关于Jacobson猜想

数学使徒（MathematicalApostle）

Jacobson猜想原表述：对于一个左诺特的幺环R，其Jacobson根J的所有幂之交J^ω=0。

但1965年该表述被证伪，目前仍然open的是以下表述：

“对于一个双边诺特的幺环R，其Jacobson根J的所有幂之交J^ω=0。”

我对该猜想比较感兴趣。

刚翻到一篇有意思的文献，声称，如果幺环R满足如下条件之一，那么J^ω=0：

1、R是左主理想区且有一个左Motria对偶；

2、R是双边诺特环，且它作为自己的左模在离散拓扑下线性紧致。

参考文献：Claudia Menini，Jacobson’s Conjecture, Morita Duality

and Related Questions，Journal of Algebra 103, 638-655 (1986)

抽象代数环论

For a ring Rwith Jacobson radical J，the nonnegative powers Jⁿ are defined by using the product of ideals.

Jacobson's conjecture：lnaright-and-left Noetherian ring，∩Jⁿ＝{0}.

n∈ℕ

3. Sᴏᴍᴇ Fᴜʀᴛʜᴇʀ Rᴇsᴜʟᴛs

In this section we will get some more results about Jacobson's conjecture and we will show that， in some particular cases，it holds.

3.1. LEMMA. Let R be α ring，J＝J(R)，Jω＝Jω(R). Suppose thαt ʀR is

I.c.d.αnd ʀJ is finitely generαted. Then

Jω＝JωJ.

Proof.Let J＝Rα₁＋· · ·＋ Rαₙ. For every left R-module M. let M⁽ⁿ⁾ denote the direct sum of n-copies of M. Define

f： ʀR⁽ⁿ⁾ → ʀR

by setting

(r₁，...，rₙ)f＝r₁α₁＋· · ·＋rₙαₙ， r₁，...，rₙ∈R.

Then f is a morphism of left R-modules and

(∩(JᵏR⁽ⁿ⁾)f＝((Jω)⁽ⁿ⁾)f＝JωJ.

ₖ

As ʀR is l.c.d.，by Satz 1 of ［L］ it is：

(∩(JᵏR⁽ⁿ⁾)f＝∩((JᵏR⁽ⁿ⁾)f＝Jω.

ₖ ₖ

3.2. Cᴏʀᴏʟʟᴀʀʏ. Let R be α ring，J＝J(R)， Jω＝Jω(R).Assume thαt ʀR is I.c.d. αnd thαt both ʀJ αnd Jωʀ αre finitely generαted. Then Jω＝0.

Proof. By Lemma 3.1 Jω＝JωJ.Apply now Nakayama’s Lemma.

Recall that a ring R is said to have a left Moritα duαlity if both ʀR and the minimal cogenerator ʀK of R-Mod are l.c.d.

3.3. Remαrk. Corollary 3.2 holds in particular when R is a noetherian ring (on both sides) having a left Morita duality. This result has been already proved， in another way，in ［J4］.

3.4.Pʀᴏᴘᴏsɪᴏɴ. Let R be α ring J＝J(R)，Jω＝Jω(R). Suppose thαt R is α locαl (i.e.，R/J is α diυision ring)，J＝Rz，ʀJω is finitely generαted αnd R hαs α left Moritα duαlity. Then there exists αn n∈ℕ such thαt JⁿJω＝0.

Proof. Let ʀK be the minimal cogenerator of R-Mod and suppose that for every n∈ℕ，there exists

eₙ∈Ann ᴋ(JⁿJω)\Ann ᴋ(Jⁿ⁻¹Jω).

For every n∈ℕ let ēₙ＝eₙ＋Ann ᴋ(Jω)∈ K/Ann ᴋ(Jω).Then the elements ēₙ yield α bαsis for α free left R/Jω module. In fact note that JωJωeₙ＝0 and assume that

ₜ

∑ rₙēₙ＝0 with rₙ∈R，rₜ ∉ Jω.

ₙ₌₁

Then rₜ eₜ ∈ Ann ᴋ(Jᵗ⁻¹Jω) and hence Jᵗ⁻¹ Jωrₜ eₜ ＝0.

Since rₜ ∉ Jω and R is local，there exist an l∈ℕ and an invertible element ε of R such that

rₜ＝εzˡ.

Then Jωrₜ＝Jωεzˡ＝Jωzˡ and， by Proposition 3.1，Jωrₜ＝Jω.Thus Jᵗ⁻¹Jωeₜ＝0. Contradiction.

Since K/Ann ᴋ(Jω) is an l.c.d. left R-module this cannot happen. Hence there exists an n such that

Ann ᴋ(JⁿJω)＝Ann ᴋ(Jⁿ⁺¹Jω).

652 ᴄʟᴀᴜᴅɪᴀ ᴍᴇɴɪɴɪ

Thus，as ʀK is a cogenerator of R-Mod，we get

JⁿJω＝Jⁿ⁺¹Jω.

Nakayama’s Lemma implies that JⁿJω＝0.

3.5. Cᴏʀᴏʟʟᴀʀʏ. Let R be α left principαl ideαl domαin with α left Moritα duαlity. Then Jω(R)＝0.

Proof.As ʀR is l.c.d.，idempotents modulo the Jacobson radical lift. Thus，since R is a domain，R must be local.Apply now Proposition 3.4.

3.6. Remαrk. The ring R in 2.10 is a left principal ring.Thus 2.11 shows that the hypothesis “domain” in Corollary 3.5 cannot be omitied.

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