$ , contradicting disjointness. Hence $a \in 𝔭$ or $b \in 𝔭$ , and $𝔭$ is prime.

Corollary—Existence of avoiding primes

Let $S$ be an $m$ -system and $𝔄$ an ideal with $𝔄 \cap S = \emptyset$ . Then there exists a prime ideal $𝔭 \supseteq 𝔄$ with $𝔭 \cap S = \emptyset$ .

Proof

Order by inclusion the set $Σ$ of ideals containing $𝔄$ and disjoint from $S$ ; it is nonempty because $𝔄 \in Σ$ . The union of a chain in $Σ$ is an ideal, contains $𝔄$ , and meets $S$ only if some member does — so it lies in $Σ$ . Zorn's Lemma supplies a maximal element $𝔭$ , which is prime by $(10.5)$ .

Corollary—Non-nilpotent elements avoid a prime

If $a \in R$ is not nilpotent, there is a prime ideal $𝔭$ of $R$ with $a \notin 𝔭$ . Indeed $S = {a^{2^{i}} : i \geq 0}$ is an $m$ -system with $0 \notin S$ , so the previous corollary applied to $𝔄 = (0)$ yields a prime $𝔭$ disjoint from $S$ ; in particular $a \notin 𝔭$ .

Remark—What maximality does not give

The ideal produced by $(10.5)$ is maximal only among ideals disjoint from $S$ ; it is generally far from a maximal ideal of $R$ . In $ℤ$ with $S = {2^{2^{i}}}$ , the ideal $(0)$ is disjoint from $S$ but not maximal in $Σ$ , whereas $(3)$ is maximal in $Σ$ — and $(9)$ , although disjoint from $S$ , is not maximal in $Σ$ and is not prime.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Negate to get closure

To convert an ideal-avoidance property into a closure property, take complements. Primeness reads as a closure condition on $R ∖ 𝔭$ ; that is the whole content of $(10.4)$ .

Move 2

Push out and catch a witness

Enlarging $𝔭$ by $(a)$ must break disjointness, delivering a concrete $s \in S$ . Maximality is used only to produce witnesses, never structurally.

Move 3

Absorb into the ideal

Expanding $(𝔭 + (a)) R (𝔭 + (b))$ , every term containing a factor from $𝔭$ is swallowed. What is left is $(a) R (b) \subseteq (a) (b)$ — the assumed inclusion.

The same three moves reappear, with $a r a$ in place of $a r b$ , in the treatment of semiprime ideals — see Semiprime Ideals and n-Systems. Learning them once covers both.

Worked Example

$M_{n} (D)$ is a prime ring, checked with matrix units

Let $D$ be a division ring and $R = M_{n} (D)$ . By $(10.4)$ it suffices to show that $R ∖ {0}$ is an $m$ -system, i.e. that $a R b \neq 0$ whenever $a, b \neq 0$ .

Choose indices with $a_{i j} \neq 0$ and $b_{k l} \neq 0$ , and take $r = e_{j k}$ . The $(i, l)$ entry of $a e_{j k} b$ is

(a e_{j k} b)_{i l} = \sum_{s, t} a_{i s} (e_{j k})_{st} b_{t l} = a_{i j} b_{k l} \neq 0,

(E.1)

Both factors are nonzero elements of a division ring, so the product is nonzero.

Hence $a e_{j k} b \neq 0$ , the complement of $(0)$ is an $m$ -system, and $(0)$ is prime. So $M_{n} (D)$ is a prime ring for every $n$ , even though for $n \geq 2$ it is very far from a domain.

Producing a prime that avoids a prescribed element

Take $R = ℤ$ and $a = 2$ , so $S = {2, 4, 16, 256, \dots} = {2^{2^{i}}}$ . An ideal $(n)$ meets $S$ exactly when $n ∣ 2^{2^{i}}$ for some $i$ , i.e. when $n$ is a power of $2$ (including $n = 1$ ). So the ideals disjoint from $S$ are $(0)$ together with all $(n)$ whose $n$ has an odd prime factor: for instance $(0), (3), (9), (5), (6), (12)$ .

Now maximise. $(9) ⊊ (3)$ and $(3)$ is still disjoint from $S$ , so $(9)$ is not maximal in $Σ$ — correctly, since $(9)$ is not prime. Likewise $(12) ⊊ (6) ⊊ (3)$ . Every ideal properly containing $(3)$ equals $ℤ$ , which meets $S$ ; hence $(3)$ is maximal in $Σ$ and $(10.5)$ certifies it prime. The maximal members of $Σ$ are precisely the ideals $(p)$ with $p$ an odd prime — exactly the primes of $ℤ$ missing every power of $2$ .

Process and Workflow

Choose what to avoidPick the element or set you want kept outside the prime, and build an

m

-system

S

containing it — the doubling powers

{a^{2^{i}}}

if you start from one element.

Check $S \cap 𝔄 = \emptyset$ For

𝔄 = (0)

this says exactly that

a

is not nilpotent. If

S

meets

𝔄

, no prime above

𝔄

can avoid

S

and the construction correctly fails.

Apply Zorn to $Σ$ Ideals containing

𝔄

and disjoint from

S

, ordered by inclusion; unions of chains stay in

Σ

Invoke (10.5)The maximal element is prime. Read off the conclusion: an element outside a prime, or a prime containing a prescribed ideal.

You need a prime ideal with a prescribed property. Which tool?

Avoid a setMake the set an

m

-system and use

(10.5)

. This is the only general existence mechanism available.

Contain an idealEvery proper ideal lies in a maximal ideal, and maximal ideals are prime; use Zorn directly with no

m

-system.

Be minimal over an idealApply Zorn downwards: intersections of descending chains of primes are prime, so minimal primes over

𝔄

exist.

Be localisableNone of the above helps — that needs the Ore condition, which

m

-systems do not supply.

Comparison and Classification

Commutative multiplicative sets versus $m$ -systems
Feature	Multiplicatively closed $S$	$m$ -system $S$
Closure condition	$a b \in S$	$a r b \in S$ for some $r \in R$
Complement of a prime	yes (commutative case)	yes (always)
Maximal disjoint ideal is prime	yes (commutative case)	yes, $(10.5)$
Contains $1$ by convention	usually assumed	not assumed
Supports localisation $S^{- 1} R$	yes	only under an Ore condition
Generated by one element	${a^{n} : n \geq 1}$	${a^{2^{i}} : i \geq 0}$ suffices

Which closure conditions a given set satisfies
	multiplicatively closed	$m$ -system	$n$ -system
$R ∖ 𝔭$ , $𝔭$ prime	partial	yes	yes
$R ∖ 𝔠$ , $𝔠$ semiprime but not prime	no	no	yes
${a^{n} : n \geq 1}$	yes	yes	yes
${a^{2^{i}} : i \geq 0}$	no	yes	yes
$U (R)$ , the units	yes	yes	yes
${e_{12}} \subseteq M_{2} (k)$	no	yes	yes

Which closure conditions a given set satisfies

The single-element set ${e_{12}}$ qualifies because $e_{12} e_{21} e_{12} = e_{12}$ : an $m$ -system may be finite, and even a singleton. The first row is marked part because $R ∖ 𝔭$ is multiplicatively closed exactly when $𝔭$ is completely prime.

Relationship Map

m

-system

S

Zorn on ideals missing

S

maximal such ideal

𝔭

𝔭

prime

Downstream, this chain is the engine of two results. Taking $S = {a^{2^{i}}}$ gives the containment $\sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄}$ , and taking arbitrary $S$ gives the hard inclusion in the theorem that $\sqrt{𝔄}$ is the intersection of the primes above $𝔄$ — both in The Radical of an Ideal as an Intersection of Primes. The $n$ -system variant, via the lemma that every $n$ -system contains an $m$ -system through any of its points, is what makes semiprime ideals intersections of primes.

Design Considerations

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

**Which $m$ -system to choose.** A larger $S$ gives a stronger avoidance conclusion but is more likely to meet the ideal you started from. The doubling-powers system is the smallest useful one built from a single element.
Whether to start from an ideal. Applying Zorn inside ${𝔅 \supseteq 𝔄}$ rather than inside all ideals is free and gives a prime above $𝔄$ ; there is never a reason not to.
Ore or not. If the real goal is a ring of fractions rather than a prime ideal, abandon $m$ -systems and check the Ore condition on a multiplicatively closed set of regular elements instead.
Sides. $m$ -systems are left-right symmetric, matching the symmetry of primeness. No side-choice is needed anywhere in this construction.

Failure Modes and Common Mistakes

Do not assume $r$ can be taken to be $1$ : that would return you to multiplicative closure and to a theory with no examples.
Do not assume the $r$ in $(10.3)$ is unique or canonical; the definition is purely existential and no choice function is implied.
Do not confuse *maximal among ideals disjoint from $S$ * with maximal ideal. The former is prime, the latter is prime, but they are rarely the same ideal.
Do not forget nonemptiness: the empty set vacuously satisfies the closure condition, and admitting it would make the improper ideal $R$ prime.
Do not conflate $m$ -systems with $n$ -systems. Every $m$ -system is an $n$ -system; the converse needs the lemma $(10.10)$ and gives only a sub- $m$ -system through a chosen point.

Quick Reference

Definition

S \neq \emptyset

, and

a, b \in S \Rightarrow a r b \in S

for some

r

Characterisation

𝔭

prime

\Leftrightarrow

R ∖ 𝔭

m

-system

Production rulemaximal ideal disjoint from

S

is prime

(10.5)

Existence

𝔄 \cap S = \emptyset

\Rightarrow

some prime

𝔭 \supseteq 𝔄

with

𝔭 \cap S = \emptyset

Standard system

{a, a^{2}, a^{4}, a^{8}, \dots}

, missing

0

iff

a

is not nilpotent

Weaker cousin

n

-system:

\forall a \exists r

with

a r a \in S

Symmetry

S

is an

m

-system in

R

iff it is one in

R^{op}

No localisation

m

-systems do not give

S^{- 1} R

without an Ore condition

Checklist for applying (10.5)
Step	What to verify	Failure mode
Nonempty	$S \neq \emptyset$	vacuous closure makes $R$ prime
Sandwich closure	$a, b \in S \Rightarrow \exists r, a r b \in S$	only ordinary products checked
Disjointness	$𝔄 \cap S = \emptyset$	$0 \in S$ ; $Σ$ empty
Chain unions	union of a chain in $Σ$ lies in $Σ$	forgetting that a union of ideals along a chain is an ideal
Conclusion	maximal element is prime	mistaking it for a maximal ideal

Frequently Asked Questions

Why is the definition of an

m

-system existential in

r

Because the corresponding condition on primes, $a R b not \subseteq 𝔭$ , is existential: it says some element of $a R b$ escapes $𝔭$ . A universal version — $a r b \in S$ for all $r$ — would be violated by taking $r = 0$ whenever $0 \notin S$ , so it would have essentially no examples.

Is every

m

-system contained in the complement of a prime?

Yes, provided $0 \notin S$ , or more generally provided some ideal is disjoint from $S$ . Apply the corollary with $𝔄 = (0)$ : there is a prime $𝔭$ disjoint from $S$ , so $S \subseteq R ∖ 𝔭$ . If $0 \in S$ no such prime exists.

Does

(10.5)

need the ring to have an identity?

The proof as given uses $(a) = R a R ∋ a$ , which is where the identity enters. Without an identity one replaces $(a)$ by the ideal generated by $a$ , namely $ℤ a + R a + a R + R a R$ , and the argument goes through with more bookkeeping. Everything on this page assumes an identity.

What replaces

m

-systems for semiprime ideals?

$n$ -systems: $𝔠$ is semiprime exactly when $R ∖ 𝔠$ is an $n$ -system. The two notions are linked by $(10.10)$ , which shows every $n$ -system contains an $m$ -system through any prescribed point — the technical heart of the proof that semiprime ideals are intersections of primes.

Can I always take the

m

-system generated by a set?

There is no canonical generated $m$ -system, because the required $r$ is not determined. One can close a set under some choice of sandwiching elements, as in the inductive construction of $(10.10)$ , but the result depends on the choices made.

How does this relate to prime avoidance in commutative algebra?

It is the opposite direction. Prime avoidance says an ideal inside a finite union of primes lies in one of them; $(10.5)$ says a set closed under sandwiched products can be avoided by a single prime. The two are used for different purposes and neither implies the other.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §10 (pp. 163–181).
N. H. McCoy, “Prime ideals in general rings”, American Journal of Mathematics 71 (1949), 823–833.
N. H. McCoy, The Theory of Rings, Macmillan, New York, 1964.
K. R. Goodearl and R. B. Warfield, Jr., An Introduction to Noncommutative Noetherian Rings, 2nd edition, Cambridge University Press, 2004, Chapters 3 and 10.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that every $n$ -system containing $a$ contains an $m$ -system containing $a$ , and identify where choice is used.
For which rings and primes does $R ∖ 𝔭$ satisfy the right Ore condition?
Give an $m$ -system in a free algebra $k ⟨ x, y ⟩$ and describe a prime ideal avoiding it.
How is the set of minimal primes over an ideal obtained by an $m$ -system argument?
Compare $m$ -systems with the multiplicative sets used in Goldie's theorem for constructing quotient rings.
Is there a version of $(10.5)$ for rings without identity, and what changes in the proof?
What is the analogue of an $m$ -system for primitive ideals rather than prime ideals?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why not just multiplicatively closed sets

The powers of a single element

06Key Results

07Proof Techniques and Method

Negate to get closure

Push out and catch a witness

Absorb into the ideal

08Worked Example

Mn(D) is a prime ring, checked with matrix units

Producing a prime that avoids a prescribed element

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Design Considerations

13Failure Modes and Common Mistakes

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

$M_{n} (D)$ is a prime ring, checked with matrix units

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