Is rad R contained in rad R[t]?

Not in general — rad Z_(p) = pZ_(p) while radZ_(p)[t] = 0. The correct statement is that rad R the nil part survives: N ⊆ rad R always, and rad R ⊆ N when rad R is nilpotent.

Amitsur’s Theorem on the Radical of a Polynomial Ring

Q: Where exactly is the ring enlarged, and why is that legitimate?

One passes from S = R[t] to S_1 = R_1[t] where R_1 adjoins a formal p-th root of unity. S_1 is free of finite rank over S on central elements including 1, so descent (5.6)(1) and ascent (5.7) combine to give S rad S_1 = rad S: nothing is lost on contraction.

Q: Does Amitsur's Theorem tell me whether rad R[t] is zero?

Yes, whenever you can rule out nonzero nil ideals in R. Semiprimitive rings, reduced rings, domains and simple rings all qualify, so their polynomial rings are semiprimitive. What the theorem cannot do is compute a nonzero answer without independent knowledge of N.

Q: How does the theorem change for the scalar extension R(T)?

The shape is identical — rad R(T) = N'(T) with N' = R rad R(T) nil — but the nilness of N' is proved differently: not by inverting 1 - at inside a polynomial ring, but through (5.15), which shows the contraction of the radical along any non-algebraic field extension is nil.

Executive Summary

Snapper's Theorem computes $rad R [T]$ for commutative $R$ as $(Nil R) [T]$ . In a noncommutative ring the nilpotent elements need not even be closed under addition, so that formula has no meaning. Amitsur's substitute is structurally identical but leaves the coefficient ideal implicit.

rad R [T] = N [T], N = R \cap rad R [T] a nil ideal of R .

(5.10)

Amitsur's Theorem, for an arbitrary ring $R$ and any nonempty set $T$ of commuting central indeterminates.

The theorem is coefficientwise: membership of a polynomial in the radical is equivalent to membership of each of its coefficients. It follows at once that if $R$ has no nonzero nil ideal — for instance if $R$ is semiprimitive or reduced — then $R [T]$ is semiprimitive. What the theorem does not do is identify $N$ . Whether $N$ is always the upper nilradical ${Nil}^{*} R$ is Problem $(5.12)$ , and it is equivalent to Köthe's conjecture.

$N [T]$

rad R [T]

NilNature of

N

1956Amitsur

OpenIs

N = {Nil}^{*} R

Overview

Two features of the commutative proof are unavailable here. There is no nilradical to extend, and there is no supply of prime quotients that are domains. Amitsur replaces both with an automorphism argument: the radical of $R [t]$ is invariant under every $R$ -algebra automorphism of $R [t]$ , and there are two useful ones — scaling $t \mapsto ζ t$ by a root of unity, and shifting $t \mapsto t + 1$ .

Scaling is not available over $R$ itself, so one enlarges $R$ to $R_{1} = R [ξ] / (1 + ξ + \dots + ξ^{p - 1})$ , a free $R$ -module of rank $p - 1$ on which a formal $p$ -th root of unity $ζ$ acts centrally. The results of Behaviour of the Radical under Ring Extensions then say that this enlargement neither loses nor gains radical: $R [t] \cap rad R_{1} [t] = rad R [t]$ .

The same architecture yields the parallel theorem $(5.13)$ for the scalar extension $R (T) = R \otimes_{k} k (T)$ of a $k$ -algebra to the rational function field, which is used in the analysis of the radical under transcendental field extensions.

Learning Objectives

State $(5.10)$ precisely, including what is and is not asserted about $N$ .
Prove $(5.10 A)$ : every element of $N$ is nilpotent, by comparing coefficients of an inverse.
Construct $R_{1}$ with its formal $p$ -th root of unity and prove the congruence $(5.11)$ .
Run the induction of $(5.10 B)$ showing $a_{i} t^{i} \in rad R [t]$ .
Apply the shift automorphism to conclude $a_{i} \in rad R [t]$ , and induct on the number of variables.
State Problem $(5.12)$ and its equivalence with Köthe's conjecture, and state $(5.13)$ .

Definitions

$N [T]$: For an ideal N of R, the set of polynomials in R[T] all of whose coefficients lie in N; equivalently the ideal N R[T].
Nil versus nilpotent: A nil ideal has all elements nilpotent with no common bound on the index; a nilpotent ideal satisfies I to the power m equals zero for a single m. Nilpotent implies nil, never conversely in general.
${Nil}^{*} R$: The upper nilradical: the sum of all nil ideals, which is itself nil because a sum of nil ideals is nil.
$R_{1}$: The ring R adjoin a formal primitive p-th root of unity, that is R with an added central element zeta satisfying the p-th cyclotomic relation; free as an R-module of rank p minus one.

Variables in $T$ are central and commute with each other; skew polynomial rings are a different problem and are not covered by this theorem.

Core Concepts

Formal roots of unity

Fix a prime $p$ and set $R_{1} = R [ξ] / (1 + ξ + \dots + ξ^{p - 1})$ , where $ξ$ is a central indeterminate. Write $ζ$ for the image of $ξ$ . Because the defining relation is monic of degree $p - 1$ , $R_{1}$ is free as a left $R$ -module with basis $1, ζ, \dots, ζ^{p - 2}$ , and $ζ$ is central. From $ζ^{p} - 1 = (ζ - 1) (1 + ζ + \dots + ζ^{p - 1}) = 0$ we get $ζ^{p} = 1$ , so $ζ$ is a unit.

Lemma(5.11)The prime is divisible by every

ζ^{j} - 1

For every integer $j$ with $0 < j < p$ we have $p \in (ζ^{j} - 1) R_{1}$ , and the witness may be taken in the central subring generated by $ζ$ .

Proof

Work modulo the central ideal $(ζ^{j} - 1) R_{1}$ , where $ζ^{j} = 1$ . Since $\gcd (j, p) = 1$ , choose integers $a, b$ with $a j + b p = 1$ ; then $ζ = ζ^{a j + b p} = (ζ^{j})^{a} (ζ^{p})^{b} = 1$ in the quotient. Substituting $ζ = 1$ into the relation $1 + ζ + \dots + ζ^{p - 1} = 0$ gives $p = 0$ there, i.e. $p \in (ζ^{j} - 1) R_{1}$ . All manipulations took place in the commutative subring generated by $ζ$ over the prime ring, so the witness is central.

Why the enlargement is harmless

Put $S = R [t]$ , $J = rad S$ , $S_{1} = R_{1} [t]$ , $J_{1} = rad S_{1}$ . Then $S_{1} = S \oplus S ζ \oplus \dots \oplus S ζ^{p - 2}$ is free of finite rank over $S$ on central elements containing $1$ . Descent $(5.6) (1)$ gives $S \cap J_{1} \subseteq J$ ; ascent $(5.7)$ gives $J \subseteq J_{1}$ , hence $J \subseteq S \cap J_{1}$ . So

S \cap J_{1} = J,

(RU)

Adjoining a formal root of unity is invisible to the radical after contraction.

f (t) \in J

f (ζ t) \in J_{1}

ζ^{n} f (t) - f (ζ t) \in J_{1}

has degree

< n

induction gives

p a_{i} t^{i} \in J

two primes give

a_{i} t^{i} \in J

The final step from $a_{i} t^{i} \in J$ to $a_{i} \in J$ uses the shift $t \mapsto t + 1$ , which converts a monomial into a polynomial whose constant term is the coefficient one wants.

Key Results

Theorem(5.10)Amitsur

Let $R$ be any ring with identity and let $T$ be a nonempty set of commuting central indeterminates. Put $S = R [T]$ , $J = rad S$ and $N = R \cap J$ . Then $N$ is a nil ideal of $R$ and $J = N [T]$ . In particular, if $R$ has no nonzero nil ideal then $R [T]$ is Jacobson semisimple.

Proposition(5.10A)

N

is nil

With the notation of $(5.10)$ , every element of $N = R \cap rad R [T]$ is nilpotent.

Proof

Let $a \in N$ and fix a variable $t \in T$ . Since $rad S$ is an ideal, $a t \in rad S$ , so $1 - a t \in U (S)$ . Applying the $R$ -algebra homomorphism $R [T] \to R [t]$ that sends every other variable to $0$ , we get an inverse inside $R [t]$ : there are $a_{0}, \dots, a_{n} \in R$ with

(1 - a t) (a_{0} + a_{1} t + \dots + a_{n} t^{n}) = 1 .

(5.10a)

Comparing coefficients of $t^{0}, t^{1}, \dots, t^{n}, t^{n + 1}$ in turn gives $a_{0} = 1$ , then $a_{i} = a a_{i - 1}$ for $1 \leq i \leq n$ , so $a_{i} = a^{i}$ , and finally $0 = a a_{n} = a^{n + 1}$ . Hence $a$ is nilpotent.

Proposition(5.10B)Monomial terms are radical

Let $R$ be any ring, $S = R [t]$ in one variable and $J = rad S$ . If $f (t) = a_{0} + a_{1} t + \dots + a_{n} t^{n} \in J$ with $a_{i} \in R$ , then $a_{i} t^{i} \in J$ for every $i$ .

Proof

Induct on $n$ , the statement being taken for all rings simultaneously. For $n = 0$ there is nothing to prove. Let $n \geq 1$ and pick a prime $p > n$ . Build $R_{1}$ , $S_{1} = R_{1} [t]$ , $J_{1} = rad S_{1}$ as above, so that $S \cap J_{1} = J$ by $(RU)$ .

The assignment $t \mapsto ζ t$ extends to an $R_{1}$ -algebra automorphism of $S_{1}$ , and the radical is invariant under automorphisms, so from $f (t) \in J \subseteq J_{1}$ we get $f (ζ t) \in J_{1}$ . Therefore

ζ^{n} f (t) - f (ζ t) = \sum_{i = 0}^{n - 1} a_{i} (ζ^{n} - ζ^{i}) t^{i} \in J_{1},

(5.10b)

The degree- $n$ terms cancel, so the induction hypothesis applies over the ring $R_{1}$ .

By the inductive hypothesis applied to $R_{1}$ , each term $a_{i} (ζ^{n} - ζ^{i}) t^{i}$ lies in $J_{1}$ for $i \leq n - 1$ . Multiplying by the central unit $ζ^{- i}$ gives $a_{i} (ζ^{n - i} - 1) t^{i} \in J_{1}$ . Now $0 < n - i \leq n < p$ , so $(5.11)$ provides a central $c$ with $p = c (ζ^{n - i} - 1)$ ; multiplying on the left by $c$ and using centrality of $ζ$ yields $p a_{i} t^{i} \in J_{1}$ , and since $p a_{i} t^{i} \in S$ we get $p a_{i} t^{i} \in S \cap J_{1} = J$ .

Repeat the argument with a second prime $q > n$ , $q \neq p$ , to obtain $q a_{i} t^{i} \in J$ . Choosing integers with $u p + v q = 1$ gives $a_{i} t^{i} = u (p a_{i} t^{i}) + v (q a_{i} t^{i}) \in J$ for all $i \leq n - 1$ . Finally $a_{n} t^{n} = f (t) - \sum_{i < n} a_{i} t^{i} \in J$ .

Proposition(5.10C)Coefficients are radical

In the notation of $(5.10 B)$ , if $f (t) = \sum_{i} a_{i} t^{i} \in J$ then $a_{i} \in J$ — hence $a_{i} \in R \cap J = N$ — for every $i$ .

Proof

By $(5.10 B)$ , $a_{i} t^{i} \in J$ . The map $t \mapsto t + 1$ is an $R$ -algebra automorphism of $R [t]$ , so it preserves $J$ , giving $a_{i} (1 + t)^{i} \in J$ . Expanding, this is a polynomial with constant term $a_{i}$ ; applying $(5.10 B)$ to it shows in particular that its degree-zero term $a_{i}$ lies in $J$ .

ProofProof of (5.10)

$N$ is nil by $(5.10 A)$ . For $J = N [T]$ : the inclusion $N [T] \subseteq J$ holds because $N \subseteq J$ and $J$ is an ideal of $S$ containing the variables' multiples. Conversely let $f \in J$ and induct on the number $m$ of variables occurring in $f$ . For $m = 0$ , $f \in R \cap J = N$ . For $m \geq 1$ choose a variable $t$ occurring in $f$ , write $T = T_{0} ⊔ {t}$ and $f = \sum_{i} a_{i} (T_{0}) t^{i}$ . Applying $(5.10 C)$ to $R [T] = R [T_{0}] [t]$ shows each $a_{i} (T_{0}) \in J$ ; each involves at most $m - 1$ variables, so induction finishes the argument.

CorollarySemiprimitivity is inherited

If $rad R = 0$ then $rad R [T] = 0$ . Indeed $N$ is a nil ideal, hence $N \subseteq rad R = 0$ by $(4.11)$ , so $J = N [T] = 0$ . The same argument applies whenever $R$ is reduced, or a domain, or a simple ring with identity.

CorollaryThe nilpotent case is completely settled

If $rad R$ is nilpotent — for instance if $R$ is left or right artinian, by $(4.12)$ — then $rad R [T] = (rad R) [T]$ .

Proof

If $(rad R)^{m} = 0$ then $((rad R) [T])^{m} = 0$ , so $(rad R) [T] \subseteq rad R [T]$ by $(4.11)$ , whence $rad R \subseteq N$ . Conversely $N$ is nil, so $N \subseteq rad R$ . Thus $N = rad R$ and $(5.10)$ gives the result.

Problem(5.12)Identifying

N

If $I$ is a nil ideal of $R$ , is $I [T] \subseteq rad R [T]$ ? Equivalently, is $N = {Nil}^{*} R$ , so that $rad R [T] = ({Nil}^{*} R) [T]$ ? For nilpotent $I$ the answer is yes, by the corollary above. For merely nil $I$ the question is open, and Krempa showed it is equivalent to Köthe's conjecture: that a ring with no nonzero nil two-sided ideal has no nonzero nil one-sided ideal.

Theorem(5.13)Amitsur, rational function field version

Let $k$ be a field, $R$ a $k$ -algebra and $T$ a nonempty set of commuting indeterminates. Let $R (T) = R \otimes_{k} k (T)$ be the scalar extension to the rational function field, put $J' = rad R (T)$ and $N' = R \cap J'$ . Then $N'$ is a nil ideal of $R$ and $J' = N' (T) = N' \otimes_{k} k (T)$ . In particular, if $R$ has no nonzero nil ideal then $R (T)$ is Jacobson semisimple.

RemarkHow (5.13) is proved

$R (T)$ is the localisation of $R [T]$ at the central multiplicative set $k [T] ∖ {0}$ , so a general element is $f (T) / g (T)$ with $f \in R [T]$ , $0 \neq g \in k [T]$ ; since $g$ is a central unit of $R (T)$ , membership of $f / g$ in $J'$ is equivalent to membership of $f$ . The one-variable coefficient statement is proved exactly as in $(5.10 B)$ – $(5.10 C)$ , and the many-variable case follows from the identification $R (T) ≅ (R (T_{0})) (t)$ for $T = T_{0} ⊔ {t}$ , viewing $R (T_{0})$ as an algebra over $k (T_{0})$ . Nilness of $N'$ is not proved as in $(5.10 A)$ ; it comes from the scalar-extension result $(5.15)$ , which shows $R \cap rad R^{K}$ is nil for every non-algebraic extension $K / k$ .

Proof Techniques and Method

The reusable moves in Amitsur's argument.

Move 1

Enlarge to gain an automorphism

If the automorphism you want does not exist over $R$ , adjoin what it needs — here a formal $p$ -th root of unity — and use the change-of-rings lemmas to show the enlargement does not change the radical after contraction.

Move 2

Difference out the top term

$ζ^{n} f (t) - f (ζ t)$ kills the degree- $n$ term and leaves a polynomial the induction can handle. Subtracting a twisted copy of an element from itself is the standard way to lower degree inside an ideal.

Move 3

Clear the integer with two primes

The root-of-unity argument yields $p a_{i} t^{i} \in J$ for every prime $p > n$ . Two coprime such integers give $a_{i} t^{i} \in J$ — a trick that replaces division by $p$ , which is unavailable in a general ring.

Move 3 is what makes the argument work in every characteristic. There is no assumption that $p$ is invertible, and indeed $R$ may have characteristic $p$ ; the conclusion is recovered by playing two primes off against each other.

Show the contraction is nilInvert

1 - a t

and read off

a^{n + 1} = 0

Adjoin a root of unityForm

R_{1}

; check freeness and centrality; conclude

S \cap J_{1} = J

Scale the variableUse

t \mapsto ζ t

and difference to drop the degree; induct.

Clear denominators with two primesGet

a_{i} t^{i} \in J

Shift the variableUse

t \mapsto t + 1

to convert monomials into coefficients.

Induct on the number of variablesWrite

R [T] = R [T_{0}] [t]

and repeat.

Worked Example

Upper triangular matrices

Let $k$ be a field and $R = T_{2} (k)$ the ring of upper triangular $2 \times 2$ matrices. Then $rad R$ is the set of strictly upper triangular matrices, and $(rad R)^{2} = 0$ . Since the radical is nilpotent, the corollary above applies:

rad (T_{2} (k) [t]) = (rad T_{2} (k)) [t] = (\begin{matrix} 0 & k [t] \\ 0 & 0 \end{matrix}) .

(E.1)

Cross-check by a different route: $T_{2} (k) [t] ≅ T_{2} (k [t])$ , and for a triangular ring the radical is computed blockwise as $(\begin{matrix} rad A & M \\ 0 & rad B \end{matrix})$ . With $A = B = k [t]$ and $rad k [t] = 0$ this gives the strictly upper triangular matrices over $k [t]$ — the same answer. Here $N = R \cap rad R [t] = rad T_{2} (k)$ , which is nilpotent, hence certainly nil, as $(5.10 A)$ requires.

A finite commutative check

For $R = ℤ / 4 ℤ$ the ideal $(2)$ is nilpotent with $(2)^{2} = 0$ , so $rad R [t] = 2 R [t]$ : a polynomial lies in the radical exactly when all its coefficients are even. Verify the unit condition directly: $(1 - 2 t) (1 + 2 t) = 1 - 4 t^{2} = 1$ in $(ℤ / 4) [t]$ . Snapper's Theorem gives the same answer since $Nil (ℤ / 4) = (2)$ — as it must, the two theorems agreeing on commutative input.

Where the theorem stops short

Suppose $R$ possesses a nil ideal $I$ that is not nilpotent — such rings exist, for example a suitable ring of infinite upper triangular matrices with entries of unbounded nilpotence index. Then $I \subseteq {Nil}^{*} R$ , and $(5.10)$ tells us $rad R [t] = N [t]$ with $N$ nil, hence $N \subseteq {Nil}^{*} R$ ; but no known argument places $I$ inside $N$ . That gap is precisely Problem $(5.12)$ .

Comparison and Classification

What $rad R [T]$ is, by hypothesis on $R$
Hypothesis on $R$	$N = R \cap rad R [T]$	$rad R [T]$	Status
Commutative	$Nil R$	$(Nil R) [T]$	theorem (5.1)
$rad R$ nilpotent (e.g. artinian)	$rad R$	$(rad R) [T]$	theorem
Semiprimitive	$0$	$0$	theorem
Reduced, or a domain	$0$	$0$	theorem
Simple with identity	$0$	$0$	theorem
Arbitrary	a nil ideal	$N [T]$	theorem (5.10)
Has a nil, non-nilpotent ideal	$\subseteq {Nil}^{*} R$ ; equality unknown	$N [T]$	open — (5.12)

Amitsur's theorem against its neighbours
	Determines the answer	Needs commutativity	Coefficientwise	Radical can grow
$(5.1)$ Snapper, $R [T]$	yes	yes	yes	no
$(5.10)$ Amitsur, $R [T]$	partial	no	yes	no
$(5.13)$ Amitsur, $R (T)$	partial	no	yes	no
$rad R [[x]]$	yes	no	no	yes
$rad M_{n} (R)$	yes	no	yes	no

Amitsur's theorem against its neighbours

Radical can grow means that $rad S$ may meet $S ∖ R -coefficient$ territory: for power series the variable itself is in the radical, which never happens for polynomials because $t \in rad R [t]$ would force $1 - t$ to be a unit.

Relationship Map

{Nil}_{*} R

N = R \cap rad R [T]

{Nil}^{*} R

rad R

The first inclusion holds because the lower nilradical is nilpotent-by-construction in the sense of being a sum of a transfinite chain of nilpotent extensions and in any case is nil; the middle inclusion is $(5.10 A)$ ; whether the middle one is an equality is $(5.12)$ .

Köthe's conjecture — no nonzero nil ideal $\Rightarrow$ no nonzero nil one-sided ideal
- equivalent to
  - $I$ nil $\Rightarrow$ $I [T] \subseteq rad R [T]$ , i.e. $N = {Nil}^{*} R$
  - the sum of two nil left ideals is nil
  - $I$ nil $\Rightarrow$ $M_{n} (I)$ nil, for all $n$
- known cases
  - $I$ nilpotent
  - $R$ with polynomial identity
  - $R$ noetherian, where nil ideals are nilpotent
- consequences if true
  - $rad R [T] = ({Nil}^{*} R) [T]$ for every ring
  - ${Nil}^{*}$ becomes computable from one-sided data

The upper nilradical and Köthe's conjecture are developed in The Upper Nilradical and the Köthe Conjecture; the scalar-extension counterpart $(5.13)$ feeds directly into The Radical under Field Extension of Scalars.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

NameAmitsur's Theorem on the radical of a polynomial ring, Amitsur 1956

Lower nilradical

{Nil}_{*} R

; also prime radical or Baer–McCoy radical

Upper nilradical

{Nil}^{*} R

; also the nil radical in some sources — check which

Variables

R [T]

for a set;

R ⟨ T ⟩

denotes noncommuting variables, a different ring

Scalar extension

R (T) = R \otimes_{k} k (T)

, not to be confused with

R [T]

ImplementationsGAP and Magma compute radicals of finite-dimensional algebras; no system computes

rad R [T]

for a general presented ring

Computational Notes

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

For a finite-dimensional algebra $A$ over a field, $rad A$ is nilpotent, so $rad A [T] = (rad A) [T]$ is computed by one radical computation in $A$ — polynomial time by the standard trace-form or Friedl–Rónyai algorithms.
For a finite ring given by a multiplication table the same reduction applies, since a finite ring is artinian.
For a general finitely presented ring the radical is not computable — the word problem is already undecidable — and $(5.10)$ gives structure without an algorithm.
The theorem is nevertheless useful computationally as a certificate shape: to prove $f \in rad R [T]$ it suffices to prove each coefficient lies in $rad R [T]$ , which reduces an infinite family of unit tests to finitely many statements about $R$ .
Reduction modulo a nilpotent ideal is the standard practical route: compute in $R / {Nil}^{*} R$ when that quotient is tractable, then lift.

Failure Modes and Common Mistakes

Do not assume $rad R [T]$ is nil merely because its coefficient ideal is: a polynomial with nilpotent coefficients need not be nilpotent when $R$ is noncommutative.
Do not apply the theorem to noncommuting variables. For the free algebra $R ⟨ x, y ⟩$ the argument breaks down at the very first step, since $R ⟨ x, y ⟩$ is not a polynomial ring in central variables.
Do not apply it to skew polynomial rings $R [x; σ]$ ; the automorphism $x \mapsto ζ x$ interacts with $σ$ and the statement changes.
Do not confuse $R [T]$ with $R (T)$ : the latter is a localisation of the former and has its own theorem, $(5.13)$ .
Do not expect $rad R \subseteq rad R [T]$ : it holds when $rad R$ is nil and $(5.12)$ is known for that ideal, and it visibly fails for $R = ℤ_{(p)}$ .

Historical Notes and Lessons Learned

1930Köthe's questionKöthe asks whether a ring with no nonzero nil two-sided ideal can have a nonzero nil one-sided ideal. The question is still open.
1950SnapperThe commutative case is settled: the radical of a polynomial ring is the extended nilradical.
1956AmitsurAmitsur proves the general structure theorem for $rad R [T]$ and the companion result for the scalar extension $R (T)$ , in the same year as his work on algebras over infinite fields.
1960sBergman's simplificationA root-of-unity argument due to Bergman streamlines the one-variable case; it is the version presented by Passman and followed by Lam.
1972KrempaKrempa proves that the identification of $N$ with the upper nilradical is equivalent to Köthe's conjecture, tying $(5.12)$ to one of the oldest open problems in ring theory.

The methodological lesson is that a theorem can be structurally complete and still uninformative. $(5.10)$ determines $rad R [T]$ up to knowing one ideal of $R$ , and seventy years of work have not identified that ideal — a reminder that reduced to is not the same as solved.

Quick Reference

$(5.10)$

rad R [T] = N [T]

N = R \cap rad R [T]

nil

$(5.10 A)$

a \in N

(1 - a t)^{- 1} = \sum a^{i} t^{i}

finite

\Rightarrow

a^{n + 1} = 0

$(5.10 B)$

f \in rad R [t] \Rightarrow a_{i} t^{i} \in rad R [t]

$(5.10 C)$

f \in rad R [t] \Rightarrow a_{i} \in rad R [t]

$(5.11)$

p \in (ζ^{j} - 1) R_{1}

for

0 < j < p

$(5.12)$ Open:

I

nil

\Rightarrow

I [T] \subseteq rad R [T]

\Leftrightarrow

Köthe

$(5.13)$

rad R (T) = N' (T)

N' = R \cap rad R (T)

nil

Artinian case

rad R [T] = (rad R) [T]

Ingredients of the proof and where they come from
Ingredient	Role	Source
$S \cap rad S_{1} = rad S$	adjoining $ζ$ is harmless	(5.6)(1), (5.7)
$t \mapsto ζ t$	lowers degree by differencing	automorphism invariance
$p \in (ζ^{j} - 1) R_{1}$	converts a $ζ$ -multiple into an integer multiple	(5.11)
Two primes $p, q > n$	clears the integer	Bézout
$t \mapsto t + 1$	monomials to coefficients	(5.10C)
Nakayama	used indirectly, through (5.7)	(4.22)

Frequently Asked Questions

Why can the commutative proof not simply be adapted?

It uses two things that fail noncommutatively: that the nilpotent elements form an ideal, and that every ring has enough prime quotients that are domains. Amitsur replaces the prime-by-prime analysis with an automorphism argument, which needs no quotients at all — only the invariance of the radical under ring automorphisms.

Where exactly is the ring enlarged, and why is that legitimate?

One passes from $S = R [t]$ to $S_{1} = R_{1} [t]$ where $R_{1}$ adjoins a formal $p$ -th root of unity. $S_{1}$ is free of finite rank over $S$ on central elements including $1$ , so descent $(5.6) (1)$ and ascent $(5.7)$ combine to give $S \cap rad S_{1} = rad S$ : nothing is lost on contraction.

Does Amitsur's Theorem tell me whether

rad R [t]

is zero?

Yes, whenever you can rule out nonzero nil ideals in $R$ . Semiprimitive rings, reduced rings, domains and simple rings all qualify, so their polynomial rings are semiprimitive. What the theorem cannot do is compute a nonzero answer without independent knowledge of $N$ .

rad R

contained in

rad R [t]

Not in general — $rad ℤ_{(p)} = p ℤ_{(p)}$ while $rad ℤ_{(p)} [t] = 0$ . The correct statement is that $rad R \cap$ the nil part survives: $N \subseteq rad R$ always, and $rad R \subseteq N$ when $rad R$ is nilpotent.

What is the connection with Köthe's conjecture, precisely?

Krempa's theorem: Köthe's conjecture holds for all rings if and only if $I [t] \subseteq rad R [t]$ for every nil ideal $I$ of every ring $R$ ; equivalently, if and only if the ideal $N$ of Amitsur's Theorem is always the upper nilradical. So $(5.12)$ is not a technical loose end but a reformulation of the main open problem about nil rings.

How does the theorem change for the scalar extension

R (T)

The shape is identical — $rad R (T) = N' (T)$ with $N' = R \cap rad R (T)$ nil — but the nilness of $N'$ is proved differently: not by inverting $1 - a t$ inside a polynomial ring, but through $(5.15)$ , which shows the contraction of the radical along any non-algebraic field extension is nil.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §5, results (5.10)–(5.13) (pp. 75–79).
S. A. Amitsur, “Radicals of polynomial rings”, Canadian Journal of Mathematics 8 (1956), 355–361.
D. S. Passman, A Course in Ring Theory, Wadsworth & Brooks/Cole, 1991, p. 192.
G. Köthe, “Die Struktur der Ringe, deren Restklassenring nach dem Radikal vollständig reduzibel ist”, Mathematische Zeitschrift 32 (1930), 161–186.
J. Krempa, “Logical connections between some open problems concerning nil rings”, Fundamenta Mathematicae 76 (1972), 121–130.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Write out the roots-of-unity argument for $n = 1$ and $n = 2$ explicitly, tracking the primes used.
Prove Krempa's equivalence between Köthe's conjecture and Problem $(5.12)$ .
Is $rad R [t]$ always a nil ideal of $R [t]$ ? What is known and what is open?
What is the correct analogue of Amitsur's Theorem for skew polynomial rings $R [t; σ]$ ?
Give an example of a ring with a nil ideal that is not nilpotent, and describe what is known about its polynomial ring.
Compare Amitsur's Theorem with the behaviour of the Levitzki and Brown–McCoy radicals under polynomial extension.
How do Bergman's and Passman's presentations of the one-variable case differ from Amitsur's original argument?

Radical of R[T]

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Formal roots of unity

Why the enlargement is harmless

Key Results

Proof Techniques and Method

Enlarge to gain an automorphism

Difference out the top term

Clear the integer with two primes

Worked Example

Upper triangular matrices

A finite commutative check

Where the theorem stops short

Comparison and Classification

Relationship Map

Standards and Notation

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

Formal roots of unity

Why the enlargement is harmless

06Key Results

07Proof Techniques and Method

Enlarge to gain an automorphism

Difference out the top term

Clear the integer with two primes

08Worked Example

Upper triangular matrices

A finite commutative check

Where the theorem stops short

09Comparison and Classification

10Relationship Map

11Standards and Notation

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

Comparison and Classification

Relationship Map

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions