Amitsur’s Theorem on the Radical of an Algebra of Small Dimension

Q: Why does the theorem compare dimension with the cardinality of the field rather than with the cardinality of R?

Because the pigeonhole is applied to a linearly independent family in a k-vector space. The obstruction is that the resolvent family, indexed by k^×, is too large to be independent — a statement about dimension. Comparing |R| with |k| would be both weaker and unrelated to the argument.

Q: Is the finite field case really necessary as a separate argument?

Yes, though it is short. For finite k the inequality _k R < |k| makes R a finite ring, and the infinite-field step |k^×| = |k| is unavailable. The conclusion then comes from the artinian theory, where the radical is even nilpotent.

Q: Does (4.20) ever give rad R = 0 on its own?

Only when R happens to have no nonzero nil ideal, which is an extra input. For CG that input is the positivity of the trace; for a domain it is immediate. The theorem itself only identifies the radical with the largest nil ideal.

Q: What is the relationship to Amitsur's theorem on polynomial rings?

They are different results by the same author. The polynomial-ring theorem computes rad(R[t]) as N[t] for a nil ideal N of R; the theorem on this page bounds the radical of an algebra by a cardinality hypothesis. Both share the moral that the radical of a large but structured object is controlled by nil ideals.

Q: Can the hypothesis be weakened to _k R ≤ |k|?

No. C[t]_(t) has _C R = |C|, is local, and its radical tR is not nil. The resolvents (a - t)^-1 in that ring really are independent, so the pigeonhole has nothing to work with.

Q: Does the result hold for algebras over a division ring rather than a field?

The argument as given uses commutativity of the base in two places: the scalars a_i must commute with r to keep everything inside k[r], and f must be an ordinary polynomial. Over a noncommutative base one is in the territory of skew polynomials and the statement requires care; it should not be quoted in that setting.

Executive Summary

Amitsur's theorem replaces every finiteness hypothesis on an algebra by a single comparison of cardinal numbers. If $\dim_{k} R < | k |$ , then $rad R$ is nil, hence — by the standard containment of nil one-sided ideals in the radical — the largest nil ideal of $R$ .

The argument is a counting argument, not a structural one. For $r \in rad R$ every scalar shift $a - r$ with $a \neq 0$ is a unit, so the algebra contains a family of resolvents $(a - r)^{- 1}$ indexed by $k^{\times}$ . If there are more of them than the dimension permits, they must be linearly dependent, and clearing denominators turns the dependence into a polynomial equation for $r$ .

$\dim_{k} R < | k |$ The whole hypothesis

(4.20)Amitsur

(4.21)Countably generated corollary

strictThe inequality cannot be relaxed

Overview

On Radical of Algebraic Algebras the radical is shown to be nil whenever every element of the algebra satisfies a polynomial. Amitsur's insight is that algebraicity need not be assumed: it is forced, for elements of the radical, as soon as the algebra is small relative to the field.

\dim_{k} R < | k | ⟹ rad R is nil ⟹ rad R = {Nil}^{*} R,

(4.20)

Cardinal arithmetic on the left; the second implication is the general fact that nil one-sided ideals lie in the radical.

The mechanism is the classical partial-fraction phenomenon. Over $ℂ$ the rational functions $1 / (a - t)$ , one for each $a \in ℂ^{\times}$ , are linearly independent — there are continuum many of them and they force the containing algebra to have continuum dimension. If the algebra is smaller than that, no element can behave like a transcendental $t$ .

The most-used consequence is $(4.21)$ : over an uncountable field, any algebra with countably many generators has nil radical. That covers finitely generated algebras over $ℂ$ , group algebras of countable groups, enveloping algebras of finite-dimensional Lie algebras, and Weyl algebras.

Learning Objectives

State $(4.20)$ precisely, with $\dim_{k} R$ and $| k |$ compared as cardinals.
Show that $a - r \in U (R)$ for every $a \in k^{\times}$ and $r \in rad R$ .
Turn a linear dependence among the resolvents into a nonzero polynomial annihilating $r$ .
Verify the nonvanishing of that polynomial by evaluating at the scalars $a_{i}$ .
Dispose of the finite field case using the artinian theory.
Exhibit an algebra with $\dim_{k} R = | k |$ whose radical is not nil.

Definitions

Definition—The resolvent family

Let $R$ be a $k$ -algebra and $r \in rad R$ . For $a \in k^{\times}$ write $a - r = a (1 - a^{- 1} r)$ . Since $a^{- 1} r \in rad R$ , the maximality property of the radical gives $1 - a^{- 1} r \in U (R)$ , so $a - r$ is a unit. The resolvent family of $r$ is ${(a - r)^{- 1} : a \in k^{\times}} \subseteq R$ .

$\dim_{k} R$: The cardinality of a $k$ -basis of $R$ as a vector space; a cardinal number, possibly infinite.
$| k |$: The cardinality of the underlying set of the field. For infinite $k$ , $| k^{\times} | = | k |$ .
Countably generated: Generated as a $k$ -algebra by a countable set; then the words in those generators span $R$ , so $\dim_{k} R \leq ℵ_{0}$ .
${Nil}^{*} R$: The upper nilradical: the largest nil two-sided ideal of $R$ .
Nil versus nilpotent: Nil is element-wise nilpotence; nilpotent asks for a single exponent killing all products. Amitsur's conclusion is nil only.

The hypothesis compares $\dim_{k} R$ with $| k |$ , not $| R |$ with $| k |$ . For infinite-dimensional $R$ over an infinite $k$ these differ, and using the wrong one makes the theorem false.

Core Concepts

Why resolvents detect transcendence

If $r$ is transcendental over $k$ then $k [r] ≅ k [t]$ and the subalgebra generated by $r$ and the resolvents is a copy of a localisation of $k [t]$ . In such a ring, partial fractions say that the elements $1 / (a - t)$ , $a$ ranging over distinct scalars, are $k$ -linearly independent: a dependence $\sum_{i} b_{i} / (a_{i} - t) = 0$ multiplied out gives a polynomial with a nonzero value at $t = a_{i}$ whenever $b_{i} \neq 0$ .

Turning that around: a dependence among the resolvents is evidence of algebraicity. Amitsur's hypothesis manufactures such a dependence by pigeonhole.

| {(a - r)^{- 1} : a \in k^{\times}} | \leq | k^{\times} | = | k | > \dim_{k} R,

(4.20a)

More vectors than the dimension allows: the family must be linearly dependent.

Everything commutes

The elements $a_{i} - r$ all lie in the commutative subalgebra $k [r]$ , so they commute with one another. An invertible element commutes with everything its underlying element commutes with, so the resolvents commute with each other and with every $a_{j} - r$ . The manipulation of clearing denominators is therefore legitimate in a noncommutative $R$ — a point worth checking rather than assuming.

Two regimes

For finite $k$ the hypothesis $\dim_{k} R < | k |$ makes $R$ a finite ring, and the conclusion follows from the artinian theory rather than from any counting. The interesting content is the infinite case, where $| k^{\times} | = | k |$ and the pigeonhole bites.

Key Results

Theorem(4.20)Amitsur

Let $k$ be a field and $R$ a $k$ -algebra with $\dim_{k} R < | k |$ as cardinal numbers. Then $rad R$ is nil, and consequently $rad R$ is the largest nil ideal of $R$ ; that is, $rad R = {Nil}^{*} R$ and every nil one-sided ideal of $R$ is contained in it.

Proof

By Lam's $(4.11)$ every nil one-sided ideal lies in $rad R$ , so it suffices to prove that $rad R$ is nil.

Finite base field. If $| k | < \infty$ then $\dim_{k} R < | k |$ is a finite cardinal, so $R$ is a finite set. A finite ring is left artinian, so $rad R$ is nilpotent by $(4.12)$ , in particular nil.

Infinite base field. By $(4.18)$ it is enough to show that every $r \in rad R$ is algebraic over $k$ . Fix such an $r$ . For each $a \in k^{\times}$ the element $a - r = a (1 - a^{- 1} r)$ is a unit, because $a^{- 1} r \in rad R$ . Since $| k^{\times} | = | k | > \dim_{k} R$ , the family ${(a - r)^{- 1} : a \in k^{\times}}$ cannot be $k$ -linearly independent. Choose distinct $a_{1}, \dots, a_{n} \in k^{\times}$ and $b_{1}, \dots, b_{n} \in k$ , not all zero, with

\sum_{i = 1}^{n} b_{i} (a_{i} - r)^{- 1} = 0 .

All the elements $a_{i} - r$ lie in the commutative subalgebra $k [r]$ , hence commute; so do their inverses. Multiplying the relation by $\prod_{j = 1}^{n} (a_{j} - r)$ therefore gives

\sum_{i = 1}^{n} b_{i} \prod_{j \neq i} (a_{j} - r) = 0,

so $r$ is a root of the polynomial $f (t) = \sum_{i = 1}^{n} b_{i} \prod_{j \neq i} (a_{j} - t) \in k [t]$ . It remains to see $f \neq 0$ . Pick $i$ with $b_{i} \neq 0$ and evaluate at $t = a_{i}$ : every summand indexed by $ℓ \neq i$ contains the factor $(a_{i} - a_{i}) = 0$ , so

f (a_{i}) = b_{i} \prod_{j \neq i} (a_{j} - a_{i}) \neq 0,

because the $a_{j}$ are distinct and $k$ is a field. Hence $f$ is a nonzero polynomial with $f (r) = 0$ , so $r$ is algebraic over $k$ , and $(4.18)$ makes $r$ nilpotent. Therefore $rad R$ is nil.

Corollary(4.21)Countably generated algebras over an uncountable field

Let $k$ be an uncountable field and let $R$ be a $k$ -algebra generated as a $k$ -algebra by a countable set. Then $rad R$ is the largest nil ideal of $R$ .

Proof

The words in a countable generating set, together with $1$ , form a countable spanning set of $R$ as a $k$ -vector space, so $\dim_{k} R \leq ℵ_{0}$ . Since $k$ is uncountable, $ℵ_{0} < | k |$ , and $(4.20)$ applies.

Corollary—Small division algebras are algebraic

Let $D$ be a division ring which is a $k$ -algebra with $\dim_{k} D < | k |$ . Then every element of $D$ is algebraic over $k$ .

Proof

If $x \in D$ were transcendental, then $k [x] ≅ k [t]$ and, $D$ being a division ring, $D$ would contain $(a - x)^{- 1}$ for every $a \in k$ . Those elements are $k$ -linearly independent, by the evaluation argument used above applied inside the field of fractions $k (x) \subseteq D$ . That would force $\dim_{k} D \geq | k |$ , a contradiction. The same counting is at work as in $(4.20)$ ; only the source of the invertibility differs — there it came from the radical, here from $D$ being a division ring.

Counterexample—The inequality must be strict

Let $k = ℂ$ and let $R = ℂ [t]_{(t)}$ be the localisation of $ℂ [t]$ at the maximal ideal $(t)$ , that is, the rational functions $p / q$ with $q (0) \neq 0$ . Then $R$ is local with $rad R = t R$ , which is not nil — indeed $R$ is a domain, so its largest nil ideal is $0 \neq rad R$ . The resolvents $(a - t)^{- 1}$ for $a \in ℂ^{\times}$ all lie in $R$ and are linearly independent, so $\dim_{ℂ} R = 2^{ℵ_{0}} = | ℂ |$ . The hypothesis of $(4.20)$ fails by exactly one notch, and the conclusion fails with it.

Proof Techniques and Method

How the proof works, and the reusable move.

Move 1

Manufacture units from the radical

For $r \in rad R$ and $a \in k^{\times}$ , $a - r = a (1 - a^{- 1} r)$ is a unit. A whole $k^{\times}$ -indexed family of invertible elements appears for free.

Move 2

Pigeonhole on dimension

A family of more than $\dim_{k} R$ vectors is dependent. Comparing an index set with a dimension is the cheapest way to produce a relation when no finiteness is available.

Move 3

Clear denominators, then evaluate

Multiply out to get a polynomial relation, then prove the polynomial is nonzero by evaluating at the very scalars that indexed the family. Distinctness of the $a_{i}$ does the rest.

Move 3 is the part that is easy to get wrong. Producing a polynomial that annihilates $r$ is worthless unless one can show it is not the zero polynomial; the evaluation at $t = a_{i}$ , where all but one summand vanishes, is the trick that makes the argument complete.

Reduce to nilnessBy

(4.11)

the only thing to prove is that

rad R

is nil.

Reduce to algebraicityBy

(4.18)

, for elements of the radical, algebraic and nilpotent are the same.

Count resolventsCompare

| k^{\times} |

with

\dim_{k} R

and extract a linear dependence.

Extract a polynomialClear denominators inside

k [r]

and verify nonvanishing by evaluation.

Worked Example

Laurent polynomials over the complex numbers

Let $R = ℂ [t, t^{- 1}] = ℂ [ℤ]$ , the group algebra of the infinite cyclic group. Then $\dim_{ℂ} R = ℵ_{0} < 2^{ℵ_{0}} = | ℂ |$ , so $(4.20)$ applies and $rad R$ is the largest nil ideal. But $R$ is a commutative domain, so its only nil ideal is $0$ :

rad ℂ [t, t^{- 1}] = 0 .

(E.1)

Direct verification: the maximal ideals of $R$ include $(t - a)$ for every $a \in ℂ^{\times}$ , and a nonzero Laurent polynomial has only finitely many roots, so it escapes some $(t - a)$ . The intersection is therefore $0$ , as $(4.20)$ predicted without any of that computation.

Group algebras of countable groups over $ℂ$

Let $G$ be a countable group and $R = ℂ G$ . Then $\dim_{ℂ} R = | G | \leq ℵ_{0} < | ℂ |$ , so by $(4.21)$ the radical $rad (ℂ G)$ is nil. Now use the positivity of the trace to show that $ℂ G$ has no nonzero nil ideal at all. For $α = \sum_{g} a_{g} g$ put $tr (α) = a_{1}$ and $α^{*} = \sum_{g} \bar{a_{g}} g^{- 1}$ , so that

tr (α α^{*}) = \sum_{g \in G} | a_{g} |^{2} > 0 whenever α \neq 0 .

(E.2)

Suppose $N$ is a nil ideal and $0 \neq α \in N$ . Put $β = α α^{*} \in N$ ; then $β^{*} = β$ and $tr (β) > 0$ , so $β \neq 0$ . Since $β$ is nilpotent, there is a least $m \geq 1$ with $β^{2^{m}} = 0$ . Set $γ = β^{2^{m - 1}} \neq 0$ ; as $β$ is self-adjoint, $γ γ^{*} = β^{2^{m}} = 0$ , whence $tr (γ γ^{*}) = 0$ , contradicting $(E . 2)$ . So $N = 0$ .

Where the hypothesis is violated

By contrast $ℂ [t]_{(t)}$ has $\dim_{ℂ} R = | ℂ |$ and radical $t R \neq 0$ containing no nonzero nilpotent, and $ℂ [[t]]$ likewise has dimension $2^{ℵ_{0}}$ with radical $(t)$ . Neither is countably generated as a $ℂ$ -algebra, even though $ℂ [[t]]$ is generated by a single element topologically — a reminder that $(4.21)$ asks for countably many algebra generators, not a countable dense subset.

Process and Workflow

How should I try to prove $rad R$ is nil?

$R$ is left artinianUse

(4.12)

: the radical is nilpotent, which is stronger. No cardinality argument is needed.

Every element is algebraic over $k$ Use

(4.19)

directly. Typical for group algebras of locally finite groups and for algebraic field extensions.

$\dim_{k} R$ is small relative to $| k |$ Use

(4.20)

. In practice this means: countably generated algebra, uncountable field, which is

(4.21)

None of theseThere is no general theorem. The radical of

k [[t]]

is not nil, and deciding nilness from a presentation is not algorithmic.

Count the generatorsIf

R

is generated by countably many elements over

k

, its dimension is at most

ℵ_{0}

Count the fieldCheck that

| k |

strictly exceeds that dimension.

ℝ

ℂ

and any uncountable field qualify against

ℵ_{0}

Conclude nilnessApply

(4.20)

rad R = {Nil}^{*} R

Look for a positivity argumentTo upgrade to

rad R = 0

, show separately that

R

has no nonzero nil ideal — a trace or involution argument if one is available.

Comparison and Classification

What each hypothesis delivers
	$rad R$ nil	$rad R$ nilpotent	$rad R = 0$	needs a chain condition
$R$ algebraic over $k$ — $(4.19)$	yes	no	no	no
$\dim_{k} R < \| k \|$ — $(4.20)$	yes	no	no	no
$k$ uncountable, $R$ countably generated — $(4.21)$	yes	no	no	no
$R$ left artinian — $(4.12)$	yes	yes	no	yes
$R$ semisimple — $(4.14)$	yes	yes	yes	yes

What each hypothesis delivers

Cardinal bookkeeping in common cases
Algebra $R$ over $k$	$\dim_{k} R$	$\| k \|$	Does $(4.20)$ apply?
$ℂ [t, t^{- 1}]$	$ℵ_{0}$	$2^{ℵ_{0}}$	yes
$ℂ G$ , $G$ countable	$\leq ℵ_{0}$	$2^{ℵ_{0}}$	yes
Weyl algebra $A_{n} (ℂ)$	$ℵ_{0}$	$2^{ℵ_{0}}$	yes
$ℚ [t, t^{- 1}]$	$ℵ_{0}$	$ℵ_{0}$	no — equality, not strict
$ℂ [t]_{(t)}$	$2^{ℵ_{0}}$	$2^{ℵ_{0}}$	no — and the conclusion is false
$ℂ [[t]]$	$2^{ℵ_{0}}$	$2^{ℵ_{0}}$	no — and the conclusion is false

The fourth row is not a counterexample: $ℚ [t, t^{- 1}]$ is in fact semiprimitive. It merely shows the theorem is silent there.

Relationship Map

Amitsur's theorem is one of two routes to the same conclusion; both funnel through $(4.18)$ .

\dim_{k} R < | k |

resolvents dependent

r

algebraic

r

nilpotent by

(4.18)

rad R = {Nil}^{*} R

$rad R = {Nil}^{*} R$ — sufficient conditions
- Element-wise hypotheses
  - $R$ algebraic over a field $k$ — $(4.19)$
  - $R$ nil — trivially
- Cardinality hypotheses
  - $\dim_{k} R < | k |$ — $(4.20)$
  - $k$ uncountable and $R$ countably generated — $(4.21)$
- Chain conditions
  - $R$ left artinian: all radicals coincide and are nilpotent — $(4.12)$
- Consequences downstream
  - the Köthe conjecture holds for all these classes
  - semiprimitivity of $ℂ G$ for countable $G$ , with the trace argument

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 rings

Semiprimitivity in characteristic zero

Combining $(4.21)$ with the trace argument gives $rad (ℂ G) = 0$ for countable $G$ . This is the model case for the whole semiprimitivity programme for group algebras, where characteristic $p$ remains open.

Enveloping algebras

Lie theory over $ℂ$

The universal enveloping algebra of a finite-dimensional complex Lie algebra is finitely generated and so has countable dimension; $(4.21)$ makes its radical nil, and since it is a domain the radical is zero.

Noncommutative Nullstellensatz

Endomorphism rings of simple modules

The same counting argument, applied to a division ring rather than to the radical, shows a division $k$ -algebra of dimension below $| k |$ is algebraic over $k$ . This is the standard route to Nullstellensatz-type statements for algebras over uncountable fields.

Symbolic computation

Certifying nilness cheaply

For a finitely presented algebra over $ℚ$ nothing follows, but base-changing to an uncountable field puts the radical inside the nil ideals, which is often enough to justify a nilpotency search in a computer algebra system.

The theorem is a tool of pure algebra. Its practical value is that it converts a hard invariant, the Jacobson radical, into a soft one, the largest nil ideal, under a hypothesis that costs nothing to verify.

Failure Modes and Common Mistakes

Do not compare $| R |$ with $| k |$ . The relevant cardinal is $\dim_{k} R$ ; over an uncountable field these can differ.
Do not forget the finite-field case, where the hypothesis silently forces $R$ to be a finite ring and the proof is a different one.
Do not extend the theorem to algebras over commutative base rings: it inherits from $(4.18)$ the requirement that the base be a field.
Do not assume the polynomial produced by clearing denominators is automatically nonzero; the evaluation at $t = a_{i}$ is a necessary step, not a formality.

Historical Notes and Lessons Learned

1945The radical for arbitrary ringsJacobson's definition makes it meaningful to ask, for a general algebra, whether the radical is nil. For Wedderburn's finite-dimensional algebras the question does not arise.
1950Rickart's positivity argumentA trace and involution argument shows that group algebras over subfields of the complex numbers closed under conjugation have no nonzero nil ideals, which is the second half of the semiprimitivity statement.
1956Amitsur on polynomial ringsAmitsur determines the radical of a polynomial ring: rad of R adjoin t equals N adjoin t for a nil ideal N of R. The cardinality technique appears in the same circle of ideas.
1959The cardinality theorem applied to group algebrasAmitsur uses the dimension-versus-cardinality comparison to prove semisimplicity results for group algebras over large fields, the argument reproduced here as (4.20).
sinceCharacteristic p remains openIn characteristic zero the semiprimitivity of group algebras is settled. Kaplansky's problem for characteristic p, where nil ideals are no longer excluded by positivity, is still unresolved in general.

The methodological lesson is that a cardinality hypothesis can substitute for a finiteness hypothesis. Nothing about $R$ is assumed to be finite, noetherian or artinian; only that it is small compared with the field it sits over. That style of argument recurs throughout the theory of infinite-dimensional algebras, in Quillen-type lemmas and in noncommutative Nullstellensätze.

Quick Reference

Hypothesis

k

a field,

R

k

-algebra,

\dim_{k} R < | k |

Conclusion

rad R = {Nil}^{*} R

, the largest nil ideal

Key units

a - r \in U (R)

for

a \in k^{\times}

r \in rad R

Counting step

| k^{\times} | = | k | > \dim_{k} R

forces dependence

Polynomial

f (t) = \sum_{i} b_{i} \prod_{j \neq i} (a_{j} - t)

, with

f (a_{i}) = b_{i} \prod_{j \neq i} (a_{j} - a_{i})

Corollary

k

uncountable,

R

countably generated

\Rightarrow

radical is nil

Sharpness

ℂ [t]_{(t)}

\dim = | k |

and radical not nil

Companiondivision

k

-algebra with

\dim_{k} D < | k |

is algebraic over

k

Proof at a glance
Step	What is used	Output
Reduce	(4.11): nil one-sided ideals lie in the radical	only nilness of $rad R$ remains
Reduce again	(4.18): in the radical, algebraic equals nilpotent	only algebraicity remains
Finite $k$	(4.12) for the finite, hence artinian, ring	radical nilpotent
Infinite $k$	pigeonhole on the resolvent family	a linear dependence
Finish	clear denominators, evaluate at $a_{i}$	a nonzero polynomial killing $r$

Frequently Asked Questions

Why does the theorem compare dimension with the cardinality of the field rather than with the cardinality of

R

Because the pigeonhole is applied to a linearly independent family in a $k$ -vector space. The obstruction is that the resolvent family, indexed by $k^{\times}$ , is too large to be independent — a statement about dimension. Comparing $| R |$ with $| k |$ would be both weaker and unrelated to the argument.

Is the finite field case really necessary as a separate argument?

Yes, though it is short. For finite $k$ the inequality $\dim_{k} R < | k |$ makes $R$ a finite ring, and the infinite-field step $| k^{\times} | = | k |$ is unavailable. The conclusion then comes from the artinian theory, where the radical is even nilpotent.

Does

(4.20)

ever give

rad R = 0

on its own?

Only when $R$ happens to have no nonzero nil ideal, which is an extra input. For $ℂ G$ that input is the positivity of the trace; for a domain it is immediate. The theorem itself only identifies the radical with the largest nil ideal.

What is the relationship to Amitsur's theorem on polynomial rings?

They are different results by the same author. The polynomial-ring theorem computes $rad (R [t])$ as $N [t]$ for a nil ideal $N$ of $R$ ; the theorem on this page bounds the radical of an algebra by a cardinality hypothesis. Both share the moral that the radical of a large but structured object is controlled by nil ideals.

Can the hypothesis be weakened to

\dim_{k} R \leq | k |

No. $ℂ [t]_{(t)}$ has $\dim_{ℂ} R = | ℂ |$ , is local, and its radical $t R$ is not nil. The resolvents $(a - t)^{- 1}$ in that ring really are independent, so the pigeonhole has nothing to work with.

Does the result hold for algebras over a division ring rather than a field?

The argument as given uses commutativity of the base in two places: the scalars $a_{i}$ must commute with $r$ to keep everything inside $k [r]$ , and $f$ must be an ordinary polynomial. Over a noncommutative base one is in the territory of skew polynomials and the statement requires care; it should not be quoted in that setting.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §4, statements (4.20) and (4.21).
S. A. Amitsur, “On the semi-simplicity of group algebras”, Michigan Mathematical Journal 6 (1959).
S. A. Amitsur, “Radicals of polynomial rings”, Canadian Journal of Mathematics 8 (1956).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977, Chapter 7.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

State and prove Amitsur's theorem on the Jacobson radical of a polynomial ring $R [t]$ .
Give an example of a countably generated algebra over $ℚ$ whose Jacobson radical is not nil.
How does the cardinality argument extend to prove Quillen's lemma about endomorphism rings of simple modules?
What is currently known about the semiprimitivity of $k G$ when $char k = p$ divides the order of elements of $G$ ?
Show that a finitely generated algebra over an uncountable field satisfies the noncommutative Nullstellensatz.
Is there an analogue of $(4.20)$ for algebras over a complete discrete valuation ring?
Compare the resolvent-counting proof with the proof of the Amitsur–Levitzki theorem; are the techniques related at all?

Amitsur’s Radical Theorem

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why resolvents detect transcendence

Everything commutes

Two regimes

Key Results

Proof Techniques and Method

Manufacture units from the radical

Pigeonhole on dimension

Clear denominators, then evaluate

Worked Example

Laurent polynomials over the complex numbers

Group algebras of countable groups over $ℂ$

Where the hypothesis is violated

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Semiprimitivity in characteristic zero

Lie theory over $ℂ$

Endomorphism rings of simple modules

Certifying nilness cheaply

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 resolvents detect transcendence

Everything commutes

Two regimes

06Key Results

07Proof Techniques and Method

Manufacture units from the radical

Pigeonhole on dimension

Clear denominators, then evaluate

08Worked Example

Laurent polynomials over the complex numbers

Group algebras of countable groups over ℂ

Where the hypothesis is violated

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Semiprimitivity in characteristic zero

Lie theory over ℂ

Endomorphism rings of simple modules

Certifying nilness cheaply

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

Group algebras of countable groups over $ℂ$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Lie theory over $ℂ$

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions