Modules over braces
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irk-123456789-1573882019-06-21T01:26:37Z Modules over braces Rump, W. 2006 Article Modules over braces / W. Rump // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 2. — С. 127–137. — Бібліогр.: 5 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 81R50. http://dspace.nbuv.gov.ua/handle/123456789/157388 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Modules over braces |
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Modules over braces |
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Modules over braces |
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Modules over braces |
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Modules over braces |
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modules over braces |
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Modules over braces / W. Rump // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 2. — С. 127–137. — Бібліогр.: 5 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT rumpw modulesoverbraces |
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2025-07-14T09:49:32Z |
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2025-07-14T09:49:32Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2006). pp. 127 – 137
c© Journal “Algebra and Discrete Mathematics”
Modules over braces
Wolfgang Rump
Communicated by Yu. A. Drozd
How to solve the QYBE? Construct a cycle set!
How to fix cycle sets? Put them into braces!
To the best of our knowledge, a brace can be used to connect different
things, to give them stability. In ancient times, bracelets, worn round the
arm, also have been precious. The mathematical structures mentioned in
the title have those two features. Like an incomplete ring, they embrace
associativity and left distributivity, refounding both properties into a
single equation. It is this equation that makes the unclosed ring precious
in turning it into a group under the circle operation.
Precisely, a brace is an abelian group A with a right distributive mul-
tiplication such that the circle operation a ◦ b := ab + a + b makes A into
a group, the adjoint group A◦ of A. With respect to a derived opera-
tion, every brace is a cycle set [4]. Cycle sets arose in connection with
set-theoretical solutions of the quantum Yang-Baxter equation (QYBE).
Such a solution can be regarded as a map R : X ×X → X ×X satisfying
the QYBE property
R12R13R23 = R23R13R12
in X × X × X, where the superscripts refer to the components where R
acts upon. Explicitly, R can be written in the form R(x, y) = (xy, xy).
If R is unitary and left non-degenerate, i. e. R21R = 1 and x 7→ xy is
invertible, the inverse map x 7→ y · x satisfies
(x · y) · (x · z) = (y · x) · (y · z)
2000 Mathematics Subject Classification: 81R50.
Key words and phrases: quantum Yang-Baxter equation, radical ring, cycle
set, brace.
128 Modules over braces
for all x, y, z ∈ X. The sets X with such a (left invertible) binary op-
eration are called cycle sets. By [4], Proposition 1, cycle sets are in
one-to-one correspondence with left non-degenerate unitary solutions R
of the QYBE. Unitary solutions R also correspond to symmetric sets in
the sense of [1]. We will show (Proposition 2) that a cycle set can be
embedded into a brace if and only if the corresponding solution R is
also right non-degenerate [4]. For the cycle set X, this means that every
element admits a unique square root.
Usually, there are many sub-cycle-sets X of a given brace A, hence
many solutions of the QYBE. If X generates A as an abelian group,
we call X a cycle base of A. There is a kind of Galois theory between
ideals of A and quotient cycle sets of X [5]. A further instance of this
correspondence is proved in Theorem 1.
Basic properties of braces are developed in §1. By an observation of
[1], a brace A can be described as a bijective 1-cocycle A◦ → A between
the group A◦ and the right A◦-module A. Every brace A contains a largest
radical subring A0, and A0 = A if and only if A is left distributive. We
show that modules over a brace form an abelian category and discuss
some delicate points in ideal theory. In §3, we consider square-free cycle
bases X of a brace A, i. e. those whose elements x satisfy x2 = 0. We
show that every element a ∈ A determines a unique element c(a) with
c(a)2 = 0, so that the order of c(a) in A◦ coincides with its order in the
abelian group A.
There is a class of Artin-Schelter regular rings of arbitrary global
dimension n, also called quantum binomial algebras [3], which give rise
to binomial semigroups [2], hence to square-free cycle sets. In [4] we
proved the conjecture [2] that the converse is true, i. e. that each finite
square-free cycle set arises in this way. Since a square-free cycle set X is
non-degenerate, we can regard it as a cycle base of a brace A. Then the
mentioned result is equivalent to the statement that the adjoint group
A◦ of A cannot be transitive on X, unless X is a singleton. If A◦ is a
p-group, we will prove that A/A2 cannot be cyclic if A2 6= 0.
1. Braces and radical rings
Let A be an abelian group together with a right distributive multiplica-
tion, that is,
(a + b)c = ac + bc (1)
for all a, b, c ∈ A. We call A a brace if the circle operation
a ◦ b := ab + a + b. (2)
W. Rump 129
makes A into a group. This group will be called the adjoint group A◦ of
A. The associativity of A◦ is easily seen to be equivalent to the equation
a(bc + b + c) = (ab)c + ab + ac (3)
which looks like a combination of the associative and the left distributive
law. In fact, if A is left distributive, then Eq. (3) boils down to the
associative law, so that A becomes a radical ring (i. e. an associative ring
which is a group under the circle operation).
Since the maps a 7→ a ◦ b are bijective, the same is true for a 7→ ab :=
ab + a. The inverse of a 7→ ab will be denoted by a 7→ b · a, i. e.
(b · a)b = b · ab = a. (4)
(To avoid confusion, we use no dot for the ordinary multiplication in A.)
If we set a = b = 0 in (1), we get 0c = 0. Inserting b = c = 0 in (3)
therefore yields a0 = (a0)0+a0+a0, whence (a0)0 = 0, and thus a0 = 0.
So we have
a0 = 0a = 0. (5)
In particular, this shows that 0 is the neutral element of the adjoint group
A◦, like in the theory of radical rings. The inverse of a ∈ A in A◦ will be
denoted by a′. Thus
aa′ + a + a′ = a′a + a + a′ = 0. (6)
Using (4), the inverse a′ can be determined explicitly:
a′ = −(a · a). (7)
In fact, (−a · a) ◦ a = −(a · a)a + a = 0.
The similarity between braces and rings can be extended to module
theory. We define a module over a brace A to be an abelian group M
together with a right operation M ×A → M such that the following hold
for x, y ∈ M and a, b ∈ A.
(x + y)a = xa + ya
x(a ◦ b) = (xa)b + xa + xb
x0 = 0.
(8)
The second equation now states that a ◦ b operates on M like ab + a + b
in End(M). This gives a new interpretation of Eq. (3) and its module
theoretic generalization. But there is yet another way to look at this
fundamental equation. As above, we may abbreviate
xa := xa + x (9)
130 Modules over braces
for x ∈ M and a ∈ A. Then the second equation in (8) takes the simple
form
(xa)b = xa◦b. (10)
This means that the adjoint group A◦ operates on M , where x0 = x is
granted by the third equation of (8). In particular, it follows that the
map x 7→ xa is bijective for all a ∈ A. So the axioms (8) just state
that M is a right module over the adjoint group A◦. By lack of the left
distributivity, there are no left modules over braces.
As a special case, a brace A is a module over itself. Using (9), Eq.
(2) can be written in the form
b − a ◦ b + ab = 0. (11)
Therefore, a brace A can be defined to be a group A◦ together with a right
A◦-module A and a bijection A◦ → A (the identical map) which satisfies
the 1-cocycle condition (11). It is natural to ask when this 1-cocycle is
a boundary. This happens if and only if there is an element a ∈ A with
b = a − ab, i. e. b = −ab, for all b ∈ A. By Eq. (6) with b := a′, this
implies that a = 0, and thus A = 0. In other words, homologically trivial
braces are trivial.
Most of the basic concepts of ring and module theory carry over to
braces without change, but they have to be handled with care. For ex-
ample, if f : A → B is a morphism between braces, i. e. a map which
respects addition and multiplication, then f(A) is a subbrace of B, i. e.
a subset which is closed under addition and multiplication. The kernel
of f is an ideal of A, that is, an additive subgroup I such that ab and ba
belong to I whenever a ∈ I and b ∈ A. For a given ideal I of A, we can
form the factor brace A/I := {a+I |a ∈ A}. To show that multiplication
is well-defined for A/I, it suffices to verify that the implication
x ∈ I ⇒ a(x + c) − ac ∈ I
is valid for a, c ∈ A. Since I is A◦-invariant, we can write x in the form
x = bc = bc + b for some b ∈ I. Then a(x + c) − ac = a(b ◦ c) − ac =
(ab)c + ab ∈ I follows.
For a brace A, morphisms between A-modules, as well as submodules
and factor modules, are defined in the obvious way. A submodule of A
is called a right ideal. Since A-modules can be regarded as A◦-modules,
they form an abelian category.
Although there are no left modules over non-associative braces, the
interaction with associative rings of left operators is of importance. Let
R be an associative ring. A module M over a brace A which is a left
W. Rump 131
R-module will be called an (R, A)-bimodule if (rx)a = r(xa) holds for all
x ∈ M , r ∈ R, and a ∈ A. The first example arises for any brace A.
Define
A0 := {x ∈ A | ∀ a, b ∈ A : x(a + b) = xa + xb}. (12)
Proposition 1. A0 is the largest radical subring of A, and A is an
(A0, A)-bimodule.
Proof. Since radical rings are left distributive, there cannot be a larger
radical subring. Eqs. (1) and (5) show that A0 is an additive subgroup.
From Eq. (3) we infer that
x(ab) = (xa)b (13)
holds for all x ∈ A0 and a, b ∈ A. For x, y ∈ A0 and a, b ∈ A, we thus
have
(xy)(a + b) = x(y(a + b)) = x(ya + yb) = x(ya) + x(yb) = (xy)a + (xy)b,
which shows that A0 is a subbrace of A. Hence A0 is is a radical ring.
The remaining assertion follows by Eq. (13). �
Note that A0 contains the center of the adjoint group A◦. We conclude
this section with two simple examples. For more sophisticated examples,
we refer to [4, 5].
Example 1. Let B be abelian group and C a right B-module. Define a
multiplication in A := B ⊕ C by
(b1 + c1)(b2 + c2) := 0 + (cb2
1 − c1). (14)
Then A is a brace with circle operation (b1 + c1) ◦ (b2 + c2) = (b1 + b2) +
(cb2
1 + c2).
Example 2. Let A be a brace, and M an A-module. Define B := M ⊕A
with
(x + a)(y + b) := xb + ab, (15)
where a, b ∈ A and x, y ∈ M . It is easily verified that B is a brace with
circle operation
(x + a) ◦ (y + b) = (xb + y) + (a ◦ b).
2. Cycle bases
Let A be a brace. Eq. (10) can be expressed in terms of the dot product
as follows. With y := xa, the equation turns into (a ◦ b) · yb = a · y.
132 Modules over braces
Here a ◦ b = ab + b. Thus if we set c := yb and replace ab by a, we get
(a + b) · c = (b · a) · (b · c). By symmetry, this yields
(a · b) · (a · c) = (b · a) · (b · c), (16)
an equation with just one operation. A set X with a left invertible mul-
tiplication satisfying (16) is said to be a cycle set [4]. Thus every brace is
a cycle set. Left invertibility means that the left multiplication y 7→ x · y
admits an inverse y 7→ yx. An A◦-invariant subset of A which generates
A as an abelian group will be called a cycle base. By Eq. (16), any cycle
base of A is a cycle set. Note that a cycle base generates the adjoint
group A◦.
A cycle set X is said to be non-degenerate [4] if the map x 7→ x · x is
bijective (i. e. every element admits a unique square root).
Proposition 2. A cycle set X is non-degenerate if and only if it arises
as a cycle base of a brace.
Proof. Assume first that X is a cycle base of a brace A. For a given
a ∈ A, the equation x · x = a can be written as x = ax or (−a) ◦ x = 0.
Hence it admits a unique solution x = (−a)′ By Eq. (7) this shows that
x · x = a ⇔ (−a) · (−a) = −x. (17)
Since −X := {−x | x ∈ X} is another cycle base of A, it follows that X
is non-degenerate. Conversely, let X be a non-degenerate cycle set. By
[4], Proposition 6, there is a unique way to make the free abelian group
Z
(X) into a cycle set with X as a sub-cycle-set such that
a · (b + c) = a · b + a · c (18)
(a + b) · c = (a · b) · (a · c) (19)
holds for a, b, c ∈ Z
(X). Therefore, the deduction of Eq. (16) at the
beginning of this section shows that Z
(X) is a brace with X as a cycle
base. �
By the absence of left distributivity, ideals of braces are not as easy
to handle as in ring theory. We will see below how cycle bases can be
used to overcome this difficulty. Let us start with a general criterion for
ideals.
Proposition 3. A right ideal I of a brace A is an ideal if and only if I
is a normal subgroup of the adjoint group A◦.
W. Rump 133
Proof. Assume first that I is an ideal. The equation
a ◦ b = ab + b (20)
shows that I is a subgroup of A◦. If a ∈ I and b ∈ A◦, then ba + a ∈ I
can be written in the form ba + a = cb + c for some c ∈ I. Hence
b ◦ a = ba + a + b = cb + c + b = c ◦ b, which shows that I is a normal
subgroup of A◦. Conversely, let I be normal in A◦. For a, b ∈ A, we have
b◦a◦b′ = (ba+b+a)◦b′ = (ba+b+a)b′+(ba+b+a)+b′ = (ba+a)b′+(ba+a),
which proves the identity
ba + a = (b ◦ a ◦ b′)b. (21)
Thus if a ∈ I and b ∈ A, Eq. (21) implies that b ◦ a ◦ b′ ∈ I. �
Proposition 3 implies that the sum of ideals is again an ideal.
Proposition 4. Let M be a module over a brace A, and let N be a
normal subgroup of A◦. Then
MN := {
n
∑
i=1
xiai | xi ∈ M, ai ∈ N}
is a submodule of M .
Proof. The proof follows from the equation
(xa)b = (xb)(b′ ◦ a ◦ b) (22)
which holds for x ∈ M and a, b ∈ A. By Eq. (10), we have
(xa)b = (xa−x)b = (xa)b−xb = (xb)b′◦a◦b−xb = (xb)(b′◦a◦b).
�
Proposition 4 implies that if I is an ideal of a brace A, then AI is
again an ideal of A. However, the product of two ideals need not be an
ideal (see Example 2 of [5]). Next we show how ideals can be understood
in terms of a cycle base.
For a set X, let Π(X) denote the set of partitions of X, i. e. sets P
of pairwise disjoint non-empty subsets of X with
⋃
P = X. If x, y ∈ X
belong to the same Y ∈ P , we write x
P
∼ y. If P ∈ Π(X) is a refinement
of P ′ ∈ Π(X), i. e. x
P
∼ y implies x
P ′
∼ y, we write P > P ′.
Every morphism f : X → Y of cycle sets gives rise to a partition
P ∈ Π(X) with x
P
∼ y if and only if f(x) = f(y). Such partitions will
be called ideals of X. They are characterized by the following properties
(∀ x, y, z ∈ X):
x
P
∼ y ⇔ z · x
P
∼ z · y (23)
x
P
∼ y ⇒ x · z
P
∼ y · z. (24)
134 Modules over braces
Let A be a brace with a cycle base X. Then every right ideal I of A
defines a partition P (I) with x
P (I)
∼ y if and only if y = xa for some
a ∈ I. This follows since I is a subgroup of A◦.
Theorem 1. Let A be a brace with a cycle base X, and let P ∈ Π(X) be
an ideal of X. The additive subgroup IP of A generated by the differences
x − y with x
P
∼ y is an ideal with P (IP ) > P . A right ideal I of A with
P (I) = P is an ideal if and only if IP ⊂ I.
Proof. The equivalence (23) implies that IP is a right ideal of A. To
show that IP is an ideal, we first remark that the conclusion of (24) can
be replaced by zx P
∼ zy. Namely, if we replace z by zx, we get z
P
∼ y · zx.
Hence (23) yields zy P
∼ zx. Assume now that x
P
∼ y. By Proposition 2,
there exists an element z ∈ X with y = z · z. Hence xz P
∼ z. Therefore,
the modified version of (24) yields tx
z P
∼ tz for all t ∈ X, and (23) gives
tx
z
◦z′ P
∼ t. By Eqs. (7) and (20), we have t(x−y) = t(x+ z′) = t(xz ◦ z′),
hence
t(x − y) = tx
z
◦z′ − t ∈ IP . (25)
Thus b(x − y) ∈ IP for all b ∈ A. By Eq. (20), every element of a ∈ IP
can be written in the form a = a1 ◦ · · · ◦ an with ai = xi − yi and xi
P
∼ yi
for all i. Therefore, Eq. (25) shows that P (IP ) > P . To prove that IP is
an ideal, it remains to verify that ba ∈ IP holds for all b ∈ A. For n = 1,
this has been proved above. For arbitrary n, the assertion follows by the
formula
b(a1 ◦ · · · ◦ an) =
n−1
∑
i=0
ba1◦···◦aiai+1 (26)
which is an immediate consequence of the equation
ba1◦···◦an − b =
n
∑
i=1
ba1◦···◦ai −
n−1
∑
i=0
ba1◦···◦ai .
Thus IP is an ideal with P (IP ) > P .
Finally, let I be a right ideal of A with P (I) = P . Then IP ⊂ I
means that xa = xa − x ∈ I holds for all x ∈ X and a ∈ I, i. e. that I is
an ideal. �
To illustrate Theorem 1, let I be an ideal of a brace A. By Propo-
sition 4, we already know that AI is an ideal. By definition, AI is gen-
erated, as an abelian group, by the products xa with x ∈ X and a ∈ A.
Since xa = xa − x, we get
AI = IP (I). (27)
W. Rump 135
By induction, this formula yields an explicit representation of the radical
series
A ⊃ A2 ⊃ A3 ⊃ · · · (28)
of A, which is defined, inductively, by A1 := A and An+1 := A(An). In
contrast to radical rings, the series (28) does not always reach zero, even
if A◦ is a finite p-group [5].
3. Square-free elements
Let A be a brace. We call an element a ∈ A square-free if a2 = 0. The
nth power of a in A◦ will be denoted by a◦n. By definition, we have
a◦2 = a + aa and a◦3 = a + (a + aa)a = a + aa + (aa)a. Similarly,
a◦4 = a + aa + (aa)a + ((aa)a)a, and in general, we obtain
a◦n =
n−1
∑
i=0
aa◦i
(29)
for n ∈ N. If the set C(a) of iterates a, aa, (aa)a, . . . is finite, they form a
cycle. Then we call C(a) the self-cycle and s(a) := |C(a)| the self-order
of a. The subgroup of A◦ generated by a operates transitively on C(a).
So if the order ord(a) of a in A◦ is finite, it satisfies an equation
ord(a) = s(a) · r(a) (30)
with r(a) ∈ N. We call r(a) the reduced order of a. Thus a is square-free
if and only if aa = a, i. e. s(a) = 1. Define
c(a) :=
∑
C(a) = a◦s(a). (31)
Proposition 5. Let A be a brace. An element a ∈ A is square-free if
and only if
a◦n = na (32)
holds for all n ∈ Z. If a ∈ A is of finite self-order in A◦, then c(a) is
square-free, and r(a) is the order of c(a) in the abelian group A.
Proof. Let s(a) be finite. Then Eq. (29) yields c(a) ◦ c(a) = a◦s(a) ◦
a◦s(a) = a◦2s(a) = 2c(a). Hence c(a) is square-free, and a is square-free
if and only if a = c(a). The remaining statements follow immediately by
Eq. (29). �
A cycle set X is said to be square-free if x ·x = x holds for all x ∈ X.
Since such an X is non-degenerate, it arises, by Proposition 2, as a cycle
base of a brace A, and the elements of X are then square-free in A.
136 Modules over braces
A partition P of a cycle set X with |P | > 2 is said to be a decompo-
sition of X if x
P
∼ y · x holds for all x, y ∈ X. If X 6= ∅ and X admits no
such partition, we call X indecomposable. In [4], we proved the following
Theorem 2. Every finite square-free cycle set X with more than one
element is decomposable.
If we regard X as a cycle base of a brace A, the theorem becomes
more plausible. For x, y ∈ X, we have xy = xy − x, i. e. x and xy
are in the same residue class modulo A2. If X is indecomposable, this
implies that A/A2 is cyclic, and the generator can be represented by an
element in X. Since x2 = 0, it should follow that X must be a singleton.
However, this reasoning is not quite correct. A precise statement is given
by the following proposition which makes use of a right radical series
A ⊃ A2 ⊃ A(3) ⊃ A(4) ⊃ · · · (33)
with A(n+1) := (A(n))A. Note that A(2) is just A2. The right powers A(n)
are right ideals, but not ideals, in general. If |A| is a prime power, it can
be shown that A(n) = 0 for some n (see [5]).
For a module M over a brace A, and x ∈ M , we abbreviate
xA :=
{
n
∑
i=1
xai
∣
∣
∣
∣
∣
a1, . . . , an ∈ A
}
. (34)
Remark. By Eq. (3), the equation (xa)b = x(a ◦ b) − xa − xb ∈ xA
holds for a, b ∈ A. Hence xA is a submodule of M .
Proposition 6. Let A be a finite brace with a square-free cycle base X,
such that A(n) = 0 for some n. If the abelian group A/A2 is cyclic, then
A2 = 0.
Proof. Since A/A2 is cyclic, we can assume that A = Zx + A2 for
some x ∈ X. Therefore, A2 = (Zx + A2)A = xA + A(3). Inductively, let
us assume that A2 = xA + A(i) has been shown for some i > 3. Then
A2 = (Zx + xA + A(i))A = xA + A(i+1). Hence A(n) = 0 implies that
A2 = xA. Now we set
X ′ := {x + xa | a ∈ A}.
Since X ′ is A◦-invariant, X ′ is a cycle set with
(x + xa)x+xb = x + x(a ◦ (x + xb)). (35)
As A = Zx + xA, the additive group of A is generated by X ′. By Eq.
(20), we have
a + b = (b · a) ◦ b
for all a, b ∈ A. Therefore, X ′ generates the adjoint group A◦, i. e. every
a ∈ A can be written in the form a = 0 ◦ x1 ◦ · · · ◦ xn with xi ∈ X ′. Thus
(35) yields
x + xa = ((xx1)x2 · · · )xn ,
which shows that X ′ is indecomposable. Since X ′ ⊂ X is square-free,
Theorem 2 implies that |X ′| = 1. Hence A2 = xA = 0. �
Proposition 6 does not hold without the nilpotency condition A(n) = 0
(see Example 3 of [5]). We conclude with the following commutation
property of square-free cycle bases.
Proposition 7. Let A be a brace with a square-free cycle base X. If
x ∈ X and a ∈ A with x + a ∈ X, then xa ∈ aA.
Proof. x + a ∈ X implies that a ◦ x = (x + a)x + (x + a) ∈ X. Hence
0 = (a ◦ x)2 = ((a ◦ x)a)x + (a ◦ x)a + (a ◦ x)x = ((ax + a)a)x + (xa)x +
(ax)a+a2+(ax+a)x. By the above remark (i. e. that xA is a submodule
of A), this shows that xa ∈ aA. �
References
[1] P. Etingof, T. Schedler, A. Soloviev: Set-theoretical solutions to the quantum
Yang-Baxter equation, Duke Math. J. 100 (1999), 169-209
[2] T. Gateva-Ivanova, M. Van den Bergh: Semigroups of I-type, J. Algebra 206
(1998), 97-112
[3] G. Laffaille: Quantum Binomial Algebras, Colloq. Homology and Representation
Theory, Vaqueŕıas, 1998, Bol. Acad. Nac. Cienc. (Cordoba) 65 (2000), 177-182
[4] W. Rump: A decomposition theorem for square-free unitary solutions of the
quantum Yang-Baxter equation, Adv. Math. 193 (2005), 40-55
[5] W. Rump: Braces, radical rings, and the quantum Yang-Baxter equation,
Preprint
Contact information
W. Rump Institut fьr Algebra und Zahlentheorie, Uni-
versitдt Stuttgart Pfaffenwaldring 57, D-
70550 Stuttgart, Germany
E-Mail: rump@mathematik.uni-stuttgart.de
Received by the editors: 31.10.2005
and in final form 30.06.2006.
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