Some Results on Semilocal Rings

Abstract

We first give simple proofs to some main results on semilocal rings of Zhang (Proc. Amer. Math. Soc. 137, 845–852, 2009). Then, as an extension, we prove that the R/J(R)-module $\overline {\textit {aR}}$ is semisimple if and only if every descending chain in $aR\;:\;a_{1}R\;\supseteq \;\dots \;\supseteq \;a_{n}R\;\supseteq \;\cdots$ with $a_{i+1}\;=\;a_{i}\;-\;a_{i}b_{i}a_{i}\;(i\;=\;1,\;2,\;\dots )$ eventually terminates.

1 Introduction

Semilocal rings are characterized in several amazing ways by Camps and Dicks [2], then based on their results, two new conceptions “hollow chain” and “hollow length” are introduced in [6]. Let R be a ring, a descending chain

$$a_{0}R\;\supsetneq\;a_{1}R\;\supsetneq\;\dots\;\supsetneq\;a_{n}R\;\supsetneq\;\dots\;\supsetneq\;a_{n+1}R\;\cdots\quad\text{ with }\;a_{n+1}\;=\;a_{n}\;-\;a_{n}b_{n}a_{n}\; \text{for some} b_{n}\;\in\; R $$

is called a hollow chain of a ₀ ∈ R. Let $r\;=\;\sup \{n\;\in \;\mathbb {Z}\;|\;a_{0}\;R\;\supsetneq \; a_{1}\;R\;\supsetneq \;\dots \;\supsetneq \;a_{n-1}\;R\;\supsetneq \;a_{n}R$ is a hollow chain of a ₀}, then r is called the right hollow length of a ₀, denoted as h.length(a ₀) = r. And it is proved in [6] that R is a semilocal ring with $u.\;\dim (R/J(R))\;=\;n$ if and only if h. length(1 _R ) = n. Several other equivalent conditions of semilocal rings are obtained in [6]. However, some proofs in [6] are not concise. Recently, Camillo and Nielsen gave a short and intuitive proof of Camps–Dicks’s following result in [1].

Theorem 1

Let R be a ring. The following conditions are equivalent.

• (1)  R is semilocal;

• (2)  There exist a ring S and a R-S-bimodule _R M _S such that M _S has finite uniform dimension, and for r ∈ R the equation ann _M (r) = (0) implies r ∈ U (R).

Motivated by the method used in [1], we first give short and intuitive proofs of some other characterizations of semilocal rings of [2, 6] in Theorem 2.

For a ring R, write $\overline {R}\;=\;R/J(R)$, we then generalized some results in [6] by discussing when will $\overline {aR}$ be a semisimple $\overline {R}$-module in Theorem 3. For a module M, write E = End(M). Generally, it is not easy to find the relations between the structure of E and the submodules of M, but based on Theorem 3, we find that for all f ∈ E, the condition $\overline {\textit {fE}}$ is a semisimple E-module is decided by chain conditions of submodules of M.

All rings are associative with 1, and modules are unital. Let R be a ring. U(R) and r. U(R) denote respectively the set of invertible and right invertible elements of R. $u.\dim (M)$ denotes the uniform dimension of module M. For more results on semilocal rings, please refer to [3, 5]. We refer to [4] for all undefined notions used in the text.

2 Main Results

The following result contains some important results of [2, Theorem 1] and [6, Corollary 10], we now provide a short and intuitive proof.

Theorem 2

Let R be a ring. The following are equivalent:

• 1)  R is semilocal;

• 2)  $h.\text {length}(1_{R})\;<\;\infty$;

• 3)  Every descending chain of principal right ideals of R

  $$a_{0}R\supseteq a_{1}R\supseteq a_{2}R\supseteq \dots\supseteq a_{n}R\supseteq \cdots\quad \text{ with }a_{i+1}=a_{i}-a_{i}b_{i}a_{i} $$

  eventually terminates.

• 4)  Every strict descending chain

  $$a_{0}R\supsetneq a_{1}R\supsetneq a_{2}R\supsetneq {\cdots} \supsetneq a_{n}R\supsetneq \cdots\quad \text{ with }a_{i+1}=a_{i}-a_{i}b_{i}a_{i} $$

  eventually terminates.

• 5)  There exists a partial order ≥ on R, satisfying the minimum conditions such that for any a, b ∈ R, if 1 − ab ∉ U (R), then a > a − aba.

• 5′)  There exists a partial order ≥ on R, satisfying the minimum conditions such that for any a, b ∈ R, if 1 − ab ∉ r.U (R), then a > a − aba.

Proof

$3)\Rightarrow 4)$, $5)\Rightarrow 5^{\prime })$ are trivial.

• 1) ⇒ 3)  By Theorem 1, there is a R-S-bimodule _R M _S satisfying condition 2) of Theorem 1. Since ann _M (a − a d a) = ann _M (a)⊕ann _M (1−d a), we know from the descending chain in condition 3) that for any n > 1, ann _M (a _n ) = ann _M (a ₀)⊕ann _M (1 − b ₀ a ₀) ⊕ ⋯ ⊕ann _M (1−b _n−1 a _n−1). So for any n > 1, $u.\dim (\text {ann}_{M}(a_{n}))\;=\;u.\dim (\text {ann}_{M}(a_{0}))\oplus u.\dim (\text {ann}_{M}(1-b_{0}a_{0}))\allowbreak +\;\cdots \;+\;u.\dim (\text {ann}_{M}(1-b_{n-1}a_{n-1}))\leq u.\dim (M)$, thus, all but finite $u.\dim (\text {ann}_{M}(1-b_{i}a_{i}))\;=\;0$. By condition 2) of Theorem 1, it means that 1 − b _i a _i ∈ U(R), thus 1 − a _i b _i ∈ U(R), so by [6, Lemma 3], all but finite a _i+1 R = a _i (1 − b _i a _i )R = a _i R. It proves 3).

• 1) ⇒ 2)  Similar to the $1)\Rightarrow 3)$, we can get $h.\text {length}(1_{R})\;\leq \;u.\dim (M)$.

  Let $A_{\overline {R}}$ be a maximal submodule of $\overline {R}_{\overline {R}}$. We will indicate in the following $2)\;\Rightarrow \;1)$ and $5^{\prime })\Rightarrow 1)$ that A is a direct summand. Since a ring is semisimple if and only if every maximal left ideal is a summand, we can get 1).

• 2) ⇒ 1)  First, for any b ∈ R, we have $h.\text {length}\;(b)\;\leq \;h.\text {length}(1)\;<\;\infty$. In fact, if b ∉ U(R), then 1 − 1(1−b) = b ∉ U(R). So by [6, Lemma 3], we get $1R\;\gneq \;(1\;-\;1(1\;-\;b)1)R\;=\;bR$, thus h.length(1) > h.length(b); If b ∈ U(R) and $bR\;\gneq \;(b\;-\;bdb)R$ for some d ∈ R, then (1 − b d) ∉ r.U(R), so $1R\;\gneq \;(1\;-\;bd)R\;=\;(1\;-\;bd)bR\;=\;(b\;-\;bdb)R$, thus h.length(1) > h.length(b − b d b), so h.length(1) ≥ h.length(b).

  Let ${\Phi }\;=\;\{b\;\in \;R| \overline {b}\;\notin \;A\}$, then 1 _R ∈ Φ. Since for any $b\;\in \;R, h.\text {length}(b)\leq h.\text {length}(1_{R})<\infty$, there exists an element b ∈ Φ such that h.length(b) = Min{h.length(b) | b ∈ Φ}. We have $\overline {\textit {bR}}\cap A\;=\;0$. In fact, for any $\overline {\textit {bx}}\in \overline {\textit {bR}}\cap \;A$, if $\overline {b\;-\;\textit {bxb}}\;\in \; A$, then $\overline {b}\;=\;\overline {(\textit {bx})b}+\overline {b\;-\;\textit {bxb}}\;\in \;A$, a contradiction. Whence $\overline {b\;-\;bxb}\;\in \;{\Phi }$. The conditions $h.\text {length}(b\;-\;\textit {bxb})\;\leqslant \; h.\text {length}(b)$ and h.length(b) = Min{h.length(b) | b∈ Φ} indicate that $\overline {\textit {bR}}\;=\;\overline {(b\;-\;\textit {bxb})R}$. [6, Lemma 3] shows 1−bx ∈ r.U(R). We can prove similarly that for any y ∈ R,1−b x y ∈ r.U(R), so b x ∈ J(R). Therefore $\overline {\textit {bR}}\cap A\;=\;0$. Since A is a maximal left ideal of $\overline {R}$, $\overline {R}\;=\;\overline {\textit {bR}}\oplus \;A$.

• 5′) ⇒ 1)  Write ${\Phi }\;=\;\{b\;\in \;R\;|\; \overline {b}\;\notin \;A\}$. By 5’), there is a minimum element b ∈ Φ. Similar to the proof of $2)\;\Rightarrow \;1)$, we can get $\overline {R}\;=\;\overline {\textit {Rb}}\;\oplus \;A$.

• 4) ⇒ 5′)  Define an order ≥ on R via a ≥ b if a = b or $aR\;\supsetneq \;bR$ and b = a − a d a for some d ∈ R. Then a routine verification shows that this order satisfying condition 5’).

• 5′) ⇒ 5)  By $5^{\prime })\;\Rightarrow \;1)$, we know that R is semilocal, thus R is direct finite, so r. U(R) = U(R).

□

Let R and S be rings and let _R M _S be a R-S-bimodule. If

• (i)  M _S has finite uniform dimension;

• (ii)  for r ∈ R, the equation ann _M (r) = (0) implies r ∈ U(R),

then we know that R is semilocal and $\dim (R/J(R))\;=\;h.\text {length}(1_{R})\leq u.\dim (M_{S})$. A natural question is:

When will $\dim (R/J(R))\;=\;u.\dim (M_{S})$?

We now give a characterization of this condition.

Proposition 1

Suppose that R is a semilocal ring with $\dim (R/J(R))\;=\;n, {}_{R}M_{S}$ is a R-S-bimodule satisfying the above conditions (i) and (ii). Let

$$a_{0}R\supsetneq a_{1}R\supsetneq a_{2}R\supsetneq {\cdots} \supsetneq a_{n}R\quad\text{ with ~}a_{i+1}=a_{i}-a_{i}b_{i}a_{i} $$

be a hollow chain of a ₀ = 1 _R .

Then $\dim (R/J(R))\;=\;u.\dim (M_{S})$ if and only if

• 1) ann _M (a _n ) is an essential submodule of M, and

• 2) for any b ∈ R, if R/bR is local, then ann _M (b) is uniform.

Proof

$``\Leftarrow "$ By [6, Proposition 8], R/(1−a _i b _i )R is local for each i. So by condition 2) and the equation $\text {ann}_{M}(a_{n})\;=\;\text {ann}_{M}(1-a_{0}b_{0}))\oplus {\cdots } \oplus \text {ann}_{M}(1-a_{n-1}b_{n-1})), u.\dim (\text {ann}_{M}(a_{n}))={\sum }_{i=0}^{n-1}u.\dim (\text {ann}_{M}(1-a_{i}b_{i}))=n,$ condition 1) shows $u.\dim (M)=u.\dim (\text {ann}_{M}(a_{n}))=n$.

$``\Rightarrow "$ a _i R≠a _i+1 R gives 1−a _i b _i ∉r.U(R)=U(R), thus 1−b _i a _i ∉U(R). Condition (ii) shows that ann _M (1−b _i a _i )≠0, whence $u.\dim (\text {ann}_{M}(1-b_{i}a_{i}))\geq 1$. So $n=u.\dim (M)\geq u.\dim (\text {ann}_{M}(a_{n}))={\sum }_{i=0}^{n-1} u.\dim (\text {ann}_{M}(1-b_{i}a_{i}))\geq n$.

Thus $u.\dim (\text {ann}_{M}(a_{n}))\;=\;n\;=\;u.\dim (M)$, whence ann _M (a _n ) is an essential submodule of M and each $u.\dim (\text {ann}_{M}(1-b_{i}a_{i}))\;=\;1$, i.e., ann _M (1−b _i a _i ) is uniform.

For any b ∈ R, suppose that R/b R is local, then b ∉ U(R). By [6, Proposition 8(1)], we know that $h.\text {length}(b)\;=\;h.\dim (R)\;-\;1\;=\;n\;-\;1$.

Therefore, we can find a hollow chain of b=b ₁

$$b_{1}R\supsetneq b_{2}R\supsetneq {\cdots} \supsetneq b_{n}R\quad \text{ with ~}b_{i+1}=b_{i}-b_{i}c_{i}b_{i}. $$

Write b ₀ = 1 _R and c ₀ = 1−b, then b ₁ = b = 1 − 1(1 − b)1 = b ₀ − b ₀ c ₀ b ₀. Since 1 − c ₀ b ₀ = b ∉ U(R), we have the following hollow chain of b ₀ = 1,

$$b_{0}R\supsetneq b_{1}R\supsetneq b_{2}R\supsetneq {\cdots} \supsetneq b_{n}R\quad \text{ with ~}b_{i+1}=b_{i}-b_{i}c_{i}b_{i}. $$

Then similar to the above proof, we can know that ann _M (1−c _i b _i ) is uniform, so ann _M (b) = ann _M (1−c ₀ b ₀) is uniform, as asserted. □

Theorem 3

Let R be a ring, a ∈ R. The following conditions are equivalent:

• 1) $\overline {aR}$ is a semisimple $\overline {R}-$ module;

• 2) $h.\text {length}(a)\;<;\infty$;

• 3) Every descending chain in aR

  $$a_{1}R\supseteq\cdots\supseteq a_{n}R\supseteq \cdots\quad \text{ with ~}a_{i+1}=a_{i}-a_{i}b_{i}a_{i}~(i=1,2,\dots) $$

  eventually terminates;

• 4) Every strict descending chain in aR

  $$a_{1}R\supsetneq {\cdots} \supsetneq a_{n}R\supsetneq \cdots\quad \text{ with ~}a_{i+1}=a_{i}-a_{i}b_{i}a_{i}~(i=1,2,\dots) $$

  eventually terminates;

• 5) Every descending chain in aR

  $$a_{1}R\supseteq \dots\supseteq a_{n}R\supseteq \cdots\quad \text{ with }~a_{1}=a ~\text{ and ~} a_{i+1}=a_{i}-a_{i}b_{i}a_{i}~(i=1,2,\dots) $$

  eventually terminates;

• 6) Every strict descending chain in aR

  $$a_{1}R\supsetneq {\cdots} \supsetneq a_{n}R\supsetneq \cdots\quad \text{with~ }a_{1}=a~ \text{ and ~} a_{i+1}=a_{i}-a_{i}b_{i}a_{i}~(i=1,2,\dots) $$

  eventually terminates.

Proof

$3)\Rightarrow 4)$, $3)\Rightarrow 5)$, $2)\Rightarrow 6)$, $4)\Rightarrow 6)$, $5)\Rightarrow 6)$ are trivial.

$1)\Rightarrow 3)$. Given a descending chain in a R

$$ aR\supseteq a_{1}R\supseteq a_{2}R\supseteq \cdots\supseteq a_{n}R\supseteq \cdots\quad \text{ with }~a_{i+1}=a_{i}-a_{i}b_{i}a_{i} $$

(1)

We have a descending chain in $\overline {aR}$

$$ \overline{a_{1}R}\supseteq \cdots\supseteq \overline{a_{n}R}\supseteq \cdots\quad \text{ with ~}\overline{a_{i+1}}=\overline{a_{i}-a_{i}b_{i}a_{i}}. $$

(2)

Since $\overline {aR}$ is semisimple and cyclic, $\overline {aR}$ is a direct sum of finite simple modules. So $\overline {aR}$ has a composition series of finite length. Thus, the descending chain (2) eventually terminates. By [6, Lemma 3], it means the chain (1) eventually terminates.

• 1) ⇒ 3)  is similar to the $1)\;\Rightarrow \;3)$.

• 6) ⇒ 1)  We first prove the following result.

Claim

If $B\;\lneq \overline {aR}$, then there exists a submodule C ≠ 0 of $\overline {aR}$ such that $B\;\cap C\;=\;0$.

Write ${\Phi }\;=\;\{b\;\in \;R\;|\; b\;=\;a\;-\;aya \text { for some } y\;\in \;R \text { and }\overline {b}\;\notin \;B\}$. a ∈ Φ shows Φ ≠ ∅.

Define an order ≤ on Φ via

$$x\leq y \text { if } x=y \text{ or }xR\supsetneq yR \text{ and }y=x-xdx \text{ for some }d\in R. $$

Then it is easy to show that “≤” is a partial order for Φ. Condition 6) shows that (Φ, ≤) is inductive, so by Zorn lemma, there is a maximum element b∈ Φ.

We have $\overline {bR}\cap B\;=\;0$. In fact, for any $\overline {bx}\in \overline {bR}\;\cap B$, we can get 1−b x ∈ r.U(R). Otherwise, we have $(b\;-\;bxb)R\;\lneq \;bR$. b ∈ Φ shows that b = a − a y a for some y ∈ R, so b − b x b = (a − a y a) − (a − a y a)x(a − a y a) = a − a[y + (1 − y a)x(1 − a y)]a. In addition, by $\overline {b}\;\notin \;B$ and $\overline {bx}\;\in \;B$, we know that $\overline {b\;-\;bxb}\;\notin \;B$, so b − b x b ∈ Φ, thus b − b x b > b, but b is a maximal element in Φ, a contradiction. For $\overline {bx}\;\in \; \overline {bR}\;\cap B$, repeating the argument, we see that 1 − b x y ∈ r.U(R) for all y ∈ R, so b x ∈ J(R), i.e., $\overline {bx}\;=\;\overline {0}$.

Since $\overline {b}\;\notin \;B$, $\overline {bR}\;\neq \;0$. Set $C\;=\;\overline {bR}\;\neq \;0$, then $B\;\cap C\;=\;0$. So the claim is proved.

Let A be a submodule of $\overline {aR}$. Suppose $A\;\lneq \;\overline {aR}$, if we can prove that A is a summand of $\overline {aR}$, then $\overline {aR}$ is semisimple.

Let X be the complement of A in $\overline {aR}$. If $A\;\oplus \;X\;\neq \;\overline {aR}$, applying the claim, we can find a $D\;\subseteq \;\overline {aR}$ such that D ≠ 0 and $(A\;\oplus \;X)\cap D\;=\;0$, and so $A\;\cap (X\oplus D)\;=\;0$, a contradiction. So $A\;\oplus X\;=\;\overline {aR}$.

□

Let S be a ring and M _S a module. Write End(M) = E and $\overline {E}\;=\; E/J(E)$. For any f ∈ E, as an application of Theorem 3, we can characterize the semisimplicity of $\overline {fE}$ by chain conditions of submodules of module M _S .

For f,g∈E, we proved in [6, Remark 4] that if f(M) = (f − f g f)(M), then 1 − f g is surjective. Besides, since $(f\;-\;fgf)(M)\;=\;f(M)\cap (1\;-\;fg)(M)$, we know that f(M) = (f − f g f)(M) if and only if 1 − f g is surjective. Since $\ker \;(f\;-\;fgf)\;=\;\ker \;f\;\oplus \ker (1-gf)$, and $\ker (1-gf)\cong \ker (1-fg)$, $\ker (f-fgf)=\ker f$ if and only if 1−f g is injective.

We can get from Theorem 3 the following

Proposition 2

Let M _S be a module and f∈E=End(M). If M is direct finite, then the following conditions are equivalent.

• (1)  $\overline {fE}$ is a semisimple $\overline {E}$ -module;

• (2)  Write f ₀ = f, for every sequence $g_{0},\;g_{1},\;g_{2},\dots$ of elements of End(M), if we set f _n+1 = f _n − f _n g _n f _n for every n ≥ 0, then the chains

  • (i)  $f_{0}(M)\supseteq f_{1}(M)\supseteq {\cdots } \supseteq f_{n}(M)\supseteq \cdots$ ,

  • (ii)  $\ker f_{0}\subseteq \ker f_{1}\subseteq {\cdots } \subseteq \ker f_{n}\subseteq \cdots$

  of submodules of M both eventually terminate.

By Theorem 3, we can get the following sufficient conditions for $\overline {fE}$ to be semisimple.

Proposition 3

Let M _S be a module and f∈E=End(M). Write f ₀ =f, for every sequence $g_{0}, g_{1}, g_{2},\dots$ of elements of End(M), setting f _n+1 =f _n −f _n g _n f _n for every n≥0, if the chains

• (i)  $f_{0}(M)\supseteq f_{1}(M)\supseteq {\cdots } \supseteq f_{n}(M)\supseteq \cdots$ ,

• (ii)  $\ker f_{0}\subseteq \ker f_{1}\subseteq {\cdots } \subseteq \ker f_{n}\subseteq \cdots$

of submodules of M both eventually terminate, then $\overline {fE}$ is a semisimple $\overline {E}$ -module.

The following result is a generalization of [6, Corollary 13], their proofs are similar.

Corollary 1

Let M _S be a module for which every monomorphism $M\rightarrow M$ splits. f ₀ =f∈E=End(M). The following conditions are equivalent

• (i)  $\overline {fE}$ is a semisimple $\overline {E}$ -module.

• (ii)  For every sequence $g_{0}, g_{1}, g_{2},\dots$ of elements of End(M), if we set f _n+1 =f _n −f _n g _n f _n for every n≥0, then the chain $\ker f_{0}\subseteq \ker f_{1}\subseteq {\cdots } \subseteq \ker f_{n}\subseteq \cdots$ of submodules of M eventually terminates.

Moreover, $h.\dim (fE)\leq \dim (M)$.

Corollary 2

Let M _S be a module for which every monomorphism in End(M) splits. Write $\overline {E}=E/J(E)$ . For any f∈E, if $M/\ker f$ has finite uniform dimension, then $\overline {fE}$ is a semisimple $\overline {E}$ -module.