Behaviour of the Radical under Ring Extensions

Q: Why does the fixed-ring case need no hypothesis on the group?

Because the argument uses uniqueness of two-sided inverses rather than averaging. If 1 - r_0 is fixed by G and s is its inverse, then g(s) is also its inverse, hence equals s. No finiteness of G, and no invertibility of |G|, is needed — which is what makes the result usable in modular situations.

Q: Is R rad S always an ideal of R?

Yes. It is the intersection of the subring R with a two-sided ideal of S, so it is closed under addition and under multiplication by elements of R on both sides. This matters: the descent proofs only produce a one-sided inverse, and the ideal property is what converts that into genuine radical membership.

Q: Can ascent hold without any finiteness?

Sometimes, but not for structural reasons. For instance rad R ⊆ rad R[[x]] holds although R[[x]] is not a finitely generated R-module — that comes from the power series geometric expansion, not from (5.7). What (5.7) asserts is a general theorem, and its counterexamples show the finiteness cannot be removed from the general statement.

Q: How is (5.7) used inside Amitsur's Theorem?

To compare S = R[t] with S_1 = R_1[t], where R_1 = R[]/(1 + + … + ^p-1) adjoins a p-th root of unity. S_1 is a free S-module of finite rank on central elements, so (5.7) gives rad S ⊆ rad S_1 and (5.6)(1) gives the reverse contraction, yielding S rad S_1 = rad S — the identity that lets the root-of-unity automorphism be exploited.

Q: Does (5.9) ever give equality?

Only accidentally. Take R = k a field and S a finite-dimensional k-algebra with rad S ≠ 0: then (rad R)S = 0 rad S. Equality holds, for example, when S = M_n(R), where rad M_n(R) = M_n(rad R) = (rad R) M_n(R).

Q: What happens for localisations?

Neither inclusion holds in general. For R = Z and S = Z_(p), descent fails; for R = Z_(p) and S = Q, ascent fails. Both hypotheses of this page are lost when elements are inverted, and no general replacement is known.

Executive Summary

Given $R \subseteq S$ , neither radical determines the other. What can be proved are one-way inclusions under specific hypotheses, and Lam isolates two of them. Descent $(5.6)$ : if $R R$ is a direct summand of $R S$ , or if $R$ is the fixed ring of a group of automorphisms of $S$ , then $R \cap rad S \subseteq rad R$ . Ascent $(5.7)$ : if $S$ is generated as a left $R$ -module by finitely many elements centralising the image of $R$ , then $i (rad R) \subseteq rad S$ .

The hypotheses are genuinely different and are needed for different reasons. Descent needs a way to project $S$ back onto $R$ ; ascent needs finiteness, because it runs on Nakayama's Lemma. The polynomial extension $R \subseteq R [t]$ satisfies the first and fails the second — which is exactly why $rad R$ can die in $R [t]$ while $R \cap rad R [t]$ stays inside $rad R$ .

$R \cap rad S \subseteq rad R$ Descent, under (5.6)

$i (rad R) \subseteq rad S$ Ascent, under (5.7)

FinitenessWhat ascent really needs

BothGives

rad R = R \cap rad S

Overview

The change-of-rings problem for the Jacobson radical has three natural formulations, and it is worth keeping them apart:

Contraction. How does $R \cap rad S$ compare with $rad R$ ?
Extension. How does $(rad R) \cdot S$ compare with $rad S$ ?
Quotient control. Does $rad S$ lie inside the ideal of $S$ generated by $R \cap rad S$ ?

No implication holds in general. The two theorems of this page each supply one direction of the first two questions; the third is much harder and is answered only in special cases, such as Amitsur's Theorem for $S = R [T]$ .

Both results are used immediately: together they prove $rad R = R \cap rad R^{K}$ for finite field extensions in The Radical under Field Extension of Scalars, and they are the two lemmas invoked in the roots-of-unity step of Amitsur's Theorem.

Learning Objectives

State both alternatives of $(5.6)$ and explain what each provides in the proof.
Prove descent in the direct summand case using the projection $S \to R$ .
Prove descent in the fixed-ring case using uniqueness of inverses.
Prove ascent $(5.7)$ by showing $rad R$ annihilates every simple left $S$ -module.
Recognise when the twisted hypothesis of $(5.8)$ applies, as for crossed products with finite groups.
Give counterexamples to each inclusion when its hypothesis is removed.

Definitions

$R \subseteq S$: A unital subring: $S$ is a ring, $R$ a subring, and the two share the same identity element.
$i : R \to S$: A unital ring homomorphism, not assumed injective. It makes $S$ an $(R, R)$ -bimodule; $r \cdot s$ means $i (r) s$ .
$R R$ a direct summand of $R S$: There is a left R-submodule T of S with S equal to the direct sum of R and T; equivalently a left R-linear projection S onto R fixing R.
$S^{G}$: The fixed ring of a group G of ring automorphisms of S: all s in S with g(s) = s for every g in G.
Skew group ring $R * G$: Free as a left R-module on units u_g indexed by G, with multiplication determined by u_g r equal to sigma_g(r) u_g for automorphisms sigma_g of R.

The radical is left-right symmetric, so all statements may be read on either side; the proofs below use whichever side is convenient.

Core Concepts

One-sided invertibility is enough

Both descent proofs produce only a one-sided inverse inside $R$ , and that is sufficient. The reason is a small lemma worth recording separately.

LemmaIdeal test for radical membership

Let $𝔄$ be a two-sided ideal of a ring $R$ such that $1 - a$ is right-invertible in $R$ for every $a \in 𝔄$ . Then $𝔄 \subseteq rad R$ .

Proof

Fix $a \in 𝔄$ . For any $x \in R$ we have $a x \in 𝔄$ , so $1 - a x$ is right-invertible. By the right-hand form of the characterisation $(4.1)$ — legitimate because $rad R = rad R^{op}$ by the symmetry $(4.3)$ – $(4.4)$ — this says precisely that $a \in rad R$ .

Note $R \cap rad S$ is always a two-sided ideal of $R$ : it is the intersection of $R$ with a two-sided ideal of $S$ . So the lemma is exactly the tool the descent proofs need.

Why ascent is harder

Membership in $rad S$ must be tested against all simple left $S$ -modules, and there is no reason for a simple $S$ -module to be simple, or even small, over $R$ . The device in $(5.7)$ is to show it is at least finitely generated over $R$ : a simple $S$ -module is cyclic, $M = S a$ , and if $S = \sum_{j} R x_{j}$ then $M = \sum_{j} R (x_{j} a)$ is generated by $n$ elements. Nakayama then forbids $(rad R) M = M$ , and simplicity upgrades that to $(rad R) M = 0$ .

M

simple over

S

M

cyclic over

S

M

finitely generated over

R

J M \neq M

by Nakayama

J M

is an

S

-submodule, so

J M = 0

The centralising hypothesis is used in exactly one place: to know $J M$ is an $S$ -submodule and not merely an $R$ -submodule. That is why it can be relaxed to a twisting condition.

Key Results

Proposition(5.6)Descent of the radical

Let $R \subseteq S$ be rings with the same identity, and assume either

as a left $R$ -module, $R R$ is a direct summand of $R S$ ; or
there is a group $G$ of ring automorphisms of $S$ with $R = S^{G} = {s \in S : g (s) = s for all g \in G}$ .

Then $R \cap rad S \subseteq rad R$ .

Proof

Write $𝔄 = R \cap rad S$ , a two-sided ideal of $R$ . By the Lemma above it suffices to show that $1 - r_{0}$ is right-invertible **in $R$ ** for every $r_{0} \in 𝔄$ . In both cases $r_{0} \in rad S$ , so $1 - r_{0} \in U (S)$ ; let $s \in S$ satisfy $(1 - r_{0}) s = 1$ .

Case (1). Write $S = R \oplus T$ with $T$ a left $R$ -submodule, and decompose $s = r + t$ with $r \in R$ , $t \in T$ . Since $1 - r_{0} \in R$ and both $R$ and $T$ are left $R$ -submodules, $(1 - r_{0}) r \in R$ and $(1 - r_{0}) t \in T$ . Then $1 = (1 - r_{0}) r + (1 - r_{0}) t$ with $1 \in R$ , and directness of the sum forces $(1 - r_{0}) r = 1$ . So $1 - r_{0}$ is right-invertible in $R$ .

Case (2). Here $1 - r_{0}$ is a unit of $S$ with two-sided inverse $s$ . For $g \in G$ , applying $g$ to $(1 - r_{0}) s = s (1 - r_{0}) = 1$ and using $g (1 - r_{0}) = 1 - r_{0}$ (as $r_{0} \in R = S^{G}$ ) shows $g (s)$ is also a two-sided inverse of $1 - r_{0}$ . Inverses are unique, so $g (s) = s$ for all $g$ , i.e. $s \in S^{G} = R$ . Hence $1 - r_{0} \in U (R)$ , and in particular it is right-invertible in $R$ .

Proposition(5.7)Ascent of the radical

Let $i : R \to S$ be a unital ring homomorphism and view $S$ as an $(R, R)$ -bimodule through $i$ . Assume there are finitely many elements $x_{1}, \dots, x_{n} \in S$ with

S = R \cdot x_{1} + \dots + R \cdot x_{n}, x_{j} i (r) = i (r) x_{j} for all r \in R, 1 \leq j \leq n .

(5.7a)

Then $i (rad R) \subseteq rad S$ , and hence $(rad R) \cdot S \subseteq rad S$ .

Proof

Put $J = rad R$ . By $(4.1)$ it is enough to show that $i (J)$ annihilates every simple left $S$ -module $M$ . Every $S$ -module is an $R$ -module via $i$ . Pick $0 \neq a \in M$ ; simplicity gives $M = S a$ , so by $(5.7 a)$

M = (\sum_{j = 1}^{n} R \cdot x_{j}) a = \sum_{j = 1}^{n} R \cdot (x_{j} a),

so $M$ is a finitely generated left $R$ -module. Next, $J \cdot M$ is an $S$ -submodule of $M$ : it is visibly an $R$ -submodule, and for each $j$ ,

x_{j} (J \cdot M) = (x_{j} J) M = (J x_{j}) M \subseteq J \cdot M,

using that $x_{j}$ centralises $i (R)$ ; since the $x_{j}$ generate $S$ over $R$ , $J \cdot M$ is stable under all of $S$ . Because $M \neq 0$ is finitely generated over $R$ , Nakayama's Lemma $(4.22)$ gives $J \cdot M \neq M$ . Simplicity of $S M$ then forces $J \cdot M = 0$ , as required. The last assertion follows because $rad S$ is an ideal of $S$ .

Remark(5.8)Twisting is allowed

The proof only used $x_{j} J \subseteq J x_{j}$ . So the centralising hypothesis in $(5.7 a)$ may be replaced by: for each $j$ there is a ring automorphism $σ_{j}$ of $R$ with $x_{j} i (r) = i (σ_{j} (r)) x_{j}$ for all $r \in R$ . Indeed $rad R$ is invariant under every automorphism of $R$ , so $x_{j} J = σ_{j} (J) x_{j} = J x_{j}$ and the argument is unchanged.

Corollary(5.9)Module-finite algebras

Let $R$ be a commutative ring and $S$ an $R$ -algebra that is finitely generated as an $R$ -module. Then $(rad R) \cdot S \subseteq rad S$ .

Proof

By definition of an $R$ -algebra over a commutative ring, the structure map sends $R$ into the centre of $S$ , so any set of $R$ -module generators of $S$ centralises the image of $R$ and $(5.7)$ applies.

CorollaryEquality for finite crossed products

Let $G$ be a finite group and $S = R * G$ a crossed product: $S$ is free as a left $R$ -module on units $u_{g}$ ( $g \in G$ , $u_{1} = 1$ ) with $u_{g} r = σ_{g} (r) u_{g}$ for automorphisms $σ_{g}$ of $R$ . Then

rad R = R \cap rad (R * G) .

(5.9a)

Indeed $S$ is free as a left $R$ -module on a basis containing $1$ , so $R R$ is a direct summand of $R S$ and $(5.6) (1)$ gives $R \cap rad S \subseteq rad R$ ; and $S = \sum_{g \in G} R u_{g}$ is generated by finitely many twisted-centralising elements, so $(5.8)$ gives $rad R \subseteq rad S$ , hence $rad R \subseteq R \cap rad S$ . For $G$ infinite only the first inclusion survives.

CounterexampleBoth inclusions can fail

**Descent fails without $(5.6)$ .** Take $R = ℤ \subseteq S = ℤ_{(p)}$ . Then $rad R = 0$ but $R \cap rad S = ℤ \cap p ℤ_{(p)} = p ℤ \neq 0$ . In particular $ℤ$ is not a direct summand of $ℤ_{(p)}$ as a $ℤ$ -module, and $ℤ$ is not a fixed ring of automorphisms of $ℤ_{(p)}$ .

Ascent fails without finiteness. Take $R = ℤ_{(p)} \subseteq S = ℚ$ . Then $rad R = p ℤ_{(p)} \neq 0$ while $rad S = 0$ . Here $S$ is generated over $R$ by central elements, but not by finitely many. The same phenomenon appears for $R \subseteq R [t]$ , where $R [t]$ is free over $R$ but of infinite rank: Snapper's Theorem shows $rad ℤ_{(p)} [t] = 0$ .

Proof Techniques and Method

The transferable moves in these proofs.

Move 1

Project the inverse back

If $S$ splits as $R \oplus T$ over $R$ , an equation $(1 - r_{0}) s = 1$ can be truncated to its $R$ -component. Any left $R$ -linear retraction $S \to R$ does the same job.

Move 2

Average, or use uniqueness

For a fixed ring one does not average — one observes that the inverse of a $G$ -fixed element is $G$ -fixed, because inverses are unique. This avoids any need for $| G |$ to be invertible.

Move 3

Make the simple module finitely generated

Cyclic over $S$ plus finitely many $R$ -module generators of $S$ gives finitely generated over $R$ , which is the hypothesis Nakayama needs. Then a proper submodule of a simple module must be zero.

Move 2 is the reason $(5.6) (2)$ has no characteristic restriction. Averaging arguments — the usual tool for fixed rings — require $| G |$ to be invertible and would fail in modular situations; uniqueness of inverses does not.

Which inclusion can I hope to prove for my extension $R \subseteq S$ ?

$S$ is free or projective over $R$ with $1$ in a basisDescent:

R \cap rad S \subseteq rad R

(5.6) (1)

. Group rings, crossed products and scalar extensions of algebras are all of this type.

$R = S^{G}$ for a group of automorphismsDescent again, by

(5.6) (2)

, with no finiteness or characteristic hypothesis on

G

$S$ is a finitely generated $R$ -module with centralising generatorsAscent:

(rad R) S \subseteq rad S

(5.7)

, or by

(5.8)

if the generators only twist

R

by automorphisms.

$S$ is a localisation, or infinitely generatedNeither, in general. Expect the failures illustrated by

ℤ \subseteq ℤ_{(p)}

and

ℤ_{(p)} \subseteq ℚ

; a special theorem such as Amitsur's is then required.

Worked Example

Group algebras of a subgroup

Let $k$ be a field, $G \leq H$ groups, and $S = k H \supseteq R = k G$ . As a left $k G$ -module, $k H$ is free on a set of right coset representatives, one of which may be taken to be $1$ ; so $(5.6) (1)$ applies and $k G \cap rad (k H) \subseteq rad (k G)$ for arbitrary $H$ . If in addition $[H : G] < \infty$ and the coset representatives centralise $k G$ — automatic when $H$ is abelian — then $(5.7)$ gives the reverse inclusion.

Take $k = 𝔽_{2}$ , $H = C_{2} \times C_{2} = {1, a, b, a b}$ and $G = {1, a}$ . Setting $x = 1 + a$ and $y = 1 + b$ , one has $x^{2} = 1 + a^{2} = 0$ and $y^{2} = 0$ in characteristic $2$ , and

k H ≅ k [x, y] / (x^{2}, y^{2}), k G ≅ k [x] / (x^{2}) .

(E.1)

Both are local rings. $rad (k H)$ is the augmentation ideal $(x, y)$ , of dimension $3$ over $𝔽_{2}$ , and $rad (k G) = (x)$ , of dimension $1$ . Intersecting,

k G \cap rad (k H) = {λ + μ a : λ + μ = 0} = {0, 1 + a} = rad (k G),

(E.2)

Descent and ascent both hold, so the two radicals match on the nose.

Sanity check on nilpotence: $rad (k H)^{3} = 0$ but $rad (k H)^{2} = (x y) \neq 0$ , consistent with $k H$ being a $4$ -dimensional local algebra.

A fixed ring

Let $k$ be a field with $char k \neq 2$ , $S = k [x] / (x^{4})$ , and let $G = {id, σ}$ where $σ (x) = - x$ . Then $R = S^{G} = k [x^{2}] / (x^{4}) ≅ k [y] / (y^{2})$ with $y = x^{2}$ . Here $rad S = (x)$ , so

R \cap rad S = (y) = rad R,

(E.3)

confirming $(5.6) (2)$ . Note that no averaging was needed in the proof, and indeed the same computation in characteristic $2$ — where $σ$ becomes the identity and $R = S$ — still satisfies the conclusion trivially.

Process and Workflow

Fix the direction you needContraction or extension? They are separate theorems with separate hypotheses; deciding first avoids proving the wrong one.

Look for a retractionIs

S

free or projective over

R

with

1

part of a basis? Then descent is available at once.

Count generatorsIs

S

a finitely generated

R

-module? If yes, check whether generators centralise or merely twist

R

; either suffices for ascent.

CombineWith both, conclude

rad R = R \cap rad S

and transport computations across the extension.

If neither holds, look for a special theoremPolynomial extensions need Amitsur; transcendental scalar extensions need

(5.15)

; localisations need direct analysis.

Comparison and Classification

Which hypothesis holds for which extension
	$R R$ summand of $R S$	$R$ a fixed ring	Finitely many centralising generators	Conclusion available
$R \subseteq R [t]$	yes	no	no	descent only
$R \subseteq R [T]$ , $T$ infinite	yes	no	no	descent only
$R \subseteq R [x] / (x^{n})$	yes	no	yes	both
$R \subseteq M_{n} (R)$	yes	no	yes	both
$R \subseteq R * G$ , $G$ finite	yes	partial	yes	both
$R \subseteq R [G]$ , $G$ infinite	yes	no	no	descent only
$ℤ \subseteq ℤ_{(p)}$	no	no	no	none — descent fails
$ℤ_{(p)} \subseteq ℚ$	no	no	no	none — ascent fails
$R \subseteq R \otimes_{k} K$ , $[K : k] < \infty$	yes	partial	yes	both

Which hypothesis holds for which extension

The two theorems side by side
	Descent $(5.6)$	Ascent $(5.7)$ – $(5.8)$
Conclusion	$R \cap rad S \subseteq rad R$	$i (rad R) \subseteq rad S$
Structural input	a left $R$ -linear retraction, or a fixed-point description	finitely many generators twisting $R$ by automorphisms
Main tool	uniqueness or truncation of an inverse	Nakayama's Lemma on a simple $S$ -module
Finiteness required	no	yes, essentially
Injectivity of $R \to S$ required	yes (it is a subring)	no
Typical failure	localisations	infinite polynomial or group extensions

Relationship Map

These two propositions are the load-bearing lemmas for the rest of the change-of-rings stream.

$(5.6)$ and $(5.7)$ — descent and ascent
- used to prove
  - $(5.14)$ : $R \cap rad R^{K} \subseteq rad R$ , with equality for algebraic $K / k$
  - the step $J_{1} \cap S = J$ inside Amitsur's Theorem $(5.10 B)$
  - $rad M_{n} (R) \supseteq M_{n} (rad R)$ via the central scalar matrices
- specialise to
  - $(5.9)$ : module-finite algebras over a commutative base
  - finite crossed products and skew group rings
  - group algebras of subgroups of finite index
- do not give
  - any control of $rad S$ in terms of $R \cap rad S$
  - equality without both hypotheses

(5.6)

(5.7)

rad R = R \cap rad S

radical computations transfer between

R

and

S

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

Restriction to subgroups

Descent along $k G \subseteq k H$ is the algebraic content of restricting modular representations: the radical of the big group algebra cannot contribute anything to the subgroup algebra beyond its own radical.

Skew polynomial constructions

Crossed products and control theory

Skew group rings and crossed products model symmetry and time-variation; $(5.8)$ says the radical behaves as in the untwisted case whenever the twisting group is finite.

Computation

Reducing to a smaller ring

Computer algebra systems compute radicals of module-finite algebras by descending to a commutative base and applying $(5.9)$ to control the answer from below.

Coding theory

Codes over extensions

Codes over a ring $S$ that is module-finite over a commutative $R$ inherit the radical filtration of $R$ through $(rad R) S \subseteq rad S$ , which is how chain-ring code constructions transfer between base and extension.

As with most radical theory, the honest summary is internal: these are the lemmas that make the later theorems provable, and they are used inside proofs far more often than they are quoted in applications.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Choose the extension to be free. When you are free to design the ambient ring — a crossed product, a scalar extension, a matrix ring — arrange it to be free as a module over the base with $1$ in the basis. That single choice buys descent for nothing.
Prefer finite index. Ascent is a finiteness statement. Modelling a symmetry by a finite group gives $rad R = R \cap rad (R * G)$ ; modelling it by an infinite group gives only half of that.
Twisting is free, commuting is not needed. $(5.8)$ means skew constructions cost nothing: $R [x; σ] / (x^{n})$ behaves like $R [x] / (x^{n})$ for these purposes.
Fixed rings versus quotients. If a construction can be presented either as a fixed ring $S^{G}$ or as a quotient, prefer the fixed-ring presentation: descent is then automatic and needs no invertibility of $| G |$ .
Watch localisation. Inverting elements is the standard way to destroy both hypotheses at once; radical information does not survive it.

Failure Modes and Common Mistakes

Do not assume $R$ and $S$ share an identity by default; if $1_{R} \neq 1_{S}$ the statement of $(5.6)$ is not the one proved here.
Do not assume a subring of a semiprimitive ring is semiprimitive: $ℤ_{(p)} \subseteq ℚ$ is a counterexample, and $(5.6)$ is exactly the hypothesis that rules it out.
Do not read $(5.9)$ as $(rad R) S = rad S$ . Take $R$ a field and $S$ any finite-dimensional algebra with nonzero radical: the left side is $0$ .
Do not apply $(5.7)$ when the generators only normalise a subring rather than twisting $R$ itself by automorphisms; the hypothesis in $(5.8)$ is about automorphisms of $R$ .
Do not forget that $R \cap rad S$ is an ideal of $R$ — the descent proofs need that, and it is what allows a one-sided inverse to suffice.

Quick Reference

$(5.6) (1)$

R R

a summand of

R S

\Rightarrow

R \cap rad S \subseteq rad R

$(5.6) (2)$

R = S^{G}

\Rightarrow

R \cap rad S \subseteq rad R

$(5.7)$

S = \sum_{j = 1}^{n} R x_{j}

with

x_{j}

centralising

i (R)

\Rightarrow

i (rad R) \subseteq rad S

$(5.8)$ Enough that

x_{j} i (r) = i (σ_{j} (r)) x_{j}

for automorphisms

σ_{j}

R

$(5.9)$

R

commutative,

S

module-finite

\Rightarrow

(rad R) S \subseteq rad S

Both together

rad R = R \cap rad S

Key tool (descent)Uniqueness of inverses; ideal test for radical membership

Key tool (ascent)Nakayama's Lemma

(4.22)

Standard extensions and what holds
Extension	Descent	Ascent
$R \subseteq M_{n} (R)$	yes	yes
$R \subseteq R * G$ , $\| G \| < \infty$	yes	yes
$R \subseteq R [T]$	yes	no
$R \subseteq R \otimes_{k} K$ , $[K : k] < \infty$	yes	yes
$R \subseteq S^{- 1} R$	no	no

Frequently Asked Questions

Why does the fixed-ring case need no hypothesis on the group?

Because the argument uses uniqueness of two-sided inverses rather than averaging. If $1 - r_{0}$ is fixed by $G$ and $s$ is its inverse, then $g (s)$ is also its inverse, hence equals $s$ . No finiteness of $G$ , and no invertibility of $| G |$ , is needed — which is what makes the result usable in modular situations.

R \cap rad S

always an ideal of

R

Yes. It is the intersection of the subring $R$ with a two-sided ideal of $S$ , so it is closed under addition and under multiplication by elements of $R$ on both sides. This matters: the descent proofs only produce a one-sided inverse, and the ideal property is what converts that into genuine radical membership.

Can ascent hold without any finiteness?

Sometimes, but not for structural reasons. For instance $rad R \subseteq rad R [[x]]$ holds although $R [[x]]$ is not a finitely generated $R$ -module — that comes from the power series geometric expansion, not from $(5.7)$ . What $(5.7)$ asserts is a general theorem, and its counterexamples show the finiteness cannot be removed from the general statement.

How is

(5.7)

used inside Amitsur's Theorem?

To compare $S = R [t]$ with $S_{1} = R_{1} [t]$ , where $R_{1} = R [ξ] / (1 + ξ + \dots + ξ^{p - 1})$ adjoins a $p$ -th root of unity. $S_{1}$ is a free $S$ -module of finite rank on central elements, so $(5.7)$ gives $rad S \subseteq rad S_{1}$ and $(5.6) (1)$ gives the reverse contraction, yielding $S \cap rad S_{1} = rad S$ — the identity that lets the root-of-unity automorphism be exploited.

Does

(5.9)

ever give equality?

Only accidentally. Take $R = k$ a field and $S$ a finite-dimensional $k$ -algebra with $rad S \neq 0$ : then $(rad R) S = 0 ⊊ rad S$ . Equality holds, for example, when $S = M_{n} (R)$ , where $rad M_{n} (R) = M_{n} (rad R) = (rad R) M_{n} (R)$ .

What happens for localisations?

Neither inclusion holds in general. For $R = ℤ$ and $S = ℤ_{(p)}$ , descent fails; for $R = ℤ_{(p)}$ and $S = ℚ$ , ascent fails. Both hypotheses of this page are lost when elements are inverted, and no general replacement is known.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §5, results (5.6)–(5.9) (pp. 74–77).
D. S. Passman, A Course in Ring Theory, Wadsworth & Brooks/Cole, 1991, Part III.
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977, Chapters 7 and 8.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter I.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that $rad R$ is invariant under every ring automorphism of $R$ , and identify where that is used in Remark $(5.8)$ .
For a finite group $G$ with $| G |$ invertible in $R$ , show that $rad (R G) = (rad R) G$ and explain what fails in the modular case.
Does $(5.6)$ hold if $R R$ is only assumed to be a direct summand of $S_{R}$ as a right module?
Give an example of rings $R \subseteq S$ with $S$ a finitely generated $R$ -module for which $R \cap rad S$ strictly contains $rad R$ .
How do these results extend to rings without identity, where the quasi-regularity definition of the radical is primary?
Work out the analogue of $(5.7)$ for the Levitzki radical and the upper nilradical.
What can be said about $rad S$ in terms of $R \cap rad S$ when $S$ is integral over a central subring $R$ ?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

One-sided invertibility is enough

Why ascent is harder

06Key Results

07Proof Techniques and Method

Project the inverse back

Average, or use uniqueness

Make the simple module finitely generated

08Worked Example

Group algebras of a subgroup

A fixed ring

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Restriction to subgroups

Crossed products and control theory

Reducing to a smaller ring

Codes over extensions

13Design Considerations

14Failure Modes and Common Mistakes

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

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions