The Lower Nilradical of Polynomial and Matrix Rings

Q: What about skew polynomial rings?

The results fail. In R[x; ] the leading coefficient of f r g is a ^n(r) ^n(b) rather than arb, and the twist can create locally nilpotent one-sided ideals in a prime ring. Lam's example, due to J. Ram, is a prime ring A[x;] with nonzero Levitzki radical — impossible for an untwisted polynomial ring over a semiprime base.

Q: Is Nil_* a Morita invariant?

The matrix statement (10.21) is the concrete face of that claim, and yes: the prime radical is preserved by Morita equivalence, as are the Jacobson radical and the class of semiprime rings. What is not Morita invariant is being a domain or being reduced, which is why those notions are the wrong ones to build a structure theory on.

Q: Can I compute Nil_* R[x] without knowing all primes of R[x]?

Yes — that is precisely the point. You need only Nil_* R, computed in the base ring, and then extend coefficientwise. The primes of R[x] are vastly more complicated than those of R even for R = Z, but their intersection is not.

Executive Summary

Two constructions dominate ring theory: adjoin commuting indeterminates, and pass to $n \times n$ matrices. This page settles how the prime radical behaves under both. The answer is the best possible one — it extends coefficientwise and entrywise, for every ring, with no chain condition, no algebra structure over a field, and no restriction on the number of variables.

The contrast with the Jacobson radical is sharp and worth carrying around. Amitsur's theorem describes $rad R [T]$ only as $N [T]$ for a nil ideal $N \subseteq R$ that is not otherwise identified; here ${Nil}_{*}$ is computed outright. The reason is structural: the prime radical is defined by ideals and multiplication, and both are visible on leading coefficients.

$({Nil}_{*} R) [T]$ Polynomial case

$M_{n} ({Nil}_{*} R)$ Matrix case

NoneHypotheses on

R

(10.19)Amitsur–McCoy

Overview

Let $T$ be a set of indeterminates that commute with one another and with the elements of $R$ ; $R$ itself is arbitrary. The programme has two steps. First show that the class of prime rings and the class of semiprime rings are stable under $R \mapsto R [T]$ and $R \mapsto M_{n} (R)$ in both directions. Then leverage that stability: the lower nilradical is characterised as the smallest ideal with semiprime quotient, so a statement about the class of semiprime rings converts directly into a computation of the radical.

{Nil}_{*} (R [T]) = ({Nil}_{*} R) [T], {Nil}_{*} (M_{n} (R)) = M_{n} ({Nil}_{*} R)

(10.19), (10.21)

Valid for every ring $R$ , every set $T$ of commuting indeterminates and every $n \geq 1$ .

The same reasoning would compute any radical defined as *the smallest ideal with quotient in a class $𝒞$ *, provided $𝒞$ is stable under the construction in both directions. It fails for the Jacobson radical because semiprimitivity is not stable under adjoining a variable in the useful direction — $k$ is semiprimitive and so is $k [x]$ , but $R$ semiprimitive does not control $rad R [x]$ tightly enough. See Amitsur's Theorem on the Radical of a Polynomial Ring.

Learning Objectives

State $(10.18)$ and run the leading-coefficient argument that proves it.
Reduce the many-variable case to the one-variable case correctly.
Prove the Amitsur–McCoy theorem ${Nil}_{*} R [T] = ({Nil}_{*} R) [T]$ using $(10.18)$ and the prime-ideal contraction $𝔭 \cap R$ .
Prove ${Nil}_{*} M_{n} (R) = M_{n} ({Nil}_{*} R)$ from the ideal correspondence for matrix rings.
Compare the polynomial behaviour of ${Nil}_{*}$ , ${Nil}^{*}$ and $rad$ , and state precisely which of the three analogues is known to fail.

Definitions

$R [T]$: Polynomials in the commuting indeterminates $t \in T$ with coefficients in $R$ ; each element involves only finitely many variables. Multiplication uses $t r = r t$ for $r \in R$ .
$𝔄 [T]$: For an ideal $𝔄 \subseteq R$ , the set of polynomials all of whose coefficients lie in $𝔄$ . It is an ideal of $R [T]$ with $R [T] / 𝔄 [T] ≅ (R / 𝔄) [T]$ .
$M_{n} (𝔄)$: Matrices with all entries in $𝔄$ ; an ideal of $M_{n} (R)$ , and every ideal of $M_{n} (R)$ is of this form.
Contraction $𝔭 \cap R$: For an ideal $𝔭 \subseteq R [T]$ , its intersection with the coefficient ring, viewed inside $R$ via the inclusion $R ↪ R [T]$ .
${Nil}_{*} R$: The lower nilradical: the intersection of all prime ideals of R, equivalently the smallest ideal whose quotient is a semiprime ring.

Rings have identities. Indeterminates are central: they commute with all coefficients. Skew polynomial rings behave quite differently and are treated on the pages devoted to them.

Core Concepts

Leading coefficients see primeness

Everything in the polynomial half of this page rests on one computation. If $f = a t^{n} + \dots$ and $g = b t^{m} + \dots$ are nonzero elements of $R [t]$ with leading coefficients $a$ and $b$ , then for $r \in R$

f r g = (a r b) t^{n + m} + (terms of degree < n + m) .

(C.1)

Because $t$ is central, the top-degree coefficient of $f r g$ is exactly $a r b$ .

So $f R [t] g = 0$ already forces $a R b = 0$ , using only the constant multipliers $r \in R$ . Primeness of $R$ then kills $a$ or $b$ , hence $f$ or $g$ . The same display with $g = f$ proves the semiprime case. The degree bookkeeping is the entire content; there is no need to control the lower coefficients.

From classes of rings to radicals

The lower nilradical has two descriptions — as an intersection of primes and as the smallest semiprime ideal — and the proofs below use them alternately. The inclusion ${Nil}_{*} R [T] \subseteq ({Nil}_{*} R) [T]$ comes from the second description; the reverse inclusion comes from the first, by showing that every prime of $R [T]$ contracts to a prime of $R$ .

Stability of the class

R

semiprime

iff R [T]

semiprime, and likewise for prime. This is

(10.18)

Upper bound on the radical

R / {Nil}_{*} R

is semiprime, so

({Nil}_{*} R) [T]

is a semiprime ideal of

R [T]

, hence contains

{Nil}_{*} R [T]

Lower bound via contractionFor any prime

𝔭 \subseteq R [T]

𝔭 \cap R

is prime in

R

, so it contains

{Nil}_{*} R

and therefore

𝔭 \supseteq ({Nil}_{*} R) [T]

IntersectTaking the intersection over all primes

𝔭

gives

{Nil}_{*} R [T] \supseteq ({Nil}_{*} R) [T]

, and the two inclusions close the argument.

Matrices: the ideal lattice does the work

For matrix rings the corresponding tool is the bijection $𝔄 \mapsto M_{n} (𝔄)$ between ideals of $R$ and ideals of $M_{n} (R)$ , which multiplies correctly: $M_{n} (𝔄) M_{n} (𝔅) = M_{n} (𝔄 𝔅)$ . Primeness and semiprimeness are conditions on products of ideals, so they transfer verbatim, and the radical follows by the same smallest-semiprime-ideal argument.

Key Results

Proposition(10.18)Polynomial rings preserve primeness and semiprimeness

Let $R$ be a ring and $T$ a set of indeterminates commuting with one another and with the elements of $R$ . Then $R [T]$ is prime if and only if $R$ is prime, and $R [T]$ is semiprime if and only if $R$ is semiprime.

Proof

**( $\Rightarrow$ ), prime case.** Suppose $A = R [T]$ is prime and $a R b = 0$ with $a, b \in R$ . Since the indeterminates are central and commute with coefficients, $a A b = a R [T] b = (a R b) [T] = 0$ , so $a = 0$ or $b = 0$ .

**( $\Leftarrow$ ), prime case.** Suppose $R$ is prime and $f A g = 0$ with $f, g \in A$ . Each of $f, g$ involves only finitely many variables, say those in a finite subset $T_{0} \subseteq T$ , and $f R [T_{0}] g = 0$ . So we may assume $T$ finite, and by induction on $| T |$ — writing $R [t_{1}, \dots, t_{m}] = (R [t_{1}, \dots, t_{m - 1}]) [t_{m}]$ — we may assume $T = {t}$ . Suppose $f, g \neq 0$ with leading coefficients $a, b$ and degrees $n, m$ . By $(C . 1)$ the coefficient of $t^{n + m}$ in $f r g$ is $a r b$ for every $r \in R$ , and $f r g = 0$ , so $a R b = 0$ . Primeness of $R$ gives $a = 0$ or $b = 0$ , contradicting the choice of leading coefficients. Hence $f = 0$ or $g = 0$ .

Semiprime case. Identical with $g = f$ : $f A f = 0$ forces $a R a = 0$ for the leading coefficient $a$ of $f$ , hence $a = 0$ and $f = 0$ ; conversely $a R a = 0$ gives $a A a = 0$ .

Theorem(10.19)Amitsur–McCoy

For any ring $R$ and any set $T$ of commuting indeterminates, ${Nil}_{*} (R [T]) = ({Nil}_{*} R) [T]$ .

Proof

Write $I = {Nil}_{*} R$ . Since $I$ is a semiprime ideal, $R / I$ is a semiprime ring, so by $(10.18)$ the ring $(R / I) [T] ≅ R [T] / I [T]$ is semiprime. Hence $I [T]$ is a semiprime ideal of $R [T]$ , and as ${Nil}_{*} R [T]$ is the smallest semiprime ideal of $R [T]$ we get ${Nil}_{*} R [T] \subseteq I [T]$ .

For the reverse inclusion it suffices to show $I [T] \subseteq 𝔭$ for every prime ideal $𝔭$ of $R [T]$ . First, $𝔭 \cap R$ is a prime ideal of $R$ : it is proper since $1 \notin 𝔭$ , and if $a, b \in R$ satisfy $a R b \subseteq 𝔭 \cap R$ , then $a R [T] b = (a R b) [T] \subseteq 𝔭$ , so primeness of $𝔭$ gives $a \in 𝔭 \cap R$ or $b \in 𝔭 \cap R$ . Consequently $I \subseteq 𝔭 \cap R \subseteq 𝔭$ , and since $𝔭$ is an ideal of $R [T]$ closed under multiplication by monomials, every polynomial with coefficients in $I$ lies in $𝔭$ . Thus $I [T] \subseteq 𝔭$ , and intersecting over all primes gives $I [T] \subseteq {Nil}_{*} R [T]$ .

Proposition(10.20)Matrix rings preserve primeness and semiprimeness

For any ring $R$ and any $n \geq 1$ : $R$ is prime if and only if $M_{n} (R)$ is prime, and $R$ is semiprime if and only if $M_{n} (R)$ is semiprime.

Proof

If $R \neq 0$ is not prime, pick nonzero ideals $𝔄, 𝔅$ with $𝔄 𝔅 = 0$ ; then $M_{n} (𝔄) M_{n} (𝔅) = M_{n} (𝔄 𝔅) = 0$ with both factors nonzero, so $M_{n} (R)$ is not prime. Conversely, if $M_{n} (R) \neq 0$ is not prime, take nonzero ideals of $M_{n} (R)$ with zero product; by the ideal correspondence for matrix rings they are $M_{n} (𝔄)$ and $M_{n} (𝔅)$ for nonzero ideals $𝔄, 𝔅 \subseteq R$ , and $M_{n} (𝔄 𝔅) = 0$ forces $𝔄 𝔅 = 0$ . The semiprime case is the same argument with $𝔅 = 𝔄$ .

Theorem(10.21)The prime radical of a matrix ring

For any ring $R$ and any $n \geq 1$ , ${Nil}_{*} M_{n} (R) = M_{n} ({Nil}_{*} R)$ .

Proof

Write $I = {Nil}_{*} R$ . Since $R / I$ is semiprime, $(10.20)$ makes $M_{n} (R / I) ≅ M_{n} (R) / M_{n} (I)$ semiprime, so $M_{n} (I)$ is a semiprime ideal of $M_{n} (R)$ and therefore contains the smallest one: ${Nil}_{*} M_{n} (R) \subseteq M_{n} (I)$ .

Conversely, by the ideal correspondence ${Nil}_{*} M_{n} (R) = M_{n} (J)$ for some ideal $J \subseteq R$ . Then $M_{n} (R) / M_{n} (J) ≅ M_{n} (R / J)$ is semiprime, being the quotient by the prime radical, so $(10.20)$ gives $R / J$ semiprime, i.e. $J$ is a semiprime ideal of $R$ . Hence $J \supseteq I$ and ${Nil}_{*} M_{n} (R) = M_{n} (J) \supseteq M_{n} (I)$ .

Remark—What the other radicals do

$rad M_{n} (R) = M_{n} (rad R)$ holds for every ring, so the matrix statement is a theorem for both ${Nil}_{*}$ and $rad$ .
For the upper nilradical, ${Nil}^{*} M_{n} (R) = M_{n} ({Nil}^{*} R)$ is equivalent to Köthe's Conjecture and is therefore open.
For polynomial rings, $({Nil}^{*} R) [T] \subseteq {Nil}^{*} R [T]$ can fail: Smoktunowicz constructed a nil ring $N$ with $N [x]$ not nil, so the Amitsur–McCoy statement has no valid analogue for ${Nil}^{*}$ .
Lam's Exercise 21 for this section asks the reader to verify that $(L - rad R) [T]$ is a nil ideal of $R [T]$ , which places the Levitzki radical between the two extremes.

Proof Techniques and Method

The reusable moves behind the proofs above.

Move 1

Test with constants only

To show $f R [t] g = 0$ is impossible, apply it to $r \in R \subseteq R [t]$ and read the top coefficient. The infinitely many polynomial multipliers are never needed.

Move 2

Sandwich the radical

Prove $\subseteq$ by exhibiting a semiprime ideal, and $\supseteq$ by showing every prime contains the candidate. The two descriptions of ${Nil}_{*}$ are dual and each supplies one inclusion cheaply.

Move 3

Contract along an inclusion

If $R \subseteq S$ and $𝔭 \subseteq S$ is prime with $a S b \subseteq 𝔭$ recoverable from $a R b$ , then $𝔭 \cap R$ is prime. This is how information travels back down from the big ring.

Move 3 is delicate and deserves a warning: contraction of a prime need not be prime for arbitrary ring extensions. It works here because $a R [T] b = (a R b) [T]$ , an identity that uses centrality of the variables. For skew polynomial rings $R [x; σ]$ the identity fails and the conclusions of this page fail with it.

Worked Example

A finite base ring

Take $R = ℤ / 12 ℤ$ , where ${Nil}_{*} R = (2) \cap (3) = (6) = {0, 6}$ . Then

{Nil}_{*} ((ℤ / 12) [x, y]) = (6) [x, y], {Nil}_{*} (M_{2} (ℤ / 12)) = M_{2} ((6)) .

(E.1)

Both can be checked by hand. Every coefficient of an element of $(6) [x, y]$ lies in ${0, 6}$ and $6 \cdot 6 = 36 = 0$ in $ℤ / 12$ , so $((6) [x, y])^{2} = 0$ : the ideal is nilpotent, hence inside the prime radical. In the other direction the quotient is $(ℤ / 6) [x, y]$ , a reduced commutative ring, hence semiprime — so nothing larger can be in the radical. Iterating gives ${Nil}_{*} M_{2} ((ℤ / 12) [x]) = M_{2} ((6) [x])$ in one step.

Where the Jacobson radical parts company

Let $R = ℤ_{(p)}$ , the localisation of $ℤ$ at a prime $p$ . It is a domain, so ${Nil}_{*} R = 0$ and Amitsur–McCoy gives ${Nil}_{*} R [x] = 0$ . But $rad R = p ℤ_{(p)} \neq 0$ , whereas $rad R [x] = 0$ because $R$ is a commutative reduced ring.

rad ℤ_{(p)} = p ℤ_{(p)} \neq 0 = rad ℤ_{(p)} [x], {Nil}_{*} ℤ_{(p)} = 0 = {Nil}_{*} ℤ_{(p)} [x] .

(E.2)

Adjoining a variable destroys the Jacobson radical here but leaves the prime radical alone. Only one of the two invariants is polynomial-stable.

A noncommutative check

For $T = T_{2} (k)$ , the upper triangular $2 \times 2$ matrices over a field, ${Nil}_{*} T = 𝔑$ , the strictly upper triangular matrices. Amitsur–McCoy gives ${Nil}_{*} T [x] = 𝔑 [x]$ , and here the Jacobson radical agrees: $T [x] / 𝔑 [x] ≅ k [x] \times k [x]$ is semiprimitive, so $rad T [x] = 𝔑 [x]$ as well. Agreement is a coincidence of this example, not a theorem.

Process and Workflow

Which radical of $R [T]$ or $M_{n} (R)$ do you need?

Lower nilradicalCompute

{Nil}_{*} R

once and extend coefficientwise or entrywise. No further work, no hypotheses.

Jacobson radical, matrix caseUse

rad M_{n} (R) = M_{n} (rad R)

, also unconditional.

Jacobson radical, polynomial caseUse Amitsur's theorem:

rad R [T] = N [T]

where

N = R \cap rad R [T]

is a nil ideal of

R

. Identifying

N

requires extra information about

R

Upper nilradicalStop. The matrix statement is equivalent to Köthe's Conjecture and the polynomial statement is false in general.

The decision matters in practice because these four questions look interchangeable and are not. Only the first two have unconditional answers.

Comparison and Classification

Does the radical extend to the construction?
	$M_{n} (R)$	$R [T]$	Hypotheses needed
${Nil}_{*}$ (prime radical)	yes	yes	none
$rad$ (Jacobson)	yes	partial	Amitsur's theorem gives only $N [T]$ with $N$ nil
${Nil}^{*}$ (upper nilradical)	open	no	matrix case equivalent to Köthe; polynomial case fails
Prime as a class	yes	yes	none
Semiprime as a class	yes	yes	none
Reduced as a class	no	yes	$M_{n}$ of a domain has nilpotents for $n \geq 2$

Does the radical extend to the construction?

Worked values
Base ring $R$	${Nil}_{*} R$	${Nil}_{*} R [x]$	${Nil}_{*} M_{2} (R)$
$ℤ$	$0$	$0$	$0$
$ℤ / 12$	$(6)$	$(6) [x]$	$M_{2} ((6))$
$ℤ / p^{m}$	$(p)$	$(p) [x]$	$M_{2} ((p))$
$T_{2} (k)$	strictly upper	strictly upper $[x]$	$M_{2} (strictly upper)$
$k [[x]]$	$0$	$0$	$0$
$M_{3} (D)$	$0$	$0$	$0$

Relationship Map

The stability statements assemble into a single picture: ${Nil}_{*}$ commutes with the two constructions that generate most of the examples in this collection, so a computation over a base ring propagates.

R

semiprime

R [T]

semiprime

M_{n} (R [T])

semiprime

M_{n} (R) [T]

semiprime

All rings

{Nil}_{*} R

defined;

R / {Nil}_{*} R

semiprime

Semiprime rings

{Nil}_{*} R = 0

; closed under

R [T]

M_{n} (-)

, products

Prime ringsclosed under

R [T]

and

M_{n} (-)

, not under products

Domainsclosed under

R [T]

, not under

M_{n} (-)

The innermost failure is the point of the whole discussion: matrix rings force the passage from domain to prime, and the prime radical is the invariant that survives that passage.

Computational Notes

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

The two theorems are reductions, and that is their computational value: the number of variables and the matrix size drop out of the problem entirely, so the cost is the cost of computing ${Nil}_{*} R$ over the base.
For a commutative noetherian ring given by generators and relations, ${Nil}_{*}$ is the radical of the zero ideal and is computed by standard Gröbner-basis radical algorithms; systems such as Singular, Macaulay2 and Sage expose it directly.
For a finite-dimensional algebra over a field, ${Nil}_{*} A = {Nil}^{*} A = rad A$ , so the radical routines of GAP and Magma answer the question; the artinian hypothesis is what collapses the four radicals.
Naively applying a radical routine to $M_{n} (R)$ or to $R [x_{1}, \dots, x_{m}]$ inflates the dimension by $n^{2}$ or unboundedly. Applying $(10.19)$ and $(10.21)$ first is the difference between feasible and infeasible.
For general finitely presented noncommutative rings no algorithm exists: the word problem is already undecidable, so membership in ${Nil}_{*} R$ is not decidable in that generality.

Failure Modes and Common Mistakes

Do not assume the contraction $𝔭 \cap R$ of a prime ideal is prime for arbitrary extensions $R \subseteq S$ ; the proof above uses the identity $a R [T] b = (a R b) [T]$ .
Do not confuse $({Nil}_{*} R) [T]$ — coefficients in the radical — with ${Nil}_{*} (R) [T]$ read as a radical of a subring; the notation is compact but the objects are different.
Do not expect $M_{n}$ of a domain to be a domain. It is prime, which is all the theory needs.
Do not use ${Nil}_{*} R = rad R$ outside the left artinian case; $ℤ_{(p)}$ is the standard counterexample.

Quick Reference

Polynomial class

R [T]

prime (semiprime)

iff R

prime (semiprime)

Matrix class

M_{n} (R)

prime (semiprime)

iff R

prime (semiprime)

Amitsur–McCoy

{Nil}_{*} R [T] = ({Nil}_{*} R) [T]

Matrix radical

{Nil}_{*} M_{n} (R) = M_{n} ({Nil}_{*} R)

Key identity

a R [T] b = (a R b) [T]

Contraction

𝔭

prime in

R [T]

\Rightarrow 𝔭 \cap R

prime in

R

Jacobson contrast

rad R [T] = N [T]

N = R \cap rad R [T]

nil

Köthe link

{Nil}^{*} M_{n} (R) = M_{n} ({Nil}^{*} R)

iff

Köthe's Conjecture

Proof obligations for each inclusion
Inclusion	Tool	Reference
${Nil}_{} R [T] \subseteq ({Nil}_{} R) [T]$	$(R / {Nil}_{*} R) [T]$ is semiprime	(10.18)
$({Nil}_{} R) [T] \subseteq {Nil}_{} R [T]$	contraction of a prime is prime	(10.19)
${Nil}_{} M_{n} (R) \subseteq M_{n} ({Nil}_{} R)$	$M_{n} (R / I)$ semiprime	(10.20)
$M_{n} ({Nil}_{} R) \subseteq {Nil}_{} M_{n} (R)$	ideal correspondence plus (10.20)	(10.21)

Frequently Asked Questions

Why is the polynomial statement for the prime radical so much stronger than Amitsur's theorem for the Jacobson radical?

Because primeness is a statement about products of ideals, and the top-degree coefficient of a product of polynomials is the product of the leading coefficients. Invertibility, which defines the Jacobson radical, has no such degree-wise behaviour: whether $1 - f g$ is invertible in $R [T]$ depends on all coefficients at once. Amitsur's theorem still says something sharp — $rad R [T] = N [T]$ for a nil ideal $N \subseteq R$ — but it does not identify $N$ .

Does the theorem hold for infinitely many variables?

Yes, with no change. Any two polynomials involve only finitely many variables between them, so every step of the argument takes place inside $R [T_{0}]$ for a finite $T_{0} \subseteq T$ . This finiteness reduction is the only role played by the size of $T$ .

What about skew polynomial rings?

The results fail. In $R [x; σ]$ the leading coefficient of $f r g$ is $a σ^{n} (r) σ^{n} (b)$ rather than $a r b$ , and the twist can create locally nilpotent one-sided ideals in a prime ring. Lam's example, due to J. Ram, is a prime ring $A [x; σ]$ with nonzero Levitzki radical — impossible for an untwisted polynomial ring over a semiprime base.

{Nil}_{*}

a Morita invariant?

The matrix statement $(10.21)$ is the concrete face of that claim, and yes: the prime radical is preserved by Morita equivalence, as are the Jacobson radical and the class of semiprime rings. What is not Morita invariant is being a domain or being reduced, which is why those notions are the wrong ones to build a structure theory on.

Can I compute

{Nil}_{*} R [x]

without knowing all primes of

R [x]

Yes — that is precisely the point. You need only ${Nil}_{*} R$ , computed in the base ring, and then extend coefficientwise. The primes of $R [x]$ are vastly more complicated than those of $R$ even for $R = ℤ$ , but their intersection is not.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §10, (10.18)–(10.21).
S. A. Amitsur, “Radicals of polynomial rings”, Canadian Journal of Mathematics 8 (1956), 355–361.
N. H. McCoy, “Prime ideals in general rings”, American Journal of Mathematics 71 (1949), 823–833.
A. Smoktunowicz, “Polynomial rings over nil rings need not be nil”, Journal of Algebra 233 (2000), 427–436.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Write out the induction that reduces the many-variable case of $(10.18)$ to one variable, and check where centrality of the variables is used.
Prove the matrix ideal correspondence $𝔄 \mapsto M_{n} (𝔄)$ and the identity $M_{n} (𝔄) M_{n} (𝔅) = M_{n} (𝔄 𝔅)$ .
What is ${Nil}_{*}$ of a Laurent polynomial ring $R [t, t^{- 1}]$ , and does the same argument apply?
Explain how Smoktunowicz's example is consistent with Köthe's Conjecture remaining open.
Compare the prime spectra of $R$ and $R [x]$ for $R$ a noncommutative noetherian ring.
Which radicals in the literature are matrix-extensible, and what general theory explains that?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Leading coefficients see primeness

From classes of rings to radicals

Matrices: the ideal lattice does the work

06Key Results

07Proof Techniques and Method

Test with constants only

Sandwich the radical

Contract along an inclusion

08Worked Example

A finite base ring

Where the Jacobson radical parts company

A noncommutative check

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Computational Notes

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

Process and Workflow

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions