T-Nilpotency

Q: Is a T-nilpotent ideal always contained in the Jacobson radical?

Yes, and much more: it lies in the lower nilradical Nil_* R by (23.15), which is contained in the Levitzki radical, the upper nilradical and hence in rad R. This is unusually strong — most nilness hypotheses on one-sided ideals give containment only in rad R.

Executive Summary

Nilpotence of an ideal ( $J^{m} = 0$ ) is a uniform condition: one exponent works for all products. Nilness ( $a^{m} = 0$ for each $a$ ) is a pointwise condition and much weaker. T-nilpotency sits strictly between: every sequence drawn from $J$ eventually multiplies to zero, but the point at which it does may depend on the sequence. The $T$ is for transfinite.

The reason it matters is $(23.16)$ . Right T-nilpotency of a right ideal $J$ is equivalent to the statement $M J = M \Rightarrow M = 0$ for arbitrary right modules $M$ — Nakayama's Lemma with the finite generation hypothesis deleted. That is precisely the property Bass isolated in defining perfect rings, and it is what makes projective covers exist.

3Grades: nilpotent, T-nilpotent, nil

${Nil}_{*} R$ Where T-nilpotent ideals live

No f.g.Nakayama without finite generation

(23.16)The characterisation theorem

Overview

Let $A$ be a subset of a ring $R$ . The two conditions differ only in the order in which a sequence is multiplied:

left T-nilpotent: \exists n, a_{1} a_{2} \dots a_{n} = 0; right T-nilpotent: \exists n, a_{n} \dots a_{2} a_{1} = 0 .

(23.13a)

For every sequence $a_{1}, a_{2}, a_{3}, \dots$ in $A$ . Note that the integer $n$ may depend on the sequence — this is what separates T-nilpotency from nilpotency.

Applying the definition to the constant sequence $a, a, a, \dots$ shows every element of a T-nilpotent set is nilpotent. Conversely, if $J^{m} = 0$ then $n = m$ works for every sequence. So for one-sided ideals:

nilpotentleft or right T-nilpotentnil

Both implications in the chain are strict, and both failures are instructive. A nil ideal that is not T-nilpotent is easy to build commutatively; a T-nilpotent ideal that is not nilpotent requires infinitely many matrix coordinates, and the standard construction also shows that right and left T-nilpotency are genuinely different conditions.

Learning Objectives

State $(23.13)$ correctly, including which side multiplies in which order.
Prove the implications of $(23.14)$ and produce counterexamples to both converses.
Prove $(23.15)$ : a right T-nilpotent one-sided ideal lies in ${Nil}_{*} R$ and is locally nilpotent.
Prove all four equivalences of $(23.16)$ , including the free-module construction for (2) $\Rightarrow$ (1).
Prove $(23.17)$ : projectives cancel modulo a right T-nilpotent right ideal.
Explain how $(23.18)$ uses T-nilpotency to define right perfect rings, and why the sides switch in Bass's theorem.

Definitions

Definition(23.13)T-nilpotency

A subset $A$ of a ring $R$ is left T-nilpotent if for every sequence $a_{1}, a_{2}, a_{3}, \dots$ of elements of $A$ there exists an integer $n \geq 1$ with $a_{1} a_{2} \dots a_{n} = 0$ . It is right T-nilpotent if for every such sequence there exists $n \geq 1$ with $a_{n} \dots a_{2} a_{1} = 0$ .

The notion is applied almost exclusively to one-sided ideals. The letter $T$ stands for transfinite: the vanishing is guaranteed along every infinite sequence, but not at a uniform stage.

Nilpotent ideal: $J^{m} = 0$ for some fixed $m$ ; equivalently every product of $m$ elements of $J$ vanishes, uniformly.
Nil ideal: Every element of $J$ is nilpotent, with no uniformity in the index.
Locally nilpotent: Every finitely generated subring (without identity) of $J$ is nilpotent; equivalently $J$ lies in the Levitzki radical.
${Nil}_{*} R$: The lower nilradical or Baer radical: the intersection of all prime ideals of $R$ , and the smallest semiprime ideal.
${ann}_{N} (J)$: For a left $R$ -module $N$ , the subgroup ${x \in N : J x = 0}$ .

Conventions for which side is which are not uniform in the literature. This page follows Lam: right T-nilpotency multiplies new elements on the left.

Core Concepts

Why a sequence and not a power

Nilpotency asks for a single $m$ that annihilates all $m$ -fold products; T-nilpotency lets $n$ depend on the sequence. The gap is exactly the gap between a uniform bound and a well-foundedness condition: T-nilpotency says the tree of nonzero products $a_{n} \dots a_{1}$ has no infinite branch, while nilpotency says it has bounded depth. König's lemma does not close the gap, because the set of available elements at each stage is generally infinite.

The general Nakayama reading

Nakayama's Lemma $(4.22)$ says: if $M$ is a finitely generated right module and $M rad R = M$ then $M = 0$ . Theorem $(23.16)$ identifies the exact hypothesis under which the finite generation can be dropped: right T-nilpotency of the ideal. This reframing is the whole point of the notion — it converts an internal nilpotence condition into a statement about the module category.

J right T-nilpotent iff (M J = M \Rightarrow M = 0) for all right R -modules M .

(23.16a)

The side switch

Condition (3) of $(23.16)$ is about left modules: every nonzero left module has a nonzero $J$ -annihilator. So a condition on right ideals is detected on left modules. This is the same switch that appears in Bass's theorem $(23.20)$ , where right perfectness is equivalent to the descending chain condition on principal left ideals. The switch is not a misprint and not removable: it reflects the fact that $J$ acts on the left of left modules, so a leftward-growing product $a_{n} \dots a_{1}$ is exactly what iterated application produces.

How strong is T-nilpotency

Much stronger than nil: by $(23.15)$ a right T-nilpotent one-sided ideal lies inside the lower nilradical ${Nil}_{*} R$ , hence inside the Levitzki radical, hence is locally nilpotent. Nil ideals satisfy none of these — whether every nil one-sided ideal lies in the upper nilradical is the Köthe conjecture, still open. So T-nilpotency lands a one-sided ideal in the smallest of the standard radicals, not the largest.

Key Results

Proposition(23.14)The three grades

For a one-sided ideal $J$ of $R$ : if $J$ is nilpotent then $J$ is both left and right T-nilpotent; if $J$ is left or right T-nilpotent then $J$ is nil.

Proof

If $J^{m} = 0$ , take $n = m$ for every sequence: any product of $m$ elements of $J$ lies in $J^{m} = 0$ , in either order. For the second implication, apply the definition to the constant sequence $a, a, a, \dots$ with $a \in J$ : it produces $n$ with $a^{n} = 0$ .

Proposition(23.15)T-nilpotent ideals lie in the lower nilradical

Let $J$ be a one-sided ideal of $R$ that is right T-nilpotent. Then $J \subseteq {Nil}_{*} R$ . In particular $J$ is locally nilpotent, since ${Nil}_{*} R \subseteq Levitzki (R)$ . The same holds for left T-nilpotent one-sided ideals, by passing to $R^{op}$ and using the left-right symmetry of ${Nil}_{*}$ .

Proof

Passing to $R / {Nil}_{*} R$ , which is semiprime, and noting that images of T-nilpotent sets are T-nilpotent, it suffices to prove: in a semiprime ring, a right T-nilpotent one-sided ideal $J$ is zero.

Suppose $0 \neq a \in J$ . Semiprimeness says $b R b \neq 0$ for every $b \neq 0$ . Build recursively $x_{1}, x_{2}, \dots \in R$ with

d_{n} : = a x_{n} a x_{n - 1} \dots a x_{1} a \neq 0 for every n \geq 1 .

(23.15a)

Indeed $d_{0} : = a \neq 0$ , and given $d_{n} \neq 0$ choose $y$ with $d_{n} y d_{n} \neq 0$ ; setting $x_{n + 1} : = x_{n} a x_{n - 1} \dots a x_{1} a y$ gives $a x_{n + 1} d_{n} = d_{n} y d_{n} \neq 0$ , which is $d_{n + 1}$ .

If $J$ is a right ideal, put $y_{0} = a$ and $y_{i} = a x_{i}$ , all in $J$ . Then $y_{n} y_{n - 1} \dots y_{1} y_{0} = d_{n} \neq 0$ for every $n$ , contradicting right T-nilpotency. If $J$ is a left ideal, put $z_{i} = x_{i} a \in J$ ; then $a \cdot (z_{n} \dots z_{2} z_{1}) = d_{n} \neq 0$ , so $z_{n} \dots z_{2} z_{1} \neq 0$ for every $n$ , the same contradiction. Hence $J = 0$ .

Theorem(23.16)Characterisations of right T-nilpotency

For any right ideal $J \subseteq R$ the following are equivalent:

$J$ is right T-nilpotent;
for every right $R$ -module $M$ , $M J = M$ implies $M = 0$ ;
for right $R$ -modules $N \subseteq M$ , $M J + N = M$ implies $N = M$ ;
for every left $R$ -module $N$ , ${ann}_{N} (J) = 0$ implies $N = 0$ .

No finite generation is assumed anywhere, which is what makes (2) a general Nakayama lemma.

Proof

**(1) $\Rightarrow$ (4).** Suppose ${ann}_{N} (J) = 0$ and $N \neq 0$ . Pick $0 \neq z \in N$ . Since $z \notin {ann}_{N} (J)$ , there is $a_{1} \in J$ with $a_{1} z \neq 0$ ; since $a_{1} z \neq 0$ and the annihilator is trivial, there is $a_{2} \in J$ with $a_{2} a_{1} z \neq 0$ ; continuing, $a_{n} \dots a_{1} z \neq 0$ for all $n$ , so $a_{n} \dots a_{1} \neq 0$ for all $n$ and $J$ is not right T-nilpotent.

**(4) $\Rightarrow$ (2).** Let $M \neq 0$ be a right module and put $A = ann (M)$ , a proper ideal. Regard $N = R / A$ as a nonzero left $R$ -module. By (4), ${ann}_{N} (J) \neq 0$ , and this annihilator is $B / A$ where $B = {b \in R : J b \subseteq A}$ . So $A ⊊ B$ , whence $M B \neq 0$ . But $J B \subseteq A$ gives $M J B \subseteq M A = 0$ . If $M J$ were equal to $M$ we would get $M B = M J B = 0$ , a contradiction. Hence $M J \neq M$ .

**(2) $\Leftrightarrow$ (3).** Apply (2) to the quotient $M / N$ , noting $(M / N) J = (M J + N) / N$ .

**(2) $\Rightarrow$ (1).** Given $a_{1}, a_{2}, \dots \in J$ , let $F = ⨁_{i \geq 0} e_{i} R$ be free on ${e_{i}}_{i \geq 0}$ and let $S \subseteq F$ be the submodule generated by $e_{i - 1} - e_{i} a_{i}$ for $i \geq 1$ . Put $M = F / S$ . In $M$ every generator satisfies ${\bar{e}}_{i - 1} = {\bar{e}}_{i} a_{i} \in M J$ , so $M = M J$ and (2) gives $M = 0$ , i.e. $S = F$ .

In particular $e_{0} \in S$ , so $e_{0} = \sum_{i = 1}^{n} (e_{i - 1} - e_{i} a_{i}) b_{i}$ for suitable $b_{i} \in R$ . Comparing coefficients of the basis elements gives $b_{1} = 1$ , then $b_{i + 1} = a_{i} b_{i}$ for $1 \leq i \leq n - 1$ , and finally $a_{n} b_{n} = 0$ . Unwinding, $b_{n} = a_{n - 1} \dots a_{2} a_{1}$ and

0 = a_{n} b_{n} = a_{n} a_{n - 1} \dots a_{2} a_{1},

(23.16b)

which is exactly right T-nilpotency for the chosen sequence.

Corollary(23.17)Cancellation for projectives

Let $J$ be a right T-nilpotent right ideal of $R$ . Then for any two projective right $R$ -modules $P$ and $Q$ , $P / P J ≅ Q / Q J$ implies $P ≅ Q$ .

Proof

Let $ψ : P / P J \to Q / Q J$ be an isomorphism. Since $P$ is projective and $Q \to Q / Q J$ is onto, $ψ$ lifts to $ϕ : P \to Q$ compatible with the projections. Surjectivity of $ψ$ gives $ϕ (P) + Q J = Q$ , so condition (3) of $(23.16)$ , applied to $ϕ (P) \subseteq Q$ , yields $ϕ (P) = Q$ .

As $Q$ is projective, $ϕ$ splits: $P = K \oplus P'$ with $K = \ker ϕ$ and $ϕ |_{P'} : P' \tilde{⟶} Q$ . Reducing modulo $J$ gives $P / P J ≅ K / K J \oplus P' / P' J$ , and $ψ$ is injective while killing the image of $K$ ; hence $K / K J = 0$ , i.e. $K = K J$ . Condition (2) of $(23.16)$ forces $K = 0$ , so $P = P' ≅ Q$ .

Definition(23.18)Perfect rings

A ring $R$ is right perfect if $R / rad R$ is semisimple and $rad R$ is right T-nilpotent; left perfect is defined with left T-nilpotency; perfect means both.

Corollary(23.19)Semiprimary rings are perfect

If $R$ is semiprimary — $R / rad R$ semisimple and $rad R$ nilpotent — then $R$ is both left and right perfect. In particular every one-sided artinian ring is perfect. Conversely a one-sided perfect ring is semiperfect: a T-nilpotent radical is nil, and idempotents lift modulo a nil ideal by $(21.28)$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Build the bad sequence

To violate T-nilpotency, do not compute a product — construct the sequence one element at a time, each chosen so that the accumulated product stays nonzero. Semiprimeness and nontriviality of an annihilator are the two hypotheses that keep such a construction going.

Move 2

Encode a sequence as a free presentation

The relations $e_{i - 1} = e_{i} a_{i}$ on a countably generated free module turn an arbitrary sequence into a module with $M = M J$ . Any hypothesis forcing $M = 0$ then forces the sequence to terminate. This is the standard way to convert a module axiom into an element statement.

Move 3

Test on $R / ann (M)$

To move between right-module and left-module statements, use the ideal $A = ann (M)$ and the left module $R / A$ . This is the bridge that makes the side switch in $(23.16)$ work.

Move 2 is the reusable one. The presentation $M = F / S$ is the module-theoretic incarnation of an infinite descending chain, and the same construction proves the harder implications of Bass's theorem, where descending chains of principal left ideals appear in place of sequences in $J$ .

Worked Example

T-nilpotent on one side only, and not nilpotent

Let $k$ be a field and let $J$ be the set of $ℕ \times ℕ$ matrices over $k$ with only finitely many nonzero entries, all strictly above the diagonal. Put $R = k \cdot 1 + J$ , where $1$ is the infinite identity matrix. Then $R$ is a ring, $J$ is an ideal of $R$ , and $R / J ≅ k$ .

$J$ is right T-nilpotent

Let $J$ act on the left of $V = ⨁_{i \geq 1} e_{i} k$ and write $V_{n} = e_{1} k \oplus \dots \oplus e_{n} k$ . Every $a \in J$ is strictly upper triangular, so $a (V_{m}) \subseteq V_{m - 1}$ for all $m$ . Given a sequence $a_{1}, a_{2}, \dots \in J$ , the matrix $a_{1}$ has finitely many nonzero columns, so $a_{1} (V) = a_{1} (V_{N}) \subseteq V_{N - 1}$ for some $N$ . Then

a_{2} a_{1} (V) \subseteq a_{2} (V_{N - 1}) \subseteq V_{N - 2}, \dots, a_{N} \dots a_{2} a_{1} (V) \subseteq V_{0} = 0,

(E.1)

so $a_{N} \dots a_{2} a_{1} = 0$ , and $J$ is right T-nilpotent.

$J$ is not left T-nilpotent and not nilpotent

Take $a_{i} = E_{i, i + 1}$ , the matrix unit. Then $a_{1} a_{2} = E_{1, 2} E_{2, 3} = E_{1, 3}$ , and inductively $a_{1} a_{2} \dots a_{n} = E_{1, n + 1} \neq 0$ for every $n$ . So $J$ fails left T-nilpotency; the same computation shows $J^{n} \neq 0$ for every $n$ , so $J$ is not nilpotent.

Nil but not T-nilpotent

Let $R = k [x_{2}, x_{3}, x_{4}, \dots] / (x_{2}^{2}, x_{3}^{3}, x_{4}^{4}, \dots)$ , a commutative ring, and let $N$ be the ideal generated by the images of all the $x_{i}$ . Every element of $N$ involves finitely many variables and has zero constant term, so it is nilpotent: $N$ is a nil ideal. But taking $a_{i} = x_{i + 1}$ gives

a_{1} a_{2} \dots a_{n} = x_{2} x_{3} \dots x_{n + 1} \neq 0,

(E.2)

since each variable appears to the first power, below its nilpotency index. So $N$ is nil but T-nilpotent on neither side — and, $R$ being commutative, the two sides agree here.

Note that $N$ is locally nilpotent, since a commutative nil ideal always is. So the containment in $(23.15)$ cannot be reversed: lying in ${Nil}_{*} R$ is necessary but far from sufficient for T-nilpotency.

Comparison and Classification

The nilpotence conditions compared
Condition on a one-sided ideal $J$	Uniform bound?	Inside ${Nil}_{*} R$ ?	Side-symmetric?
$J$ nilpotent, $J^{m} = 0$	yes	yes	yes
$J$ right T-nilpotent	no	yes, (23.15)	no
$J$ locally nilpotent	on finite subsets only	not in general	yes
$J$ nil	no	open in general (Köthe)	yes

Which examples satisfy which condition
	Nilpotent	Right T-nilp.	Left T-nilp.	Nil
Strictly upper triangular in $M_{n} (k)$	yes	yes	yes	yes
$J$ in the infinite-matrix ring above	no	yes	no	yes
$N$ in $k [x_{2}, x_{3}, \dots] / (x_{i}^{i})$	no	no	no	yes
$rad k [[x]] = (x)$	no	no	no	no
$rad R$ for $R$ left artinian	yes	yes	yes	yes

Which examples satisfy which condition

The last row is why T-nilpotency is invisible in classical finite-dimensional theory: over a left artinian ring the radical is nilpotent, so all four columns collapse. The notion earns its keep only outside chain conditions.

Relationship Map

$J$ right T-nilpotent right ideal — the hypothesis
- implies about $J$
  - $J$ is nil $(23.14)$
  - $J \subseteq {Nil}_{*} R \subseteq rad R$ $(23.15)$
  - $J$ is locally nilpotent
- implies about modules
  - $M J = M \Rightarrow M = 0$ for all right $M$
  - ${ann}_{N} (J) \neq 0$ for all nonzero left $N$
  - projectives cancel modulo $J$ $(23.17)$
- with $R / rad R$ semisimple and $J = rad R$
  - $R$ is right perfect $(23.18)$
  - $R$ is semiperfect $(23.19)$
  - DCC on principal left ideals — Bass $(23.20)$
  - every right module has a projective cover

Reading the branches together: T-nilpotency is the exact hypothesis that lets radical-quotient arguments run for arbitrary modules rather than finitely generated ones, and every homological consequence of perfectness — projective covers, cancellation, flat implies projective — traces back to that single substitution.

Applications and Industry Use

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

Homological algebra

Projective covers

A ring is right perfect exactly when every right module has a projective cover. The existence proof is a lifting argument that needs $M J = M \Rightarrow M = 0$ for arbitrary $M$ , so T-nilpotency of the radical is not a convenience but the precise hypothesis.

Module theory

Flat equals projective

Over a right perfect ring every flat right module is projective. This is one of Bass's equivalent characterisations and is the statement most often invoked when perfectness is assumed in homological arguments.

Representation theory

Infinite-dimensional algebras

Algebras of infinite dimension over a field can have a radical that is T-nilpotent but not nilpotent. Perfectness is then the right substitute for semiprimarity when transferring finite-dimensional techniques such as Krull-Schmidt and Cartan matrices.

Radical theory

Locating one-sided ideals

$(23.15)$ is a rare tool that places a one-sided ideal inside the lower nilradical. Most nilness hypotheses only place ideals in the upper nilradical, and whether nil one-sided ideals do so at all is the Köthe conjecture.

The honest summary is that this is a technical hypothesis with a large homological payoff. Nobody studies T-nilpotency for its own sake; it is studied because it is exactly what perfect rings need.

Standards and Notation

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

This page (Lam)right T-nilpotent:

a_{n} \dots a_{2} a_{1} = 0

Also written

T

-nilpotent, T-nilpotent, or transfinitely nilpotent

Radicals

{Nil}_{*} R

lower,

{Nil}^{*} R

upper,

Levitzki (R)

locally nilpotent

PerfectRight perfect:

R / rad R

semisimple and

rad R

right T-nilpotent

OriginBass, Trans. AMS 95 (1960); condition

T

in his numbering

ImplementationsNone expose T-nilpotency; GAP, Magma and Sage handle only the nilpotent case for finite-dimensional algebras

Computational Notes

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

T-nilpotency is a condition on all infinite sequences, so it is not a finitary property and admits no direct finite verification procedure for a general presented ring.
For a finite-dimensional algebra $A$ over a field, $rad A$ is nilpotent, so T-nilpotency is automatic and carries no information; any computation reduces to finding the nilpotency index, which is at most $\dim_{k} A$ .
For a ring given by a finite presentation, nilness of the radical is already undecidable in general, and T-nilpotency is a stronger condition; certificates in practice are structural, as in the infinite-matrix example, rather than computed.
The usable finite test is the negative one: exhibiting a sequence $a_{1}, a_{2}, \dots$ with all products $a_{n} \dots a_{1}$ nonzero refutes right T-nilpotency, and such sequences are usually periodic or given by a simple pattern such as $a_{i} = E_{i, i + 1}$ .
Computer algebra systems expose IsNilpotent and radical computations for finite-dimensional algebras but have no predicate for T-nilpotency, precisely because it is not decidable at that level of generality.

Failure Modes and Common Mistakes

Do not read $(23.16) (2)$ as Nakayama's Lemma for finitely generated modules — the content is precisely that no finiteness hypothesis is needed, and the equivalence fails if one is imposed.
Do not assume the integer $n$ in the definition is uniform; if it were, the condition would be nilpotency.
Do not expect the converse of $(23.15)$ : lying in ${Nil}_{*} R$ , or even being locally nilpotent, does not give T-nilpotency.
Do not conflate right T-nilpotency of $rad R$ with a chain condition on right ideals: Bass's theorem trades it for DCC on principal left ideals.

Quick Reference

Right T-nilpotent

\forall (a_{i}) \subseteq J \exists n : a_{n} \dots a_{1} = 0

Left T-nilpotent

\forall (a_{i}) \subseteq J \exists n : a_{1} \dots a_{n} = 0

Gradesnilpotent

\Rightarrow

T-nilpotent

\Rightarrow

nil, both strict

Location

J \subseteq {Nil}_{*} R

, hence locally nilpotent

Module test

M J = M \Rightarrow M = 0

for all right

M

Left-module test

{ann}_{N} (J) \neq 0

for all nonzero left

N

Cancellation

P / P J ≅ Q / Q J \Rightarrow P ≅ Q

for

P, Q

projective

Perfect

R / rad R

semisimple and

rad R

right T-nilpotent

Statements and their hypotheses
Statement	Hypotheses	Reference
nilpotent $\Rightarrow$ T-nilpotent $\Rightarrow$ nil	$J$ a one-sided ideal	(23.14)
$J \subseteq {Nil}_{*} R$	$J$ a right T-nilpotent one-sided ideal	(23.15)
General Nakayama: $M J = M \Rightarrow M = 0$	$J$ a right ideal, right T-nilpotent	(23.16)
${ann}_{N} (J) = 0 \Rightarrow N = 0$	$N$ a left module, $J$ right T-nilpotent	(23.16)(3)
$P / P J ≅ Q / Q J \Rightarrow P ≅ Q$	$P, Q$ projective, $J$ right T-nilpotent	(23.17)
Semiprimary $\Rightarrow$ perfect $\Rightarrow$ semiperfect	one-sided perfect suffices for the second	(23.19)

Frequently Asked Questions

What does the T stand for?

Transfinite. The point is that the vanishing of products is guaranteed along every infinite sequence, but the stage at which it happens is not bounded in advance — the condition is about well-foundedness rather than about a fixed exponent.

Why is T-nilpotency stated for sequences rather than for products of a fixed length?

Because fixing the length gives back nilpotency. The whole content of the notion is that the length may depend on the sequence. The infinite strictly-upper-triangular matrix ideal is the standard witness: every sequence terminates, but no single exponent works for all of them.

Is a T-nilpotent ideal always contained in the Jacobson radical?

Yes, and much more: it lies in the lower nilradical ${Nil}_{*} R$ by $(23.15)$ , which is contained in the Levitzki radical, the upper nilradical and hence in $rad R$ . This is unusually strong — most nilness hypotheses on one-sided ideals give containment only in $rad R$ .

Why does a condition on a right ideal get characterised by left modules?

Because a right T-nilpotent right ideal builds its products by multiplying new elements on the left, which is exactly how $J$ acts by iteration on a left module. Condition (3) of $(23.16)$ says the iteration must terminate, and this is the same side switch that appears in Bass's theorem, where right perfectness corresponds to DCC on principal left ideals.

Does right perfect imply left perfect?

No. Lam's infinite-matrix example is a local ring whose radical is right T-nilpotent but not left T-nilpotent, so it is right perfect and not left perfect. This distinguishes perfectness from semiperfectness, which is a two-sided notion.

Where does T-nilpotency actually get used?

In the homological characterisations of perfect rings: existence of projective covers for all modules, flat modules being projective, and descending chain conditions on principal one-sided ideals. All of these need the Nakayama statement $M J = M \Rightarrow M = 0$ for arbitrary $M$ , and $(23.16)$ says that statement is T-nilpotency.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §23, results (23.13)–(23.22).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §28.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2 (radicals and nilpotence conditions).
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter I.

AI Suggested Questions

Prove that a right T-nilpotent right ideal $J$ satisfies ${ann}_{N} (J) \neq 0$ for every nonzero left module $N$ , and interpret this as a socle statement.
Work through Bass's proof that right perfect implies DCC on principal left ideals.
Construct a ring whose radical is left and right T-nilpotent but not nilpotent.
How does T-nilpotency of the radical give projective covers for arbitrary modules?
Compare the Köthe conjecture with the containment $(23.15)$ : why is the T-nilpotent case so much easier than the nil case?
Is there a transfinite hierarchy of nilpotency conditions between T-nilpotent and nil, indexed by ordinals?
For which group rings $k G$ with $G$ infinite is the augmentation ideal T-nilpotent?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why a sequence and not a power

The general Nakayama reading

The side switch

How strong is T-nilpotency

06Key Results

07Proof Techniques and Method

Build the bad sequence

Encode a sequence as a free presentation

Test on R/ann(M)

08Worked Example

T-nilpotent on one side only, and not nilpotent

J is right T-nilpotent

J is not left T-nilpotent and not nilpotent

Nil but not T-nilpotent

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Projective covers

Flat equals projective

Infinite-dimensional algebras

Locating one-sided ideals

12Standards and Notation

13Computational Notes

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

Test on $R / ann (M)$

Worked Example

$J$ is right T-nilpotent

$J$ is not left T-nilpotent and not nilpotent

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

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