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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 R-algebra which is a direct limit of module finite R-algebras
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 R-algebra which is locally
a direct limit of module-finite algebras, is strongly clean if R is a π-regular
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 module-finite
R-algebra 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 R-algebra A which is a direct limit of module-finite algebras and
for each integer n ≥ 1, the matrix ring Mn(A) is strongly clean;
(3) Each R-algebra A which is a direct limit of module-finite 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.
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2 FRANC ¸OIS COUCHOT
Proof. (1) ⇒ (2). Let A be a direct limit of module-finite R-algebras and
a ∈ Mn(A). Then R[a] is a commutative module-finite R-algebra. 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 module-finite R-algebra 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 semi-perfect. 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 R-algebra A which is locally a direct limit of module-finite algebras
and for each integer n ≥ 1, Mn(A) is strongly clean;
(3) Each R-algebra A which is locally a direct limit of module-finite algebras is
clean.
Proof. (1) ⇒ (2). Let A be an R-algebra which is locally a direct limit of
module-finite 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 module-finite 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 module-finite R-algebra. It follows that A is
clean. By Theorem 1 RP is Henselian.
?
A ring R is said to be strongly π-regular 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 π-regular commutative ring. Then, for each
R-algebra A which is locally a direct limit of module-finite 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 π-regular 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 π-regular 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 π-regular commutative ring. Then, for each
R-algebra A which is locally a direct limit of module-finite algebras and for each
integer n ≥ 1, the matrix ring Mn(A) is strongly π-regular.
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STRONG CLEANNESS OF MATRIX RINGS3
Proof. Let S = Mn(A) and s ∈ S. Then R[s] is locally a module-finite algebra.
It is easy to prove that each prime ideal of R[s] is maximal. Consequently R[s] is
strongly π-regular. So, S is strongly π-regular 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 R-module 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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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
E-mail address: couchot@math.unicaen.fr
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