CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

arXiv:1511.00936v1 [math.RA] 3 Nov 2015

LI GUO AND HOUYI YU
Abstract. The characteristic is a simple yet important invariant of an algebra. In this paper,
we study the characteristic of a Rota-Baxter algebra, called the Rota-Baxter characteristic. We
introduce an invariant, called the ascent set, of a Rota-Baxter characteristic. By studying its
properties, we classify Rota-Baxter characteristics in the homogenous case and relate Rota-Baxter
characteristics in general to the homogeneous case through initial ideals. We show that the RotaBaxter quotients of Rota-Baxter characteristics have the same underlying sets as those in the
homogeneous case. We also give a more detailed study of Rota-Baxter characteristics with special
base rings. In particular, we determine the prime characteristics of Rota-Baxter rings.

Contents
1. Introduction
2. Characteristics of Rota-Baxter algebras
3. Classification of the characteristics of Rota-Baxter algebras
3.1. The ascent set of a Rota-Baxter characteristic
3.2. Classification of homogeneous Rota-Baxter characteristics
3.3. Rota-Baxter characteristics and their quotients
4. Characteristics of Rota-Baxter rings
References

1
2
4
4
6
9
11
15

1. Introduction
A Rota-Baxter algebra is an associative algebra together with a linear endomorphism that is
an algebraic analogue of the integral operator. This concept has it origin from a work of G.
Baxter [5] in probability theory. In the late 1960s, Rota [16] studied the subject from an algebraic
and combinatorial perspective and suggested that they are closely related to hypergeometric
functions, incidence algebras and symmetric functions [17, 18]. Since then, these algebras have
been investigated by mathematicians and mathematical physicists with various motivations. For
example, a Rota-Baxter operator on a Lie algebra is closely related to the operator form of the
classical Yang-Baxter equation. Here the Baxter is the physicist R. Baxter [3, 19]. In recent
years Rota-Baxter operators have found applications to many areas, such as number theory [15],
combinatorics [10, 16, 17], operads [1, 4] and quantum field theory [6]. See [11, 13] and the
references therein for further details.
For the theoretic study of this important algebraic structure, it is useful to generalize the study
of characteristics of algebras to Rota-Baxter algebras. The characteristic of a (unitary) ring R is
(the nonnegative generator of) the kernel of the structure map Z → R sending 1 to the identity
Date: January 20, 2018.
2010 Mathematics Subject Classification. 13C05, 13E05, 16W99.
Key words and phrases. Rota-Baxter algebra, Rota-Baxter characteristic, initial object, initial ideal, prime ideal.
1

2

LI GUO AND HOUYI YU

element of R. More generally, the characteristic of an algebra R over a commutative ring k
is the kernel of the structure map k → R. To generalize this concept to the context of RotaBaxter algebras, we note that the structure map k → R comes from the fact that k is the initial
object in the category of k-algebras, as the free k-algebra on the empty set. Then to study the
characteristics of Rota-Baxter k-algebras, we consider the initial object in the category of RotaBaxter k-algebras, as the free Rota-Baxter algebra on the empty set. Thus every Rota-Baxter
k-algebra comes with a (unique) structure map from the initial object to this Rota-Baxter algebra.
Then the kernel of the structure map should be the characteristic of the Rota-Baxter algebra.
In this paper, we study the characteristics of Rota-Baxter algebras. More precisely, we study
the Rota-Baxter ideals, and their corresponding quotients, of the initial Rota-Baxter k-algebra. In
particular, we study the Rota-Baxter ideals of the initial Rota-Baxter ring. The initial Rota-Baxter
algebra is a generalization of the divided power algebra, and the polynomial algebra k[x] when k
is taken to contain Q. So some our results are naturally related to results of these algebras.
The construction of the initial object will be reviewed at the beginning of Section 2 and will
be applied to give the concept of Rota-Baxter characteristics. In Section 3, we first classify
homogeneous Rota-Baxter characteristics and their quotients (Theorem 3.6). We then show that
the Rota-Baxter quotients of general Rota-Baxter characteristics have the same underlying sets
as those in the homogeneous case (Theorem 3.9). Finally specializing to the case of k = Z
in Section 4, we give the structures of Rota-Baxter characteristics of Rota-Baxter rings, and
determine the prime Rota-Baxter characteristics (Theorem 4.3).
2. Characteristics of Rota-Baxter algebras
Notations. Unless otherwise specified, an algebra in this paper is assumed to be unitary associative defined over a unitary commutative ring k. Let N and P denote the set of nonnegative and
positive integers respectively. For n ∈ N, denote [n] := {1, · · · , n}. For notational convenience,
we also denote [∞] = P. Let Zn denote the set of integers modulo n.
Before giving the definition of the characteristic of a Rota-Baxter algebra, we first provide
some background and preliminary results on Rota-Baxter algebras. See [9, 11, 14, ] for details.
Let λ ∈ k be given. A Rota-Baxter k-algebra of weight λ is a k-algebra R paired with a linear
operator P on R that satisfies the identity
(1)

P(x)P(y) = P(xP(y)) + P(P(x)y) + λP(xy)

for all x, y ∈ R. When k is Z, then the pair (R, P) is called a Rota-Baxter ring.
A Rota-Baxter ideal of a Rota-Baxter algebra (R, P) is an ideal I of R such that P(I) ⊆ I. Then
we denote I 6 R. The concepts of Rota-Baxter subalgebras, quotient Rota-Baxter algebras and
homomorphisms of Rota-Baxter algebras can be similarly defined.
We recall that the free Rota-Baxter algebra on a set X is a Rota-Baxter algebra (F RB (X), PX )
together with a set map iX : X → F RB (X) characterized by the universal property that, for any
Rota-Baxter algebra (R, P) and set map f : X → R, there is a unique Rota-Baxter algebra
homomorphism f˜ : F RB(X) → R such that iX ◦ f˜ = f . The free Rota-Baxter algebra on the
empty set is also the free Rota-Baxter k-algebra F RB (k) on the k-algebra k [14, ], characterized
by the universal property that, for any Rota-Baxter algebra (R, P) and k-algebra homomorphism
f : k → R, there is a unique Rota-Baxter algebra homomorphism f˜ : F RB (k) → R such that
ik ◦ f˜ = f , where ik is the structure map ik : k → F RB (k). Since there is only one k-algebra
homomorphism k → R, this universal property shows that F RB (k) is the initial object in the
category of Rota-Baxter k-algebras.

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

3

We refer the reader to [14], as well as [7, 16], for the general constructions of free Rota-Baxter
algebras, but will focus on a simple construction of F RB (k) following [2], where it is denoted by
Xk (k). This free Rota-Baxter k-algebra not only provides the simplest example of free RotaBaxter algebras but also establishes the connection between Rota-Baxter algebra and some wellknown concepts such as divided powers and Stirling numbers [2, 12, ]. As a k-module, Xk (k) is
given by the free k-module
∞
M
Xk (k) =
kam
m=0

on the basis {am | m > 0}. For a given λ ∈ k, the product ⋄ = ⋄λ on Xk (k) is defined by
min(m,n)
X m + n − i ! m!
(2)
λi am+n−i ,
m, n ∈ N.
am ⋄λ an =
i
m
i=0
Thus when λ = 0, Xk (k) is the divided power algebra. Note that ⋄ is an extension of the product
on k viewed as ka0 . Thus there should no confusion if the notation ⋄ is suppressed, as we often
do. Define the k-linear operator P = Pk on Xk (k) by assigning Pk (am ) = am+1 , m > 0 and
extending by additivity.
By [14, ], when k contains Q and λ = 0, we have Xk (k)  k[x] as a k-algebra.
Theorem 2.1. The pair (Xk (k), Pk ) is the initial object in the category of Rota-Baxter k-algebras.
More precisely, for any Rota-Baxter k-algebra (R, P), there is a unique Rota-Baxter k-algebra
homomorphism ϕ : (Xk (k), Pk ) → (R, P).
Proof. Suppose a k-algebra A has a linear basis X. By [9, 11], the free Rota-Baxter algebra F RB (A)
on A (denoted by XNC
k (A)) has a linear basis consisting of Rota-Baxter words in the alphabet set
X. By definition, a Rota-Baxter word in X is a bracketed word in the alphabet set X in which
there are no adjacent pairs of brackets. Taking A = k, then X = {1}. Thus a Rota-Baxter word in
X can only be of the form
ur := ⌊· · · ⌊ 1 ⌋ · · ·⌋
|{z}
|{z}
r iterations

r iterations

(namely applying the bracket operator ⌊ ⌋ to 1 r times), for r > 1, together with u0 := 1. Thus
X
F RB(k) =
kur ,
r>0

on which the Rota-Baxter operator Qk is given by Qk (ur ) = ⌊ur ⌋ = ur+1 . By the universal property
of F RB (k), the natural inclusion map f : k → Xk (k) sending 1 → a0 (that is, the structure map)
extends to a Rota-Baxter algebra homomorphism
f˜ : F RB (k) → Xk (k).
As such, we obtain f˜(u0 ) = a0 and recursively,
f˜(ur+1 ) = f˜(Qk (ur )) = Pk ( f˜(ur )) = Pk (ar ) = ar+1 , r > 0.
Therefore f˜ is a linear isomorphism and thus a Rota-Baxter algebra isomorphism, showing that
Xk (k) is the free Rota-Baxter algebra on k and hence the initial object in the category of RotaBaxter k-algebras.

Thus Xk (k) plays the same role in the category of Rota-Baxter k-algebras as the role played
by k in the category of k-algebras.

4

LI GUO AND HOUYI YU

Definition 2.2. Let (R, P) be a Rota-Baxter k-algebra and let ϕ = ϕ(R,P) : (Xk (k), Pk ) → (R, P) be
the unique Rota-Baxter algebra homomorphism from the initial object (Xk (k), Pk ) in the category
of Rota-Baxter k-algebras to (R, P). The kernel of the structure map ϕ is called the Rota-Baxter
characteristic of (R, P).
In view of Theorem 2.1, the characteristic of a Rota-Baxter k-algebra R must be a Rota-Baxter
ideal of Xk (k). Conversely, any Rota-Baxter ideal I of Xk (k) is the characteristic of some RotaBaxter algebra, for example of Xk (k)/I.
Remark 2.3. Based on the above observation, we will use the terminology Rota-Baxter characteristics in exchange with Rota-Baxter ideals of Xk (k) in the rest of the paper.
3. Classification of the characteristics of Rota-Baxter algebras
In this section, we study Rota-Baxter characteristics and their quotients. First in Section 3.1,
we introduce an invariant, called the ascent set, of a Rota-Baxter characteristic. We then apply
the invariants to classify all homogeneous Rota-Baxter characteristics in Section 3.2. Finally in
Section 3.3 we relate a Rota-Baxter characteristic to a homogeneous Rota-Baxter characteristic
by taking the initial terms and show that the quotients of the two Rota-Baxter characteristics share
the same underlying sets, but not the same k-modules. This approach is motivated from the study
in the polynomial algebra k[x1 , · · · , xn ] when k is a field. There the quotient modulo an ideal and
the quotient modulo the corresponding initial ideal are known to share the same basis [8, § 5.3,
Proposition 4].
3.1. The ascent set of a Rota-Baxter characteristic. As recalled in the last section, the RotaBaxter algebra Xk (k), as the initial object in the category of Rota-Baxter algebras, is the direct
sum
M
Xk (k) =
kam
m>0

on the basis {am | m > 0} and hence is an N-graded k–module. Any nonzero element f of Xk (k)
n
P
ci ai with cn , 0. Then n is called the degree of f and cn an
can be uniquely written as f =
i=0

is called the initial term of f , denoted by deg f and in( f ) respectively. In addition, we define
deg 0 = −∞. We call the set supp( f ) := {ci ai |ci , 0} the support of f .
Let I be a Rota-Baxter characteristic, namely an ideal of Xk (k). We introduce an invariant of
I. For each j ∈ N, we denote


j


X




Ω j := Ω j (I) := 
b
∈
k
(∃
f
∈
I)
f
=
b
a
.

j
i
i




i=0

Equivalently,

o
n
Ω j = {0} ∪ b j ∈ k (∃ f ∈ I) in( f ) = b j a j .

The smallest index j such that Ω j , 0 is called the starting point of I, denoted by st(I), that is,
o
n
st(I) = min j ∈ N Ω j , 0 .

Lemma 3.1. Let I be a Rota-Baxter ideal of Xk (k). Then for each j ∈ N, Ω j is an ideal of k and
Ω j ⊆ Ω j+1 .

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

Proof. For any b j , c j ∈ Ω j , there exist f and g ∈ I such that f =

Pj

5

bi ai and g =

Pj

ci ai . Hence

i=0

i=0

f +g=

Pj

(bi + ci )ai is in I. So b j + c j ∈ Ω j . On the other hand, for any c ∈ k, c f =

Pj

cbi ai ∈ I.

i=0

i=0

This yields cb j ∈ Ω j . Thus Ω j is an ideal of k.
Since I is a Rota-Baxter ideal of Xk (k), we have P( f ) =

Pj

bi ai+1 ∈ I. So b j ∈ Ω j+1 . Hence

i=0

Ω j ⊆ Ω j+1 , as required.



For a given Rota-Baxter ideal I of Xk (k), Lemma 3.1 shows that the ideals Ω j ⊆ k, j ∈ N,
form a non-decreasing sequence of ideals of k. Thus the sequence is controlled by the locations
and extents where the increases occur. This motivates us to introduce the following notions.
Let s1 < s2 < · · · be the integers t ∈ N such that Ωt−1 ( Ωt with the convention that Ω−1 = {0}.
We will use the notation si , i ∈ [NI ] where N = NI is either in N or ∞ with the convention adopted
at the beginning of the last section. The integers si , i ∈ [NI ], are called the ascending points of I
and the ideals Ωsi , i ∈ [NI ] are called the ascending levels of I. In view of Lemma 3.1, we have
s1 = st(I).
Thus for a given Rota-Baxter characteristic I, the set of pairs
o
n
(3)
A(I) := (s j , Ωs j ) j ∈ [NI ] ,
called the ascent set, is uniquely determined by I. By the definition of Ωs j , we have s j < s j+1 and
Ωs j = Ωs j +1 = · · · = Ωs j+1 −1 ( Ωs j+1

(4)

for all j ∈ [NI ], as illustrated by Figure 1.
Ω
✻

Ω s3 r

Ω s2 r
Ω s1 r
0

r

r

r

s1

s2

s3

✲

N

Figure 1.
We next study how the information from A(I) can be used to recover I. Denote
o
n
(5)
A := {(s j , Ωs j )} j∈[N] s j ∈ N, Ωs j 6 k, s j < s j+1 , Ωs j ( Ωs j+1 , j ∈ [N], 1 6 N 6 ∞ ,

called the set of ascending pairs. So A consists of pairs of sequences of the same lengths
with one sequence of strictly increasing nonnegative integers and a second sequence of strictly
increasing ideals of k. Let I = I(k) denote the set of Rota-Baxter characteristics, namely the set
of Rota-Baxter ideals of Xk (k). Then taking the ascent set of a Rota-Baxter characteristic defines

6

LI GUO AND HOUYI YU

a map
Φ : I → A,

(6)

I 7→ A(I), I 6 Xk (k).

In the rest of the paper, we study the property of this map, including its surjectivity and fibers,
that is, inverse images. In Theorem 3.6, we show that the restriction of Φ to the subset of
homogeneous Rota-Baxter characteristics gives a bijection to A, proving the surjectivity of Φ.
In Proposition 3.8, we show that two Rota-Baxter characteristics are in the same fiber if and only
if they have the same initial ideal.
n
o
Lemma 3.2. Let I be a Rota-Baxter ideal of Xk (k) with A(I) = (s j , Ωs j ) | j ∈ [NI ] . For each
o
n
j ∈ [NI ], we let Θ j := ω s j ,ℓ ℓ ∈ [N j ] be a set of generators of the ideal Ωs j . Here N j is either a
positive integer or ∞. For each ω s j ,ℓ ∈ Θ j , let f s j ,ℓ be an element of I whose initial term is ω s j ,ℓ as j .
Then the set
[ n
o
f s j ,ℓ ℓ ∈ [N j ]
j∈[NI ]

is a generating set of the Rota-Baxter ideal I.
o
S n
We call the above-mentioned set
f s j ,ℓ ℓ ∈ [N j ] an ascent generating set of the Rotaj∈[NI ]

Baxter ideal I.

Proof. Let I ′ be the Rota-Baxter ideal generated by the set

S n

j∈[NI ]

o
f s j ,ℓ ℓ ∈ [N j ] . Since I ′ ⊆ I is

trivial, it suffices to show that each element f of I belongs to I ′ .
So take an f ∈ I. Then f is in I ′ if f = 0. We next assume that f , 0. Then we have
in( f ) = badeg f for some 0 , b ∈ k. We proceed by induction on deg f . Clearly, deg f > st(I) = s1 .
If deg f = s1 , then b ∈ P
Ωs1 . So there exist cs1 ,ℓ ∈ k, ℓ ∈ [N1 ], all but finitely many of which
cs1 ,ℓ ω s1 ,ℓ . Thus the element
being zero, such that b =
ℓ∈[N1 ]

g= f−

X

cs1 ,ℓ f s1 ,ℓ

ℓ∈[N1 ]

is in I. But now the degree
P of g is less than′ deg f . It follows from deg f = st(I) that g must be 0,
cs1 ,ℓ f s1 ,ℓ is in I and we are done. For a given n > st(I), assume that
which means that f =
ℓ∈[N1 ]

all f ∈ I with deg f 6 n are in I ′ and take f ∈ I with deg f = n + 1. Then there exists r ∈ [NI ]
such that sr 6 n + 1 < P
sr+1 with the convention that sNI +1 = ∞ if NI is finite. By Eq. (4), we have
csr ,ℓ ω sr ,ℓ where csr ,ℓ ∈ k, ℓ ∈ [Nr ], with all but a finite number of csr ,ℓ
Ωn+1 = Ωsr . So b =
ℓ∈[Nr ]

being zero. Put

h= f −

X

ℓ∈[Nr ]


csr ,ℓ Pn+1−sr f sr ,ℓ .

Then h ∈ I with deg h < deg f . So we can apply the induction hypothesis to obtain that h is in I ′
and hence f is in I ′ , as required.

3.2. Classification of homogeneous Rota-Baxter characteristics. We now apply ascent sets to
classify homogeneous Rota-Baxter characteristics.
Definition 3.3. A Rota-Baxter ideal I of Xk (k) is called a homogeneous Rota-Baxter ideal
if I is a Rota-Baxter ideal generated by a set of homogeneous elements. Then the Rota-Baxter
characteristic I is called homogeneous.

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

7

We next show that, for a Rota-Baxter ideal of Xk (k), its homogeneity as a Rota-Baxter ideal
is equivalent to its homogeneity as an ideal. For this purpose, we first give a general relation
between generating a Rota-Baxter ideal and generating an ideal in a Rota-Baxter algebra.
Lemma 3.4. Let (R, P) be a Rota-Baxter k-algebra of weight λ, and S a subset of R. Then the
Rota-Baxter ideal (S )RB generated by S is the ideal of R generated by the set
[
S RB :=
(7)
(◦mi=1 P xi )(a)|a ∈ S , xi ∈ R, 1 6 i 6 m ,
m∈N

where (◦mi=1 P xi )(a) := P(xm P(xm−1 P(· · · P(x1 a)))) with the convention that (◦0i=1 P xi )(a) := a.
If R = Xk (k), then it suffices to take xi to be the homogeneous elements in Eq. (7).
Proof. Let I be the ideal of R generated by S RB. Then an element of I is a sum of elements of the
form r(◦mi=1 P xi )(a) with r ∈ R. Then we have
P(r(◦mi=1 P xi )(a)) = (◦m+1
i=1 P xi )(a),
where xm+1 = r. So, by the additivity of P, we obtain P(I) ⊆ I, and hence I is the Rota-Baxter
ideal of R generated by S RB , that is, I = (S RB )RB . Notice that S ⊆ S RB ⊆ (S )RB , so (S )RB = (S RB )RB
whence I = (S )RB , as required. The second statement follows since any element in Xk (k) is a
linear combination of homogeneous elements.

Now we can give the following characterization of homogeneous Rota-Baxter ideals in Xk (k).
Proposition 3.5. Let I ⊆ Xk (k) be a Rota-Baxter ideal. Then I is a homogeneous Rota-Baxter
ideal if only if I is a homogeneous ideal.
Proof. (⇐=) If a set of homogeneous elements generates I as an ideal, then it does so as a RotaBaxter ideal.
(=⇒) Suppose that I is a homogeneous Rota-Baxter ideal. Then we can assume that there are a
subset Λ of N and numbers ni ∈ P (i ∈ Λ) or ni = ∞ such that I is generated by the following set
of homogeneous elements:
G := {bi j ai ∈ kai | j ∈ [ni ], i ∈ Λ}.
Then, by Lemma 3.4, I is the ideal generated by the set
n
o
G′ := (◦mk=1 P xk )(bi j ai ) | bi j ai ∈ G, xk ∈ Xk (k) homogeneous, 1 6 k 6 m, m > 0

with the convention that (◦0i=1 P xi )(a) := a. Hence I is contained in the homogeneous ideal
generated by the supports of elements in G′ . Thus to complete the proof, we just need to prove
that for each u ∈ G′ , we have supp(u) ⊂ I. We will prove this for u in the form (◦mk=1 P xk )(bi j ai ),
where bi j ai ∈ G, by induction on m > 0. When m = 0, we have u = bi j ai ∈ G which is assumed
to be in I. Suppose that the statement is true for m = k > 0 and consider

m
u = (◦m+1
k=1 P xk )(bi j ai ) = P xm+1 (◦k=1 P xk )(bi j ai ) .

By the induction hypothesis, the support of (◦mk=1 P xk )(bi j ai ), that is, the set of homogenous components of it, is contained in I. Let ba p ∈ I, b ∈ k, be such a homogenous component and denote
xm+1 = can , c ∈ k. Then by Eq. (2), we have
min(n,p)
 min(n,p)
 X n + p − i! p!
 X n + p − i! p!

i

P xm+1 (ba p ) = P (can )(ba p ) = P 
λ bcan+p−i  =
λi cPn−i+1 (ba p ).
i
p
i
p
i=0
i=0
Since Pn−i+1 (ba p ) is in I, the homogeneous components of P xm+1 (ba p ) are in I. Hence the homogeneous components of u are in I. This completes the induction.


8

LI GUO AND HOUYI YU

Theorem 3.6.
(a) Let I be a homogeneous Rota-Baxter
characteristic,
that is, a homogeneous
o
n
Rota-Baxter ideal of Xk (k), with A(I) = (s j , Ωs j ) j ∈ [NI ] . Then

j+1 −1
∞ sL

L

s −1
 


Ωs j ai ,
NI = ∞,
NI  M
∞
 

M
M
 j+1


j=1 i=s j






(8)
I=
Ωi ai =
Ωs j ai  = 

j+1 −1
∞
I −1 s L


 L  L
NL



i=s1
i=s j
j=1
 , NI < ∞,




Ω
a
Ω
a



s
i
s
i
NI
j




j=1

i=sNI

i=s j

as the direct sum of the k-modules Ωs j ai .
(b) The quotient Xk (k)/I is the direct sum of the k-modules (k/Ωs j )ai , that is,
 −1

NI  sM

M
 j+1


Xk (k)/I 
(k/Ω
)a
s j i


i=s j
j=0
 s −1


!
sL
j+1
∞ L

1 −1

L L




 ,

ka
(k/Ω
)a
NI = ∞,
i
s
i

j


 i=0
j=1 i=s j




!
= 
(9)

j+1 −1
sL
∞
I −1 s L
1 −1

 L  L
NL

L


 , NI < ∞,




(k/Ω
)a
(k/Ω
)a
ka




s NI i 
s j i



 i=0 i
i=sN
j=1 i=s j
I

with the convention that s0 = 0, Ωs0 = 0 and sNI +1 = ∞ if NI is finite.
(c) For any ascent pair {(s j , Ωs j )} j∈[N] in A defined in Eq. (5), consisting of a positive integer
N or N = ∞, strictly increasing nonnegative integers si and strictly increasing ideals
Ωs j , j ∈ [N], there is a unique homogeneous Rota-Baxter ideal I of Xk (k) such that A(I) =
{(s j , Ωs j )} j∈[N] . In other words, the restriction of the map Φ : I → A in Eq. (6) to the set
of homogeneous Rota-Baxter characteristics is a bijection.

Proof. (a) By Proposition 3.5, the k-module I is the direct sum of the k-modules Ωi ai .
(b) Since I is a graded submodule of Xk (k), Item (b) is proved.
(c) For a given ascending pair {(s j , Ωs j )} j∈[N] ∈ A, define I ⊆ Xk (k) as in Eq. (8) with NI replaced
by N. We next show that I is a homogeneous Rota-Baxter ideal.
It follows from s1 < s2 < · · · and Ωs1 ( Ωs2 ( · · · that P(I) ⊆ I. I is a k-submodule since
Ωs j is an ideal of k for each j ∈ [N]. Further by its definition, I is homogeneous as a k-module.
We next prove that I(Xk (k)) ⊆ I. Since each element of Xk (k) can be written as the summand
of finitely many homogeneous components, it suffices to show that (ban )(ca p ) ∈ I, where ban ∈ I
and ca p ∈ Xk (k). By Eq. (2), we have
X n + p − i ! p!
  min(n,p)
(ban ) ca p =
bcλi an+p−i .
i
p
i=0
Since ban ∈ I, we have b ∈ Ωsr where r ∈ [N] is such that sr 6 n < sr+1 . For any given i with
0 6 i 6 min(n, p), let t be the integer in [N] such that st 6 n + p − i < st+1 . Note that we always
have n + p − i > n > sr , so sr 6 st . Thus Ωsr ⊆ Ωst whence b ∈ Ωst . Then for each
 iwith
n+p−i p
i
0 6 i 6 min(n, p), the element p i bcλ an+p−i is in Ωst an+p−i . This shows that (ban ) ca p ∈ I.
Thus I is a homogeneous Rota-Baxter ideal of Xk (k).
Thus we obtain a map Ψ from A to the set of homogeneous Rota-Baxter ideals of Xk (k). It is
direct to check that the left and right compositions of Ψ with the restriction of Φ to homogeneous

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

9

Rota-Baxter ideals of Xk (k) are the identities. Thus the restriction of Φ to the set of homogeneous
Rota-Baxter ideals is bijective.

3.3. Rota-Baxter characteristics and their quotients. Now we relate a Rota-Baxter characteristic to the homogeneous case and describe the structure of the quotient of Xk (k) modulo a
Rota-Baxter characteristic. We will see that such a quotient has the underlying set of the form
defined by Eq. (9). However, the same underlying set may be shared by different Rota-Baxter
characteristics, as can be seen from the following theorem. As mentioned at the beginning of this
section, this theorem is a property of Xk (k) coming as an analog to polynomial algebras.
We first relate a Rota-Baxter ideal of Xk (k) to a suitable homogeneous Rota-Baxter ideal.
Definition 3.7. The initial Rota-Baxter ideal of a Rota-Baxter ideal I of Xk (k) is the RotaBaxter ideal in(I) generated by {in( f )| f ∈ I}.
Thus in(I) is a homogeneous ideal.
Proposition 3.8. Let I be a Rota-Baxter ideal of Xk (k). Then for its ascent set we have A(I) =
A(in(I)). In other words, two Rota-Baxter characteristics are in the same fiber under the map Φ
in Eq. (6) if and only if they have the same initial ideal.
o
n
Proof. Assume that A(I) := (s j , Ωs j ) j ∈ [NI ] . We only need to show that Ωi (I) = Ωi (in(I))
for all i ∈ N. If i < st(I), then there is no element f ∈ I with degree lower than i, so Ωi (I) =
Ωi (in(I)) = ∅. Next we assume that i > st(I). Take a nonzero element b ∈ k. Then b ∈ Ωi (I) if and
only if there exists an element f in I such that in( f ) = bai , which is equivalent to b ∈ Ωi (in(I)).
Thus A(I) = A(in(I)), as required.

Now we can determine the underlying set of the quotient of a Rota-Baxter characteristic.
Theorem 3.9. Let I be a Rota-Baxter ideal of Xk (k). Then Xk (k)/I has the same underlying
set as Xk (k)/in(I) in Eq. (9). More precisely, for each j with j ∈ {0} ∪ [NI ], fix a complete set
T j ⊆ k, such that 0 ∈ T j , of representatives of k modulo Ωs j , with the convention that s0 = 0,
Ωs0 = 0 and sNI +1 = ∞ if NI is finite. Then Xk (k)/I has a complete set of representatives given
∞
L
by the following subset of Xk (k) =
kam :
m=0

(10)

 s −1


!
sL
j+1
∞ L

1 −1

L L


 s −1
 

 ,

ka
T
a
NI = ∞,
i
j
i

N
j+1
I M
 
M



i=s
i=0
j=1
j







!
T :=
T j ai  = 


sL
j+1 −1
∞
1 −1
I −1 s L
 L  L


L NL


i=s j
j=0




ka
T
a
T NI ai  , NI < ∞.




i
j i



 i=0
i=sN
j=1 i=s j
I

n
o
Proof. Let I be a Rota-Baxter ideal of Xk (k) with its ascent set A(I) = (s j , Ωs j ) | j ∈ [NI ] . By
Proposition 3.8, we have A(I) = A(in(I)). Thus it suffices to show that the underlying set of
Xk (k)/I is T in light of Theorem 3.6(b).
By convention, we take s0 = 0 and Ωs0 = 0. Then T 0 = k and
s1 −1
M
i=0

Note that this term does not exist if s1 = 0.

T 0 ai =

s1 −1
M
i=s0

kai .

10

LI GUO AND HOUYI YU

Let f ∈ Xk (k) be a nonzero element with in( f ) = bn an for some bn ∈ k. So deg f = n > 0. We
first show that
there is a unique
f ′ in T such that f − f ′ is in I.
o
S n
Let
f s j ,ℓ ℓ ∈ [N j ] be an ascent generating set of I, we may assume that for each j ∈
j∈[NI ]

[N
o j ] we have in( f s j ,ℓ ) = ω s j ,ℓ as j and hence Ωs j is generated by the set Θ j :=
n I ] and ℓ ∈ [N
ω s j ,ℓ ℓ ∈ [N j ] by Lemma 3.2.
sL
sL
1 −1
1 −1
T 0 ai since s0 = 0, Ωs0 = 0. Thus, it is
If deg f < s1 , then f ∈
kai , which equals to
i=0

i=s0

enough to take f ′ = f .
Next we assume that deg f > s1 . We prove that there exists an element f ′ ∈ T with deg f ′ 6
deg f such that f − f ′ ∈ I by induction on deg f . If deg f = s1 , then in( f ) = b s1 as1 for some
b s1 ∈ k. Since T 1 ⊆ k is a complete set of representatives of k modulo Ωs1 , there exists a unique
b′s1 ∈ T 1 such that b s1 − b′s1 ∈ Ωs1 . Note that Ωs1 is generated by Θ1 , so there P
exist cs1 ,ℓ in k,
cs1 ,ℓ ω s1 ,ℓ . If
ℓ ∈ [N1 ], with all but a finite number of cs1 ,ℓ being zero, such that b s1 − b′s1 =
ℓ∈[N1 ]
P
cs1 ,ℓ f s1 ,ℓ , then f − f ′ ∈ I. It is clear that f ′ ∈ T if f ′ = 0. If f ′ , 0, then
we take f ′ = f −
ℓ∈[N1 ]

f ′ = b′s1 as1 + g′ for some g′ ∈ Xk (k) with deg g′ < deg f = s1 . Thus g′ ∈

sL
1 −1
i=0

kai =

sL
1 −1

T 0 ai so

i=s0

that f ′ is also in T .
Now we assume that the claim has been proved for f with deg f 6 n − 1 for a given n > 1 and
show that the claim is true when deg f = n. Assume that in( f ) = bn an . Let r ∈ {0} ∪ [NI ] be such
that sr 6 n < sr+1 . So Ωn = Ωsr by Eq. (4), and hence bn − b′n ∈ Ωsr for some b′n ∈ T r .PThen there
exist cn,ℓ ∈ k, ℓ ∈ [Nr ], all but finitely many of which being zero, such that bn − b′n =
cn,ℓ ω sr ,ℓ .
ℓ∈[Nr ]
P
cn,ℓ f sr ,ℓ ), then we have g = b′n an + g′ for some g′ ∈ Xk (k) with
Taking g = f − Pn−sr (
ℓ∈[Nr ]

deg g′ < deg f . If g′ = 0, then put f ′ = b′n an so that f ′ ∈ T and we have f − f ′ ∈ I. If g′ , 0,
then deg g′ < deg f = n. By the induction hypothesis, there exists g′′ ∈ T with deg g′′ 6 deg g′
such that g′ − g′′ ∈ I. If we take f ′ = b′n an + g′′ , then f − f ′ ∈ I. Since deg g′′ 6 deg g′ , we have
deg g′′ < deg f = n, so that f ′ is an element of T .
Suppose that there is another element h′ in T such that h′ , f ′ and f − h′ ∈ I. Then f ′ − h′ ∈ I.
m
k
P
P
bi ai , where bi , ci ∈ T r if sr 6 i < sr+1 for some r ∈ {0} ∪ [NI ].
ci ai , h′ =
Let f ′ =
i=0

i=0

By symmetry, we may assume that k 6 m and put ci = 0 for all i with k < i 6 m. Thus
m
P
h′ − f ′ = (bi − ci )ai . Since h′ , f ′, there is a nonnegative integer p with 0 6 p 6 m such that
i=0

b p − c p , 0 and in(h′ − f ′ ) = (b p − c p )a p . Let t ∈ {0} ∪ [NI ] be such that st 6 p < st+1 . Then
b p ∈ T t , c p ∈ T t and b p − c p ∈ Ωst . But T t is a complete set of representatives of k modulo Ωst , so
b p = c p , a contradiction.
It remains to show that every element of T represents an element of Xk (k)/I. For any f ′ in T
we take f = f ′ in Xk (k), then f − f ′ ∈ I whence f ′ ∈ T is corresponding to the element f + I in
Xk (k)/I. Thus, the underlying set of Xk (k)/I is T and the statement follows at once.


We remark that the underlying k-module T in Eq. (10) is usually not the direct sum of these
k-modules (k/Ωs j )ai for a nonhomogeneous Rota-Baxter ideal I, even though we write it in the
form of a direct sum. The scalar multiplication by k should be defined according to the k-module
Xk (k)/I. We illustrate this by the following example.

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

11

Example 3.10. Let XZ (Z) be the free Rota-Baxter Z-algebra on Z of weight 1. Let I be the RotaBaxter ideals of XZ (Z) generated by the element f = 2(a1 + a0 ). Then XZ (Z)/I and XZ (Z)/in(I)
are in bijection as sets, but are not isomorphic as Z-modules (that is, abelian groups).
L
Proof. Since in( f ) = 2a1 , we have 2am+1 = Pm (2a1 ) ∈ in(I) for m > 0 whence in(I) ⊇
2Zai .
i>1

We note from Lemma 3.4 that all elements of I can be obtained from iterated operations on
f by the Rota-Baxter operator P, the scalar product, the multiplication and the addition of the
algebra XZ (Z). On one hand, we have Pm ( f ) = 2(am+1 + am ). On the other hand, it follows from
Eq. (2) that
am f = 2(m + 1)(am+1 + am ).

Thus, any element of I must be of the form
2c1 (a1 + a0 ) + 2c2 (a2 + a1 ) + · · · + 2cn+1 (an+1 + an ),
L
L
where n ∈ N, ci ∈ Z, 1 6 i 6 n + 1. Consequently, in(I) ⊆
2Zai and hence in(I) =
2Zai .
(11)

i>1

i>1

Clearly, A(in(I)) = {(1, 2Z)}, whence both the underlying sets of XZ (Z)/in(I) and XZ (Z)/I are


M M



Za0
Z
a
2
i

i>1

by Theorems 3.6 and 3.9.
For the Z-module XZ (Z)/in(I), it follows from 2a1 ∈ in(I) that 2a1 = 0. However, for the
Z-module XZ (Z)/I, we have 2a1 + I = −2a0 + I so that 2a1 = −2a0 , which is clearly not zero
since −2a0 is not in I by Eq. (11). Therefore, the Rota-Baxter ideals I and in(I) share the same
quotient sets, but not Z-modules. From the fact that 2a1 , 0 in XZ (Z)/I we also see that XZ (Z)/I
is not the direct sum of the Z-modules Za0 and Z2 ai , i > 1.

The following example shows that a similar phenomenon can already be found in the polynomial algebra k[x].
Example 3.11. Let I be the ideal of Z[x] generated by the polynomial 2x + 2. Then the initial
ideal in6 (I) is generated by 2x. Thus the ideal I and its initial ideal in(I) have the same quotient
set


M M


n
 .
Z
Z
x
2

n>1

However, the two ideals do not have isomorphic quotient groups. For instance, 2x + I = −2 + I is
not zero since −2 is not in I, but 2x + in6 (I) = 0 since 2x ∈ in6 (I). Furthermore, 2x + I , 0 shows
that Z[x]/I is not the direct sum of the abelian groups Z and Z2 xn , n > 1.
4. Characteristics of Rota-Baxter rings
We next focus on the case when k = Z and classify the Rota-Baxter ideals and prime RotaBaxter ideal of XZ (Z). Note that XZ (Z) is the initial object in the category of unitary Rota-Baxter
rings (that is, Rota-Baxter Z-algebras). So we are talking about characteristics of Rota-Baxter
rings.

12

LI GUO AND HOUYI YU

Theorem 4.1. Let I be a Rota-Baxter ideal of XZ (Z). Then there exist a positive integer m and
m pairs (s1 , ω1), · · · , (sm , ωm ) ∈ N × P with s j < s j+1 , ω j+1 , ω j and ω j+1 |ω j , 1 6 j 6 m − 1, such
that A(I) = {(s1 , ω1 Z), · · · , (sm , ωm Z)}. Thus if I is homogeneous then
 −1

m  sM

M
 j+1


I=
ω
Za
sj
i


j=1

i=s j

and the quotient XZ (Z)/I is isomorphic to


 −1
m  sM

M
 j+1
 ,

Z
ω j ai 



(12)

j=0

i=s j

with the convention that s0 = 0, ω s0 = 0 and sm+1 = ∞. If I is nonhomogeneous then XZ (Z)/I
has the same underlying set as Xk (k)/in(I) in Eq. (12).
Conversely, for any positive integer m and m pairs (s1 , ω1 ), · · · , (sm , ωm ) ∈ N × P with s j < s j+1 ,
ω j+1 , ω j and ω j+1 |ω j , 1 6 j 6 m − 1, there is a Rota-Baxter ideal I of XZ (Z) such that the
underlying set of XZ (Z)/I is the construction defined by Eq. (12).
Proof. Let I be a Rota-Baxter ideal of XZ (Z). Since Z satisfies the ascending chain condition
on ideals, I has only finite ascend steps, say s1 , · · · , sm , where m is a positive integer. Since Z
is a PID, we have Ωs j = ω j Z for some positive integer ω j , j = 1, · · · , m. Then, by Lemma 3.1,
we have ω j+1 , ω j and ω j+1 |ω j , 1 6 j 6 m − 1. It follows from the definition of A(I) that
A(I) = {(s1 , ω1 Z), · · · , (sm , ωm Z)}. Consequently, in view of Theorems 3.6 and 3.9, the first part
of the theorem follows.
Conversely, take Ωs j = ω j Z. Then it follows from s j < s j+1 , ω j+1 , ω j and ω j+1 |ω j , 1 6 j 6
m − 1 that Ωs1 ( Ωs2 ( · · · ( Ωsm . So by Theorem 3.9, there is a Rota-Baxter ideal I of XZ (Z)

such that the underlying set of XZ (Z)/I is the construction defined by Eq. (12).
Let k be an integral domain with characteristic 0. The next lemma tells us that if Ωt = kωt is a
principal ideal for t = st(I), then the element f ∈ I with in( f ) = ωt at is completely determined by
ωt and the number of nonzero terms in f .
Proposition 4.2. Let k be an integral domain with characteristic 0 and Xk (k) the free RotaBaxter algebra of weight λ, and let I ⊆ Xk (k) be a nonzero Rota-Baxter ideal. Suppose that
st(I) = t and Ωt = kωt .
t
P
(a) If f = ci ai ∈ I with ct = ωt and cr , 0, then
i=r
!
t − r t−i
λ ct ,
r 6 i 6 t − 1.
ci =
t−i
L
kai .
(b) If λ = 0, then I ⊆
i>t
L
kai .
(c) If λ = 0 and k is a field, then I =
i>t

Proof. (a) Since I is a Rota-Baxter ideal and f =

t
P

ci ai is in I, we have g1 := (t+1)P( f )−a1 f ∈ I.

i=r

Now from Eq. (2) it follows that
a1 f = (t + 1)ct at+1 +

t
X

i=r+1

i(ci−1 + λci )ai + λrcr ar ,

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

13

which together with
(t + 1)P( f ) = (t + 1)

t
X

ci ai+1 = (t + 1)

i=r

t+1
X

ci−1 ai

i=r+1

implies that
g1 =

t
X

((t + 1 − i)ci−1 − λici ) ai − λrcr ar .

i=r+1

Thus, the coefficient of at is in Ωt , that is, ct−1 − λtct is in kct so that ct−1 is in kct . Hence there
exists b ∈ k such that ct−1 = bct . Then
t
X
((t + 1 − i)ci−1 + (λ(t − i) − b)ci ) ai + (λ(t − r) − b)cr ar
g2 :=g1 − (b − λt) f =
i=r+1

is in I. But the coefficient of at in g2 is 0, so the fact that st(I) = t yields that g2 = 0, which is
equivalent to (λ(t − r) − b)cr = 0 and
(13)

(t + 1 − i)ci−1 + (λ(t − i) − b)ci = 0,

r + 1 6 i 6 t.

By the hypotheses that k is an integral domain and cr , 0, it follows from (λ(t − r) − b)cr = 0 that
b = λ(t − r). Consequently, Eq. (13) becomes
(t + 1 − i)ci−1 = λ(i − r)ci ,

r + 1 6 i 6 t.

So for each r 6 i 6 t − 1, we have
t 
t 
Y
 Y

(t + 1 − j)c j−1 =
λ( j − r)c j ,
j=i+1

j=i+1

which is equivalent to
(t − i)!

t
Y

j=i+1

c j−1

t
− r)! Y
=λ
c j.
(i − r)! j=i+1
t−i (t

Since k is an integral domain with characteristic 0, we obtain
!
(t − r)!
t − r t−i
t−i
λ ct ,
ci = λ
ct =
t−i
(t − i)!(i − r)!
This completes the proof of part (a) of this lemma.
(b) Take arbitrary f ∈ I. Since st(I) = t, we may assume that

r 6 i 6 t − 1.

f = bm am + bm−1 am−1 + · · · + bt at + f0 ∈ I,
where m > t and deg f0 6 t − 1. Note that Ωt = kωt . So there exists an element g in I with ωt at as
the initial term. Since λ = 0, by part (a), we must have that g = ωt at is in I. So ωt am = Pm−t (ωt at )
is in I. Then h := ωt f − bm ωt am is in I, that is,
(14)

h = bm−1 ωt am−1 + · · · + bt ωt at + ωt f0

is in I. Since P(I) ⊆ I, we have ωt aℓ = Pℓ−t (ωt at ) ∈ I for all t 6 ℓ 6 m − 1, so that bm−1 ωt am−1 +
· · · + bt ωt at ∈ I. Then Eq. (14) implies that ωt f0 ∈ I. Note that deg f0 6 t − 1, which
Ltogether
kai and
with st(I) = t yields ωt f0 = 0. But k is an integral domain, so f0 = 0. Therefore, I ⊆
i>t

we are done.

14

LI GUO AND HOUYI YU

(c) Since λ = 0, we know that ωt at is in I. But k is a field, so Ωt = k, whence
L
kai .
together with part (b) gives I =

L

kai ⊆ I, which

i>t



i>t

We finally give a classification for the prime Rota-Baxter ideals of XZ (Z) when λ = 0. Recall
that a Rota-Baxter ideal I of a Rota-Baxter algebra (R, P) is said to be prime if it is a prime ideal
with P(I) ⊆ I. If λ = 0, then for any m, n ∈ N, it follows from Eq. (2) that
!
m+n
(15)
am+n .
am an =
m
Theorem 4.3. Let XZ (Z) be the free Rota-Baxter Z-algebra on Z of weight 0 and let I be a proper
nonzero Rota-Baxter ideal of XZ (Z). Then I is prime if and only if


M M 


I = pZa0
Zai  ,
i>1

where p is either 0 or a prime number of Z.

By Theorems 4.1 and 4.3, the quotient of a prime characteristic of a Rota-Baxter ring is either
Z or Z/pZ for a prime number p, as in the case of prime characteristic of a ring.
Proof. Let I be a prime Rota-Baxter ideal. Denote t = st(I) and Ωt = ωZ for some positive integer
ω. Then, by Proposition 4.2(a), the element in I of the form
ωat + lower degree terms
must be ωat , that is, ωat ∈ I.
If t > 1, then ωat−1 < I since st(I) = t. From Eq. (15) we know that a1 (ωat−1 ) = tωat ∈ I. Since
I is prime, we have a1 ∈ I, which means that t = st(I) 6 1 and hence t = 1. Now a1 ∈ I gives
1 ∈ Ω1 so that Ω1 = Z. By Lemma 3.1, Ω j = Z for all positive integer j. Therefore,
M
I=
Zai .
i>1

If t = 0, then ωa0 ∈ I. Since I is a prime ideal and a0 is the identity of XZ (Z), ω must be a
prime number. Let ω = p. Then p > 2 and pa0 ∈ I. So a1p = p!a p = (p − 1)!P p (pa0 ) ∈ I which
means that a1 is in the prime ideal I. Hence I is generated by pa0 and a1 . Therefore,


M M 

I = pZa0
Zai  .

i>1

Conversely, I is a homogeneous Rota-Baxter ideal generated by either {a1 } or {pa0 , a1 }, where
p is a prime number. By Theorem 4.1, if I is generated by {a1 }, then XZ (Z)/I is isomorphic to
Z. If I is generated by {pa0 , a1 }, then XZ (Z)/I is isomorphic to Z p . In either case, I is a prime
ideal.

Acknowledgements: This work was supported by NSFC grant 11426183, 11501467, Chongqing
Research Program of Application Foundation and Advanced Technology (No. cstc2014jcyjA00028).
H. Yu would like to thank the NYU Polytechnic School of Engineering for hospitality and support.

CHARACTERISTICS OF ROTA-BAXTER ALGEBRAS

15

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Department of Mathematics and Computer Science, Rutgers University, Newark, NJ 07102, USA
E-mail address: liguo@rutgers.edu
School of Mathematics and Statistics, Southwest University, Chongqing, China
E-mail address: yuhouyi@swu.edu.cn

