The Radical of an Algebraic Algebra

Q: Does (4.18) say that algebraic elements are nilpotent?

No. It says that inside rad R the two conditions coincide. Outside the radical algebraic elements are everywhere and are usually not nilpotent — every element of a finite field extension of k inside R is algebraic and none of the nonzero ones is nilpotent.

Q: Why can the radical of an algebraic algebra fail to be nilpotent?

Because nilpotence is a uniform statement — one exponent that works for all products — while nil is element-wise. In k[x_1, x_2, …]/(x_i^p) every element dies, but the exponent needed grows with the number of variables involved, and x_1 x_2 … x_N ≠ 0 shows no uniform exponent exists.

Q: Is every algebraic algebra locally finite?

No. Golod's 1964 construction gives finitely generated nil algebras of infinite dimension over any field; a nil algebra is algebraic, since each element satisfies t^m = 0. This answers the Kurosh problem in the negative and shows the class of algebraic algebras is strictly larger than the locally finite one.

Q: Does (4.19) identify the radical with the lower nilradical too?

No, only with the upper nilradical Nil^* R, the largest nil ideal. The lower nilradical Nil_* R is the intersection of the prime ideals and can be strictly smaller; the two agree under stronger hypotheses such as the artinian condition.

Q: What happens if the base is a commutative ring rather than a field?

The proof breaks at the point where it divides by the lowest nonzero coefficient. There is no useful replacement statement: integrality over a central subring does not force elements of the radical to be nilpotent unless that subring is a field.

Q: How does this compare with Amitsur's cardinality theorem?

Both conclude that rad R is the largest nil ideal, and both route through (4.18). The difference is how algebraicity is obtained: here it is assumed of every element, whereas Amitsur derives it from the inequality _k R < |k| by a linear dependence argument among the inverses (a - r)^-1.

Executive Summary

For a general ring the Jacobson radical need not be nil: $rad k [[t]] = (t)$ has no nonzero nilpotent element. Lam's $(4.18)$ isolates exactly what goes wrong. Inside the radical of a $k$ -algebra, algebraic and nilpotent are the same condition, so the radical fails to be nil only by containing transcendental elements.

Restricting to algebraic algebras removes that possibility. For such an algebra $rad R$ is nil, and since every nil one-sided ideal already lies in the radical, $rad R$ is the largest nil ideal — it coincides with the upper nilradical ${Nil}^{*} R$ . Two nontrivial consequences follow: the radical need not be nilpotent, and the Köthe conjecture is a theorem in this class.

(4.18)Algebraic = nilpotent in the radical

(4.19)Radical = largest nil ideal

${Nil}^{*} R$ What the radical becomes

nil, not nilpotentBest possible conclusion

Overview

Let $k$ be a field and $R$ a $k$ -algebra with identity, so $k$ maps into the centre of $R$ . An element $x \in R$ is **algebraic over $k$ ** if $f (x) = 0$ for some nonzero $f \in k [t]$ ; $R$ is an algebraic algebra if every element is algebraic.

Every finite-dimensional algebra is algebraic, since $1, x, x^{2}, \dots$ cannot be independent. But the class is much wider: infinite algebraic field extensions, group algebras of locally finite groups, and — by Golod's construction — finitely generated infinite-dimensional nil algebras all qualify.

x \in rad R and x algebraic over k ⟺ x \in rad R and x nilpotent

(4.18)

The equivalence is false outside the radical: $x = 1$ is algebraic and not nilpotent.

The result is the bridge between the unit-theoretic radical of Jacobson Radical Definition and Characterisations and the nil-theoretic radicals discussed under Upper Nilradical and the Köthe Conjecture. For algebraic algebras the two theories agree.

Learning Objectives

State $(4.18)$ with the hypothesis that $k$ is a field and $R$ a $k$ -algebra.
Prove that an algebraic element of $rad R$ is nilpotent by factoring out the lowest power.
Combine $(4.18)$ with $(4.11)$ to obtain $rad R = {Nil}^{*} R$ for algebraic algebras.
Separate the classes finite-dimensional, locally finite and algebraic, and cite the separating examples.
Build an algebraic algebra whose radical is nil but not nilpotent.
Explain why $(4.19)$ resolves the Köthe conjecture for algebraic algebras.

Definitions

Definition—Algebraic algebra

Let $k$ be a field. A $k$ -algebra $R$ is algebraic if every $x \in R$ satisfies a nonzero polynomial $f \in k [t]$ . Equivalently, the subalgebra $k [x] \subseteq R$ is finite-dimensional for every $x$ , so that $R$ is the union of its finite-dimensional subalgebras $k [x]$ .

Nil ideal: A one-sided or two-sided ideal all of whose elements are nilpotent.
Nilpotent ideal: An ideal $𝔄$ with $𝔄^{n} = 0$ for some $n$ , meaning every product of $n$ elements vanishes. Strictly stronger than nil.
${Nil}^{*} R$: The upper nilradical: the sum of all nil ideals of $R$ , itself nil, hence the largest nil ideal.
${Nil}_{*} R$: The lower nilradical or prime radical: the intersection of the prime ideals of $R$ , equal to the smallest semiprime ideal.
Locally finite algebra: Every finitely generated subalgebra is finite-dimensional over $k$ . Strictly between finite-dimensional and algebraic.

The base ring must be a field. Over a commutative base ring that is not a field the leading coefficient $a_{r}$ need not be invertible and $(4.18)$ fails.

Core Concepts

Why the lowest term, not the leading term

Given a polynomial relation for $x$ , one is tempted to divide by the leading coefficient and solve for the top power. That is the wrong move here. Write the relation in ascending order and let $r$ be the least index with $a_{r} \neq 0$ :

a_{r} x^{r} + a_{r + 1} x^{r + 1} + \dots + a_{n} x^{n} = 0, a_{r} \neq 0,

(4.18a)

Factoring $x^{r}$ out on the left leaves $x^{r} (a_{r} + a_{r + 1} x + \dots + a_{n} x^{n - r}) = 0$ , and dividing by the scalar $a_{r}$ turns the bracket into $1 + u$ where $u = a_{r}^{- 1} a_{r + 1} x + \dots + a_{r}^{- 1} a_{n} x^{n - r}$ . Since $x \in rad R$ and $rad R$ is a two-sided ideal, $u \in rad R$ , hence $1 + u \in U (R)$ and $x^{r} = 0$ .

From elements to ideals

Nilpotence of every element of $rad R$ says $rad R$ is a nil ideal. The reverse containment is Lam's $(4.11)$ : if a one-sided ideal $𝔄$ is nil then for $y \in 𝔄$ and any $x$ , the element $x y$ is nilpotent, so $1 - x y$ has the geometric-series inverse $1 + x y + (x y)^{2} + \dots$ , a finite sum. Hence $𝔄 \subseteq rad R$ .

every element algebraic

rad R

nil

rad R \subseteq {Nil}^{*} R

{Nil}^{*} R \subseteq rad R

(4.11)

equality

Nil is as far as it goes

One cannot upgrade nil to nilpotent in $(4.19)$ . The obstruction is genuine and appears already for commutative algebraic algebras — the worked example below is a group algebra whose radical is nil of unbounded index. Nilpotence of the radical requires a chain condition, as in Lam's $(4.12)$ for left artinian rings.

Key Results

Proposition(4.18)Algebraic elements of the radical

Let $k$ be a field, $R$ a $k$ -algebra, and $x \in rad R$ . Then $x$ is algebraic over $k$ if and only if $x$ is nilpotent.

Proof

**( $\Leftarrow$ )** If $x^{m} = 0$ then $x$ satisfies $t^{m}$ , a nonzero polynomial over $k$ .

**( $\Rightarrow$ )** Assume $R \neq 0$ and let $f \in k [t]$ be nonzero with $f (x) = 0$ . Write $f$ in ascending order, $f (t) = a_{r} t^{r} + a_{r + 1} t^{r + 1} + \dots + a_{n} t^{n}$ with $a_{r} \neq 0$ . Then

0 = f (x) = x^{r} (a_{r} + a_{r + 1} x + \dots + a_{n} x^{n - r}) = a_{r} x^{r} (1 + u), u : = \sum_{i = r + 1}^{n} a_{r}^{- 1} a_{i} x^{i - r} .

Every term of $u$ has a factor $x \in rad R$ , and $rad R$ is a two-sided ideal, so $u \in rad R$ and therefore $1 + u \in U (R)$ . Multiplying on the right by $(1 + u)^{- 1}$ and by the scalar $a_{r}^{- 1}$ gives $x^{r} = 0$ . Finally $r \geq 1$ , since $r = 0$ would give $1 = 0$ ; hence $x$ is nilpotent.

Corollary(4.19)The radical of an algebraic algebra

Let $R$ be an algebraic algebra over a field $k$ . Then $rad R$ is the largest nil ideal of $R$ ; it contains every nil one-sided ideal, and $rad R = {Nil}^{*} R$ .

Proof

Every $x \in rad R$ is algebraic by hypothesis, hence nilpotent by $(4.18)$ ; so $rad R$ is a nil ideal. Conversely, by $(4.11)$ every nil one-sided ideal of any ring is contained in $rad R$ . So $rad R$ is a nil ideal containing all nil one-sided ideals, which is exactly the assertion that it is the largest nil ideal, and by definition of the upper nilradical, $rad R = {Nil}^{*} R$ .

Corollary(4.19a)Köthe's conjecture for algebraic algebras

If $R$ is an algebraic algebra over a field, then every nil one-sided ideal of $R$ is contained in a nil two-sided ideal, and the sum of two nil left ideals of $R$ is nil. That is, the Köthe conjecture holds for $R$ .

Proof

By $(4.19)$ , every nil one-sided ideal lies in ${Nil}^{*} R = rad R$ , which is itself nil and two-sided. A sum of two nil left ideals therefore lies in ${Nil}^{*} R$ and so is nil. The general conjecture asserts precisely this for arbitrary rings and remains open.

Corollary(4.19b)Finite-dimensional algebras

If $\dim_{k} R < \infty$ then $R$ is algebraic, and moreover $R$ is left artinian, so $rad R$ is the largest nilpotent ideal of $R$ and $R / rad R$ is semisimple. The nilpotence, unlike the nil property, comes from the chain condition and not from $(4.19)$ .

Counterexample—Why the base must be a field

Let $A = ℤ_{(p)}$ and regard $R = ℤ_{(p)}$ as an algebra over itself. Then $rad R = p R$ , and the element $x = p$ lies in the radical and satisfies the nonzero polynomial $t - p \in A [t]$ , so it is *algebraic over $A$ *. It is not nilpotent. The step that fails is the division by the lowest nonzero coefficient: here that coefficient is $- p$ , which is not invertible in $A$ . Over a field this cannot happen, which is why $(4.18)$ is stated for a field base and admits no straightforward generalisation to commutative base rings.

Proof Techniques and Method

How the proof works, and the reusable move.

One trick, used twice.

Move 1

Factor out the lowest power, invert the rest

From $x^{r} v = 0$ with $v \in a_{r} (1 + rad R)$ conclude $x^{r} = 0$ . This converts any polynomial relation into a nilpotence statement, provided the lowest coefficient is invertible in the base.

Move 2

Sandwich the radical between nil ideals

To prove $rad R$ equals the largest nil ideal, show (a) $rad R$ is nil, and (b) every nil one-sided ideal is inside it. Half (b) is free from $(4.11)$ for every ring; only (a) needs hypotheses.

The second move is the standard template for identifying $rad R$ with a nil-type radical, and it recurs on Amitsur's Radical Theorem, where hypothesis (a) is obtained from a cardinality argument instead of an algebraicity one.

Take $x \in rad R$ The goal is a bound on the order of nilpotence, or at least nilpotence itself.

Produce a polynomial relationFrom algebraicity, or from a cardinality or dimension count.

Normalise itPush the relation into the form

x^{r} (1 + u) = 0

with

u \in rad R

Invert and conclude

1 + u \in U (R)

gives

x^{r} = 0

; then

(4.11)

closes the loop to

{Nil}^{*} R

Worked Example

A finite example first

Let $k$ be a field with $char k = p$ and $G = ℤ / p$ generated by $g$ . Then $k G ≅ k [t] / (t^{p} - 1)$ , and in characteristic $p$ we have $t^{p} - 1 = (t - 1)^{p}$ , so with $x = t - 1$ ,

k G ≅ k [x] / (x^{p}) .

(E.1)

This is a local $k$ -algebra of dimension $p$ with $rad (k G) = (x)$ , of $k$ -dimension $p - 1$ and nilpotency index exactly $p$ . It is algebraic (finite-dimensional), and $(4.19)$ is confirmed: $(x)$ is the largest nil ideal.

An infinite-dimensional algebraic algebra whose radical is not nilpotent

Keep $char k = p$ and take $G = ⨁_{i \geq 1} ℤ / p$ , an infinite elementary abelian $p$ -group. $G$ is locally finite, so $k G$ is an algebraic $k$ -algebra. Writing $x_{i} = g_{i} - 1$ for the generators, the computation above tensors up to

k G ≅ k [x_{1}, x_{2}, x_{3}, \dots] / (x_{1}^{p}, x_{2}^{p}, x_{3}^{p}, \dots) .

(E.2)

A commutative local $k$ -algebra of countably infinite dimension.

Let $𝔪 = (x_{1}, x_{2}, \dots)$ be the augmentation ideal. Every element of $k G$ outside $𝔪$ has nonzero constant term and lies in some finite-dimensional local subalgebra $k [x_{1}, \dots, x_{N}] / (x_{i}^{p})$ , where it is a unit; so $k G$ is local with $rad (k G) = 𝔪$ .

** $𝔪$ is nil.** Any $f \in 𝔪$ involves finitely many $x_{i}$ , say $x_{1}, \dots, x_{N}$ , so $f$ lies in $k [x_{1}, \dots, x_{N}] / (x_{i}^{p})$ whose maximal ideal satisfies $𝔪_{N}^{N (p - 1) + 1} = 0$ . Hence $f^{N (p - 1) + 1} = 0$ .
** $𝔪$ is not nilpotent.** For every $N$ the product $x_{1} x_{2} \dots x_{N}$ is a nonzero monomial of $k G$ lying in $𝔪^{N}$ , so $𝔪^{N} \neq 0$ for all $N$ .
Conclusion. $rad (k G) = {Nil}^{*} (k G) = 𝔪$ is nil of unbounded index. $(4.19)$ is sharp: nil cannot be improved to nilpotent.

A non-example that explains the hypothesis

Let $R = k [[t]]$ . Then $rad R = (t)$ but $t$ is not nilpotent. By $(4.18)$ , read contrapositively, $t$ cannot be algebraic over $k$ — and it is not. So $k [[t]]$ is not an algebraic algebra, and $(4.19)$ does not apply; $rad R$ here is not even nil, while the largest nil ideal of $R$ is $0$ .

Frameworks and Models

The classes of algebras involved are nested, and each containment is strict.

All $k$ -algebras

rad R

can be non-nil, e.g.

k [[t]]

Algebraic algebrasevery element satisfies a polynomial over

k

;

rad R = {Nil}^{*} R

(4.19)

Locally finite algebrasevery finitely generated subalgebra is finite-dimensional

Finite-dimensional algebras

rad R

is nilpotent and

R / rad R

is semisimple

Sources of algebraic algebras
- Finite-dimensional
  - $M_{n} (D)$ for $D$ finite-dimensional over $k$
  - $k G$ for $G$ a finite group
  - path algebras of finite quivers without oriented cycles
- Locally finite but infinite-dimensional
  - algebraic field extensions $K / k$ with $[K : k] = \infty$ , such as an algebraic closure
  - $k G$ for $G$ locally finite, in particular any abelian torsion group
  - unions of ascending chains of finite-dimensional subalgebras
- Algebraic but not locally finite
  - Golod's finitely generated infinite-dimensional nil algebras (1964), which answer the Kurosh problem negatively

Relationship Map

Four radicals are in play. The containments on the left hold in every ring; the collapses on the right are what hypotheses buy.

{Nil}_{*} R

Levitzki (R)

{Nil}^{*} R

rad R

When the chain collapses
Hypothesis on $R$	What becomes equal	Reason
none	nothing in general	$k [[t]]$ has ${Nil}^{*} R = 0 ⊊ (t) = rad R$
algebraic over a field $k$	${Nil}^{*} R = rad R$	$(4.18)$ plus $(4.11)$
$\dim_{k} R < \| k \|$	${Nil}^{*} R = rad R$	Amitsur's theorem $(4.20)$ , via $(4.18)$
left artinian	all four are equal and nilpotent	$(4.12)$ : the radical is the largest nilpotent ideal
commutative	${Nil}_{} R = {Nil}^{} R =$ the nilradical	nilpotents form an ideal; $rad R$ can still be larger

The middle two rows are the point of this page and of Amitsur's Radical Theorem: two quite different hypotheses — algebraicity of elements, smallness of dimension — deliver the same collapse, and both do it through $(4.18)$ .

Applications and Industry Use

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

Group representation theory

Locally finite groups

For $G$ locally finite and $k$ any field, $k G$ is algebraic, so $rad (k G)$ is the largest nil ideal. This is the starting point for the modular representation theory of infinite locally finite groups, where no artinian hypothesis is available.

Radical theory

A test class for Köthe

The Köthe conjecture is open in general but a theorem for algebraic algebras by $(4.19 a)$ . Any counterexample must therefore contain a transcendental element in its radical, which sharply constrains where to look.

Field theory

Algebraic extensions

An algebraic field extension $K / k$ of infinite degree is an algebraic algebra with zero radical. The content of $(4.19)$ is vacuous here, but the class is the reason algebraic algebras are not merely finite-dimensional algebras in disguise.

Symbolic computation

Recognising nilpotence

In a computer algebra system, testing whether a radical element is nilpotent reduces to computing its minimal polynomial when the algebra is algebraic. $(4.18)$ is the theorem that licenses replacing a nilpotence test by a linear algebra computation in $k [x]$ .

The honest summary: this is internal machinery. Its value lies in extending radical theory beyond the artinian world that most computational and representation-theoretic applications inhabit.

Computational Notes

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

For a single element $x$ of a finite-dimensional algebra $A$ with $\dim_{k} A = n$ , the minimal polynomial of the left multiplication operator $L_{x}$ is computable in $O (n^{3})$ field operations, and $x$ is nilpotent exactly when that polynomial is a power of $t$ .
$(4.18)$ turns the question *is $x \in rad A$ nilpotent?* into *is $x$ algebraic?*, which is automatic in finite dimension. The proposition therefore has no algorithmic content for finite-dimensional input; it earns its keep on infinite-dimensional algebras.
For an algebra presented by generators and relations, algebraicity of a given element is undecidable in general, because the word problem for finitely presented associative algebras is undecidable. No procedure can decide membership in ${Nil}^{*} R$ from a presentation.
Group algebras of locally finite groups are handled by direct limits: computations are performed in a finite subgroup algebra $k H$ and the answer is stable under enlarging $H$ only for element-wise questions such as nilpotence, not for ideal-theoretic ones such as nilpotency index.

Failure Modes and Common Mistakes

Do not read $(4.18)$ as algebraic implies nilpotent. The hypothesis $x \in rad R$ is essential; $1$ is algebraic and is not nilpotent.
Do not assume ${Nil}_{*} R = {Nil}^{*} R$ for algebraic algebras. $(4.19)$ identifies the radical with the upper nilradical only; the lower nilradical can be strictly smaller.
Do not conclude that an algebraic algebra is left artinian or even left noetherian. Neither follows, as $(E . 2)$ shows.
Do not apply $(4.19)$ to a ring that is merely integral over a central subring which is not a field; the argument is genuinely about fields.

Historical Notes and Lessons Learned

1908Wedderburn's nilpotent radicalFor finite-dimensional algebras the radical is defined as the largest nilpotent ideal. Nil and nilpotent coincide in that setting, and the distinction is invisible.
1930Koethe's conjectureKoethe asks whether a ring with no nonzero nil two-sided ideal can have a nonzero nil one-sided ideal. Equivalently, whether every nil one-sided ideal lies in the upper nilradical. It is still open.
1941The Kurosh problemKurosh asks whether every finitely generated algebraic algebra is finite-dimensional, the algebraic analogue of the Burnside problem for groups.
1945Jacobson's radicalThe radical becomes available for arbitrary rings and is no longer defined by nilpotence, which makes the comparison with nil ideals a genuine question.
1964Golod and ShafarevichGolod constructs finitely generated infinite-dimensional nil algebras, answering Kurosh negatively and separating algebraic from locally finite. The same machinery produces infinite finitely generated torsion groups.

The lesson is about the right level of generality. Wedderburn's nilpotence and Jacobson's unit-theoretic radical agree on finite-dimensional algebras, diverge on general rings, and are reconciled on algebraic algebras — a class defined element by element rather than by any finiteness of the whole. Element-wise hypotheses are often the ones that survive the passage to infinite dimension.

Quick Reference

$(4.18)$

x \in rad R

: algebraic over

k

iff

nilpotent

$(4.19)$

R

algebraic

\Rightarrow

rad R = {Nil}^{*} R

, the largest nil ideal

Key inequality

1 + rad R \subseteq U (R)

— the only fact used

Always truenil one-sided ideal

\subseteq rad R

, by

(4.11)

Not impliednilpotence of

rad R

; that needs a chain condition

Class inclusionsfinite-dimensional

⊊

locally finite

⊊

algebraic

Kötheholds for algebraic algebras; open in general

Standard example

k [x_{1}, x_{2}, \dots] / (x_{i}^{p})

: radical nil, not nilpotent

What the hypotheses give
Hypothesis	Conclusion about $rad R$	Reference
$k$ -algebra, $x \in rad R$ algebraic	$x$ is nilpotent	(4.18)
$R$ an algebraic $k$ -algebra	$rad R$ is the largest nil ideal	(4.19)
$\dim_{k} R < \infty$	$rad R$ is the largest nilpotent ideal	(4.12)
$R$ left artinian	$rad R$ nilpotent, $R / rad R$ semisimple	(4.12), (4.14)
none	$rad R$ need not be nil	$k [[t]]$

Frequently Asked Questions

Does

(4.18)

say that algebraic elements are nilpotent?

No. It says that inside $rad R$ the two conditions coincide. Outside the radical algebraic elements are everywhere and are usually not nilpotent — every element of a finite field extension of $k$ inside $R$ is algebraic and none of the nonzero ones is nilpotent.

Why can the radical of an algebraic algebra fail to be nilpotent?

Because nilpotence is a uniform statement — one exponent that works for all products — while nil is element-wise. In $k [x_{1}, x_{2}, \dots] / (x_{i}^{p})$ every element dies, but the exponent needed grows with the number of variables involved, and $x_{1} x_{2} \dots x_{N} \neq 0$ shows no uniform exponent exists.

Is every algebraic algebra locally finite?

No. Golod's 1964 construction gives finitely generated nil algebras of infinite dimension over any field; a nil algebra is algebraic, since each element satisfies $t^{m} = 0$ . This answers the Kurosh problem in the negative and shows the class of algebraic algebras is strictly larger than the locally finite one.

Does

(4.19)

identify the radical with the lower nilradical too?

No, only with the upper nilradical ${Nil}^{*} R$ , the largest nil ideal. The lower nilradical ${Nil}_{*} R$ is the intersection of the prime ideals and can be strictly smaller; the two agree under stronger hypotheses such as the artinian condition.

What happens if the base is a commutative ring rather than a field?

The proof breaks at the point where it divides by the lowest nonzero coefficient. There is no useful replacement statement: integrality over a central subring does not force elements of the radical to be nilpotent unless that subring is a field.

How does this compare with Amitsur's cardinality theorem?

Both conclude that $rad R$ is the largest nil ideal, and both route through $(4.18)$ . The difference is how algebraicity is obtained: here it is assumed of every element, whereas Amitsur derives it from the inequality $\dim_{k} R < | k |$ by a linear dependence argument among the inverses $(a - r)^{- 1}$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §4, statements (4.18) and (4.19).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
I. N. Herstein, Noncommutative Rings, Carus Mathematical Monographs 15, Mathematical Association of America, 1968.
E. S. Golod, “On nil-algebras and finitely approximable p-groups”, Izvestiya Akademii Nauk SSSR, Seriya Matematicheskaya 28 (1964).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Sketch Golod's construction of a finitely generated infinite-dimensional nil algebra and explain the role of the Golod–Shafarevich inequality.
Give an algebraic algebra whose lower nilradical is strictly smaller than its upper nilradical.
Is the tensor product of two algebraic algebras over a field again algebraic?
For which locally finite groups $G$ and fields $k$ is $rad (k G)$ nonzero?
State the equivalent formulations of the Köthe conjecture and identify which one $(4.19 a)$ verifies.
What is known about the radical of an algebra that is integral over its centre rather than algebraic over a field?
Can an algebraic algebra over a field be left noetherian without being finite-dimensional?

Radical of Algebraic Algebras

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why the lowest term, not the leading term

From elements to ideals

Nil is as far as it goes

Key Results

Proof Techniques and Method

Factor out the lowest power, invert the rest

Sandwich the radical between nil ideals

Worked Example

A finite example first

An infinite-dimensional algebraic algebra whose radical is not nilpotent

A non-example that explains the hypothesis

Frameworks and Models

Relationship Map

Applications and Industry Use

Locally finite groups

A test class for Köthe

Algebraic extensions

Recognising nilpotence

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the lowest term, not the leading term

From elements to ideals

Nil is as far as it goes

06Key Results

07Proof Techniques and Method

Factor out the lowest power, invert the rest

Sandwich the radical between nil ideals

08Worked Example

A finite example first

An infinite-dimensional algebraic algebra whose radical is not nilpotent

A non-example that explains the hypothesis

09Frameworks and Models

10Relationship Map

11Applications and Industry Use

Locally finite groups

A test class for Köthe

Algebraic extensions

Recognising nilpotence

12Computational Notes

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Frameworks and Models

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions