Is A always nil modulo A?

Yes: (10.6) gives A ⊆ s : s^n ∈ A, so every element of A is nilpotent in R/A. The converse fails, and A need not be nilpotent modulo A — nil and nilpotent differ without chain conditions.

Does commute with quotients?

Yes in the form that matters: for A ⊆ B, B/A = B/A computed in R/A, because the primes above B correspond to the primes of R/A above B/A. The special case B = A gives A/A = Nil_*(R/A).

The Radical of an Ideal as an Intersection of Primes

Q: Why is A an ideal when its definition mentions no ring operations?

Because (10.7) identifies it with an intersection of prime ideals, and any intersection of two-sided ideals is a two-sided ideal. There is no direct proof from (10.6) that avoids the prime characterisation, which is precisely why Lam proves the theorem before drawing any consequence.

Q: What happens if no prime ideal contains A?

That occurs only for A = R, since any proper ideal lies in a maximal ideal and maximal ideals are prime. In that case the intersection is empty and equals R by convention, matching R = R from (10.6).

Q: Can I use only the minimal primes above A?

Yes. Every prime above A contains one that is minimal over A, because a descending chain of primes has prime intersection and Zorn applies. So the intersection over minimal primes gives the same ideal, and it is the form used computationally.

Q: Why is the notation a square root?

Historical inheritance from commutative algebra, where A is literally obtained by extracting roots of elements. In the noncommutative theory nothing is extracted and the symbol is pure notation; some authors write rad(A) or N(A) instead, at the cost of a clash with the Jacobson radical.

Executive Summary

In a commutative ring, $\sqrt{𝔄} = {s : s^{n} \in 𝔄}$ and this happens to be both an ideal and the intersection of the primes above $𝔄$ . In a noncommutative ring the element description fails on both counts — it is usually not an ideal — so the definition is taken from the $m$ -system side instead.

Definition $(10.6)$ declares $s \in \sqrt{𝔄}$ when every $m$ -system through $s$ meets $𝔄$ . Theorem $(10.7)$ then identifies this set with $⋂ {𝔭 \supseteq 𝔄 : 𝔭 prime}$ , which settles at one stroke that it is an ideal, that it is intrinsic, and that it generalises the commutative radical correctly.

$⋂ 𝔭$ What it equals

$m$ -systemsHow it is defined

$\subseteq$ nilRelation to powers

(10.7)The identification

Overview

Fix a ring $R$ with identity and an ideal $𝔄 \subseteq R$ .

\sqrt{𝔄} : = {s \in R : every m -system containing s meets 𝔄} \subseteq {s \in R : s^{n} \in 𝔄 for some n \geq 1}

(10.6)

The definition, together with the containment that justifies the radical notation. The containment can be strict.

The containment is immediate from the standard $m$ -system ${s, s^{2}, s^{4}, s^{8}, \dots}$ : it contains $s$ , so if $s \in \sqrt{𝔄}$ it must meet $𝔄$ , giving $s^{2^{i}} \in 𝔄$ for some $i$ .

When $𝔄 = R$ there are no primes above $𝔄$ and the empty intersection is $R$ ; consistently, $\sqrt{R} = R$ , since every $m$ -system is a nonempty subset of $R$ and so meets $R$ .

Learning Objectives

Parse $(10.6)$ : which quantifier is universal, which existential, and why.
Prove $\sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄}$ using the doubling-powers $m$ -system.
Prove both inclusions of $(10.7)$ , identifying where $(10.4)$ and $(10.5)$ are used.
Deduce that $\sqrt{𝔄}$ is an ideal, that $𝔄 \subseteq \sqrt{𝔄}$ , and that $\sqrt{\sqrt{𝔄}} = \sqrt{𝔄}$ .
Show that in the commutative case the containment in $(10.6)$ is an equality.
Compute $\sqrt{M_{2} (12 ℤ)}$ inside $M_{2} (ℤ)$ and exhibit the failure of the element description.

Definitions

Definition(10.6)Radical of an ideal

For an ideal $𝔄$ of a ring $R$ , set $\sqrt{𝔄} = {s \in R : every m -system S with s \in S satisfies S \cap 𝔄 \neq \emptyset}$ .

$\sqrt{(0)}$: The lower nilradical ${Nil}_{*} R$ , the intersection of all prime ideals of $R$ .
Radical ideal: An ideal with $𝔄 = \sqrt{𝔄}$ ; in the noncommutative setting these are exactly the semiprime ideals.
${s : s^{n} \in 𝔄}$: The set of elements nilpotent modulo $𝔄$ . In general only a subset-containing set, not an ideal, and strictly larger than $\sqrt{𝔄}$ .
Minimal prime over $𝔄$: A prime $𝔭 \supseteq 𝔄$ minimal with that property; the intersection in $(10.7)$ may be restricted to these.

The square-root notation is inherited from commutative algebra, where the radical really is defined by extracting roots. Here it is a name, not a description.

Core Concepts

Why the element description is abandoned

In $M_{2} (k)$ the elements $e_{12}$ and $e_{21}$ both square to zero, so both are nilpotent modulo $(0)$ . Their sum satisfies $(e_{12} + e_{21})^{2} = e_{11} + e_{22} = 1$ , so it is a unit. The set of nilpotent elements of a noncommutative ring is therefore not closed under addition and cannot serve as a radical.

Definition $(10.6)$ sidesteps this by asking a question about all $m$ -systems through $s$ , a condition strong enough to be stable under the ring operations — though only the identification $(10.7)$ makes that stability visible.

How the two inclusions are proved

Both directions of $(10.7)$ are one-line applications of the previous page. That $\sqrt{𝔄}$ lies in each prime $𝔭 \supseteq 𝔄$ uses $(10.4)$ : the complement $R ∖ 𝔭$ is an $m$ -system that misses $𝔄$ . That every element outside $\sqrt{𝔄}$ escapes some prime uses $(10.5)$ : an $m$ -system witnessing $s \notin \sqrt{𝔄}$ can be avoided by a prime above $𝔄$ .

s \notin \sqrt{𝔄}

m

-system

S ∋ s

S \cap 𝔄 = \emptyset

prime

𝔭 \supseteq 𝔄

𝔭 \cap S = \emptyset

s \notin 𝔭

The commutative case, recovered

When $R$ is commutative the containment of $(10.6)$ is an equality. Suppose $s^{n} \in 𝔄$ and let $S$ be any $m$ -system with $s \in S$ . Applying the defining property repeatedly produces elements $s^{2} r_{1}, s^{3} r_{1} r_{2}, \dots$ of $S$ — commutativity lets the ring elements be collected on the right — so some $s^{n} r \in S$ . Since $s^{n} \in 𝔄$ and $𝔄$ is an ideal, $s^{n} r \in 𝔄$ , so $S$ meets $𝔄$ and $s \in \sqrt{𝔄}$ .

Key Results

Theorem(10.7)Radical as an intersection of primes

Let $R$ be a ring with identity and $𝔄 \subseteq R$ an ideal. Then

\sqrt{𝔄} = ⋂_{substack 𝔭 \supseteq 𝔄}

In particular $\sqrt{𝔄}$ is a two-sided ideal of $R$ containing $𝔄$ .

Proof

** $\subseteq$ .** Let $s \in \sqrt{𝔄}$ and let $𝔭 \supseteq 𝔄$ be prime. By $(10.4)$ the complement $S = R ∖ 𝔭$ is an $m$ -system, and $S \cap 𝔄 \subseteq S \cap 𝔭 = \emptyset$ . If $s$ belonged to $S$ , then $S$ would be an $m$ -system containing $s$ that misses $𝔄$ , contradicting $s \in \sqrt{𝔄}$ . Hence $s \in 𝔭$ .

** $\supseteq$ .** Suppose $s \notin \sqrt{𝔄}$ . By definition there is an $m$ -system $S$ with $s \in S$ and $S \cap 𝔄 = \emptyset$ . Order by inclusion the ideals containing $𝔄$ and disjoint from $S$ ; the set is nonempty (it contains $𝔄$ ) and closed under unions of chains, so Zorn's Lemma gives a maximal element $𝔭$ . By $(10.5)$ , $𝔭$ is prime, and $𝔭 \supseteq 𝔄$ with $𝔭 \cap S = \emptyset$ ; since $s \in S$ we get $s \notin 𝔭$ . So $s$ is missing from the intersection.

Both inclusions give the equality. The intersection of two-sided ideals is a two-sided ideal, and every prime above $𝔄$ contains $𝔄$ , so $𝔄 \subseteq \sqrt{𝔄}$ .

Corollary—Formal properties

For ideals $𝔄, 𝔅$ of a ring $R$ :

$𝔄 \subseteq 𝔅$ implies $\sqrt{𝔄} \subseteq \sqrt{𝔅}$ ;
$\sqrt{\sqrt{𝔄}} = \sqrt{𝔄}$ ;
$\sqrt{𝔄 \cap 𝔅} = \sqrt{𝔄 𝔅} = \sqrt{𝔄} \cap \sqrt{𝔅}$ ;
$\sqrt{𝔄} / 𝔄 = {Nil}_{*} (R / 𝔄)$ .

Proof

(1) A prime containing $𝔅$ contains $𝔄$ , so the intersection defining $\sqrt{𝔅}$ runs over a subfamily of that defining $\sqrt{𝔄}$ .

(2) The primes above $\sqrt{𝔄}$ are exactly the primes above $𝔄$ : one direction is $𝔄 \subseteq \sqrt{𝔄}$ , the other is $(10.7)$ , which puts $\sqrt{𝔄}$ inside every prime above $𝔄$ . Equal families give equal intersections.

(3) A prime $𝔭$ contains $𝔄 𝔅$ iff it contains $𝔄$ or $𝔅$ , by $(10.1)$ ; and $𝔄 𝔅 \subseteq 𝔄 \cap 𝔅 \subseteq 𝔄$ shows the same family of primes arises from $𝔄 \cap 𝔅$ . Intersecting over the union of the two families gives $\sqrt{𝔄} \cap \sqrt{𝔅}$ .

(4) The primes of $R / 𝔄$ are the images of the primes of $R$ containing $𝔄$ , and intersection commutes with the quotient map on ideals containing $𝔄$ .

Proposition—The commutative case

If $R$ is commutative then $\sqrt{𝔄} = {s \in R : s^{n} \in 𝔄 for some n \geq 1}$ , so $(10.6)$ recovers the classical radical and $(10.7)$ recovers the classical theorem that a radical ideal is an intersection of primes.

Proof

The inclusion $\subseteq$ holds in any ring by $(10.6)$ . For $\supseteq$ , let $s^{n} \in 𝔄$ and let $S$ be an $m$ -system containing $s$ . Build elements of $S$ inductively: $t_{1} = s$ , and given $t_{i} \in S$ choose $r_{i}$ with $t_{i + 1} = s r_{i} t_{i} \in S$ . By commutativity $t_{i} = s^{i} r_{1} \dots r_{i - 1}$ , so $t_{n} \in s^{n} R \subseteq 𝔄$ . Hence $S$ meets $𝔄$ , and as $S$ was arbitrary, $s \in \sqrt{𝔄}$ .

Remark—Only minimal primes matter

Every prime above $𝔄$ contains a prime minimal over $𝔄$ — a descending chain of primes has prime intersection, so Zorn applies downwards. Hence the intersection in $(10.7)$ may be taken over the minimal primes above $𝔄$ alone, which is how radicals are computed in practice.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Define by a universal condition

When a set has no obvious algebraic structure, define it by a condition quantified over all objects of an auxiliary type — here all $m$ -systems — and recover the structure by a separate identification theorem.

Move 2

Test membership by escaping

To show $s$ is not in an intersection of primes, produce one prime missing $s$ . Existence comes from Zorn plus $(10.5)$ ; this converts a negative statement into a construction.

Move 3

Reduce to the quotient

$\sqrt{𝔄}$ is the preimage of ${Nil}_{*} (R / 𝔄)$ . Any question about radicals of ideals may be moved to a question about the lower nilradical of a quotient ring, and usually should be.

Move 3 explains the architecture of this section: Lam proves everything for a general ideal $𝔄$ first, then specialises to $𝔄 = (0)$ to define ${Nil}_{*} R$ . Nothing new is needed at the second step.

Worked Example

Computing $\sqrt{M_{2} (12 ℤ)}$

Let $R = M_{2} (ℤ)$ and $𝔄 = M_{2} (12 ℤ)$ . Every ideal of $M_{n} (S)$ has the form $M_{n} (I)$ for a unique ideal $I \subseteq S$ , and $M_{n} (I)$ is prime exactly when $I$ is prime. The prime ideals of $ℤ$ containing $12 ℤ$ are $2 ℤ$ and $3 ℤ$ , so the primes of $R$ above $𝔄$ are $M_{2} (2 ℤ)$ and $M_{2} (3 ℤ)$ . Therefore

\sqrt{M_{2} (12 ℤ)} = M_{2} (2 ℤ) \cap M_{2} (3 ℤ) = M_{2} (6 ℤ) .

(E.1)

Equivalently ${Nil}_{*} (M_{2} (ℤ / 12 ℤ)) = M_{2} (6 ℤ / 12 ℤ)$ , by property (4) of the corollary.

The element description fails here

Consider $s = e_{12} = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ . Then $s^{2} = 0 \in 𝔄$ , so $s$ is nilpotent modulo $𝔄$ ; but $s \notin M_{2} (6 ℤ)$ because its $(1, 2)$ entry is $1$ . Hence

\sqrt{𝔄} = M_{2} (6 ℤ) ⊊ {s \in R : s^{n} \in 𝔄 for some n},

(E.2)

The containment of (10.6) is strict, and the larger set is not even closed under addition.

Indeed $e_{12}$ and $e_{21}$ both square to $0 \in 𝔄$ , while $(e_{12} + e_{21})^{2} = e_{11} + e_{22} = 1$ and no power of $1$ lies in $𝔄$ . So the nilpotent-modulo set is not closed under addition and cannot be an ideal, whereas $\sqrt{𝔄}$ always is.

Comparison and Classification

The radical: commutative versus general
Feature	Commutative $R$	General $R$
Definition used	${s : s^{n} \in 𝔄}$	$m$ -system condition $(10.6)$
Is it an ideal?	yes, directly	yes, but only via $(10.7)$
Intersection of primes	yes	yes
Element description	exact	only an upper bound, often strict
Fixed points	radical ideals	semiprime ideals
$\sqrt{(0)}$	$Nil R$	${Nil}_{*} R \subseteq Nil R$

Radicals of some explicit ideals
	$\sqrt{𝔄}$	equals the nilpotent-modulo set?	semiprime?
$𝔄 = 12 ℤ \subseteq ℤ$	$6 ℤ$	yes	yes
$𝔄 = M_{2} (12 ℤ) \subseteq M_{2} (ℤ)$	$M_{2} (6 ℤ)$	no	yes
$𝔄 = (0) \subseteq M_{2} (k)$	$0$	no	yes
$𝔄 = (0) \subseteq T_{2} (k)$	$k e_{12}$	yes	yes
$𝔄 = (x^{3}) \subseteq k [x]$	$(x)$	yes	yes
$𝔄 = R$	$R$	yes	yes

Radicals of some explicit ideals

Relationship Map

$R$ the whole ring

${s : s^{n} \in 𝔄}$ elements nilpotent modulo

𝔄

— a set, generally not an ideal

$\sqrt{𝔄}$ the intersection of all primes above

𝔄

— the smallest semiprime ideal containing

𝔄

$𝔄$ the ideal you started with

Reading outward: $𝔄 \subseteq \sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄} \subseteq R$ , with the middle containment an equality precisely when the nilpotent-modulo set happens to be an ideal — automatic in the commutative case, rare otherwise.

𝔄

any ideal

\sqrt{𝔄}

semiprime

\sqrt{𝔄} / 𝔄 = {Nil}_{*} (R / 𝔄)

R / \sqrt{𝔄}

semiprime

Computational Notes

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

Commutative, finitely generated over a field. $\sqrt{𝔄}$ is computable: Gröbner-basis algorithms (Krick–Logar, Eisenbud–Huneke–Vasconcelos) are implemented as radical in Singular and Macaulay2. Worst-case cost is doubly exponential in the number of variables, and characteristic $p$ needs separate handling because of inseparability.
Finite-dimensional algebras over a field. Here ${Nil}_{*} A = {Nil}^{*} A = rad A$ , so the radical of an ideal reduces to a Jacobson radical computation in the quotient — polynomial time by the trace form in characteristic $0$ and by the Friedl–Rónyai algorithm in characteristic $p$ .
Finitely presented noncommutative algebras. No general algorithm exists: the word problem is already undecidable, so membership in $\sqrt{𝔄}$ cannot be decided in general. Non-commutative Gröbner bases (Plural, GBNP) terminate only under favourable term orders and degree bounds.
Practical shortcut. Because $\sqrt{𝔄}$ is the intersection of the minimal primes above $𝔄$ , any algorithm that enumerates minimal primes computes the radical; in the commutative world this is exactly what primary decomposition packages do.

Failure Modes and Common Mistakes

Do not assume $\sqrt{𝔄 𝔅} = \sqrt{𝔄} \sqrt{𝔅}$ — the correct identity is $\sqrt{𝔄 𝔅} = \sqrt{𝔄} \cap \sqrt{𝔅}$ .
Do not assume the intersection in $(10.7)$ is over finitely many primes; it is finite only under strong hypotheses such as Noetherian conditions.
Do not apply the radical to one-sided ideals. $(10.6)$ and $(10.7)$ are statements about two-sided ideals, and the prime ideals involved are two-sided.
Do not confuse $\sqrt{(0)} = {Nil}_{*} R$ with the Jacobson radical. The containment ${Nil}_{*} R \subseteq rad R$ is usually strict — ${Nil}_{*} k [[x]] = 0$ but $rad k [[x]] = (x)$ .
Do not forget the degenerate case: $\sqrt{R} = R$ , and $R$ is not prime, so the empty intersection convention is doing real work.

Quick Reference

Definition

s \in \sqrt{𝔄}

iff every

m

-system containing

s

meets

𝔄

Identification

\sqrt{𝔄} = ⋂ {𝔭 prime : 𝔭 \supseteq 𝔄}

Bound

\sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄}

, strict in general

Commutativeequality; the classical radical

Idempotence

\sqrt{\sqrt{𝔄}} = \sqrt{𝔄}

Products

\sqrt{𝔄 𝔅} = \sqrt{𝔄 \cap 𝔅} = \sqrt{𝔄} \cap \sqrt{𝔅}

Quotient

\sqrt{𝔄} / 𝔄 = {Nil}_{*} (R / 𝔄)

Fixed points

𝔄 = \sqrt{𝔄}

iff

𝔄

is semiprime

Proving membership and non-membership
Goal	Cheapest route	Reference
$s \in \sqrt{𝔄}$	show $s$ lies in every prime above $𝔄$	(10.7)
$s \notin \sqrt{𝔄}$	exhibit one $m$ -system through $s$ missing $𝔄$	(10.6)
$\sqrt{𝔄} = 𝔄$	show $𝔄$ is semiprime	(10.11)
compute $\sqrt{𝔄}$	intersect the minimal primes above $𝔄$	(10.7) plus Zorn
$\sqrt{(0)}$	identify ${Nil}_{*} R$	(10.13)

Frequently Asked Questions

Why is

\sqrt{𝔄}

an ideal when its definition mentions no ring operations?

Because $(10.7)$ identifies it with an intersection of prime ideals, and any intersection of two-sided ideals is a two-sided ideal. There is no direct proof from $(10.6)$ that avoids the prime characterisation, which is precisely why Lam proves the theorem before drawing any consequence.

\sqrt{𝔄}

always nil modulo

𝔄

Yes: $(10.6)$ gives $\sqrt{𝔄} \subseteq {s : s^{n} \in 𝔄}$ , so every element of $\sqrt{𝔄}$ is nilpotent in $R / 𝔄$ . The converse fails, and $\sqrt{𝔄}$ need not be nilpotent modulo $𝔄$ — nil and nilpotent differ without chain conditions.

What happens if no prime ideal contains

𝔄

That occurs only for $𝔄 = R$ , since any proper ideal lies in a maximal ideal and maximal ideals are prime. In that case the intersection is empty and equals $R$ by convention, matching $\sqrt{R} = R$ from $(10.6)$ .

Does

\sqrt{}

commute with quotients?

Yes in the form that matters: for $𝔄 \subseteq 𝔅$ , $\sqrt{𝔅} / 𝔄 = \sqrt{𝔅 / 𝔄}$ computed in $R / 𝔄$ , because the primes above $𝔅$ correspond to the primes of $R / 𝔄$ above $𝔅 / 𝔄$ . The special case $𝔅 = 𝔄$ gives $\sqrt{𝔄} / 𝔄 = {Nil}_{*} (R / 𝔄)$ .

Can I use only the minimal primes above

𝔄

Yes. Every prime above $𝔄$ contains one that is minimal over $𝔄$ , because a descending chain of primes has prime intersection and Zorn applies. So the intersection over minimal primes gives the same ideal, and it is the form used computationally.

Why is the notation a square root?

Historical inheritance from commutative algebra, where $\sqrt{𝔄}$ is literally obtained by extracting roots of elements. In the noncommutative theory nothing is extracted and the symbol is pure notation; some authors write $rad (𝔄)$ or $N (𝔄)$ instead, at the cost of a clash with the Jacobson radical.

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.
R. Baer, “Radical ideals”, American Journal of Mathematics 65 (1943), 537–568.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.
D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry, Graduate Texts in Mathematics 150, Springer-Verlag, 1995, Chapter 4 (for the commutative comparison).

AI Suggested Questions

Show that a descending chain of prime ideals has prime intersection, and deduce the existence of minimal primes over an ideal.
Find a ring and an ideal for which the set of elements nilpotent modulo the ideal is not closed under multiplication either.
How is $\sqrt{𝔄}$ computed for an ideal of a finite-dimensional algebra given by structure constants?
Is there a noncommutative analogue of primary decomposition that refines the intersection in (10.7)?
Compare $\sqrt{(0)}$ with the set of nilpotent elements for the free algebra and for the Weyl algebra.
What is the radical of an ideal in a ring without identity, and which steps of the proof of (10.7) change?
Explain how (10.7) specialises to the commutative Nullstellensatz correspondence between radical ideals and varieties.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the element description is abandoned

How the two inclusions are proved

The commutative case, recovered

06Key Results

07Proof Techniques and Method

Define by a universal condition

Test membership by escaping

Reduce to the quotient

08Worked Example

Computing M2(12ℤ)

The element description fails here

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Quick Reference

14Frequently Asked Questions

15Related KEVOS Topics

16References

17AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Computing $\sqrt{M_{2} (12 ℤ)}$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions