Semiprime Ideals and n-Systems

Q: Why is R allowed to be a semiprime ideal when it is not allowed to be prime?

So that (10.11) is true without exceptions. R is the intersection of the empty family of primes, and if it were excluded the statement *semiprime = intersection of primes* would fail for rings whose only ideal is R — and (10.12) would have no closure operator for c = R.

Q: Why does the proof of (10.11) need the lemma about n-systems?

Because semiprimeness is a statement about n-systems while the radical is defined by m-systems. To show c ⊆ c you must produce an m-system missing c through a given point outside c, and the complement only gives you an n-system. (10.10) bridges the gap.

Q: Does a semiprime ring have no nilpotent elements?

No — that is the *reduced* condition. A semiprime ring has no nonzero nilpotent one-sided ideals, but may have plenty of nilpotent elements: M_2(k) is semiprime, indeed simple, and e_12^2 = 0.

Executive Summary

Prime ideals are the noncommutative analogue of primes; semiprime ideals are the analogue of radical ideals. The definition is the squaring condition $𝔄^{2} \subseteq 𝔠 \Rightarrow 𝔄 \subseteq 𝔠$ , and $(10.9)$ reduces it to the single-element test $a R a \subseteq 𝔠 \Rightarrow a \in 𝔠$ .

The main theorem $(10.11)$ closes the loop opened in the previous two pages: semiprime, radical-fixed and intersection-of-primes are the same condition. The technical bridge is $(10.10)$ , a small inductive lemma that finds an $m$ -system inside any $n$ -system.

$𝔄^{2} \subseteq 𝔠$ Defining test

$a R a$ Element form

3Equivalent global descriptions (10.11)

$n$ -systemComplement structure

Overview

In a commutative ring, an ideal is radical when $a^{n} \in 𝔠$ forces $a \in 𝔠$ , and radical ideals are precisely the intersections of primes. Both halves generalise, but as with primeness the element condition must be sandwiched.

𝔄^{2} \subseteq 𝔠 ⟹ 𝔄 \subseteq 𝔠 for every ideal 𝔄 \subseteq R

(10.8)

The definition of a semiprime ideal. Unlike $(10.1)$ there is no properness requirement, so $𝔠 = R$ is semiprime — the empty intersection of primes.

Every prime ideal is semiprime, since $𝔄 𝔄 \subseteq 𝔠$ is a special case of $(10.1)$ . The converse fails badly: in the upper triangular ring $T_{2} (k)$ the ideal $k e_{12}$ is semiprime but not prime, being the intersection of the two maximal ideals.

Learning Objectives

State $(10.8)$ and derive the element test $(10.9) (3)$ .
Prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$ of $(10.9)$ , including the right-handed variant.
Show that a proper ideal is semiprime iff its complement is an $n$ -system.
Prove $(10.10)$ by constructing $a_{i + 1} = a_{i} r_{i} a_{i}$ and verifying the $m$ -system property.
Prove $(10.11)$ and deduce $(10.12)$ : $\sqrt{𝔠}$ is the smallest semiprime ideal above $𝔠$ .
Show that $𝔄^{n} \subseteq 𝔠$ implies $𝔄 \subseteq 𝔠$ for semiprime $𝔠$ .

Definitions

Definition(10.8)Semiprime ideal

An ideal $𝔠$ of a ring $R$ is semiprime if for every ideal $𝔄 \subseteq R$ , $𝔄^{2} \subseteq 𝔠$ implies $𝔄 \subseteq 𝔠$ .

Definition—n-System

A nonempty subset $S \subseteq R$ is an ** $n$ -system** if for every $a \in S$ there exists $r \in R$ with $a r a \in S$ . Setting $b = a$ in $(10.3)$ shows that every $m$ -system is an $n$ -system.

Completely semiprime: $a^{2} \in 𝔠 \Rightarrow a \in 𝔠$ ; equivalently $R / 𝔠$ is reduced. Implies semiprime; the converse fails, as $(0) \subseteq M_{2} (k)$ shows.
Nilpotent ideal: $𝔄^{n} = 0$ for some $n$ . A semiprime ideal contains every ideal nilpotent modulo it.
Semiprime closure: $\sqrt{𝔠}$ , the smallest semiprime ideal containing $𝔠$ , by $(10.12)$ .
Semiprime ring: $(0)$ is semiprime; equivalently there is no nonzero nilpotent ideal, equivalently no nonzero nilpotent left ideal.

The improper ideal R is semiprime by convention and by the definition, matching the convention that R is the intersection of the empty family of primes.

Core Concepts

Squaring is enough

The definition only mentions $𝔄^{2}$ , but it controls all powers. If $𝔠$ is semiprime and $𝔄^{n} \subseteq 𝔠$ , choose $m$ with $2 m \geq n$ ; then $(𝔄^{m})^{2} = 𝔄^{2 m} \subseteq 𝔄^{n} \subseteq 𝔠$ , so $𝔄^{m} \subseteq 𝔠$ , and descending gives $𝔄 \subseteq 𝔠$ . In particular a semiprime ideal absorbs every ideal that is nilpotent modulo it.

Why $a R a$ and not $a^{2}$

The condition $a^{2} \in 𝔠 \Rightarrow a \in 𝔠$ defines completely semiprime ideals, whose quotients are reduced rings. That is too strong: $M_{2} (k)$ is a simple ring, so $(0)$ must be semiprime in any usable theory, yet $e_{12}^{2} = 0$ with $e_{12} \neq 0$ . Inserting the ring — $e_{12} R e_{12} = 0$ is false, since $e_{12} e_{21} e_{12} = e_{12}$ — restores the correct verdict.

$n$ -systems and the missing link

Complements again convert the ideal condition into a closure condition: a proper ideal $𝔠$ is semiprime exactly when $R ∖ 𝔠$ is an $n$ -system. But the radical $\sqrt{}$ was defined by $m$ -systems, not $n$ -systems, so a comparison is needed. That is $(10.10)$ : any $n$ -system contains an $m$ -system through any prescribed element. Its proof is a two-line induction whose only subtlety is the bookkeeping of indices.

𝔠

semiprime

R ∖ 𝔠

n

-systemcontains an

m

-system through each point

\sqrt{𝔠} \subseteq 𝔠

Key Results

Proposition(10.9)Characterisations of semiprimeness

Let $R$ be a ring with identity and $𝔠 \subseteq R$ an ideal. The following are equivalent:

$𝔠$ is semiprime;
for $a \in R$ , $(a)^{2} \subseteq 𝔠$ implies $a \in 𝔠$ ;
for $a \in R$ , $a R a \subseteq 𝔠$ implies $a \in 𝔠$ ;
for every left ideal $𝔄$ of $R$ , $𝔄^{2} \subseteq 𝔠$ implies $𝔄 \subseteq 𝔠$ ;
for every right ideal $𝔄$ of $R$ , $𝔄^{2} \subseteq 𝔠$ implies $𝔄 \subseteq 𝔠$ .

Proof

**(1) $\Rightarrow$ (2).** $(a) = R a R$ is an ideal, so this is the definition applied to $𝔄 = (a)$ .

**(2) $\Rightarrow$ (3).** If $a R a \subseteq 𝔠$ then $(a)^{2} = (R a R) (R a R) = R a (R R) a R \subseteq R (a R a) R \subseteq 𝔠$ , since $𝔠$ is an ideal. Apply (2).

**(3) $\Rightarrow$ (4).** Let $𝔄$ be a left ideal with $𝔄^{2} \subseteq 𝔠$ and let $a \in 𝔄$ . Then $R a \subseteq 𝔄$ , so $a R a \subseteq a 𝔄 \subseteq 𝔄 𝔄 \subseteq 𝔠$ . By (3), $a \in 𝔠$ ; hence $𝔄 \subseteq 𝔠$ .

**(4) $\Rightarrow$ (1).** Every ideal is a left ideal.

For (5), argue on the other side: if $𝔄$ is a right ideal with $𝔄^{2} \subseteq 𝔠$ and $a \in 𝔄$ , then $a R \subseteq 𝔄$ gives $a R a \subseteq 𝔄 a \subseteq 𝔄^{2} \subseteq 𝔠$ , so (3) $\Rightarrow$ (5) $\Rightarrow$ (1) as well.

Corollary—Complements are n-systems

A proper ideal $𝔠 ⊊ R$ is semiprime if and only if $R ∖ 𝔠$ is an $n$ -system. Indeed, $(10.9) (3)$ says exactly that whenever $a \notin 𝔠$ there is $r \in R$ with $a r a \notin 𝔠$ . (For $𝔠 = R$ the complement is empty; $R$ is semiprime by convention.)

Lemma(10.10)An m-system inside an n-system

Let $N$ be an $n$ -system in a ring $R$ and let $a \in N$ . Then there is an $m$ -system $M$ with $a \in M \subseteq N$ .

Proof

Define elements of $N$ recursively: $a_{1} = a$ , and having chosen $a_{i} \in N$ , use the $n$ -system property to pick $r_{i} \in R$ with $a_{i + 1} = a_{i} r_{i} a_{i} \in N$ . Put $M = {a_{1}, a_{2}, a_{3}, \dots} \subseteq N$ ; clearly $a \in M$ .

First observe that $a_{j} \in a_{i} R a_{i}$ whenever $j > i$ . This holds for $j = i + 1$ by construction, and if $a_{j} \in a_{i} R a_{i}$ then $a_{j + 1} = a_{j} r_{j} a_{j} \in (a_{i} R a_{i}) r_{j} (a_{i} R a_{i}) \subseteq a_{i} R a_{i}$ .

Now take any $a_{i}, a_{j} \in M$ . If $i \leq j$ then $a_{i} R a_{j} \supseteq a_{j} R a_{j} ∋ a_{j + 1}$ , using $a_{j} \in a_{i} R a_{i} \subseteq a_{i} R$ for $i < j$ and the trivial case $i = j$ . If $i > j$ then $a_{i} R a_{j} \supseteq a_{i} R a_{i} ∋ a_{i + 1}$ , using $a_{i} \in a_{j} R a_{j} \subseteq R a_{j}$ . Either way $a_{i} R a_{j}$ meets $M$ , so $M$ is an $m$ -system.

Theorem(10.11)Semiprime equals intersection of primes

For an ideal $𝔠$ of a ring $R$ the following are equivalent:

$𝔠$ is a semiprime ideal;
$𝔠$ is an intersection of prime ideals of $R$ ;
$𝔠 = \sqrt{𝔠}$ .

In the commutative case this says that semiprime ideals are exactly the radical ideals.

Proof

**(3) $\Rightarrow$ (2).** By $(10.7)$ , $\sqrt{𝔠}$ is the intersection of the primes containing $𝔠$ .

**(2) $\Rightarrow$ (1).** Let $𝔠 = ⋂_{i} 𝔭_{i}$ with each $𝔭_{i}$ prime, and let $𝔄^{2} \subseteq 𝔠$ . For each $i$ , $𝔄 𝔄 \subseteq 𝔭_{i}$ gives $𝔄 \subseteq 𝔭_{i}$ by $(10.1)$ ; intersecting, $𝔄 \subseteq 𝔠$ .

**(1) $\Rightarrow$ (3).** Always $𝔠 \subseteq \sqrt{𝔠}$ , so it suffices to prove $\sqrt{𝔠} \subseteq 𝔠$ ; for $𝔠 = R$ this is trivial, so assume $𝔠$ proper. Let $a \notin 𝔠$ . Then $N = R ∖ 𝔠$ is an $n$ -system containing $a$ , by the corollary above. By $(10.10)$ there is an $m$ -system $M$ with $a \in M \subseteq N$ , and $M \cap 𝔠 = \emptyset$ because $M \subseteq R ∖ 𝔠$ . So $M$ is an $m$ -system containing $a$ and missing $𝔠$ , whence $a \notin \sqrt{𝔠}$ by $(10.6)$ .

Corollary(10.12)Semiprime closure

For any ideal $𝔠 \subseteq R$ , $\sqrt{𝔠}$ is the smallest semiprime ideal of $R$ containing $𝔠$ .

Proof

$\sqrt{𝔠}$ is an intersection of primes by $(10.7)$ , hence semiprime by $(10.11)$ , and it contains $𝔠$ . If $𝔡 \supseteq 𝔠$ is semiprime, monotonicity of the radical and $(10.11)$ give $\sqrt{𝔠} \subseteq \sqrt{𝔡} = 𝔡$ .

Corollary—Absorbing nilpotent ideals

If $𝔠$ is semiprime and $𝔄$ is a left, right or two-sided ideal with $𝔄^{n} \subseteq 𝔠$ for some $n \geq 1$ , then $𝔄 \subseteq 𝔠$ . Choose $m$ with $2 m \geq n$ : then $(𝔄^{m})^{2} \subseteq 𝔄^{n} \subseteq 𝔠$ , so $𝔄^{m} \subseteq 𝔠$ by $(10.9)$ , and repeating halves the exponent until it reaches $1$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Set $b = a$

Every statement about semiprime ideals is the corresponding statement about prime ideals with the two elements identified. Proofs transfer verbatim; only the conclusion weakens.

Move 2

Iterate the witness

In $(10.10)$ the $n$ -system property is applied to its own output: $a_{i + 1} = a_{i} r_{i} a_{i}$ . The resulting sequence is automatically nested inside every earlier $a_{i} R a_{i}$ , which is what upgrades $n$ to $m$ .

Move 3

Halve the exponent

To pass from $𝔄^{n} \subseteq 𝔠$ to $𝔄 \subseteq 𝔠$ , apply the squaring condition to $𝔄^{lceil n / 2 rceil}$ . Repeated halving reaches exponent $1$ in $\log_{2} n$ steps.

Move 2 is the only genuinely new idea in the section. It is worth remembering as a template: to strengthen a closure property, close the witnessing construction under itself and check that the resulting set is nested.

Worked Example

Which ideals of $T_{2} (k)$ are semiprime?

Let $R = T_{2} (k)$ , upper triangular $2 \times 2$ matrices over a field. Its ideals are $0$ , $𝔍 = k e_{12}$ , $ℑ_{1} = k e_{11} \oplus k e_{12}$ , $ℑ_{2} = k e_{12} \oplus k e_{22}$ and $R$ , with $ℑ_{1}, ℑ_{2}$ maximal and $𝔍 = ℑ_{1} \cap ℑ_{2}$ .

$0$ is not semiprime

Apply $(10.9) (3)$ with $a = e_{12}$ : for $X = (\begin{matrix} x & y \\ 0 & z \end{matrix})$ we get $e_{12} X e_{12} = z e_{12} e_{12} = 0$ , so $e_{12} R e_{12} = 0$ while $e_{12} \neq 0$ . Equivalently $𝔍^{2} = 0$ exhibits a nonzero nilpotent ideal.

$𝔍$ is semiprime

Direct verification with the $n$ -system criterion. Let $a = (\begin{matrix} x & y \\ 0 & z \end{matrix}) \notin 𝔍$ , so $x \neq 0$ or $z \neq 0$ . If $x \neq 0$ , then $a e_{11} a = (\begin{matrix} x^{2} & x y \\ 0 & 0 \end{matrix}) \notin 𝔍$ ; if $z \neq 0$ , then $a e_{22} a = (\begin{matrix} 0 & y z \\ 0 & z^{2} \end{matrix}) \notin 𝔍$ . So $R ∖ 𝔍$ is an $n$ -system and $𝔍$ is semiprime — as it must be, being $ℑ_{1} \cap ℑ_{2}$ , an intersection of primes.

\sqrt{(0)} = ℑ_{1} \cap ℑ_{2} = k e_{12} = {Nil}_{*} T_{2} (k) .

(E.1)

By $(10.12)$ this is the smallest semiprime ideal of $T_{2} (k)$ ; the only smaller ideal, $0$ , fails.

An arithmetic check of (10.12)

In $ℤ$ the ideal $12 ℤ$ is not semiprime: $(6 ℤ)^{2} = 36 ℤ \subseteq 12 ℤ$ but $6 ℤ not \subseteq 12 ℤ$ . The primes above it are $2 ℤ$ and $3 ℤ$ , so $\sqrt{12 ℤ} = 6 ℤ$ , and $6 ℤ$ is indeed the smallest semiprime ideal containing $12 ℤ$ — the intermediate ideals $12 ℤ$ and $4 ℤ$ both fail the squaring test.

Frameworks and Models

Ideals of a ring — classified by the multiplicative conditions they satisfy
- Maximal
  - always prime
  - quotient is a simple ring
- Completely prime
  - $a b \in 𝔭 \Rightarrow a$ or $b \in 𝔭$
  - quotient is a domain
  - implies prime
- Prime
  - $a R b \subseteq 𝔭 \Rightarrow a$ or $b \in 𝔭$
  - quotient is a prime ring
  - implies semiprime
- Completely semiprime
  - $a^{2} \in 𝔠 \Rightarrow a \in 𝔠$
  - quotient is reduced
  - implies semiprime
- Semiprime
  - $a R a \subseteq 𝔠 \Rightarrow a \in 𝔠$
  - quotient has no nonzero nilpotent ideals
  - intersection of primes

The four implications *maximal $\Rightarrow$ prime*, *completely prime $\Rightarrow$ prime*, *completely semiprime $\Rightarrow$ semiprime* and *prime $\Rightarrow$ semiprime* are all strict. Witnesses: $(0) \subseteq ℤ$ is prime and not maximal; $(0) \subseteq M_{2} (k)$ is prime and not completely prime, and semiprime and not completely semiprime; $k e_{12} \subseteq T_{2} (k)$ is semiprime and not prime.

Comparison and Classification

Prime and semiprime side by side
Aspect	Prime ideal $𝔭$	Semiprime ideal $𝔠$
Ideal test	$𝔄 𝔅 \subseteq 𝔭$	$𝔄^{2} \subseteq 𝔠$
Element test	$a R b \subseteq 𝔭$	$a R a \subseteq 𝔠$
Complement	$m$ -system	$n$ -system
Properness	$𝔭 \neq R$ required	$𝔠 = R$ allowed
Global form	—	intersection of primes $(10.11)$
Quotient ring	prime ring	semiprime ring
Commutative analogue	prime ideal	radical ideal
Closure operator	—	$𝔠 \mapsto \sqrt{𝔠}$ , $(10.12)$

Status of some explicit ideals
	prime	semiprime	completely semiprime
$(0) \subseteq M_{2} (k)$	yes	yes	no
$(0) \subseteq T_{2} (k)$	no	no	no
$k e_{12} \subseteq T_{2} (k)$	no	yes	yes
$ℑ_{1} \subseteq T_{2} (k)$	yes	yes	yes
$12 ℤ \subseteq ℤ$	no	no	no
$6 ℤ \subseteq ℤ$	no	yes	yes
$3 ℤ \subseteq ℤ$	yes	yes	yes

Status of some explicit ideals

The third row is the one to remember: semiprime without prime — and note that it is completely semiprime, since $T_{2} (k) / k e_{12} ≅ k \times k$ is reduced, so the two strengthenings are independent. The first row is the other lesson: prime without completely semiprime, because $e_{12}^{2} = 0$ .

Relationship Map

𝔠

semiprime

𝔠 = \sqrt{𝔠}

𝔠 = ⋂ 𝔭_{i}

R / 𝔠

semiprime ring

All four conditions are interchangeable. The last is the bridge to Prime and Semiprime Rings, where semiprime rings are characterised by the absence of nonzero nilpotent left ideals; specialising $𝔠 = (0)$ turns every statement here into a statement about rings.

All idealsno condition

Semiprime idealsclosed under arbitrary intersections; fixed points of

\sqrt{}

Prime idealsnot closed under intersection

Maximal idealsthe primes with simple quotient

The middle band is a closure system: arbitrary intersections of semiprime ideals are semiprime, and $\sqrt{}$ is the associated closure operator. Prime ideals form no such system — the intersection of two primes is usually only semiprime, which is exactly the content of the $T_{2} (k)$ example.

Failure Modes and Common Mistakes

Do not require $𝔠 \neq R$ : unlike primeness, the definition of semiprime deliberately admits the whole ring, so that every ideal has a semiprime closure.
Do not confuse $n$ -systems with $m$ -systems. The inclusion is one-way, and $(10.10)$ recovers only an $m$ -system through one chosen point, not the whole set.
Do not assume a semiprime ideal contains no nilpotent elements — it contains no ideal nilpotent modulo it, which is a much weaker statement about elements.
Do not use the squaring test on arbitrary additive subgroups: $(10.9) (4)$ applies to one-sided ideals, and the proof genuinely uses $R 𝔄 \subseteq 𝔄$ .
Do not expect $\sqrt{𝔠}$ to be nilpotent modulo $𝔠$ ; it is only nil modulo $𝔠$ , and even that needs $(10.6)$ .

Best Practices

To prove an ideal semiprime, look first for a presentation as an intersection of primes — it is usually shorter than the $a R a$ verification.
To prove an ideal not semiprime, exhibit one nonzero ideal $𝔄$ with $𝔄^{2} \subseteq 𝔠$ ; a single square-zero ideal settles it.
When a proof needs a semiprime hypothesis, record which of the five forms of $(10.9)$ you are using — the one-sided versions $(4)$ and $(5)$ are what make arguments about nilpotent left ideals work.
Pass to $R / 𝔠$ early. Every statement about a semiprime ideal is a statement about a semiprime ring, where the literature is far richer.

Quick Reference

Definition

𝔄^{2} \subseteq 𝔠 \Rightarrow 𝔄 \subseteq 𝔠

Element test

a R a \subseteq 𝔠 \Rightarrow a \in 𝔠

One-sided formsame test for left ideals, and for right ideals

Complementproper

𝔠

semiprime iff

R ∖ 𝔠

is an

n

-system

Global formsemiprime

\Leftrightarrow

intersection of primes

\Leftrightarrow

𝔠 = \sqrt{𝔠}

Closure

\sqrt{𝔠}

= smallest semiprime ideal

\supseteq 𝔠

Powers

𝔄^{n} \subseteq 𝔠 \Rightarrow 𝔄 \subseteq 𝔠

Stronger notioncompletely semiprime:

a^{2} \in 𝔠 \Rightarrow a \in 𝔠

The prime / semiprime dictionary
Prime side	Semiprime side	Reference
$𝔄 𝔅 \subseteq 𝔭$	$𝔄^{2} \subseteq 𝔠$	(10.1), (10.8)
$a R b$ test	$a R a$ test	(10.2), (10.9)
$m$ -system	$n$ -system	(10.3), (10.9)
maximal disjoint ideal is prime	$n$ -system contains an $m$ -system	(10.5), (10.10)
$\sqrt{𝔄} = ⋂ 𝔭$	$𝔠 = \sqrt{𝔠}$	(10.7), (10.11)

Frequently Asked Questions

Why is

R

allowed to be a semiprime ideal when it is not allowed to be prime?

So that $(10.11)$ is true without exceptions. $R$ is the intersection of the empty family of primes, and if it were excluded the statement semiprime = intersection of primes would fail for rings whose only ideal is $R$ — and $(10.12)$ would have no closure operator for $𝔠 = R$ .

Is a semiprime ideal the same as a radical ideal?

In the commutative case, yes — that is the parenthetical remark after $(10.11)$ . In general, semiprime means $𝔠 = \sqrt{𝔠}$ with the $m$ -system radical, whereas the naive element-wise radical condition defines the strictly stronger notion of a completely semiprime ideal.

Why does the proof of

(10.11)

need the lemma about

n

-systems?

Because semiprimeness is a statement about $n$ -systems while the radical is defined by $m$ -systems. To show $\sqrt{𝔠} \subseteq 𝔠$ you must produce an $m$ -system missing $𝔠$ through a given point outside $𝔠$ , and the complement only gives you an $n$ -system. $(10.10)$ bridges the gap.

Does a semiprime ring have no nilpotent elements?

No — that is the reduced condition. A semiprime ring has no nonzero nilpotent one-sided ideals, but may have plenty of nilpotent elements: $M_{2} (k)$ is semiprime, indeed simple, and $e_{12}^{2} = 0$ .

Are intersections and sums of semiprime ideals semiprime?

Intersections always are, immediately from the definition or from $(10.11)$ . Sums need not be: in $k [x, y]$ the ideals $(y)$ and $(y - x^{2})$ are prime, hence semiprime, but their sum $(y, x^{2})$ is not semiprime, since $(x)^{2} \subseteq (y, x^{2})$ while $x \notin (y, x^{2})$ . Semiprime ideals form a closure system under intersection only.

How do I recognise the semiprime ideals of a small ring quickly?

List the ideals, find the maximal ones (automatically prime), and take all possible intersections of primes. By $(10.11)$ that is exactly the set of semiprime ideals, provided you have found all the primes — checking the $a R a$ test on the remaining ideals confirms it.

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.
J. Levitzki, “Prime ideals and the lower radical”, American Journal of Mathematics 73 (1951), 25–29.
K. R. Goodearl and R. B. Warfield, Jr., An Introduction to Noncommutative Noetherian Rings, 2nd edition, Cambridge University Press, 2004, Chapter 3.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Show that the sum of two semiprime ideals need not be semiprime, with an explicit noncommutative example.
Prove directly from (10.9) that a semiprime ring has no nonzero nilpotent left ideals.
Which rings have the property that every semiprime ideal is completely semiprime?
Describe the semiprime ideals of the Weyl algebra and of the free algebra on two generators.
How does (10.10) change for rings without identity, and does (10.11) survive?
Explain the role of semiprimeness in Goldie's theorem and why it cannot be weakened.
Give an algorithm that finds all semiprime ideals of a finite-dimensional algebra over a finite field.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Squaring is enough

Why aRa and not a2

n-systems and the missing link

06Key Results

07Proof Techniques and Method

Set b=a

Iterate the witness

Halve the exponent

08Worked Example

Which ideals of T2(k) are semiprime?

0 is not semiprime

𝔍 is semiprime

An arithmetic check of (10.12)

09Frameworks and Models

10Comparison and Classification

11Relationship Map

12Failure Modes and Common Mistakes

13Best Practices

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why $a R a$ and not $a^{2}$

$n$ -systems and the missing link

Key Results

Proof Techniques and Method

Set $b = a$

Worked Example

Which ideals of $T_{2} (k)$ are semiprime?

$0$ is not semiprime

$𝔍$ is semiprime

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