Strong cleanness of matrix rings over commutative rings
ABSTRACT Let $R$ be a commutative local ring. It is proved that $R$ is Henselian if and only if each $R$algebra which is a direct limit of module finite $R$algebras is strongly clean. So, the matrix ring $\mathbb{M}_n(R)$ is strongly clean for each integer $n>0$ if $R$ is Henselian and we show that the converse holds if either the residue class field of $R$ is algebraically closed or $R$ is an integrally closed domain or $R$ is a valuation ring. It is also shown that each $R$algebra which is locally a direct limit of modulefinite algebras, is strongly clean if $R$ is a $\pi$regular commutative ring.

Article: A Survey of Strongly Clean Rings
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ABSTRACT: Let R be an associative ring with identity. An element a∈R is called clean if a=e+u with e an idempotent and u a unit of R and a is called strongly clean if, in addition, eu=ue. Aring R is called clean if every element of R is clean and R is strongly clean if every element of R is strongly clean. In the paper [Nicholson and Zhou, Clean rings: a survey, Advances in Ring Theory, 181–198, World Sci. Pub., Hackensack, NJ, 2005], the authors brought out an up to date account of the results in the study of clean rings. Here, we give an account of the results on strongly clean rings.Acta Applicandae Mathematicae 01/2009; 108(1):157173. · 0.99 Impact Factor
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arXiv:0804.1221v1 [math.RA] 8 Apr 2008
STRONG CLEANNESS OF MATRIX RINGS OVER
COMMUTATIVE RINGS
FRANC ¸OIS COUCHOT
Abstract. Let R be a commutative local ring. It is proved that R is Henselian
if and only if each Ralgebra which is a direct limit of module finite Ralgebras
is strongly clean. So, the matrix ring Mn(R) is strongly clean for each integer
n > 0 if R is Henselian and we show that the converse holds if either the residue
class field of R is algebraically closed or R is an integrally closed domain or
R is a valuation ring. It is also shown that each Ralgebra which is locally
a direct limit of modulefinite algebras, is strongly clean if R is a πregular
commutative ring.
As in [10] a ring R is called clean if each element of R is the sum of an idempotent
and a unit. In [8] Han and Nicholson proved that a ring R is clean if and only if
Mn(R) is clean for every integer n ≥ 1. It is easy to check that each local ring
is clean and consequently every matrix ring over a local ring is clean. On the
other hand a ring R is called strongly clean if each element of R is the sum
of an idempotent and a unit that commute. Recently, in [12], Chen and Wang
gave an example of a commutative local ring R with M2(R) not strongly clean.
This motivates the following interesting question: what are the commutative local
rings R for which Mn(R) is strongly clean for each integer n ≥ 1? In [4], Chen,
Yang and Zhou gave a complete characterization of commutative local rings R with
M2(R) strongly clean. So, from their results and their examples, it is reasonable
to conjecture that the Henselian rings are the only commutative local rings R with
Mn(R) strongly clean for each integer n ≥ 1. In this note we give a partial answer
to this problem. If R is Henselian then Mn(R) is strongly clean for each integer
n ≥ 1 and the converse holds if R is an integrally closed domain, a valuation ring
or if its residue class field is algebraically closed.
All rings in this paper are associative with unity. By [11, Chapitre I] a com
mutative local ring R is said to be Henselian if each commutative modulefinite
Ralgebra is a finite product of local rings. It was G. Azumaya ([1]) who first
studied this property which was then developed by M. Nagata ([9]). The following
theorem gives a new characterization of Henselian rings.
Theorem 1. Let R be a commutative local ring. Then the following conditions are
equivalent:
(1) R is Henselian;
(2) For each Ralgebra A which is a direct limit of modulefinite algebras and
for each integer n ≥ 1, the matrix ring Mn(A) is strongly clean;
(3) Each Ralgebra A which is a direct limit of modulefinite algebras is clean.
2000 Mathematics Subject Classification. Primary 13H99, 16U99.
Key words and phrases. clean ring, strongly clean ring, local ring, Henselian ring, matrix ring,
valuation ring.
1
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2FRANC ¸OIS COUCHOT
Proof. (1) ⇒ (2). Let A be a direct limit of modulefinite Ralgebras and
a ∈ Mn(A). Then R[a] is a commutative modulefinite Ralgebra. Since R is
Henselian, R[a] is a finite direct product of local rings. So R[a] is clean. Hence a is
a sum of an idempotent and a unit that commute.
It is obvious that (2) ⇒ (3).
(3) ⇒ (1). Let A be a commutative modulefinite Ralgebra and let J(A) be its
Jacobson radical. Since J(R)A ⊆ J(A), where J(R) is the Jacobson radical of R,
we deduce that A/J(A) is semisimple artinian. By [10, Propositions 1.8 and 1.5]
idempotents can be lifted modulo J(A). Hence A is semiperfect. It follows that A
is a finite product of local rings, whence R is Henselian.
?
Let P be a ring property. We say that an algebra A over a commutative ring R
is locally P if AP satisfies P for each maximal ideal P of R.
Corollary 2. Let R be a commutative ring. Then the following conditions are
equivalent:
(1) R is clean and locally Henselian;
(2) For each Ralgebra A which is locally a direct limit of modulefinite algebras
and for each integer n ≥ 1, Mn(A) is strongly clean;
(3) Each Ralgebra A which is locally a direct limit of modulefinite algebras is
clean.
Proof. (1) ⇒ (2). Let A be an Ralgebra which is locally a direct limit of
modulefinite algebras and a ∈ Mn(A). Consider the following polynomial equa
tions: E + U = a, E2= E, UV = 1, V U = 1, EU = UE. By Theorem 1 these
equations have a solution in Mn(AP), for each maximal ideal P of R. So, by [5,
Theorem I.1] they have a solution in Mn(A) too.
It is obvious that (2) ⇒ (3).
(3) ⇒ (1). Let P be a maximal ideal of R and let A be a modulefinite RP
algebra. Since R is clean, the natural map R → RP is surjective by [5, Theorem
I.1 and Proposition III.1]. So A is a modulefinite Ralgebra. It follows that A is
clean. By Theorem 1 RP is Henselian.
?
A ring R is said to be strongly πregular if, for each r ∈ R, there exist s ∈ R
and an integer q ≥ 1 such that rq= rq+1s.
Corollary 3. Let R be a strongly πregular commutative ring. Then, for each
Ralgebra A which is locally a direct limit of modulefinite algebras and for each
integer n ≥ 1, the matrix ring Mn(A) is strongly clean.
Proof. It is known that R is clean and that each prime ideal is maximal. So, for
every maximal P, PRP is a nilideal of RP. Hence RP is Henselian. We conclude
by Corollary 2.
?
By [6, Th´ eor` eme 1] each strongly πregular R satisfies the following condition:
for each r ∈ R, there exist s ∈ R and an integer q ≥ 1 such that rq= srq+1.
Moreover, by [3, Proposition 2.6.iii)] each strongly πregular ring is strongly clean.
So, Corollary 3 is also a consequence of the following proposition. (Probably, this
proposition is already known).
Proposition 4. Let R be a strongly πregular commutative ring. Then, for each
Ralgebra A which is locally a direct limit of modulefinite algebras and for each
integer n ≥ 1, the matrix ring Mn(A) is strongly πregular.
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STRONG CLEANNESS OF MATRIX RINGS3
Proof. Let S = Mn(A) and s ∈ S. Then R[s] is locally a modulefinite algebra.
It is easy to prove that each prime ideal of R[s] is maximal. Consequently R[s] is
strongly πregular. So, S is strongly πregular too.
?
The following lemma will be useful in the sequel.
Lemma 5. Let R be a commutative local ring with maximal ideal P. Let n be
an integer > 1 such that Mn(R) is strongly clean. Let f be a monic polynomial of
degree n with coefficients in R such that f(0) ∈ P and f(a) ∈ P for some a ∈ R\P.
Then f is reducible.
Proof. Let A ∈ Mn(R) such that its characteristic polynomial is f, i.e. f =
det(XIn− A), where In is the unit element of Mn(R). Then A = E + U where
E is idempotent, U is invertible and EU = UE. First we assume that a = 1. So,
0 and 1 are eigenvalues of A the reduction of A modulo P. Consequently A and
A − Inare not invertible. It follows that E ?= Inand E ?= 0n,nwhere 0p,q is the
p × q matrix whose coefficients are 0. Let F be a free Rmodule of rank n and let
ǫ be the endomorphism of F for which E is the matrix associated with respect to
some basis. Then F = Im ǫ ⊕ Ker ǫ. Moreover Im ǫ and Ker ǫ are free because R
is local. Consequently there exists a n × n invertible matrix Q such that:
QEQ−1= B =
?Ip
0q,p
0p,q
0q,q
?
where p is an integer such that 0 < p < n and q = n−p. Since E and A commute,
then B and QAQ−1commute too. So, QAQ−1is of the form:
QAQ−1=
?C
0q,p
0p,q
D
?
where C is a p×p matrix and D is a q×q matrix. We deduce that f is the product of
the characteristic polynomial g of C with the characteristic polynomial h of D. Let
us observe that (C−Ip) and D are invertible. So, g(1) / ∈ P, h(0) / ∈ P, g(0) ∈ P and
h(1) ∈ P. Now suppose that a ?= 1. Then a−nf(X) = g(Y ) where Y = a−1X and
g is a monic polynomial of degree n. We easily check that g(1) ∈ P and g(0) ∈ P.
It follows that g is reducible, whence f is reducible too.
?
A commutative ring R is a valuation ring (respectively arithmetic) if its
lattice of ideals is totally ordered by inclusion (respectively distributive).
Theorem 6. Let R be a local commutative ring with maximal ideal P and with
residue class field k. Consider the following two conditions:
(1) R is Henselian;
(2) the matrix ring Mn(R) is strongly clean ∀n ∈ N∗.
Then (1) ⇒ (2) and the converse holds if R satisfies one of the following properties:
(a) k is algebraically closed;
(b) R is an integrally closed domain;
(c) R is a valuation ring.
Proof. By Theorem 1 it remains to prove that (2) implies (1) when one of
(a), (b) or (c) is valid. We will use [2, Theorem 1.4] and [7, Theorem II.7.3.(iv)].
Consider the polynomial f = Xn+ cn−1Xn−1+ ··· + c1X + c0and assume that
∃m, 1 ≤ m < n such that cm / ∈ P and ci∈ P, ∀i < m. Since c0∈ P, we see that
f(0) ∈ P.
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4FRANC ¸OIS COUCHOT
Hence, if k is algebraically closed, ∃a ∈ R \ P such that f(a) ∈ P. By Lemma 5
f is reducible. So, by [2, Theorem 1.4] R is Henselian.
If R is an integrally closed domain, we take m = n−1 for proving the condition
(iv) of [7, Theorem II.7.3]. In this case f(−cn−1) ∈ P. By Lemma 5 (possibly
applied several times) f satisfies the condition (iv) of [7, Theorem II.7.3]. Hence R
is Henselian.
Assume that R is a valuation ring. Let N be the nilradical of R and let R′=
R/N. We know that R is Henselian if and only if R′is Henselian too. For each
n ∈ N∗, Mn(R′) is strongly clean. Since R′is a valuation domain, R′is integrally
closed. It follows that R′and R are Henselian.
?
Corollary 7. Let R be an arithmetic commutative ring. Then the following con
ditions are equivalent:
(1) R is clean and locally Henselian;
(2) the matrix ring Mn(R) is strongly clean ∀n ∈ N∗.
Proof. By Corollary 2 it remains to show (2) ⇒ (1). Let P be a maximal ideal
of R. Since R is clean the natural map R → RP is surjective by [5, Theorem I.1
and Proposition III.1]. So, Mn(RP) is strongly clean ∀n ∈ N∗. Theorem 6 can be
applied because RP is a valuation ring. We conclude that RP is Henselian.
?
The following generalization of [4, Theorem 8] holds even if the properties
(a),(b),(c) of Theorem 6 are not satisfied.
Theorem 8. Let R be a local commutative ring with maximal ideal P and with
residue class field k. Let p be an integer such that 2 ≤ p ≤ 5. Then the following
conditions are equivalent:
(1) Mn(R) is strongly clean ∀n, 2 ≤ n ≤ p;
(2) each monic polynomial f of degree n, 2 ≤ n ≤ p, for which f(0) ∈ P and
f(1) ∈ P, is reducible.
Proof. By Lemma 5 it remains to prove that (2) ⇒ (1). Let A ∈ Mn(R).
We denote by f the characteristic polynomial of A. If A is invertible then A =
0n,n+ A. If A − Inis invertible then A = In+ (A − In). So, we may assume that
A and (A − In) are not invertible. It follows that f(0) ∈ P and f(1) ∈ P. Then,
f = gh where g and h are monic polynomials of degree ≥ 1. We may assume that
g(0) ∈ P, g(1) / ∈ P, h(0) / ∈ P and h(1) ∈ P (possibly by applying condition (2)
several times). We denote by¯f, ¯ g,¯h the images of f, g, h by the natural map
R[X] → k[X]. If ¯ g and¯h have a common factor of degree ≥ 1 then this factor
is of degree 1 because n ≤ 5. In this case ∃a ∈ R \ P such that g(a) ∈ P and
h(a) ∈ P. As in the proof of Lemma 5 we show that g is reducible. Hence, after
changing g and h, we get that ¯ g and¯h have no common divisor of degree ≥ 1. It
follows that there exist two polynomials u and v with coefficients in R such that
¯ u¯ g + ¯ v¯h = 1. Since PR[A] is contained in the Jacobson radical of R[A], we may
assume that u(A)g(A)+v(A)h(A) = In. We put e = vh. Then we easily check that
e(A) is idempotent. It remains to show that (A − e(A)) is invertible. It is enough
to prove that (¯ A − ¯ e(¯A)) is invertible because PMn(R) is the Jacobson radical of
Mn(R). Let V be a vector space of dimension n over k and let B be a basis of
V . Let α be the endomorphism of V for which¯A is the matrix associated with
respect to B. We put ǫ = ¯ e(α). Since V has finite dimension, it is sufficient to
show that (α − ǫ) is injective. Let w ∈ V such that α(w) = ǫ(w). It follows that
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STRONG CLEANNESS OF MATRIX RINGS5
α(ǫ(w)) = ǫ(α(w)) = ǫ2(w) = ǫ(w). Since ¯ e is divisible by (X −¯1) we get that
ǫ(w) = 0. So, α(w) = 0. We deduce that ǫ(w) = w because ¯ e −¯1 is divisible by X.
Hence w = 0.
?
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Laboratoire de Math´ ematiques Nicolas Oresme, CNRS UMR 6139, D´ epartement de
math´ ematiques et m´ ecanique, 14032 Caen cedex, France
Email address: couchot@math.unicaen.fr