Prime and Semiprime Rings

Q: Is a semiprime ring the same as a ring with zero Jacobson radical?

No, and the implication goes only one way. Nil_* R ⊆ rad R always, so rad R = 0 forces semiprimeness. The converse fails: Z_(p) and k[[x]] are domains, hence prime and semiprime, but have nonzero Jacobson radical. The two conditions coincide when R is left artinian.

Q: Does semiprimeness pass to subrings?

Not in general. T_2(k) is a subring of M_2(k); the larger ring is semiprime and the smaller one is not. What does pass are the constructions that preserve the ideal lattice in a controlled way: matrix rings, polynomial rings, direct products, and centres in the prime case.

Q: How do I recognise a nonzero nilpotent left ideal in practice?

Look for an element a with aRa = 0; then Ra is a left ideal with (Ra)^2 = RaRa = 0. This turns a global search over left ideals into an element-wise condition, which is why (10.16) is stated with condition (4) and proved with condition (3).

Q: What is the relationship between prime rings and prime ideals?

They are the same information viewed twice. An ideal p R is prime exactly when R/p is a prime ring, and the primes of R are therefore the kernels of the surjections from R onto prime rings. The intersection of all of them is Nil_* R.

Q: Is there a version of Passman's criterion over a field?

Yes, and it is the form most often quoted: for a field k, the group ring kG is semiprime if char k = 0, with no condition on G; and for char k = p > 0, kG is semiprime exactly when G has no finite normal subgroup whose order is divisible by p.

Executive Summary

Commutative algebra has two ring-level conditions that make everything work: being a domain and being reduced. Neither survives the passage to noncommutative rings — $M_{2} (k)$ is as well-behaved as a ring can be and is neither. The right replacements are obtained by demanding that the zero ideal be prime, respectively semiprime, in the ideal-theoretic sense of Prime Ideals in Noncommutative Rings.

What results is a pair of classes closed under matrix rings, polynomial rings and (for semiprimeness) arbitrary direct products, containing all simple rings, all domains and all semiprimitive rings, and characterised by transparent element tests. Semiprimeness has an extra face that primeness does not: it is exactly the vanishing of every nilpotent one-sided ideal, which is why it is the hypothesis under which Wedderburn-style arguments restart.

$a R b = 0$ Prime test

$a R a = 0$ Semiprime test

${Nil}_{*} R = 0$ Semiprime, radically

(10.15)Lam's definition

Overview

Recall the ideal-theoretic definitions. An ideal $𝔭 ⊊ R$ is prime if $𝔄 𝔅 \subseteq 𝔭$ implies $𝔄 \subseteq 𝔭$ or $𝔅 \subseteq 𝔭$ for ideals $𝔄, 𝔅$ ; an ideal $ℭ$ is semiprime if $𝔄^{2} \subseteq ℭ$ implies $𝔄 \subseteq ℭ$ . Applying these to $𝔭 = ℭ = (0)$ gives the two ring classes of this page.

R prime iff (𝔄 𝔅 = 0 \Rightarrow 𝔄 = 0 or 𝔅 = 0), R semiprime iff (𝔄^{2} = 0 \Rightarrow 𝔄 = 0)

(10.15)

Ideals $𝔄, 𝔅$ range over two-sided ideals; by $(10.2)$ and $(10.9)$ one may equally let them range over left ideals or over right ideals.

The definitions transfer to quotients without friction: for an ideal $𝔄 \subseteq R$ , the ring $R / 𝔄$ is prime exactly when $𝔄$ is a prime ideal, and semiprime exactly when $𝔄$ is a semiprime ideal. Everything proved about prime ideals therefore becomes a statement about prime rings, and conversely. In particular $R / {Nil}_{*} R$ is always semiprime, since ${Nil}_{*} R$ is the smallest semiprime ideal — see The Lower Nilradical (Baer Radical).

In the commutative category the two classes are the familiar ones: prime rings are the integral domains, semiprime rings are the reduced rings, and ${Nil}_{*} R$ collapses to the ideal $Nil (R)$ of nilpotent elements. Noncommutatively all three statements fail as stated, and the failures are instructive rather than pathological.

Learning Objectives

State Definition $(10.15)$ and translate it into the element tests $a R b = 0$ and $a R a = 0$ .
Prove the equivalence of semiprimeness with ${Nil}_{*} R = 0$ and with the absence of nonzero nilpotent left ideals.
Explain why $M_{n} (D)$ is prime but not a domain, and semiprime but not reduced.
Show that a nonzero central element of a prime ring is a non-zero-divisor.
Deduce that a left artinian prime ring is simple artinian.
State the Connell and Passman criteria for $k G$ to be prime, respectively semiprime.

Definitions

Definition(10.15)Prime and semiprime rings

A ring $R$ is called a prime ring if the zero ideal is a prime ideal of $R$ , and a semiprime ring if the zero ideal is a semiprime ideal of $R$ . Since a prime ideal is by definition proper, a prime ring is nonzero; the zero ring is semiprime by the empty-intersection convention.

$(a)$: The two-sided ideal generated by $a$ ; in a ring with identity, $(a) = R a R$ .
$𝔄^{n}$: The additive subgroup generated by all products $a_{1} a_{2} \dots a_{n}$ with $a_{i} \in 𝔄$ . Nilpotent means $𝔄^{n} = 0$ for some $n$ ; nil means every element of $𝔄$ is nilpotent.
Domain: A nonzero ring in which $a b = 0$ implies $a = 0$ or $b = 0$ . Strictly stronger than prime.
Reduced: No nonzero nilpotent elements. Strictly stronger than semiprime.
${Nil}_{*} R$: The lower nilradical, equal to $\sqrt{(0)}$ and to the intersection of all prime ideals of $R$ .

Rings have an identity and modules are unital. Ideal means two-sided ideal unless the word left or right appears.

Core Concepts

Why the element test is not the naive one

In a commutative ring, $𝔭$ is prime iff $a b \in 𝔭$ forces $a \in 𝔭$ or $b \in 𝔭$ . Transporting that verbatim to $𝔭 = (0)$ would define domain, and the class of noncommutative domains is far too small: it excludes every matrix ring. The correct test inserts the whole ring between the two elements.

a R b = 0 ⟹ a = 0 or b = 0

The prime test. It says $(a) (b) = R a R b R = 0$ forces $a = 0$ or $b = 0$ , so it is exactly primeness of $(0)$ read on principal ideals.

The gap between $a b = 0$ and $a R b = 0$ is precisely the room in which matrix rings live: in $M_{2} (k)$ one has $E_{11} E_{22} = 0$ , but $E_{11} M_{2} (k) E_{22} ∋ E_{11} E_{12} E_{22} = E_{12} \neq 0$ .

Semiprimeness is the absence of nilpotence, at the ideal level

Semiprimeness admits a formulation that mentions no test elements at all: $R$ is semiprime iff it has no nonzero nilpotent ideal, and — this is the part that takes an argument — iff it has no nonzero nilpotent one-sided ideal. That upgrade is what makes semiprimeness usable in module theory, where left ideals are the natural objects.

No nilpotent left idealNo nilpotent ideal

(0)

semiprime

{Nil}_{*} R = 0

Key Results

Proposition(10.2), (10.9)Element characterisations

For a nonzero ring $R$ the following are equivalent: (1) $R$ is prime; (2) $𝔄 𝔅 = 0$ for ideals implies $𝔄 = 0$ or $𝔅 = 0$ ; (3) $a R b = 0$ for $a, b \in R$ implies $a = 0$ or $b = 0$ ; (4) $𝔄 𝔅 = 0$ for left ideals implies $𝔄 = 0$ or $𝔅 = 0$ ; (4 $'$ ) the same for right ideals.

For any ring $R$ the following are equivalent: (1) $R$ is semiprime; (2) $(a)^{2} = 0$ implies $a = 0$ ; (3) $a R a = 0$ implies $a = 0$ ; (4) $𝔄^{2} = 0$ for a left ideal $𝔄$ implies $𝔄 = 0$ ; (4 $'$ ) the same for right ideals.

These are the specialisations to $𝔭 = (0)$ and $ℭ = (0)$ of the general characterisations of prime and semiprime ideals.

Proposition(10.16)Semiprimeness and nilpotent left ideals

For any ring $R$ the following are equivalent:

$R$ is a semiprime ring;
${Nil}_{*} R = 0$ ;
$R$ has no nonzero nilpotent ideal;
$R$ has no nonzero nilpotent left ideal.

Proof

**(1) $\Leftrightarrow$ (2).** By definition ${Nil}_{*} R = \sqrt{(0)}$ is the smallest semiprime ideal of $R$ , so it is zero exactly when $(0)$ is already semiprime.

**(4) $\Rightarrow$ (3)** is trivial, an ideal being in particular a left ideal, and **(3) $\Rightarrow$ (1)** is immediate from the definition of a semiprime ideal applied to $𝔄^{2} = 0$ .

**(1) $\Rightarrow$ (4).** Let $𝔄$ be a nilpotent left ideal and choose $n \geq 1$ minimal with $𝔄^{n} = 0$ . Suppose $n > 1$ . Then $(𝔄^{n - 1})^{2} = 𝔄^{2 n - 2} \subseteq 𝔄^{n} = 0$ , because $2 n - 2 \geq n$ for $n \geq 2$ . Since $𝔄^{n - 1}$ is again a left ideal and $R$ is semiprime, characterisation (4) of semiprime ideals gives $𝔄^{n - 1} = 0$ , contradicting minimality of $n$ . Hence $n = 1$ and $𝔄 = 0$ .

Corollary—Central elements of a prime ring

If $R$ is a prime ring, every nonzero element of the centre $Z (R)$ is a non-zero-divisor in $R$ . In particular $Z (R)$ is an integral domain.

Proof

Let $0 \neq z \in Z (R)$ and suppose $z a = 0$ for some $a \in R$ . For every $r \in R$ we have $z r a = r z a = 0$ , so $z R a = 0$ ; primeness forces $z = 0$ or $a = 0$ , hence $a = 0$ . The same computation with $a z = 0$ gives $a R z = 0$ and $a = 0$ . Restricting to $Z (R)$ shows $Z (R)$ has no zero divisors, and $Z (R) \neq 0$ since $1 \in Z (R)$ .

Corollary—Artinian prime rings are simple

A left artinian prime ring $R$ is simple artinian, hence $R ≅ M_{n} (D)$ for a division ring $D$ and an integer $n \geq 1$ .

Proof

A prime ring is semiprime. Since $R$ is left artinian, $rad R$ is a nilpotent ideal, and by $(10.16)$ a semiprime ring has no nonzero nilpotent ideal, so $rad R = 0$ . A left artinian ring with zero radical is semisimple, hence a finite product $M_{n_{1}} (D_{1}) \times \dots \times M_{n_{r}} (D_{r})$ by Wedderburn–Artin. If $r \geq 2$ , the two nonzero ideals given by the first factor and by the product of the remaining factors multiply to zero, contradicting primeness. So $r = 1$ .

Example(10.17)The standard list

Every domain is prime, and every reduced ring is semiprime; neither converse holds.
Every simple ring is prime, because a maximal ideal is always prime: if $𝔄, 𝔅 not \subseteq 𝔪$ then $𝔪 + 𝔄 = R = 𝔪 + 𝔅$ , so $R = (𝔪 + 𝔄) (𝔪 + 𝔅) \subseteq 𝔪 + 𝔄 𝔅$ and $𝔄 𝔅 not \subseteq 𝔪$ .
$R / {Nil}_{*} R$ is semiprime for every $R$ , and a surjection $f : R \to S$ satisfies $f ({Nil}_{*} R) \subseteq {Nil}_{*} S$ , so it descends to a surjection of semiprime rings.
$rad R = 0$ implies ${Nil}_{*} R = 0$ , since ${Nil}_{*} R \subseteq rad R$ . Hence semiprimitive rings — in particular semisimple rings and von Neumann regular rings — are semiprime.
Any direct product of semiprime rings is semiprime. By contrast a direct product of two or more nonzero rings is never prime: the ideals $R \times 0$ and $0 \times S$ are nonzero with product zero.

Theorem(10.17g)Connell's criterion

Let $k$ be a ring and $G$ a group. The group ring $k G$ is prime if and only if $k$ is prime and $G$ has no finite normal subgroup other than ${1}$ .

The easy direction is instructive. If $H ⊴ G$ is finite, let $𝔄$ be the ideal of $k G$ generated by all $h - 1$ with $h \in H$ and $𝔅$ the ideal generated by $\sum_{h \in H} h$ . Normality of $H$ makes both two-sided, and $(h - 1) \sum_{h' \in H} h' = 0$ gives $𝔄 𝔅 = 0$ . Primeness forces $𝔄 = 0$ , i.e. $H = {1}$ . The converse rests on the structure of the f.c. centre $Δ (G)$ : Dietzmann's Lemma makes $Δ (G)$ torsion-free, hence abelian, and the argument used for the group-ring zero-divisor analysis then shows $γ k G γ' = 0$ only for $γ = γ' = 0$ .

Theorem(10.17h)Passman's criterion

Let $k$ be a ring and $G$ a group. The group ring $k G$ is semiprime if and only if $k$ is semiprime and, for every finite normal subgroup $H ⊴ G$ , the integer $| H |$ is not a zero divisor in $k$ .

With $𝔄, 𝔅$ as in Connell's proof one has $(𝔄 \cap 𝔅)^{2} \subseteq 𝔄 𝔅 = 0$ , so semiprimeness gives $𝔄 \cap 𝔅 = 0$ ; if $a \in k$ satisfies $| H | a = 0$ then $a \sum_{h \in H} h$ lies in $𝔄 \cap 𝔅$ and hence vanishes, forcing $a = 0$ . Over a field this reads: $k G$ is semiprime iff $char k = 0$ , or $char k = p > 0$ and $G$ has no finite normal subgroup of order divisible by $p$ .

Proof Techniques and Method

The reusable moves behind the proofs above.

Move 1

Sandwich the ring

Replace a product $a b$ by the set $a R b$ . Every commutative primeness argument transports if you make this substitution, and the resulting condition is automatically side-symmetric.

Move 2

Take a minimal nilpotency index

Given $𝔄^{n} = 0$ with $n$ minimal, square $𝔄^{n - 1}$ : the exponent $2 n - 2$ already exceeds $n$ , so semiprimeness collapses $𝔄^{n - 1}$ and contradicts minimality. This one-line trick is what promotes ideals to one-sided ideals.

Move 3

Build an annihilating pair

To defeat primeness, exhibit two nonzero ideals with zero product. For group rings the pair is the augmentation-type ideal of a finite normal subgroup and the ideal generated by its element sum.

Move 2 explains a recurring asymmetry in the literature: statements about nilpotent one-sided ideals are elementary, while the corresponding statements about nil one-sided ideals are open. Nothing in Move 2 survives when $𝔄$ is merely nil, because there is no index to minimise.

Worked Example

Three rings on $ℤ / 12 ℤ$ and its quotients

Take $R = ℤ / 12 ℤ$ . Its prime ideals are $(2)$ and $(3)$ , both maximal, so

{Nil}_{*} R = (2) \cap (3) = {0, 6} = (6),

(E.1)

and $(6)^{2} = (36) = 0$ , so $R$ carries a nonzero nilpotent ideal and is not semiprime — consistent with $(10.16)$ . Passing to the quotient, $R / (6) ≅ ℤ / 6 ℤ ≅ ℤ / 2 \times ℤ / 3$ is reduced, hence semiprime, but it is not prime: the nonzero ideals $(2)$ and $(3)$ of $ℤ / 6$ satisfy $(2) (3) = (6) = 0$ .

A prime ring that is not a domain

Let $R = M_{2} (ℤ)$ . Every ideal of $R$ is $M_{2} (n ℤ)$ for some $n \geq 0$ , and $M_{2} (m ℤ) \cdot M_{2} (n ℤ) = M_{2} (m n ℤ)$ , which is nonzero whenever $m, n \neq 0$ . So $R$ is prime. It is not a domain: $E_{11} E_{22} = 0$ . It is not reduced either, since $E_{12}^{2} = 0$ , yet it is semiprime — a single nilpotent element is not a nilpotent ideal, and indeed $E_{12} R E_{12} ∋ E_{12} E_{21} E_{12} = E_{12} \neq 0$ .

A ring that fails semiprimeness for structural reasons

Let $k$ be a field and $T = T_{2} (k)$ the upper triangular $2 \times 2$ matrices. The strictly upper triangular matrices form an ideal $𝔑$ with $𝔑^{2} = 0$ , so $T$ is not semiprime, and ${Nil}_{*} T = 𝔑$ because $T / 𝔑 ≅ k \times k$ is reduced.

𝔑 = (\begin{matrix} 0 & k \\ 0 & 0 \end{matrix}), 𝔑^{2} = 0, T / 𝔑 ≅ k \times k semiprime, not prime .

(E.2)

For $T_{2} (k)$ the lower nilradical, upper nilradical, Levitzki radical and Jacobson radical all coincide with $𝔑$ .

Comparison and Classification

Prime and semiprime status of standard rings
Ring	Prime?	Semiprime?	${Nil}_{*} R$
Division ring $D$	yes	yes	$0$
$M_{n} (D)$ , $n \geq 2$	yes (not a domain)	yes (not reduced)	$0$
$ℤ$	yes	yes	$0$
$ℤ / 6 ℤ$	no	yes	$0$
$ℤ / 12 ℤ$	no	no	$(6)$
$T_{2} (k)$ , $k$ a field	no	no	strictly upper triangular
$k [[x]]$	yes	yes	$0$ (but $rad = (x)$ )
$R \times S$ , both nonzero	no	iff both are	${Nil}_{} R \times {Nil}_{} S$
$k G$ , $G$ finite of order $n$	no for $\| G \| > 1$	iff $k$ is semiprime and each $\| H \|$ , $H ⊴ G$ , is a non-zero-divisor in $k$	depends on $char k$

Which implications hold in which direction
	Prime	Semiprime	Domain	Reduced
Domain	yes	yes	—	yes
Reduced	no	yes	no	—
Simple	yes	yes	no	no
Semisimple	no	yes	no	no
Semiprimitive	no	yes	no	no
Left primitive	yes	yes	no	no
Von Neumann regular	no	yes	no	no

Which implications hold in which direction

Read the table as: does the row class force the column class? The entry prime and reduced is worth isolating — a ring is a domain precisely when it is both.

Relationship Map

SimpleLeft primitivePrimeSemiprime

Each arrow is strict. $M_{2} (k [x])$ is prime but not primitive over a field $k$ ; ${End}_{D} (V)$ for infinite-dimensional $V$ is primitive but not simple; $ℤ / 6$ is semiprime but not prime. The full picture is drawn in Primitive, Simple, Prime and Semiprimitive: How the Classes Relate.

Semiprime rings — ${Nil}_{*} R = 0$
- contain
  - all prime rings
  - all reduced rings, in particular all domains
  - all semiprimitive rings, hence all semisimple and all von Neumann regular rings
  - arbitrary direct products of semiprime rings
- exclude
  - any ring with a nonzero nilpotent left ideal
  - $ℤ / n$ for $n$ not squarefree
  - triangular rings $T_{n} (k)$ for $n \geq 2$
- are closed under
  - matrix rings $M_{n} (-)$
  - polynomial rings $R [T]$
  - direct products
  - passing to $R / {Nil}_{*} R$ from any ring

The closure properties in the last branch are the content of The Lower Nilradical of Polynomial and Matrix Rings; they fail for the Jacobson radical, which is one reason the prime radical is the more computable invariant.

Applications and Industry Use

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

Rings of quotients

Goldie's theorem

A ring has a semisimple artinian classical left ring of quotients exactly when it is semiprime left Goldie; it has a simple artinian one exactly when it is prime left Goldie. Semiprimeness is the hypothesis that makes noncommutative localisation behave.

Group algebras

Connell and Passman

The prime and semiprime questions for $k G$ are completely settled, in contrast to the semiprimitivity question, which remains open. Modular representation theory uses the semiprime criterion to decide when $k G$ has no nilpotent ideals.

Operator algebras

Prime C*-algebras

A C*-algebra is semiprime automatically, and primeness of the algebra corresponds to the primitive-ideal space being irreducible. The ideal-theoretic language on this page is the algebraic shadow of that topology.

Noncommutative geometry

Spectra and supports

Prime ideals are the points of the noncommutative spectrum; semiprime ideals are the closed sets. Computer algebra systems that manipulate ideals in Weyl algebras and enveloping algebras use exactly this correspondence.

Honestly stated, prime and semiprime rings are infrastructure. They are where the structure theory of noncommutative rings begins once artinian hypotheses are dropped, and their industrial reach is via the objects built on them — quotient rings, PI theory, and the ideal-theoretic engines inside symbolic computation systems.

Failure Modes and Common Mistakes

Do not conclude *nil one-sided ideal $= 0$ * from semiprimeness alone; that implication is Köthe's Conjecture and is available only under extra hypotheses such as ACC on right annihilators.
Do not assume a product of prime rings is prime — it never is once two factors are nonzero, although the product of semiprime rings is semiprime.
Do not read Passman's criterion as a statement about $char k$ alone; over a general coefficient ring the condition is that $| H |$ is a non-zero-divisor, which is finer.
Do not forget that a prime ring is nonzero by convention, since a prime ideal must be proper.

Best Practices

Verify primeness with $a R b = 0$ and semiprimeness with $a R a = 0$ ; the ideal-theoretic definitions are for proofs, the element tests are for computation.
When a ring is presented as a quotient $R / 𝔄$ , decide primeness of the ideal rather than of the ring; the two questions are the same and the ideal is usually the concrete object.
Record the coefficient ring's own status before applying Connell or Passman: both criteria have a condition on $k$ and a condition on $G$ .
When you need both no-nilpotents and no-zero-divisors, say domain; the words prime and semiprime should be reserved for their technical meanings.

Quick Reference

Prime ring

(0)

is a prime ideal;

a R b = 0 \Rightarrow a = 0

b = 0

Semiprime ring

(0)

is a semiprime ideal;

a R a = 0 \Rightarrow a = 0

Radical form

R

semiprime

iff {Nil}_{*} R = 0

Nilpotence form

R

semiprime

iff

no nonzero nilpotent left ideal

Commutative caseprime

=

domain, semiprime

=

reduced

Quotients

R / 𝔄

prime (semiprime)

iff 𝔄

prime (semiprime)

Closure

M_{n} (R)

and

R [T]

inherit both properties from

R

Group ringsConnell

(10.17 g)

for prime; Passman

(10.17 h)

for semiprime

Tests at a glance
Property	Test	Reference
Prime	$a R b = 0 \Rightarrow a = 0$ or $b = 0$	(10.2)(3)
Prime	product of two nonzero left ideals is nonzero	(10.2)(4)
Semiprime	$a R a = 0 \Rightarrow a = 0$	(10.9)(3)
Semiprime	no nonzero nilpotent left ideal	(10.16)(4)
Semiprime	${Nil}_{*} R = 0$	(10.16)(2)
Domain	prime and reduced	Exercise 10.3

Frequently Asked Questions

Why not simply define a prime ring as one with no zero divisors?

That defines a domain, and the class of noncommutative domains is far too restrictive to carry a structure theory: it excludes every matrix ring $M_{n} (D)$ with $n \geq 2$ , hence every simple artinian ring except division rings. Primeness weakens the test from $a b = 0$ to $a R b = 0$ , which keeps matrix rings inside the class while retaining the property that makes prime ideals useful — an ideal-theoretic irreducibility.

Is a semiprime ring the same as a ring with zero Jacobson radical?

No, and the implication goes only one way. ${Nil}_{*} R \subseteq rad R$ always, so $rad R = 0$ forces semiprimeness. The converse fails: $ℤ_{(p)}$ and $k [[x]]$ are domains, hence prime and semiprime, but have nonzero Jacobson radical. The two conditions coincide when $R$ is left artinian.

Does semiprimeness pass to subrings?

Not in general. $T_{2} (k)$ is a subring of $M_{2} (k)$ ; the larger ring is semiprime and the smaller one is not. What does pass are the constructions that preserve the ideal lattice in a controlled way: matrix rings, polynomial rings, direct products, and centres in the prime case.

How do I recognise a nonzero nilpotent left ideal in practice?

Look for an element $a$ with $a R a = 0$ ; then $R a$ is a left ideal with $(R a)^{2} = R a R a = 0$ . This turns a global search over left ideals into an element-wise condition, which is why $(10.16)$ is stated with condition (4) and proved with condition (3).

What is the relationship between prime rings and prime ideals?

They are the same information viewed twice. An ideal $𝔭 ⊊ R$ is prime exactly when $R / 𝔭$ is a prime ring, and the primes of $R$ are therefore the kernels of the surjections from $R$ onto prime rings. The intersection of all of them is ${Nil}_{*} R$ .

Is there a version of Passman's criterion over a field?

Yes, and it is the form most often quoted: for a field $k$ , the group ring $k G$ is semiprime if $char k = 0$ , with no condition on $G$ ; and for $char k = p > 0$ , $k G$ is semiprime exactly when $G$ has no finite normal subgroup whose order is divisible by $p$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §10, especially (10.15)–(10.17).
N. H. McCoy, “Prime ideals in general rings”, American Journal of Mathematics 71 (1949), 823–833.
I. G. Connell, “On the group ring”, Canadian Journal of Mathematics 15 (1963), 650–685.
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977, Chapter 4.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, Chapter 4 (Goldie's theorem).

AI Suggested Questions

Give an example of a prime ring that is not left primitive, and explain what obstructs primitivity.
How does Goldie's theorem use semiprimeness, and what fails for rings that are merely nonsingular?
Prove that the centre of a prime ring is an integral domain and that the ring embeds in a ring over its central quotient field.
Describe the minimal prime ideals of $ℤ / n ℤ$ and of $T_{n} (k)$ , and check that their intersection is the lower nilradical.
Why is the semiprimitivity problem for $k G$ still open when the prime and semiprime problems are solved?
Compare completely prime ideals with prime ideals in the Weyl algebra $A_{1} (k)$ .
Show that a ring is a domain if and only if it is prime and reduced.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the element test is not the naive one

Semiprimeness is the absence of nilpotence, at the ideal level

06Key Results

07Proof Techniques and Method

Sandwich the ring

Take a minimal nilpotency index

Build an annihilating pair

08Worked Example

Three rings on ℤ/12ℤ and its quotients

A prime ring that is not a domain

A ring that fails semiprimeness for structural reasons

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Goldie's theorem

Connell and Passman

Prime C*-algebras

Spectra and supports

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

Key Results

Proof Techniques and Method

Worked Example

Three rings on $ℤ / 12 ℤ$ and its quotients

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions