Factorization of elements in noncommutative rings, I
We extend the classical theory of factorization in noncommutative integral domains to the more general classes of right saturated rings and right cyclically complete rings. Our attention is focused, in particular, on the factorizations of right regular elements into left irreducible elements. We stu...
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irk-123456789-1557402019-06-18T01:26:14Z Factorization of elements in noncommutative rings, I Facchini, A. Fassina, M. We extend the classical theory of factorization in noncommutative integral domains to the more general classes of right saturated rings and right cyclically complete rings. Our attention is focused, in particular, on the factorizations of right regular elements into left irreducible elements. We study the connections among such factorizations, right similar elements, cyclically presented modules of Euler characteristic 0 and their series of submodules. Finally, we consider factorizations as a product of idempotents. 2016 Article Factorization of elements in noncommutative rings, I / A. Facchini, M. Fassina // Algebra and Discrete Mathematics. — 2016. — Vol. 22, № 2. — С. 209-232. — Бібліогр.: 21 назв. — англ. 1726-3255 2010 MSC:16U30, 16B99. http://dspace.nbuv.gov.ua/handle/123456789/155740 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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We extend the classical theory of factorization in noncommutative integral domains to the more general classes of right saturated rings and right cyclically complete rings. Our attention is focused, in particular, on the factorizations of right regular elements into left irreducible elements. We study the connections among such factorizations, right similar elements, cyclically presented modules of Euler characteristic 0 and their series of submodules. Finally, we consider factorizations as a product of idempotents. |
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Facchini, A. Fassina, M. |
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Facchini, A. Fassina, M. Factorization of elements in noncommutative rings, I Algebra and Discrete Mathematics |
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Facchini, A. Fassina, M. |
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Facchini, A. |
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Factorization of elements in noncommutative rings, I |
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Factorization of elements in noncommutative rings, I |
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Factorization of elements in noncommutative rings, I |
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Factorization of elements in noncommutative rings, I |
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Factorization of elements in noncommutative rings, I |
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factorization of elements in noncommutative rings, i |
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Інститут прикладної математики і механіки НАН України |
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Factorization of elements in noncommutative rings, I / A. Facchini, M. Fassina // Algebra and Discrete Mathematics. — 2016. — Vol. 22, № 2. — С. 209-232. — Бібліогр.: 21 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT facchinia factorizationofelementsinnoncommutativeringsi AT fassinam factorizationofelementsinnoncommutativeringsi |
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2025-07-14T07:58:54Z |
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1837608400894033920 |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 22 (2016). Number 2, pp. 209–232
© Journal “Algebra and Discrete Mathematics”
Factorization of elements
in noncommutative rings, I
Alberto Facchini∗ and Martino Fassina
Communicated by R. Wisbauer
Abstract. We extend the classical theory of factorization in
noncommutative integral domains to the more general classes of right
saturated rings and right cyclically complete rings. Our attention is
focused, in particular, on the factorizations of right regular elements
into left irreducible elements. We study the connections among such
factorizations, right similar elements, cyclically presented modules
of Euler characteristic 0 and their series of submodules. Finally, we
consider factorizations as a product of idempotents.
1. Introduction
The study of factorizations has always given strong impulses to algebra
in its history. Modern commutative algebra was practically born in 1847,
when Gabriel Lamé announced at the Paris Académie des Sciences his
solution to Fermat’s Problem [18]. In his proof, he claimed that the ring
Z[ζp] is a UFD for every prime integer p, where ζp denotes a primitive
p-th root of the unity. He didn’t know that Ernst Kummer had proved
four years before that Z[ζ23] is not a UFD. This mistake made apparent
the necessity of a rigorous study of the subrings of C that are UFDs, that
is, the necessity of a rigorous foundation of commutative algebra.
∗The first author is partially supported by Università di Padova (Progetto ex 60%
“Anelli e categorie di moduli”) and Fondazione Cassa di Risparmio di Padova e Rovigo
(Progetto di Eccellenza “Algebraic structures and their applications”.)
2010 MSC: 16U30, 16B99.
Key words and phrases: Divisibility; Factorization; Right irreducible element.
210 Factorization of elements in noncommutative rings
Richard Dedekind, Gauss’s last student, said to his collaborators
around 1855 that the goal of number theory was to do for the general ring
of integers of an algebraic number field what Kummer had done for the
particular case of Z[ζp]. Dedekind completely succeeded in his programme
in 1871, and one of his main result was that each proper ideal of the ring
of integers of an algebraic number field can be factored in an essentially
unique way as the product of prime ideals.
In the classical noncommutative setting, the study of factorizations of
elements into irreducible elements in a noncommutative integral domain
and the theory of noncommutative UFDs [5], started by Asano and
Jacobson [16, pp. 33–36] for noncommutative PIDs, lead P. M. Cohn to
the discovery of the theory of free ideal rings and the study of factorization
in rings of noncommutative polynomials (see [6]).
The Auslander-Reiten theory and its almost split sequences were born
studying the factorizations of morphisms into irreducible morphisms [1,3].
(Recall that a morphism h between indecomposable modules is irreducible
if it is not invertible and, in every factorization h = βα of h, either α
is left invertible or β is right invertible. Irreducible morphisms are the
arrows of the Auslander-Reiten quiver.)
It is therefore natural to wonder if Cohn’s factorization theory can be
also extended to the case of any noncommutative ring R, not necessarily
an integral domain. This is what we begin to do in this paper.
Over an arbitrary ring R, a great simplification takes place when we
limit ourselves to the study of right regular elements a ∈ R (i.e., such
that a 6= 0 and a is not a left zerodivisor) and we restrict our study to the
factorizations of the right regular element a as a product of right regular
elements. In this paper we deal with this case, postponing the general
case of a not necessarily a right regular element to a further paper.
The study of factorizations of elements in noncommutative rings can
proceed in a number of different directions. Our attention is focused,
in particular, on the factorizations of right regular elements into left
irreducible elements. We study the connections among such factorizations,
right similar elements, cyclically presented modules of Euler characteristic
0 and their series of submodules. We finally consider factorizations as a
product of idempotents. Our main example is the case of factorization
of elements in the ring Mn(k) of n× n matrices with entries in a ring k.
When k is a division ring, the right regular elements of the ring Mn(k)
are the invertible matrices. Thus the study of factorizations of regular
elements in the ring Mn(k) is the noncommutative analogue of the study
of factorizations of invertible elements in a division ring k, for which all
A. Facchini, M. Fassina 211
factorizations are trivial. As far as singular matrices are concerned, every
such matrix is a product of idempotent matrices [17].
In the study of factorization in a noncommutative domain R, it is
a natural step to replace an element a ∈ R by the cyclically presented
right R-module R/aR. When R is commutative, R/aR is isomorphic to
R/bR if and only if a and b are associated. Also, when R is a domain,
the right R-modules R/aR and R/bR are isomorphic if and only if the
left R-modules R/Ra and R/Rb are isomorphic, so that the condition
is right-left symmetric. We generalize this point of view to the case of
an element a of an arbitrary ring R, considering, instead of a left factor
b ∈ R of a, the cyclic right R-module Rb/Ra.
In the continuation of this paper, we will essentially follow Cohn’s
“lattice method" [6, Section 5]. The factorizations of an element a of a
ring R will be described by the partially ordered set of all principal right
ideals of R between aR and RR itself.
For any subset X of a ring R, the left annihilator l. annR(X) is the
set of all r ∈ R such that rx = 0 for every x ∈ X. Similarly for the right
annihilator r. annR(X). We denote by U(R) the group of all invertible
elements of a ring R, that is, all the elements c ∈ R for which there exists
d ∈ R with cd = dc = 1.
2. Right regular elements, left irreducible elements
Let R be an associative ring with an identity 1 6= 0. We consider
factorizations of an element a ∈ R. If xy = 1, that is, if x ∈ R is right
invertible and y ∈ R is left invertible, then a has always the trivial
factorizations a = (ax)y and a = x(ya). If a, b are elements of R, we say
that a is a left divisor of b (in R), and write a|lb, if there exists an element
x ∈ R with ax = b. Similarly for right divisors, in which case we use the
symbol |r. Right invertible elements are left divisors of all elements of
R and left invertible elements are right divisors of all elements of R. An
element u ∈ R is right invertible if and only if u|l1.
For any ring R, the relation |l is a preorder on R, that is, a relation
on R that is reflexive and transitive. If a, b ∈ R, we will say that a and
b are left associates, and write a ∼l b, if a|lb and b|la. Clearly, ∼l is an
equivalence relation on the set R and the preorder |l on R induces a
partial order on the quotient set R/∼l. The equivalence class [1] of 1 ∈ R
in R/∼l is the least element of R/∼l and consists of all right invertible
212 Factorization of elements in noncommutative rings
elements of R. The equivalence class [0] of 0 ∈ R in R/∼l is the greatest
element of R/∼l and consists only of 0.
One has that a|lb if and only bR ⊆ aR, that is, if and only if the
principal right ideal generated by b is contained in the principal right ideal
generated by a. Thus the quotient setR/∼l is in one-to-one correspondence
with the set Lp(RR) of all principal right ideals of R. If R/∼l is partially
ordered by the partial order induced by |l as above and Lp(RR) is partially
ordered by set inclusion ⊆, then the one-to-one correspondence R/∼l →
Lp(RR) turns out to be an anti-isomorphism of partially ordered sets.
In particular, two elements of R are left associates if and only if they
generate the same principal right ideal of R.
Every element a ∈ R always has, among its left divisors, all right
invertible elements of R and all left associates with a. These are called
the trivial left divisors of a. If these are all the left divisors of a, a 6= 0
and a is not right invertible, then a is said to be a left irreducible element
of R.
Lemma 1. The following conditions are equivalent for an element a ∈ R:
(1) a is a left irreducible element.
(2) The right ideal aR is nonzero and is a maximal element in the set
Lp(RR) \ {0, R} of all nonzero proper principal right ideals of R.
Recall that an element a of a ring R is a left zerodivisor if it is nonzero
and there exists b ∈ R, b 6= 0 such that ab = 0, and is right regular if it
is 6= 0 and is not a left zerodivisor. Thus a ∈ R is right regular if and
only if ax = 0 implies x = 0 for every x ∈ R. For any element a ∈ R, left
multiplication by a is a right R-module homomorphism λa : RR → RR,
which is a monomorphism if and only if a is right regular. Notice that a
right ideal I of a ring R is isomorphic to RR if and only if it is a principal
right ideal of R generated by a right regular element of R. Similarly,
we define right zerodivisors and left regular elements. An element is a
zerodivisor if it is either a right zerodivisor or a left zerodivisor. An element
is regular if it is both right regular and left regular.
Lemma 2. A right regular element is right invertible if and only if it is
invertible.
Proof. Let a ∈ R be a right regular right invertible element. Then there
exists b ∈ R such that ab = 1. Then 0 = (ab − 1)a = a(ba − 1). Now a
right regular implies ba = 1, so a is also left invertible.
Lemma 3. Let R be a ring and S the set of all right regular elements
of R. Then:
A. Facchini, M. Fassina 213
(1) If a, b ∈ S, then ab ∈ S.
(2) If a, b ∈ R and ab ∈ S, then b ∈ S.
(3) Suppose that every principal right ideal of R generated by a right
regular element of R is essential in RR. Then a, b ∈ R and ab ∈ S
imply that a ∈ S and b ∈ S.
(4) If a ∈ S and I ⊆ J are right ideals of R, then J/I and aJ/aI are
isomorphic right R-modules.
Proof. The proofs of (1) and (2) are elementary.
For (3), assume that every principal right ideal of R generated by
a right regular element of R is essential in RR, and that a, b ∈ R and
ab ∈ S. Then b ∈ S by (2). Let us show that r. annR(a) ∩ bR = 0. If
bx ∈ R and bx ∈ r. annR(a), then abx = 0, so that x = 0. It follows that
r. annR(a)∩bR = 0. Since bR is essential in RR, we get that r. annR(a) = 0.
Equivalently, a ∈ S.
Part (4) follows immediately from the fact that left multiplication by
a ∈ S is a right R-module isomorphism RR → aR.
We will say that a ring R is a right saturated ring if every left divisor
of a right regular element is right regular, that is, if for every a, b ∈ R,
ab ∈ S implies a ∈ S.
Examples 1. (1) Every (not necessarily commutative) integral domain
is a right saturated ring. Every commutative ring is a (right) saturated
ring.
(2) By Lemma 3(3), if R is a ring and every principal right ideal of R
generated by a right regular element of R is essential in RR, then R is a
right saturated ring.
(3) Recall that a ring R is directly finite (or Dedekind finite) if every
right invertible element is invertible (equivalently, if every left invertible
element is invertible). Every right saturated ring is directly finite.
(4) Let R be a ring and suppose that the right R-module RR has
finite Goldie dimension. Then every principal right ideal of R generated
by a right regular element of R is essential in RR [21, Lemma II.2.3], so
that R is a right saturated ring (Example (2)).
(5) If R is a right nonsingular ring and every principal right ideal of R
generated by a right regular element ofR is essential in RR, then every right
regular element of R is left regular. In fact, let R be a right nonsingular
ring and suppose that every principal right ideal of R generated by a
right regular element of R is essential in RR. Let a ∈ R be a right regular
element. Suppose that xa = 0 for some x ∈ R. Then left multiplication
214 Factorization of elements in noncommutative rings
λx : RR → RR by x induces a right R-module morphism R/aR → RR.
Now aR is essential, so R/aR is singular [14, Proposition 1.20(b)], and
RR is nonsingular. Thus the right R-module morphism R/aR → RR is
the zero morphism. Hence x = 0, so that a is left regular.
The preorder |l on R induces a preorder on the set S of all right
regular elements of R. The set S contains all left invertible elements.
Lemma 4. Let a, b be two right regular elements of R. Then a ∼l b if
and only if b = au for some invertible element u ∈ R.
Proof. If a ∼l b, there exist u, v ∈ R such that au = b and bv = a. Thus
a(1 − uv) = 0. Since a is right regular, we get that u is right invertible
and v is left invertible. By symmetry, since b is also right regular, we get
that v is right invertible and u is left invertible. Thus u is invertible. The
converse is clear.
For the case where a, b are not right regular elements, we need a
further definition. Let a be an element of a ring R and I a subgroup of
the additive group R. We say that u is right invertible modulo I if there
exists v ∈ R such that uv − 1 ∈ I. Similarly, u is left invertible modulo I
if there exists v ∈ R such that vu− 1 ∈ I.
Notice that if u is right invertible modulo I and I is a right ideal, then
all the elements of the coset u+ I are also right invertible modulo I.
Lemma 5. Let a, b be elements of a ring R. Then a ∼l b if and only if
b = au for some element u ∈ R right invertible modulo r. annR(a).
Proof. If a ∼l b, there exist u, v ∈ R such that au = b and bv = a. Thus
a(1 − uv) = 0, so that u is right invertible modulo r. annR(a). Conversely,
suppose b = au for element u ∈ R right invertible modulo r. annR(a).
Then bR = auR ⊆ aR and there exists v ∈ R with a(1 − uv) = 0. Thus
aR = auvR ⊆ auR = bR, so aR = bR and a ∼l b.
Lemma 6. Let R be a right saturated ring. The following conditions are
equivalent for a right regular element a ∈ R that is not right invertible:
(1) a is left irreducible.
(2) For every factorization a = bc of the element a (b, c ∈ R), either b
is invertible or c is invertible.
Proof. (1) =⇒ (2) If a is right regular left irreducible and a = bc is a
factorization of a, then b is either right invertible or left associate with a.
Moreover, b and c are right regular (Lemma 3(3)). If b is right invertible,
A. Facchini, M. Fassina 215
then it is invertible by Lemma 2. If b is left associate with a, then there
exists an invertible element d ∈ R such that b = ad (Lemma 4), so that
a = bc = adc, hence 1 = dc. Thus c is invertible.
(2) =⇒ (1) Since a is right regular, we must have a 6= 0. Suppose
that (2) holds. In order to prove that a is left irreducible, we must show
that every left divisor b of a is trivial. Now, if b is a left divisor of a, there
is a factorization a = bc. By (2), either b is invertible, hence it is a trivial
divisor of a, or c is invertible, in which case b and a are left associates.
The preorder |l on the set S of all right regular elements induces a
partial order on the quotient set S/∼l. The equivalence class [1R] of 1R
in S/∼l is the least element of S/∼l. By Lemma 4, the equivalence class
[1R] of 1R in S/∼l consists of all invertible elements of R. Thus we can
consider the subset S∗ := S \ [1R] consisting of all right regular elements
of R that are not invertible in R.
The maximal elements in S∗/∼l are exactly the equivalence classes
modulo ∼l of the left irreducible elements of R modulo ∼l. Two elements
a, b ∈ S∗ are equivalent modulo ∼l if and only if a = bu for some invertible
element u ∈ R (Lemma 4).
3. Cyclically presented modules, right similar elements
We now pass from right regular elements of a ring R to cyclically
presented right modules over R. The proof of the following lemma is
elementary.
Lemma 7. The following conditions are equivalent for a right
R-module MR:
(1) There exists a right regular element a ∈ R such that MR
∼= RR/aR.
(2) There exists a short exact sequence of the form 0 → RR → RR →
MR → 0.
We call the modules satisfying the equivalent conditions of Lemma 7
cyclically presented modules of Euler characteristic 0. When R is an
IBN ring, so that the Euler characteristic of a module with a finite free
resolution is well defined, the modules satisfying the conditions of Lemma 7
are exactly the cyclic modules with a finite free resolution of length 1 of
Euler characteristic χ(MR) equal to 0.
Obviously, cyclically presented modules of Euler characteristic 0 are
cyclic modules. Take any other epimorphism RR → MR of such a module
216 Factorization of elements in noncommutative rings
MR, that is, fix any other generator of MR. Then the kernel of the epi-
morphism is the annihilator of the new generator of MR, and is a right
ideal I of R.
Lemma 8. Let R be a ring, a ∈ R a right regular element, and I a right
ideal of R such that R/aR ∼= R/I. Then:
(1) RR ⊕RR
∼= RR ⊕ I.
(2) I is a projective right ideal of R that can be generated by two
elements.
(3) I is isomorphic to the kernel of an epimorphism RR ⊕RR → RR.
(4) There exist elements b, c ∈ R such that bR = cR and IR is isomor-
phic to the submodule of RR ⊕ RR consisting of all pairs (x, y) ∈
RR ⊕RR such that bx = cy.
(5) There exists b ∈ R such that I = (aR : b) := {x ∈ R | bx ∈ aR }
and aR + bR = R. Moreover, r.annR(b) ⊆ I, and the two right
R-modules I/r.annR(b) and aR ∩ bR are isomorphic. Conversely,
if a, b ∈ R, a is right regular and aR+ bR = R, then R/(aR : b) is
a cyclically presented right R-module of Euler characteristic 0.
(Statement (5) appears in Cohn [7, Proposition 3.2.1]).
Proof. From R/aR ∼= R/I, it follows thatRR⊕RR
∼= RR⊕I by Schanuel’s
Lemma [1, p. 214]. This proves (1). Statements (2) and (3) follow imme-
diately from (1).
(4) By (3), I is isomorphic to the kernel of an epimorphism RR⊕RR →
RR. Every such an epimorphism is of the form (x, y) 7→ bx + cy with
bR+ cR = R. We can substitute −c for c, getting that every epimorphism
RR ⊕RR → RR is of the form (x, y) 7→ bx− cy, where b, c are elements
of R and bR + cR = R. The kernel of this epimorphism consists of all
pairs (x, y) ∈ RR ⊕RR such that bx = cy.
(5) Let ϕ : R/I → R/aR be an isomorphism and assume that ϕ(1 +
I) = b + aR. Surjectivity of ϕ gives aR + bR = R. Injectivity gives
that I = {x ∈ R | bx ∈ aR }. This proves the first part of (5). For
the second part, notice that r.annR(b) ⊆ (aR : b) = I. Finally, from
I = {x ∈ R | bx ∈ aR }, it follows that the mapping I → aR ∩ bR,
x → bx, is a well defined epimorphism with kernel r.annR(b), so that
I/r.annR(b) ∼= aR ∩ bR.
For the converse, let a, b be elements of R, with a right regular and
aR + bR = R. Then left multiplication by b induces an isomorphism
R/(aR : b) → R/aR.
A. Facchini, M. Fassina 217
The reason why we have introduced cyclically presented R-modules
in the study of factorizations of elements of R is the following. Suppose
that a = a1a2 . . . an is a factorization in R, where a, a1, . . . , an ∈ R.
Then aR = a1a2 . . . anR ⊆ a1a2 . . . an−1R ⊆ · · · ⊆ a1R ⊆ R is a se-
ries of principal right ideals of R, so that aR/aR = a1a2 . . . anR/aR ⊆
a1a2 . . . an−1R/aR ⊆ · · · ⊆ a1R/aR ⊆ R/aR is a series of submodules
of the cyclically presented right R-module R/aR. Similarly, Ra/Ra =
Ra1a2 . . . an/Ra ⊆ Ra2a3 . . . an/Ra ⊆ · · · ⊆ Ran/Ra ⊆ R/Ra is a series
of submodules of the cyclically presented left R-module R/Ra.
We would like the submodules a1a2 . . . aiR/aR in these series and the
factor modules
(a1a2 . . . ai−1R/aR)/(a1a2 . . . aiR/aR) ∼= a1a2 . . . ai−1R/a1a2 . . . aiR
to be cyclically presented modules as well. Unluckily, the situation is the
following:
Lemma 9. Let x, y be elements of a ring R. Then the right R-module
xR/xyR is cyclically presented if and only if there exist z, w ∈ R such
that xyR+ xwR = xR and, for every t ∈ R, one has xwt ∈ xyR if and
only if t ∈ zR.
Proof. The right R-module xR/xyR is cyclically presented if and only
if there exists z ∈ R with R/zR ∼= xR/xyR, i.e., if and only if there
exist z, w ∈ R such that left multiplication λxw : RR → xR/xyR is an
epimorphism with kernel zR. Now it is easy to conclude.
In the previous lemma, the situation is much simpler when x is right
regular. In this case, it suffices to take z := y and w := 1, and xR/xyR ∼=
R/yR always turns out to be cyclically presented.
Therefore, we will suppose that all the elements a1, a2, . . . , an ∈ R
are right regular, in which case a = a1a2 . . . an is also right regular
(Lemma 3(1)). Under this hypothesis, the right R-modules aR/aR =
a1a2 . . . anR/aR ⊆ a1a2 . . . an−1R/aR ⊆ · · · ⊆ a1R/aR ⊂ R/aR of the
series are cyclically presented of Euler characteristic 0, and so are the
factors a1a2 . . . ai−1R/a1a2 . . . aiR of the series.
Two elements a, b of an arbitrary ring R are said to be right similar if
the cyclically presented right R-modules R/aR,R/bR are isomorphic. If
two elements a, b of R are left associates, then they generate the same
principal right ideals aR, bR of R, hence they are clearly right similar.
218 Factorization of elements in noncommutative rings
4. Factorizations of right regular elements and right cycli-
cally complete rings
Let R be a ring and U(R) its group of units, that is, the group of all
elements with a (two-sided) inverse. The direct product U(R)n−1 of n− 1
copies of U(R) acts on the set of factorizations Fn(a) of length n > 1 of
an element a into right regular elements. Here a is an element of R, the
set of factorizations of length n of a into right regular elements is the set
of all n-tuples Fn(a) := { (a1, a2, . . . , an) | ai ∈ S, a1a2 . . . an = a }. In
particular, Fn(a) is empty if a is not right regular. The direct product
U(R)n−1 acts on the set Fn(a) via (u1, u2, . . . , un−1) · (a1, a2, . . . , an) =
(a1u1, u
−1
1 a2u2, u
−1
2 a3u3, . . . , u
−1
n−1an). We say that two factorizations of
a of length n,m respectively, are equivalent if n = m and the two factor-
izations are in the same orbit under the action of U(R)n−1.
In Section 3, we have associated with every factorization (a1, . . . , an)
of length n of an element a ∈ R into right regular elements, the series
aR/aR = a1 . . . anR/aR ⊆ a1 . . . an−1R/aR ⊆ · · · ⊆ a1R/aR ⊆ R/aR
of cyclically presented submodules of Euler characteristic 0 of the cyclically
presented right R-module R/aR. In this series, the factors of the series
are cyclically presented modules of Euler characteristic 0 as well.
Proposition 1. Let a be a right regular element of a ring R. Two fac-
torizations
(a1, a2, . . . , an) and (b1, b2, . . . , bm)
of a into right regular elements a1, . . . an, b1, . . . , bm are equivalent if and
only if their associated series of submodules of the cyclically presented
right R-module R/aR are equal.
(Two series 0 = A0 6 A1 6 A2 6 · · · 6 An = R/aR, 0 = B0 6 B1 6
B2 6 · · · 6 Bm = R/aR of R/aR are equal if n = m and Ai = Bi for
every i = 0, 1, 2, . . . , n.)
Proof. The two series
aR/aR = a1 . . . anR/aR ⊆ a1 . . . an−1R/aR ⊆ · · · ⊆ a1R/aR ⊆ R/aR
and
aR/aR = b1 . . . bmR/aR ⊆ b1 . . . bn−1R/aR ⊆ · · · ⊆ b1R/aR ⊆ R/aR
A. Facchini, M. Fassina 219
are equal if and only if n = m and a1a2 . . . aiR = b1b2 . . . biR for every
i = 1, 2, . . . , n− 1, that is, if and only if a1a2 . . . ai ∼l b1b2 . . . bi for every
i = 1, 2, . . . , n− 1.
Suppose a1a2 . . . ai ∼l b1b2 . . . bi for every i = 1, 2, . . . , n − 1. Then
there exist invertible elements u1, u2, . . . , un−1 ∈ R such that b1b2 . . . bi =
a1a2 . . . aiui (Lemma 4). Set u0 = un = 1. Let us prove by induction
on i that bi = u−1
i−1aiui for every i = 1, 2, . . . , n. The case i = 1 is
obvious. Suppose that (*) bi = u−1
i−1aiui for every i = 1, 2, . . . , j with
j < n. Let us show that bj+1 = u−1
j aj+1uj+1. Replacing in b1b2 . . . bj+1 =
a1a2 . . . aj+1uj+1 the equalities (*), which we assume to hold, we get that
u−1
0 a1u1u
−1
1 a2u2 . . . u
−1
j−1ajujbj+1 = a1a2 . . . aj+1uj+1,
so that ujbj+1 = aj+1uj+1. It follows that bj+1 = u−1
j aj+1uj+1, as desired.
This concludes the proof by induction. It is now clear that the two
factorizations (a1, a2, . . . , an), (b1, b2, . . . , bn) are in the same orbit.
Conversely, suppose that the factorizations (a1, . . . , an), (b1, . . . , bn) are
equivalent. Then there exist invertible elements u1, u2, . . . , un−1 ∈ R such
that bi = u−1
i−1aiui for every i = 1, 2, . . . , n, where u0 = un = 1. It follows
that b1b2 . . . biR = u−1
0 a1u1u
−1
1 a2u2 . . . u
−1
i−1aiuiR = a1a2 . . . aiR, and the
two series of cyclically presented submodules of R/aR coincide.
Remark 1. We have defined two factorizations of a into right regular
elements to be equivalent if one differs from the other by insertion of units.
Now suppose that the ring R is not directly finite, that is, there exist
noninvertible elements c, d ∈ R with cd = 1. Suppose that (a1, a2, . . . , an)
is a factorization of an element a ∈ R into right regular elements. Then
a = a1a2 . . . an = (a1c)(da2)a3 . . . an, but in this second factorization
the element a1c is not right regular, otherwise c would be right regular
(Lemma 3(2)). But c is right invertible, hence it would be invertible by
Lemma 2, contradiction. Thus, it is natural to define two factorizations
of a into right regular elements to be equivalent if one differs from the
other by insertion of units, because insertion by elements invertible only
on one side is not sufficient.
In the following, it will be often convenient to restrict our attention to
the rings for which the projective right ideals I of R, studied in Lemma 8,
are all necessarily principal right ideals generated by a right regular
element. We will call them right cyclically complete rings. Thus a ring R
is right cyclically complete if for every a, b ∈ R with a right regular and
aR+bR = R, the right ideal (aR : b) is a principal right ideal generated by
220 Factorization of elements in noncommutative rings
a right regular element. Equivalently, a ring R is right cyclically complete
if and only if, for every right regular element a ∈ R, the right annihilator
of any generator of R/aR is a principal right ideal generated by a right
regular element.
Examples 2. (1) If R is a projective free (that is, a ring for which every
projective ideal is free) IBN ring, then R is right cyclically complete. In
fact, if a, b ∈ R, a is right regular and aR+ bR = R, then R/(aR : b) is a
cyclically presented right R-module of Euler characteristic 0 (Lemma 8),
so that R/(aR : b) ∼= R/rR for some right regular element r ∈ R. By
Schanuel’s Lemma, R ⊕ (aR : b) ∼= R ⊕ rR ∼= R2
R, so that (aR : b)
is a finitely generated projective right ideal. But R is projective free,
hence (aR : b) ∼= Rn
R for some nonnegative integer n. The isomorphism
R ⊕ (aR : b) ∼= R2
R and R IBN imply that (aR : b) ∼= RR. Therefore
(aR : b) is a principal right ideal generated by a right regular element.
(2) Assume that RR cancels from direct sums, that is, if AR, BR are
arbitrary right R-modules and RR ⊕ AR
∼= RR ⊕ BR, then AR
∼= BR.
Then R is right cyclically complete. In fact, we can argue as in Example
(1), as follows. If a, b ∈ R, a is right regular and aR + bR = R, then
R/(aR : b) ∼= R/rR for some right regular element r ∈ R, so that
R⊕(aR : b) ∼= R2
R. ButRR cancels from direct sums, so that (aR : b) ∼= RR
is a principal right ideal generated by a right regular element.
(3) If R is any ring of stable range 1, then R is right cyclically complete.
In fact, if R is a ring of stable range 1, then RR cancels from direct sums
([10, Theorem 2] or [11, Theorem 4.5]), and we conclude by Example (2).
(4) If R is any semilocal ring, then R is right cyclically complete. In
fact, semilocal rings are of stable range 1 ([4] or [11, Theorem 4.4]). Thus
this Example (4) is a special case of Example (3). In particular, local
rings are right cyclically complete.
(5) Commutative Dedekind rings are right cyclically complete rings.
To see this, we can argue as in Examples (1) and (2). Let R be a Dedekind
ring and let a, b be elements of R, with a right regular and aR+ bR = R.
Then R⊕ (aR : b) ∼= R2
R. By [20, Lemma 6.18], we get that (aR : b) ∼= RR
is a principal right ideal generated by a right regular element.
(6) Right Bézout domains, that is, the (not necessarily commutative)
integral domains in which every finitely generated right ideal is principal,
are right cyclically complete rings. To prove this, let R be a right Bézout
domain and a, b be elements of R, with a right regular and aR+ bR = R.
Then, as in the previous examples, R⊕ (aR : b) ∼= R2
R, so that (aR : b) is
a finitely generated right ideal. But R is right Bézout, so that (aR : b)
is a principal right ideal of R. If (aR : b) = 0, then RR
∼= R2
R, so that
A. Facchini, M. Fassina 221
R has a nontrivial idempotent. But R is an integral domain, and this is
a contradiction, which proves that (aR : b) is a nonzero principal right
ideal of R. As R is an integral domain, the nonzero principal right ideal
(aR : b) of R is generated by a right regular element.
(7) Unit-regular rings are of stable range 1, and are therefore right
cyclically complete (Example (3)).
(8) Every commutative ring is right cyclically complete. In fact, if
a, b ∈ R, a is right regular and aR+ bR = R, then R/(aR : b) ∼= R/aR, so
that R/(aR : b) and R/aR have the same annihilators, that is, (aR : b) =
aR. The same proof shows that every right duo ring is right cyclically
complete.
(9) We will now give an example of a ring R that is not right cyclically
complete. Let Vk be a vector space of countable dimension over a field
k, and let v0, v1, v2, . . . be a basis of Vk. Set R := End(Vk). Notice that
an element a ∈ R is a right regular element of R if and only if it is
an injective endomorphism of Vk. For instance, the element a ∈ R such
that a : vi 7→ v2i for every i > 0 is a right regular element of R. Apply
the exact functor Hom(RVk,−) : Mod-k → Mod-R to the split exact
sequence 0 → Vk
a
−→ Vk → Vk → 0, getting a split exact sequence
0 → RR
λa−→ RR → RR → 0. Then R/aR ∼= RR. To show that R is not
right cyclically complete, it suffices to prove that RR has a generator
whose right annihilator is not a principal right ideal generated by a right
regular element. As a generator of RR, consider the element x ∈ R such
that x : vi 7→ vi−1 for every i > 1 and x : v0 7→ 0. The endomorphism x of
Vk is surjective, hence split surjective, i.e., right invertible, so that it is a
generator of RR. Its right annihilator is r. annR(x) = { y ∈ R | xy = 0 } =
{ y ∈ R | im(y) ⊆ ker(x) = v0k } = Hom(Vk, v0k). This is a projective
cyclic right ideal of R, which contains no right regular elements of R,
because no element of Hom(Vk, v0k) is an injective mapping. This proves
that R is not a right cyclically complete ring.
Lemma 10. A ring R is right cyclically complete if and only if, for
every right ideal I of R with R/I a cyclically presented module of Euler
characteristic 0, I is a principal right ideal generated by a right regular
element.
Lemma 11. Let 0 → A → B → C → 0 be a short exact sequence of right
modules over a right cyclically complete ring R. If A and C are cyclically
presented modules of Euler characteristic 0 and B is cyclic, then B is a
cyclically presented module of Euler characteristic 0.
222 Factorization of elements in noncommutative rings
Proof. Since B is cyclic, we have that B ∼= R/I for some right ideal
I of R. Then there exists a right ideal J of R with I ⊆ J , C ∼= R/J
and A ∼= J/I. Now C ∼= R/J is a cyclically presented module of Euler
characteristic 0 and R is cyclically complete, so that J = rR is the
principal right ideal generated by a right regular element r of R. Thus
left multiplication λr : RR → RR by r is a monomorphism, which induces
an isomorphism λr : RR → rR = J . If K is the inverse image of I ⊆ J
via this isomorphism, then K is a right ideal of R and rK = I. Then
A ∼= J/I = rR/rK ∼= R/K. But A is a cyclically presented module of
Euler characteristic 0 and R is cyclically complete, so K = sR is the
principal right ideal generated by a right regular element s of R. Then
I = rK = rsR, so that B ∼= R/I = R/rsR is a cyclically presented
module of Euler characteristic 0.
The following Proposition is motivated by [7, Proposition 0.6.1].
Proposition 2. Let R be a right cyclically complete ring. Let
0 = M0 ⊆ M1 ⊆ . · · · ⊆ Mn = MR (1)
be a finite series of submodules of a right R-module MR. Assume that all
the factors M/Mi (i = 0, 1, 2, . . . , n) are cyclically presented R-modules
of Euler characteristic 0. Let a ∈ R be any right regular element such
that M ∼= RR/aR. Then there exists a factorization a = a1a2 . . . an of
a with a1, a2, . . . , an ∈ R right regular elements, such that Mj/Mi
∼=
R/an−j+1an−j+2 . . . an−i−1an−i−2R for every 0 6 i < j 6 n. In partic-
ular, all the modules Mj/Mi (0 6 i < j 6 n) are cyclically presented
R-module MR of Euler characteristic 0.
Proof. Let a ∈ R be a right regular element such that M ∼= RR/aR, so
that there exists an epimorphism ϕ : RR → MR with kerϕ = aR. By
the Correspondence Theorem applied to the epimorphism ϕ, there is
a one-to-one correspondence between the set L of all right ideals of R
containing aR and the set L′ of all submodules ofMR. The correspondence
is given by IR 7→ ϕ(IR) for every IR ∈ L and its inverse is given by
NR 7→ ϕ−1(NR) for every NR ∈ L′. Thus if Ki := ϕ−1(Mi), then the
finite series of right ideals of R corresponding to the series (1) is the series
aR = kerϕ = K0 ⊆ K1 ⊆ . · · · ⊆ Kn = RR. If ϕi := ϕ|Ki
: Ki → Mi
denotes the restriction of ϕ, then ϕi is an epimorphism with kernel aR,
so that Mi
∼= Ki/aR for every i = 0, 1, 2, . . . , n. If 0 6 i 6 j 6 n,
the composite mapping of ϕj : Kj → Mj and the canonical projection
A. Facchini, M. Fassina 223
Mj → Mj/Mi is an epimorphism Kj → Mj/Mi with kernel Ki, so that
Kj/Ki
∼= Mj/Mi. (2)
For j = n, we get in particular that RR/Ki
∼= M/Mi for every i =
0, 1, . . . , n. Thus, the modules RR/Ki are cyclically presented of Euler
characteristic 0, so that the right ideals Ki are of our type, so that they
are isomorphic to RR because R is right cyclically complete. Thus we
have right regular elements c0 := a, c1, c2 . . . , cn−1, cn := 1 of R such
that Ki = ciR for every i = 0, 1, . . . , n. Now K0 ⊆ K1 ⊆ . · · · ⊆ Kn =
RR, so that ci = ci+1bi+1 for suitable elements b1, b2 . . . , bn ∈ R. These
elements bi are right regular by Lemma 3(2). Then, for 0 6 i < j 6 n,
we get that ci = ci+1bi+1 = ci+2bi+2bi+1 = ci+3bi+3bi+2bi+1 = · · · =
cjbjbj−1 . . . bi+2bi+1. In particular, for i = 0 and j = n, we have that
a = c0 = cnbnbn−1 . . . b2b1 = bnbn−1 . . . b2b1.
Finally, from (2), we get that
Mj/Mi
∼= Kj/Ki = cjR/ciR = cjR/cjbjbj−1 . . . bi+2bi+1R
∼= R/bjbj−1 . . . bi+2bi+1R
by Lemma 3(4). Now change the notation, orderly substituting the se-
quence a1, a2, . . . , an for the sequence bn, bn−1, . . . , b1.
5. Correspondence between factorizations and series of
submodules
Let R be a ring and S be the set of all right regular elements of R.
Let S := ˙⋃
n>1S
n be the disjoint union of the sets Sn of all n-tuples of
elements of S, i.e., the cartesian product of n copies of S. The set S can
be seen as the set of all factorizations of finite length into right regular
elements. Let MR be the class of all series of finite length of cyclically
presented right R-modules of Euler characteristic 0:
MR := { (M1,M2, . . . ,Mn) | n > 1, 0 = M0 6 M1 6 M2 6 · · · 6 Mn,
and the modulesMi andMi/Mi−1 are cyclically presented rightR-modules
of Euler characteristic 0 for every i = 1, 2, . . . , n}.
Two series (M1,M2, . . . ,Mn), (M ′
1,M
′
2, . . . ,M
′
m) ∈ MR of finite length
are isomorphic if n = m and there is a right R-module isomorphism
ϕ : Mn → M ′
m such that ϕ(Mi) = M ′
i for every i = 1, 2, . . . , n − 1. In
this case, we will write (M1,M2, . . . ,Mn) ∼= (M ′
1,M
′
2, . . . ,M
′
m), so that ∼=
224 Factorization of elements in noncommutative rings
turns out to be an equivalence relation on the class MR, and the quotient
class MR/∼= has a set of representatives modulo ∼=.
In the first paragraph of Section 4, we have considered, for any a ∈ R,
the set Fn(a) := { (a1, a2, . . . , an) | ai ∈ S, a1a2 . . . an = a } of factoriza-
tions of length n of a into right regular elements. For every fixed right
regular element a ∈ R, let S(a) be the disjoint union of the sets Fn(a),
n > 1, so that S(a) is a subset of S. Similarly, we can consider the set
MR(R/aR) := { (M1,M2, . . . ,Mn) ∈ MR | Mn = R/aR }.
It is the set of all finite series of submodules of R/aR. There is a mapping
f : S(a) → MR(R/aR)
defined by
f(a1, a2, . . . , an) = (a1a2 . . . an−1R/aR, . . . , a1R/aR,R/aR)
for every (a1, a2, . . . , an) ∈ S(a).
Proposition 3. Let a be a right regular element of a ring R.
(1) If R is right cyclically complete, then the mapping f : S(a) →
MR(R/aR) is surjective.
(2) Two factorizations in S(a) are mapped to the same element of
MR(R/aR) via f if and only if they are equivalent factorizations
of a.
Proof. (1) Let (M1,M2, . . . ,Mn) be an element of MR(R/aR). Then
M := Mn = R/aR, and all the modules M/Mi are cyclically presented
modules of Euler characteristic 0 by Lemma 11. Thus we can apply Propo-
sition 2. Following its proof, we can take as ϕ : RR → M the canonical
projection, so that Mi = Ki/aR, Ki = ciR for suitable right regular
elements ci, with ci = ci+1an−i, where a = a1a2 . . . an and a1, . . . , an ∈ R
are right regular elements of R. Thus f(a1, . . . , an) = (M1, . . . ,Mn).
(2) is Proposition 1.
If a and b are right similar right regular elements of a right cyclically
complete ring R, there is an isomorphism f : R/aR → R/bR, which is
defined by left multiplication by an element c ∈ R. This bijection f = λc
induces a bijection MR(R/aR) → MR(R/bR), in which every series of
submodules of R/aR is mapped to an isomorphic series of submodules of
R/bR. Since R is right cyclically complete, this bijection MR(R/aR) →
A. Facchini, M. Fassina 225
MR(R/bR) induces a bijection S(a)/∼= → S(b)/∼= (Proposition 3), where,
for two factorizations (a1, a2, . . . , an) and (a′
1, a
′
2, . . . , a
′
n) in S(a), we write
(a1, a2, . . . , an) ∼= (a′
1, a
′
2, . . . , a
′
n) if the two factorizations are equivalent
according to the definition given in the first paragraph of Section 4. Fix
(a1, a2, . . . , an) ∈ S(a). The factorization a = a1a2 . . . an of a corresponds
to the series of submodules
0 =
aR
aR
6
a1 . . . an−1R
aR
6
a1 . . . an−2R
aR
6 · · · 6
a1R
aR
6
R
aR
of R/aR, which is mapped isomorphically by λc to the series
0 =
caR+ bR
bR
6
ca1 . . . an−1R+ bR
bR
6
ca1 . . . an−2R+ bR
bR
6 . . .
· · · 6
ca1R+ bR
bR
6
cR+ bR
bR
=
R
bR
of submodules of R/bR. As far as the factors are concerned, it follows
that R/(ca1 . . . aiR + bR) ∼= R/a1 . . . aiR. Set d0 := 1 and dn := b. For
i = 1, . . . n− 1, let di be a right regular element such that ca1 . . . aiR +
bR = diR. Such a di exists because R is right cyclically complete. Then
a1 . . . aiR ⊆ a1 . . . ai−1R, so that 0 6= diR ⊆ di−1R. Hence there exists
bi ∈ R with di = di−1bi, and bi is right regular, i = 1, 2, . . . , n. Then b =
dn = dn−1bn = dn−2bn−1bn = dn−3bn−2bn−1bn = · · · = d0b1b2 . . . bn =
b1b2 . . . bn is the factorization of b corresponding to the factorization
a = a1a2 . . . an of a, up to equivalence of factorizations.
The set S(a) can obviously be partially ordered by the refinement
relation 6. The partially ordered set (S(a),6) has no maximal elements,
because
(a1, a2, . . . , an) 6 (a1, a2, . . . , ai, 1, ai+1, . . . , an).
If we want to relate maximal elements of S(a) with respect to the refine-
ment relation 6 and factorizations of a as a product of left irreducible
right invertible elements, we must exclude from the factorizations into
right regular elements left invertible factors (recall that left invertible
elements are always right regular). Thus, let T be the set of all elements
of R that are right regular but not left invertible, so that T ⊆ S. Set
T (a) := { (a1, a2, . . . , an) | n > 1, ai ∈ T, a1a2 . . . an = a } ⊆ S(a). The
set T (a) is also partially ordered by the refinement relation 6. We leave
to the reader the proof of the following easy lemma.
226 Factorization of elements in noncommutative rings
Lemma 12. Let a be an element of a right saturated ring. An element
(a1, a2, . . . , an) ∈ T (a) is a maximal element of T (a) with respect to
the refinement relation 6 if and only if ai is left irreducible for every
i = 1, 2, . . . , n.
In particular, if R is a right saturated ring and a ∈ R, then T (a) has a
maximal element if and only if a has a factorization as a product of finitely
many right regular left irreducible elements that are not left invertible.
Such a factorization is very rarely unique up to equivalence, that is,
T (a) modulo equivalence has very rarely a unique maximal element. For
instance, if R is a commutative ring, (a1, . . . , an) is a maximal element of
T (a) and σ ∈ Sn is a permutation, then (aσ(1), . . . , aσ(n)) is a maximal
element of T (a) as well.
Recall that a nonzero element a of an integral domain R is rigid
[8, p. 43] if a = bc = b′c′ implies b ∈ b′R or b′ ∈ bR. More generally,
this definition can be extended to any element a of any ring R. Thus
we say that an element a of a ring R is right rigid if the principal right
ideals between aR and R form a chain under inclusion. When the ring
R is a right cyclically complete ring and a ∈ R is right regular, then
the principal right ideals between aR and R form a finite chain under
inclusion if and only if a has a unique factorization into right regular left
irreducible elements up to equivalence of factorizations.
6. Idempotents
Particularly interesting, in study of factorizations in rings with zero-
divisors and the uniqueness of such factorizations, is the case of idempo-
tents. In fact, consider the example of matrices. In [9], J. A. Erdos showed
that singular matrices over commutative fields factorize as a product of
idempotent matrices. This result was later extended to matrices over
euclidean rings and division rings [17]. Fountain [13] considered the case
of commutative Hermite rings, using techniques of semigroup theory. In-
spired by this paper, Ruitenburg [19] studied the case of noncommutative
Hermite rings. Fountain and Ruitenberg determined a clear connection
between product decompositions of singular matrices into idempotents
and product decompositions of invertible matrices into elementary ones.
If an element x ∈ R is a product x = e1 . . . en of idempotents ei ∈ R,
then x is annihilated both by left multiplication by 1 − e1 and by right
multiplication by 1 − en [12, Section 2], so that, whenever x ∈ R is a
product of finitely many idempotents, we must necessarily have that either
x = 1, or both l. ann(x) 6= 0 and r. ann(x) 6= 0.
A. Facchini, M. Fassina 227
If e ∈ R is an idempotent and we consider the factorizations e = xy
of e as a product of two elements x, y ∈ R, then e always has the trivial
factorizations e = u(ve), e = (eu)v and e = (eu)(e), where u, v ∈ R
are any two elements with uv = 1. The composition factors of these
factorizations e = xy, i.e., the composition factors R/xR and xR/xyR of
the series eR = xyR ⊆ xR ⊆ R are 0 and R/eR in all these three types
of factorizations. We will say that e is an irreducible idempotent if these
are the only factorizations of e as a product of two elements x and y in
R and e 6= 1. For example:
Proposition 4. Let k be a division ring, Vk a right vector space over k of
finite dimension n > 2, R := End(Vk) its endomorphism ring and ϕ ∈ R
an idempotent endomorphism. Then ϕ is is an irreducible idempotent of
R if and only if dim(kerϕ) = 1.
Proof. If dim(kerϕ) = 0, then ϕ is the identity, so that it is not an
irreducible idempotent.
If dim(kerϕ) > 2, then there is a nontrivial direct-sum decomposition
kerϕ = A ⊕ B, so that Vk = A ⊕ B ⊕ im(ϕ). Then ϕ is the composite
mapping ϕ = ψω of the endomorphism ω of Vk that is zero on A and the
identity on B ⊕ im(ϕ) and the endomorphism ψ that is zero on B and
the identity on A⊕ im(ϕ).
Now suppose dim(kerϕ) = 1 and that ϕ decomposes as ϕ = ψω. Then
kerω ⊆ kerϕ and imϕ ⊆ imψ, so that dim(kerω) 6 1 and dim(imψ) >
n− 1. If dim(kerω) = 0, then ω is an automorphism and the factorization
ϕ = ψω is trivial. If dim(imψ) = n, then ψ is an automorphism and
the factorization ϕ = ψω is also trivial. Hence it remains to consider
the factorizations ϕ = ψω with dim(kerω) = 1 and dim(imψ) = n − 1.
Now dim(kerω) = dim(kerϕ) = 1 implies that kerω = ker(ψω), so that
kerψ ∩ imω = 0. As dim(imω) = n − 1 and dim(kerψ) = 1, we get
that Vk = kerψ ⊕ imω. In matrix notation, it follows that ω : Vk =
kerϕ ⊕ imϕ → Vk = kerψ ⊕ imω is of the form ω =
(0 0
0 f
)
, where
f : imϕ → imω is an isomorphism, and ψ : Vk = kerψ ⊕ imω → Vk =
kerϕ ⊕ imϕ is of the form ψ =
(0 0
0 g
)
, where g is the inverse of f . Let
u : Vk = kerϕ⊕ imϕ → Vk = kerψ ⊕ imω be the isomorphism u =
(h 0
0 f
)
,
where h is any isomorphism between the one-dimensional vector spaces
kerϕ and kerψ. Then ω = uϕ and ψ = ϕu−1, so that the factorization
ϕ = ψω is also trivial in this case.
Laffey proved that every singular n× n matrix with entries in a di-
vision ring k can be expressed as a product of idempotents over k [17].
228 Factorization of elements in noncommutative rings
Clearly, every idempotent ε in R := End(Vk) is a product of t irreducible
idempotents, where t = dim(ker ε). It follows that every noninvertible ele-
ment of R is a product of finitely many irreducible idempotents. As far as
uniqueness of such a decomposition is concerned, assume that ϕ ∈ R can
be written as ϕ = ε1 . . . εm = e1 . . . et, where the εi and the ej are all irre-
ducible idempotents. To determine the factors ε1 . . . εi−1R/ε1 . . . εiR and
e1 . . . ej−1R/e1 . . . ejR of the series of principal right ideals corresponding
to the two factorizations of ϕ, we need the following lemma.
Lemma 13. Let k be a division ring, Vk a right vector space over k
of finite dimension n, R := End(Vk) its endomorphism ring, g ∈ R an
endomorphism of Vk and e ∈ R an irreducible idempotent. Then exactly
one of the following two cases occurs:
(a) dim(ker(ge)) = dim(ker g) and gR/geR = 0; or
(b) dim(ker(ge)) = dim(ker g) + 1 and gR/geR is a simple right
R-module.
Notice that R is a simple artinian ring, so that all its simple right
R-modules are isomorphic.
Proof. We have that dim(im(ge)) = dim(g(eV )) = dim(g(eV + ker g)).
Now eV +ker g contains eV , which has dimension n−1 and is contained in
V , that has dimension n. Hence either eV +ker g has dimension n−1, and
in this case eV + ker g = eV and ker g ⊆ eV , or eV + ker g has dimension
n, in which case eV + ker g = V and ker g 6⊆ eV . Let us distinguish these
two cases:
(a) Case eV + ker g = V and ker g 6⊆ eV . In this case, dim(im(ge)) =
dim(g(eV + ker g)) = dim(g(V )) = dim(im(g)), so that dim(ker(ge)) =
dim(ker g).
(b) Case ker g ⊆ eV . Since g induces a lattice isomorphism be-
tween the lattice of all subspaces of Vk that contain ker g and the lat-
tice of all subspaces of im(g) and for every subspace W of Vk with
W ⊇ ker g we have dim g(W ) = dim(W ) − dim(ker g), it follows that
dim(im(ge)) = dim(g(eV )) = dim(eV ) − dim(ker g) = n− 1 − dim(ker g),
so that dim(ker(ge)) = dim(ker g) + 1.
Now in both cases, left multiplication by g is a right R-module epi-
morphism RR → gR, which induces a right R-module epimorphism
R/eR → gR/geR. But R/eR is a simple right R-module, so that ei-
ther gR/geR = 0 or gR/geR is simple. Hence, in order to conclude the
proof of the lemma, it suffices to show that gR/geR = 0 if and only if
dim(ker(ge)) = dim(ker g).
A. Facchini, M. Fassina 229
Now if gR/geR = 0, then gR = geR, so that g ∈ geR. Hence
there exists h ∈ R with g = geh. Then im(g) = im(geh) ⊆ im(ge) ⊆
im(g). Thus im(ge) = im(g), from which dim(im(ge)) = dim(im(g)), and
dim(ker(ge)) = dim(ker g).
Conversely, suppose dim(ker(ge)) = dim(ker g). Then dim(im(ge)) =
dim(im(g)) and im(ge) ⊆ im(g), so im(ge) = im(g). It follows that eV +
ker g = V . Let x be an element in ker g not in eV , so that eV ⊕ xk = V .
With respect to this direct-sum decomposition the matrices of e and g
are, respectively,
g =
(
g11 0
g21 0
)
and g =
(
1 e12
0 e22
)
.
Let h ∈ R be the endomorphism of V with matrix
h =
(
1 −e12
0 1
)
.
It is easily seen that g = geh, so that gR = geR, and gR/geR = 0.
Proposition 5. Let k be a division ring, Vk a right vector space over k of
finite dimension n,R := End(Vk) its endomorphism ring and f = e1 . . . em
a product decomposition of an endomorphism f ∈ R into irreducible
idempotents e1, . . . , em of R. Let t be the dimension of the kernel of f .
Then t of the factors e1 . . . ei−1R/e1 . . . eiR are simple R-modules, and
the other m− t are zero.
Proof. Induction of m. If m = 1, then the kernel of f = e1 is 1, so that
t = m = 1, and the unique factor R/e1R is a simple right R-module.
Suppose m > 1 and that the proposition is true for endomorphisms
that are products of m − 1 irreducible idempotents. Suppose that the
dimension of the kernel of f = e1 . . . em is t. Set g := e1 . . . em−1, so
that the m factors e1 . . . ei−1R/e1 . . . eiR relative to the factorization f =
e1 . . . em are the m−1 factors relative to the factorization g := e1 . . . em−1
plus the module gR/gemR. By the previous lemma, we have one of the
following two cases:
(a) dim(ker(ge)) = dim(ker g) and gR/geR = 0. In this case, t =
dim(ker g), so that, by the inductive hypothesis, t factors relative to the
factorization g := e1 . . . em−1 are simple, and all the other m− t factors
relative to factorization f = e1 . . . em are zero.
(b) dim(ker(ge)) = dim(ker g) + 1 and gR/geR is a simple right
R-module. In this case, dim(ker g) = t − 1, so that by the inductive
230 Factorization of elements in noncommutative rings
hypothesis t − 1 factors relative to the factorization g := e1 . . . em−1
are simple and the other m − t are zero. The other factor relative to
factorization f = e1 . . . em is the simple module gR/geR.
Thus if ϕ = ε1 . . . εm = e1 . . . et are two factorizations with the idem-
potents εi and ej irreducible idempotents, then the series of principal
right ideals associated to the two factorizations have the same number of
simple factors and all the other factors are zero. This remark justifies the
following definitions.
Let R be any ring and a = a1 . . . an a factorization in R, where
a, a1, . . . , an ∈ R. We call
aR = a1 . . . anR ⊆ a1 . . . an−1R ⊆ · · · ⊆ a1R ⊆ R
the series of principal right ideals of R associated to the factorization and
R/a1R, a1R/a1a2R, . . . , a1a2 . . . an−1R/a1a2 . . . anR
the factors of the series. We call right length of the factorization the
number of nonzero factors. Similarly, we can consider the series
Ra = Ra1a2 . . . an ⊆ Ra2a3 . . . an ⊆ · · · ⊆ Ran ⊆ R
of principal left ideals of R associated to the factorization, and define its
factors Rai . . . an/Rai−1 . . . an and the left length of the factorization. In
our last example of this first paper, we show that the right length and
the left length of a factorization can be different.
Example 1. Let R be the ring
(
Q 0
R R
)
. For every element r =
(
q 0
α β
)
, the
principal right ideal generated by r is
rR =
(
q 0
α β
)(
1 0
0 0
)
R+
(
q 0
α β
)(
0 0
0 1
)
R =
=
(
q 0
α 0
)
R+
(
0 0
0 β
)
R =
(
q 0
α 0
)
Q + β
(
0 0
R R
)
so that rR is the improper right ideal R if and only if qβ 6= 0, and rR
is maximal among the proper principal right ideals if and only if either
β 6= 0 and q = 0, or β = 0 and q 6= 0. Thus the elements r with qβ 6= 0
are right invertible, and those with q = 0 or β = 0 but not both q and
β equal to zero are the left irreducible elements of R. For the elements
r with q = β = 0 and α 6= 0, the series rR ⊂
(
0 0
R R
)
⊂ R is a series of
A. Facchini, M. Fassina 231
principal right ideals that cannot be properly refined, so that the proper
left divisors of r are exactly the generators of the principal right ideal
(
0 0
R R
)
, that is, the elements s =
(
0 0
α′ β′
)
with β′ 6= 0. The factorizations
of r in this case are
r =
(
0 0
α 0
)
=
(
0 0
α′ β′
)(
a 0
b 0
)
, (3)
where a ∈ Q and b ∈ R are any two numbers with α′a + β′b = α. The
factorization (3) has right length 2 and its right factors are the simple
module R/
(
0 0
R R
)
and the nonsimple module
(
0 0
R R
)
/
(
0 0
αQ 0
)
.
Let us compute the left length of the factorization (3). It is easily seen
that, for r =
(
q 0
α β
)
, the principal left ideal generated by r is
Rr = q
(
Q 0
R 0
)
+ R
(
0 0
α β
)
.
The series of left principal ideals associated to the factorization (3) of
r = ( 0 0
α 0 ) is
Rr ⊆ R
(
a 0
b 0
)
⊆ R.
Now Rr = R ( 0 0
α 0 ) =
(
0 0
R 0
)
and R
(
a 0
b 0
)
= a
(
Q 0
R 0
)
+ R
(
0 0
b 0
)
, so that we
have two cases according to when a = 0 or a 6= 0:
(a) If a = 0, then b 6= 0, so that R
(
a 0
b 0
)
=
(
0 0
R 0
)
= Rr, and the
factorization (3) of r has left length 1.
(b) If a 6= 0, then R
(
a 0
b 0
)
=
(
Q 0
R 0
)
⊃ Rr, so that the factorization (3)
of r has left length 2.
In our next paper, we will treat the factorizations of arbitrary elements
of a ring, including the case of zerodivisors.
References
[1] F. W. Anderson and K. R. Fuller, “Rings and Categories of Modules”, Second
Edition, GTM 13, Springer-Verlag, New York, 1992.
[2] M. Auslander, Almost split sequences, I, Proceedings of the International Conference
on Representations of Algebras (Carleton Univ., Ottawa, Ont., 1974), Paper No. 1,
8 pp., Carleton Math. Lecture Notes, No. 9, Carleton Univ., Ottawa, Ont., 1974.
[3] M. Auslander and I. Reiten, Almost split sequences, II, Proceedings of the Interna-
tional Conference on Representations of Algebras (Carleton Univ., Ottawa, Ont.,
1974), Paper No. 2, 13 pp., Carleton Math. Lecture Notes, No. 9, Carleton Univ.,
Ottawa, Ont., 1974.
232 Factorization of elements in noncommutative rings
[4] H. Bass, “K-theory and stable algebra”, Publ. Math. I. H. E. S. 22 (1964), 5–60.
[5] P. M. Cohn, Noncommutative unique factorization domains, Trans. Amer. Math.
Soc. 109 (1963), 313–331.
[6] P. M. Cohn, Unique factorization domains, Amer. Math. Monthly 80 (1973), 1–18.
[7] P. M. Cohn, “Free rings and their relations”, Second Edition, London Math. Soc.
Monographs 19, Academic Press, London-New York, 1985.
[8] P. M. Cohn, “Free ideal rings and localization in general rings”, New Mathematical
Monographs 3, Cambridge Univ. Press, Cambridge, 2006.
[9] J. A. Erdos, On products of idempotent matrices, Glasg. Math. J. 8 (1967), 118–122.
[10] E. G. Evans, Jr., Krull-Schmidt and cancellation over local rings, Pacific J. Math.
46 (1973), 115–121.
[11] A. Facchini, “Module Theory. Endomorphism rings and direct sum decompositions
in some classes of modules”, Progress in Mathematics 167, Birkäuser Verlag, Basel,
1998. Reprinted in Modern Birkhäuser Classics, Birkhäuser Verlag, Basel, 2010.
[12] A. Facchini and A. Leroy, Elementary matrices and products of idempotents, Linear
Multilinear Algebra 64 (2016), 1916–1935.
[13] J. Fountain, Products of idempotent integer matrices, Math. Cambridge Philos.
Soc. 110 (1991), 431–441.
[14] K. Goodearl, “Ring theory. Nonsingular rings and modules", Marcel Dekker, Inc.,
New York-Basel, 1976.
[15] K. R. Goodearl, “von Neumann regular rings”, Second edition, Robert E. Krieger
Publishing Co., Inc., Malabar, FL, 1991.
[16] N. Jacobson, “The Theory of Rings”, Amer. Mat. Soc., New York, 1943.
[17] T. J. Laffey, Products of idempotent matrices, Linear and Multilinear Algebra 14
(1983), 309–314.
[18] G. Lamé, Demonstration generale du Theoreme de Fermat, sur l’impossibilite,
en nombres entier, de l’equation xn + yn = zn, Compte rendue des Séances de
L’Académie des Sciences, Séance du lundi 1er mars 1847.
[19] W. Ruitenburg, Products of idempotent matrices over Hermite domains, Semigroup
Forum 46 (1993), 371–378.
[20] D. W. Sharpe and P. Vámos, “Injective modules", Cambridge Univ. Press, Cam-
bridge, 1972.
[21] B. Stenström, “Rings of quotients”, Springer-Verlag, New York-Heidelberg, 1975.
Contact information
A. Facchini Dipartimento di Matematica, Universitá di
Padova, 35121 Padova, Italy
E-Mail(s): facchini@math.unipd.it
M. Fassina Department of Mathematics, University of Illi-
nois at Urbana-Champaign, 61801 Urbana, IL,
USA
E-Mail(s): fassina2@illinois.edu
Received by the editors: 11.01.2016
and in final form 09.02.2016.
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