Gr\"obner--Shirshov bases for commutative dialgebras

TL;DR

This paper develops a Gr"obner--Shirshov bases theory for commutative dialgebras, providing normal forms, solvability of the word problem, and basis construction methods, extending algebraic computational tools to this class of algebras.

Contribution

It introduces a unique reduced Gr"obner--Shirshov basis theory for commutative dialgebras, including finite basis existence and applications to word problems and ideal equality.

Findings

Unique reduced Gr"obner--Shirshov bases exist for ideals in commutative dialgebras.

The word problem for finitely presented commutative dialgebras is solvable.

Deciding ideal equality is solvable when the generating set is finite.

Abstract

We establish Gr\"obner--Shirshov bases theory for commutative dialgebras. We show that for any ideal $I$ of $Di[X]$, $I$ has a unique reduced Gr\"obner--Shirshov basis, where $Di[X]$ is the free commutative dialgebra generated by a set $X$, in particular, $I$ has a finite Gr\"obner--Shirshov basis if $X$ is finite. As applications, we give normal forms of elements of an arbitrary commutative disemigroup, prove that the word problem for finitely presented commutative dialgebras (disemigroups) is solvable, and show that if $X$ is finite, then the problem whether two ideals of $Di[X]$ are identical is solvable. We construct a Gr\"obner--Shirshov basis in associative dialgebra $Di\langle X\rangle$ by lifting a Gr\"obner--Shirshov basis in $Di[X]$.

Equations

$$⊢:D\times D\to D,⊣:D\times D\to D,$$

$$\{\begin{array}{l}a⊣(b⊢c)=a⊣(b⊣c),\\ (a⊣b)⊢c=(a⊢b)⊢c,\\ a⊢(b⊣c)=(a⊢b)⊣c.\end{array}$$

$$\lfloor X^{+}\rfloor :=\{\lfloor x_{i_1}{x}_{i_2}\cdots x_{i_n}\rfloor \mid i_1,\dots ,i_n\in I,x_{i_1}\le x_{i_2}\le \cdots \le x_{i_n},n\in {\mathbb{Z}}^{+}\},$$

$$\lfloor X^{\ast }\rfloor :=\lfloor X^{+}\rfloor \cup \{\epsilon \},$$

$$\lfloor u\rfloor =\lfloor x_{j_1}{x}_{j_2}\cdots x_{j_n}\rfloor :=\lfloor x_{i_1}{x}_{i_2}\cdots x_{i_n}\rfloor ,x_{i_1}\le x_{i_2}\le \cdots \le x_{i_n},$$

$$\lfloor X^{+}{\rfloor }_{{}_1}:$$

$$\lfloor XX{\rfloor }_{{}_2}:$$

$$Disgp[X]:=(\lfloor X^{+}{\rfloor }_{{}_1}\cup \lfloor XX{\rfloor }_{{}_2},⊢,⊣).$$

$$\lfloor v{\rfloor }_{p_2}⊢\lfloor u{\rfloor }_{p_1}=\lfloor u{\rfloor }_{p_1}⊢\lfloor v{\rfloor }_{p_2}=\lfloor u{\rfloor }_{p_1}⊣\lfloor v{\rfloor }_{p_2}=\lfloor v{\rfloor }_{p_2}⊣\lfloor u{\rfloor }_{p_1}=\lfloor uv{\rfloor }_1,$$

$$x⊣x^{\prime }=x^{\prime }⊣x=\lfloor x{x}^{\prime }{\rfloor }_1,$$

$$x⊢x^{\prime }=x^{\prime }⊢x=\lfloor x{x}^{\prime }{\rfloor }_2.$$

$$\lfloor u{\rfloor }_1:={\dot{x}}_{i_1}{x}_{i_2}\cdots x_{i_n}=x_{i_1}⊣x_{i_2}⊣\cdots ⊣x_{i_n},\lfloor v{\rfloor }_2:=x_{j_1}{\dot{x}}_{j_2}=x_{j_1}⊢x_{j_2}.$$

$$\lfloor u\rfloor >\lfloor v\rfloor \iff (|u|,x_{i_1},x_{i_2},\cdots ,x_{i_n})>(|v|,x_{j_1},x_{j_2},\cdots ,x_{j_m})\text{mbox}lexicographically.$$

$$\lfloor u\rfloor >\lfloor v\rfloor \Rightarrow \lfloor uw\rfloor >\lfloor vw\rfloor .$$

$$\lfloor u{\rfloor }_m{>}_d\lfloor v{\rfloor }_n\text{mbox}if(\lfloor u\rfloor ,m)>(\lfloor v\rfloor ,n)\text{mbox}lexicographically.$$

$$f={\alpha }_1\lfloor u_1{\rfloor }_{m_1}+\cdots +{\alpha }_n\lfloor u_n{\rfloor }_{m_n},$$

$$f={\alpha }_1\lfloor x{x}^{\prime }{\rfloor }_2+{\alpha }_2\lfloor x{x}^{\prime }{\rfloor }_1+g,$$

$$\lfloor u{\rfloor }_m>\lfloor v{\rfloor }_n,\lfloor u\rfloor >\lfloor v\rfloor$$

$$\lfloor u{\rfloor }_m>\lfloor v{\rfloor }_n,\lfloor u\rfloor =\lfloor v\rfloor$$

$$\overline{(f⊢\lfloor u{\rfloor }_m)}\le \overline{f}⊢\lfloor u{\rfloor }_m,\overline{(f⊣\lfloor u{\rfloor }_m)}\le \overline{f}⊣\lfloor u{\rfloor }_m.$$

$$\overline{f}=\lfloor x_2{x}_3{\rfloor }_2,lt(f)=2\lfloor x_2{x}_3{\rfloor }_2,lc(f)=2,\tilde{f}=\lfloor x_2{x}_3\rfloor ,r{}_{{}_f}=-2\lfloor x_2{x}_3{\rfloor }_1+3\lfloor x_1{x}_3{\rfloor }_2.$$

$$\overline{(f⊣x_1)}=\overline{(f⊢x_1)}$$

$$\overline{(r{}_{{}_f}⊣x_1)}=\overline{(r{}_{{}_f}⊢x_1)}$$

$$g=s⊢x,\text{mbox}wheres\in S,|\tilde{s}|=1,x\in X,$$

$$g=s⊣\lfloor a{\rfloor }_1,\text{mbox}wheres\in S,s\text{mbox}isstrong,\lfloor a\rfloor \in \lfloor X^{+}\rfloor .$$

$$\overline{g}=\{\begin{array}{ll}\overline{s}& \text{if g=s,}\\ \lfloor \tilde{s}x{\rfloor }_2& \text{if g=s⊢x,}\\ \lfloor \tilde{s}a{\rfloor }_1& \text{if g=s⊣⌊a⌋1.}\end{array}$$

$$\tilde{s}>\lfloor \tilde{s}c\rfloor >\lfloor \tilde{s}{c}^2\rfloor >\lfloor \tilde{s}{c}^3\rfloor >\cdots ,$$

$$\lfloor sb{\rfloor }_m=\{\begin{array}{ll}s& \text{if b=ε,}\\ s⊢b& \text{if ∣s∣=1 and b∈X,}\\ s⊣\lfloor b{\rfloor }_1& \text{if s is strong and ⌊b⌋∈⌊X+⌋.}\end{array}$$

$$f\equiv g\mathrm{mod}(S,\lfloor w{\rfloor }_m)$$

$$Irr(S):=\{\lfloor u{\rfloor }_n\in \lfloor X^{+}{\rfloor }_{{}_1}\cup \lfloor XX{\rfloor }_{{}_2}\mid \lfloor u{\rfloor }_n\ne \overline{\lfloor sb{\rfloor }_m}\text{mbox}foranynormalS\text{mbox}-polynomial\lfloor sb{\rfloor }_m\}$$

$$s,s⊣\lfloor a{\rfloor }_m,s⊢\lfloor a{\rfloor }_m,$$

$$f=\sum _i{\alpha }_i\lfloor u_i{\rfloor }_{n_i}+\sum _j{\beta }_j\lfloor s_j{b}_j{\rfloor }_{m_j},$$

$$\lfloor s_1{b}_1{\rfloor }_{m_1}-\lfloor s_2{b}_2{\rfloor }_{m_2}\equiv 0\mathrm{mod}(S,\lfloor w{\rfloor }_m).$$

Full text

1 Introduction

Recently the study of algebraic properties of dialgebras has attracted considerable attention. Dialgebras are vector spaces over a field equipped with two binary bilinear associative operations satisfying some axioms. Moreover, if operations of a dialgebra coincide, we obtain an associative algebra and so, dialgebras are a generalization of associative algebras. The class of dialgebras is rather interesting despite the lack of simple examples distinct from the associative algebras. It is well-known that for Lie algebras there is a notion of a universal enveloping associative algebra. By the Poincar$\acute{e}$-Birkhoff-Witt theorem, to a given Lie algebra $L$ there exists an associative algebra $A$ such that $L$ is isomorphic to a subalgebra of the Lie algebra $A^{(-)}$. The universal enveloping algebra for Leibniz algebras, which are a non-commutative variation of Lie algebras, was found in Loday (1993, 1995). Dialgebras serve as these enveloping algebras. In Bokut et al. (2010), a Composition-Diamond lemma for dialgebras was given to obtain normal forms for some dialgebras including the universal enveloping algebra for Leibniz algebras. Pozhidaev (2009) studied the connection of Rota-Baxter algebras and dialgebras with associative bar-unity. Kolesnikov (2008) proved that each dialgebra may be obtained in turn from an associative conformal algebra. Lately normal forms of free commutative dialgebras were found by Zhuchok (2010), Zhang and Chen (2017).

Gröbner bases and Gröbner–Shirshov bases were invented independently by A.I. Shirshov for ideals of free (commutative, anti-commutative) non-associative algebras in Shirshov (1962, 2009), free Lie algebras in Shirshov (2009) and implicitly free associative algebras in Shirshov (2009) (see also Bergman (1978); Bokut (1976)), by Hironaka (1964) for ideals of the power series algebras (both formal and convergent), and by Buchberger (1970) for ideals of the polynomial algebras. Gröbner bases and Gröbner–Shirshov bases theories have been proved to be very useful in different branches of mathematics, including commutative algebra and combinatorial algebra. It is a powerful tool to solve the following classical problems: normal form; word problem; conjugacy problem; rewriting system; automaton; embedding theorem; PBW theorem; extension; homology; growth function; Dehn function; complexity; etc. See, for example, the books by Adams and Loustaunau (1994); Bokut and Kukin (1994); Buchberger et al. (1982); Buchberger and Winkler (1998); Cox et al. (2015); Eisenbud (1995) and the surveys by Bokut and Chen (2008, 2014); Bokut et al. (2000); Bokut and Kolesnikov (2000, 2004); Bokut and Shum (2005).

In Gröbner–Shirshov bases theory for a category of algebras, a key part is to establish “Composition-Diamond lemma” for such algebras. The name “Composition-Diamond lemma” combines the Diamond Lemma in Newman (1942), the Composition Lemma in Shirshov (1962) and the Diamond Lemma in Bergman (1978).

In this paper, we establish Gröbner–Shirshov bases theory for commutative dialgebras. A Composition Diamond lemma for commutative dialgebras is given, see Theorem 14. We show that for a given monomial-center ordering, each ideal of $Di[X]$ has a unique reduced Gröbner–Shirshov basis; if $X$ is finite, then $Di[X]$ is Noetherian and each ideal of $Di[X]$ has a finite Gröbner–Shirshov basis, and an algorithm is given to find such a finite (reduced) Gröbner–Shirshov basis. As applications, we give normal forms of elements of an arbitrary commutative disemigroup and prove that for finitely presented commutative dialgebras (disemigroups), the word problem and the problem whether two ideals of $Di[X]$ are identical are solvable. These results will be applied to computer algebra systems, which are referring to the common use of the Buchberger approach to polynomials. Moreover, we prove a theorem on the pair of algebras ($Di[X],Di\langle X\rangle$) following the spirit of Eisenbud-Peeva-Sturmfels’ theorem on the pair ($k[X],k\langle X\rangle$) in Eisenbud et al. (1998). Namely, we construct a Gröbner–Shirshov basis in associative dialgebra $Di\langle X\rangle$ by lifting a given Gröbner–Shirshov basis $S$ in commutative dialgebra $Di[X]$.

The paper is organized as follows. In section 2, we review the free commutative dialgebra $Di[X]$ generated by $X$ over a field $\mathbf{k}$. In section 3, we establish a Composition-Diamond lemma for commutative dialgebras. In section 4, it is shown that each ideal of a finitely generated polynomial dialgebra has a finite Gröbner–Shirshov basis. Section 5 gives normal forms of commutative disemigroups and shows the word problem for finitely presented commutative dialgebras (disemigroups) is solvable. The main results in this section are similar to ones in Gröbner bases theory for commutative algebras (Buchberger, 1965, 1970; Buchberger et al., 1982; Buchberger and Winkler, 1998). Section 6 provides a method by which we can lift commutative Gröbner–Shirshov bases to associative ones.

2 Free commutative dialgebras

Throughout the paper, we fix a field $\mathbf{k}$. $\mathbb{Z}^{+}$ stands for the set of positive integers.

Definition 1**.**

(Loday et al., 2001)* A disemigroup (dialgebra) is a set ($\mathbf{k}$-linear space) $D$ equipped with two maps*

[TABLE]

where $\vdash$ and $\dashv$ are associative and satisfy the following identities: for all $a,b,c\in D,$

[TABLE]

A disemigroup (dialgebra) $(D,\vdash,\dashv)$ is commutative if both $\vdash$ and $\dashv$ are commutative.

Let $X=\{x_{i}|i\in I\}$ be a total-ordered set, $X^{+}\ (X^{*})$ the free semigroup (monoid) generated by $X$,

[TABLE]

the free commutative semigroup without unit generated by $X$, and

[TABLE]

the free commutative monoid generated by $X$, where $\varepsilon$ is the empty word.

For any $u=x_{j_{1}}x_{j_{2}}\cdots x_{j_{n}}\in X^{+},\ x_{j_{k}}\in X$, we define

[TABLE]

where $\{x_{i_{1}},x_{i_{2}},\dots,x_{i_{n}}\}=\{x_{j_{1}},x_{j_{2}},\dots,x_{j_{n}}\}$ as multisets.

Write

[TABLE]

where $|v|$ is the length of $v$. For any $h=\lfloor u\rfloor_{p}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$, we call $\lfloor u\rfloor$ the associative (commutative) word of $h$. For convenience, we denote $\lfloor u\rfloor_{1}=u$ if $u\in X$.

Lemma 2**.**

(Zhuchok, 2010; Zhang and Chen, 2017)* Let $X=\{x_{i}\mid i\in I\}$ be a total-ordered set and*

[TABLE]

Then $Disgp[X]$ is the free commutative disemigroup generated by $X$, where the operations $\vdash$ and $\dashv$ are as follows: for any $x,x^{\prime}\in X,\ \lfloor u\rfloor_{p_{1}},\ \lfloor v\rfloor_{p_{2}}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ with $|uv|>2$,

[TABLE]

Let $Di[X]$ be the $\mathbf{k}$-linear space with a $\mathbf{k}$-basis $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then $(Di[X],\ \vdash,\dashv)$ is the free commutative dialgebra generated by $X$.

For example, if $\lfloor u\rfloor,\lfloor v\rfloor\in\lfloor X^{+}\rfloor,\ \lfloor u\rfloor=\lfloor x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\rfloor,\lfloor v\rfloor=\lfloor x_{j_{1}}x_{j_{2}}\rfloor$, $x_{i_{l}},x_{j_{k}}\in X$, then with the notation as in Bokut et al. (2010); Zhang and Chen (2017),

[TABLE]

Let $X$ be a well-ordered set. We define the deg-lex ordering on $\lfloor X^{+}\rfloor$ by the following: for any $\lfloor u\rfloor=\lfloor x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\rfloor,\lfloor v\rfloor=\lfloor x_{j_{1}}x_{j_{2}}\cdots x_{j_{m}}\rfloor\in\lfloor X^{+}\rfloor$, where $x_{i_{l}},x_{j_{t}}\in X$,

[TABLE]

An ordering $>$ on $\lfloor X^{+}\rfloor$ is said to be monomial if $>$ is a well ordering and for any $\lfloor u\rfloor,\lfloor v\rfloor,\lfloor w\rfloor\in\lfloor X^{+}\rfloor$,

[TABLE]

Clearly, the deg-lex ordering is monomial.

3 Composition-Diamond lemma for commutative dialgebras

Let $>$ be a monomial ordering on $\lfloor X^{+}\rfloor$. We define the monomial-center ordering $>_{d}$ on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ as follows. For any $\lfloor u\rfloor_{m},\lfloor v\rfloor_{n}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$,

[TABLE]

In particular, if $>$ is the deg-lex ordering on $\lfloor X^{+}\rfloor$, we call the ordering defined by $(\ref{equ0})$ the deg-lex-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. For simplicity of notation, we write $>$ instead of $>_{d}$ when no confusion can arise. It is clear that a monomial-center ordering is a well ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Such an ordering plays an important role in the sequel.

Here and subsequently, $>$ is a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ unless otherwise stated.

For any nonzero polynomial $f\in Di[X]$,

[TABLE]

where each $0\neq\alpha_{i}\in\mathbf{k},\ \lfloor u_{i}\rfloor_{m_{i}}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ and $\lfloor u_{1}\rfloor_{m_{1}}>\cdots>\lfloor u_{n}\rfloor_{m_{n}}$. We write

$\bullet$ $supp(f):=\{\lfloor u_{1}\rfloor_{m_{1}},\cdots,\lfloor u_{n}\rfloor_{m_{n}}\}$;

$\bullet$ $\overline{f}:=\lfloor u_{1}\rfloor_{m_{1}}$, the leading monomial of $f$;

$\bullet$ $lt(f):=\alpha_{1}\lfloor u_{1}\rfloor_{m_{1}}$, the leading term of $f$;

$\bullet$ $lc(f):=\alpha_{1}$, the coefficient of $\overline{f}$;

$\bullet$ $\widetilde{f}:=\lfloor u_{1}\rfloor$, the associative word of $\overline{f}$;

$\bullet$ $r\!_{{}_{f}}:=f-lt(f)$.

Note that $\widetilde{f}\in\lfloor X^{+}\rfloor$. $f$ is called monic if $lc(f)=1$. For any nonempty subset $S$ of $Di[X]$, $S$ is* monic* if $s$ is monic for all $s\in S$.

For convenience we assume that $\lfloor u\rfloor>0$ for any $\lfloor u\rfloor\in\lfloor X^{+}\rfloor$ and $\widetilde{0}=0$, and $\lfloor u\rfloor_{m}>0$ for any $\lfloor u\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$.

Definition 3**.**

A nonzero polynomial $f\in Di[X]$ is strong if $\widetilde{f}>\widetilde{r\!_{{}_{f}}}$.

The proof of the following proposition follows from the Definition 3.

Proposition 4**.**

A nonzero polynomial $f\in Di[X]$ is not strong if and only if

[TABLE]

where $0\neq\alpha_{1},\alpha_{2}\in\mathbf{k},\ x,x^{\prime}\in X,\ \ g\in Di[X]$ and $\overline{g}<\lfloor xx^{\prime}\rfloor_{1}$.

It is easy to check that $>$ on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ is compatible with operations $\vdash$ and $\dashv$ in the following sense: for any $\lfloor u\rfloor_{m},\lfloor v\rfloor_{n},\lfloor w\rfloor_{k}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$,

[TABLE]

From this, it follows that

Lemma 5**.**

Let $0\neq f\in Di[X]$ and $\lfloor u\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then

[TABLE]

In particular, if $f$ is strong, then $\overline{(f\vdash\lfloor u\rfloor_{m})}=\overline{f}\vdash\lfloor u\rfloor_{m},$ and $\overline{(f\dashv\lfloor u\rfloor_{m})}=\overline{f}\dashv\lfloor u\rfloor_{m}$.

Example 6**.**

Let $X=\{x_{1},x_{2},x_{3}\}$, $x_{3}>x_{2}>x_{1}$, $Char\mathbf{k}\neq 2,3$ and $>$ be the deg-lex-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Let $f=2\lfloor x_{2}x_{3}\rfloor_{2}-2\lfloor x_{2}x_{3}\rfloor_{1}+3\lfloor x_{1}x_{3}\rfloor_{2}$. Then

[TABLE]

The polynomial $f$ is not strong since $\widetilde{f}=\lfloor x_{2}x_{3}\rfloor=\widetilde{r\!_{{}_{f}}}$ and

[TABLE]

Here and subsequently, $S$ denotes a monic subset of $Di[X]$ unless otherwise stated.

Definition 7**.**

Let $S$ be a monic subset of $Di[X]$. A polynomial $g\in Di[X]$ is called a normal $S$-polynomial in $Di[X]$ if either $g\in S$ or $g$ is one of the following:

[TABLE]

or

[TABLE]

If this is so, we also call $g$ a normal $s$-polynomial.

Lemma 8**.**

Let $s\in S$ and $g$ be a normal $s$-polynomial. Then

[TABLE]

In particular, $\overline{g}\geq\overline{s}$, and $\overline{g}=\overline{s}$ if and only if $g=s$.

Proof.

The proof of (3) is straightforward. It remains to prove that $\overline{g}>\overline{s}$ if $\overline{g}\neq\overline{s}$. Suppose that $\overline{g}\neq\overline{s}$. Then $\widetilde{g}=\lfloor\widetilde{s}c\rfloor$ where $\lfloor c\rfloor\in\lfloor X^{+}\rfloor$. We claim that $\widetilde{g}>\widetilde{s}$. Otherwise, $\widetilde{s}>\widetilde{g}=\lfloor\widetilde{s}c\rfloor$. We have an infinite descending chain

[TABLE]

which contradicts the fact that $>$ is a monomial ordering on $\lfloor X^{+}\rfloor$. This clearly forces $\overline{g}>\overline{s}$. ∎

Note that if $g$ is a normal $s$-polynomial, then $\overline{g}=\lfloor\widetilde{s}b\rfloor_{m}$ for some $\lfloor b\rfloor\in\lfloor X^{*}\rfloor$ and $m\in\{1,2\}$.

From now on, we use $\lfloor sb\rfloor_{m}$ to present a normal $S$​-polynomial, where $s\in S,\ \lfloor b\rfloor\in\lfloor X^{*}\rfloor$ and $m\in\{1,2\}$, i.e.

[TABLE]

It follows immediately that $\overline{\lfloor sb\rfloor_{m}}=\lfloor\widetilde{s}b\rfloor_{m}$.

For $\lfloor a\rfloor,\lfloor b\rfloor\in\lfloor X^{+}\rfloor$, we denote the least common multiple of $\lfloor a\rfloor$ and $\lfloor b\rfloor$ by $lcm\{\lfloor a\rfloor,\lfloor b\rfloor\}$.

Definition 9**.**

Let $f,g$ be two monic polynomials in $Di[X]$.

1. (i)

If $f$ is not strong, then for any $x\in X$, we call both $f\vdash x$ and $f\dashv x$ the multiplication compositions of $f$. 2. (ii)

If $f$ is strong, $|\widetilde{f}|>1$ and $supp(f)\cap X\neq\emptyset$, then for any $x\in X$, we call $(f\vdash x)-(f\dashv x)$ the special composition of $f$. 3. (iii)

If $f\neq g$ and $\overline{f}=\overline{g}$, then we call $(f,g)_{\overline{f}}=f-g$ the equal composition of $f$ and $g$. 4. (iv)

Suppose that $|\widetilde{f}|=2$ and $|\widetilde{g}|=1$. If there exists a normal $g$-polynomial $\lfloor gx\rfloor_{m}$ for some $x\in X$ such that $\overline{f}=\overline{\lfloor gx\rfloor_{m}}$, then we call $(f,g)_{\overline{f}}=f-\lfloor gx\rfloor_{m}$ the short intersection composition of $f$ and $g$. 5. (v)

Suppose that $f,g$ are strong, $|\widetilde{f}|+|\widetilde{g}|>3$ and $\overline{f}\neq\overline{g}$. If $\widetilde{f}=\widetilde{g}$, then for any $x\in X$, we call $(f,g)_{\lfloor\widetilde{f}x\rfloor_{1}}=\lfloor fx\rfloor_{1}-\lfloor gx\rfloor_{1}$ the equal multiplication composition of $f$ and $g$; if $\widetilde{f}\neq\widetilde{g}$ and $lcm\{\widetilde{f},\widetilde{g}\}=\lfloor w\rfloor=\lfloor\widetilde{f}a\rfloor=\lfloor\widetilde{g}b\rfloor$ for some $\lfloor a\rfloor\in\lfloor X^{*}\rfloor,\lfloor b\rfloor\in\lfloor X^{+}\rfloor$ and $|w|<|\widetilde{f}|+|\widetilde{g}|$, then we call $(f,g)_{\lfloor w\rfloor_{1}}=\lfloor fa\rfloor_{1}-\lfloor gb\rfloor_{1}$ the long intersection composition of $f$ and $g$.

Definition 10**.**

Let $S$ be a monic subset of $Di[X]$. A polynomial $h\in Di[X]$ is called trivial modulo $S$, if $h=\sum_{i}\alpha_{i}\lfloor s_{i}b_{i}\rfloor_{m_{i}}$, where each $\alpha_{i}\in\mathbf{k},s_{i}\in S,\lfloor b_{i}\rfloor\in\lfloor X^{*}\rfloor$, and $\overline{\lfloor s_{i}b_{i}\rfloor_{m_{i}}}\leq\overline{h}$ if $\alpha_{i}\neq 0$.

A monic set $S$ is called a Gröbner–Shirshov basis in $Di[X]$ if any composition of polynomials in $S$ is trivial modulo $S$.

$S$* is said to be closed under the multiplication and special compositions if any multiplication and special composition of polynomials in $S$ is trivial modulo $S$.*

For convenience, for any $f,g\in Di[X]$ and $\lfloor w\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$, we write

[TABLE]

which means that$f-g=\sum_{i}\alpha_{i}\lfloor s_{i}b_{i}\rfloor_{m_{i}}$, where each $\alpha_{i}\in\mathbf{k},s_{i}\in S,\lfloor b_{i}\rfloor\in\lfloor X^{*}\rfloor$, and $\overline{\lfloor s_{i}b_{i}\rfloor_{m_{i}}}<\lfloor w\rfloor_{m}$ if $\alpha_{i}\neq 0$.

We set

[TABLE]

and use $Id(S)$ to denote the ideal of $Di[X]$ generated by $S$.

Lemma 11**.**

Let $S$ be closed under the multiplication and special compositions, $\lfloor sb\rfloor_{n}$ a normal $S$​-polynomial, $\lfloor a\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ and $f\in Id(S)$. Then

1. (i)

$\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{m}$* is trivial modulo $S$;* 2. (ii)

$\lfloor sb\rfloor_{n}\vdash\lfloor a\rfloor_{m}$* is trivial modulo $S$;* 3. (iii)

$f=\sum_{i}\alpha_{i}\lfloor s_{i}b_{i}\rfloor_{m_{i}},$* where each $\alpha_{i}\in\mathbf{k},s_{i}\in S,\lfloor b_{i}\rfloor\in\lfloor X^{*}\rfloor$.*

Proof.

$(i)$ It suffices to show that $\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{1}$ is trivial modulo $S$. The proof is by induction on $|a|$. Suppose that $|a|=1$. The result holds trivially if $s$ is strong. Assume that $s$ is not strong. Then $\lfloor b\rfloor$ is empty and we are done by the triviality of multiplication composition. Now, let $|a|>1$ and write $\lfloor a\rfloor=y\lfloor a_{1}\rfloor$ in $X^{*}$, where $y\in X,\lfloor a_{1}\rfloor\in\lfloor X^{+}\rfloor$. Then, by the above arguments, $\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{1}=(\lfloor sb\rfloor_{n}\dashv y)\dashv\lfloor a_{1}\rfloor_{1}$ is a linear combination of polynomials of the form $\lfloor s^{\prime}c\rfloor_{l}\dashv\lfloor a_{1}\rfloor_{1}$, where $s^{\prime}\in S,\lfloor c\rfloor\in\lfloor X^{*}\rfloor$ and $\overline{\lfloor s^{\prime}c\rfloor_{l}}\leq\overline{\lfloor sb\rfloor_{n}\dashv y}$. By induction, $\lfloor s^{\prime}c\rfloor_{l}\dashv\lfloor a_{1}\rfloor_{1}$ is a linear combination of normal $S$​-polynomials $\lfloor s_{i}b_{i}\rfloor_{m_{i}}$ and $\overline{\lfloor s_{i}b_{i}\rfloor_{m_{i}}}\leq\overline{\lfloor s^{\prime}c\rfloor_{l}\dashv\lfloor a_{1}\rfloor_{1}}\leq\overline{\lfloor s^{\prime}c\rfloor_{l}}\dashv\lfloor a_{1}\rfloor_{1}\leq(\overline{\lfloor sb\rfloor_{n}\dashv y})\dashv\lfloor a_{1}\rfloor_{1}=\overline{\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{1}}$, which implies that $\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{1}$ is trivial modulo $S$.

$(ii)$ If $b\neq\varepsilon$ or $|a|>1$, then $\lfloor sb\rfloor_{n}\vdash\lfloor a\rfloor_{m}=\lfloor sb\rfloor_{n}\dashv\lfloor a\rfloor_{1}$ and we have done by $(i)$. It remains to prove that $s\vdash a$ is trivial modulo $S$, where $a\in X$. Note that $s\vdash a$ is a normal $S$​-polynomial if $|\widetilde{s}|=1$. Let $|\widetilde{s}|>1$. If $s$ is not strong, then we are done by the triviality of multiplication composition. Otherwise, $s\dashv a$ is a normal $S$​-polynomial and $\overline{s\dashv a}=\overline{s\vdash a}$. For $supp(s)\cap X=\emptyset$ it is easy to see that $s\vdash a=s\dashv a$ and we are done. For $supp(s)\cap X\neq\emptyset$, by the triviality of special composition, $s\vdash a-s\dashv a$ is a linear combination of normal $S$​-polynomials $\lfloor s_{i}b_{i}\rfloor_{m_{i}}$ and $\overline{\lfloor s_{i}b_{i}\rfloor_{m_{i}}}\leq\overline{s\vdash a-s\dashv a}<\overline{s\vdash a}$. Therefore the result holds.

$(iii)$ It is clear that $f$ is a linear combination of polynomials of the forms

[TABLE]

where $s\in S,\lfloor a\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then the result follows from (i) and (ii). ∎

Lemma 12**.**

Let $S$ be a monic subset of $Di[X]$. Then for any nonzero $f\in Di[X]$,

[TABLE]

where each $\lfloor u_{i}\rfloor_{n_{i}}\in Irr(S),\ \alpha_{i},\beta_{j}\in\mathbf{k},\ s_{j}\in S,\lfloor b_{j}\rfloor\in\lfloor X^{*}\rfloor$, $\lfloor u_{i}\rfloor_{n_{i}}\leq\overline{f}$ and $\overline{\lfloor s_{j}b_{j}\rfloor_{m_{j}}}\leq\overline{f}$.

Proof.

If $\overline{f}\in Irr(S)$, then take $\lfloor u\rfloor_{n}=\overline{f}$ and $f_{1}=f-lc(f)\lfloor u\rfloor_{n}$. If $\overline{f}\notin Irr(S)$, then $\overline{f}=\overline{\lfloor sb\rfloor_{m}}$ for some normal $S$​-polynomial $\lfloor sb\rfloor_{m}$ and take $f_{1}=f-lc(f)\lfloor sb\rfloor_{m}$. In both cases, we have $\overline{f_{1}}<\overline{f}$ and the result follows from induction on $\overline{f}$. ∎

Lemma 13**.**

Let $S$ be a Gröbner–Shirshov basis in $Di[X]$, $\lfloor s_{1}b_{1}\rfloor_{m_{1}}$ and $\lfloor s_{2}b_{2}\rfloor_{m_{2}}$ normal $S$​-polynomials. If $\lfloor w\rfloor_{m}=\overline{\lfloor s_{1}b_{1}\rfloor_{m_{1}}}=\overline{\lfloor s_{2}b_{2}\rfloor_{m_{2}}}$, then

[TABLE]

Proof.

Since $\lfloor w\rfloor_{m}=\overline{\lfloor s_{1}b_{1}\rfloor_{m_{1}}}=\overline{\lfloor s_{2}b_{2}\rfloor_{m_{2}}}$, it follows that $\lfloor w\rfloor=\lfloor\widetilde{s_{1}}b_{1}\rfloor=\lfloor\widetilde{s_{2}}b_{2}\rfloor$ and $m=m_{1}=m_{2}$.

If $b_{1}=b_{2}=\varepsilon$, then $\overline{s_{1}}=\overline{s_{2}}$ and we are done by the triviality of equal composition.

Suppose only one of $b_{1}$ and $b_{2}$ is empty, say, $b_{1}=\varepsilon$ and $b_{2}\neq\varepsilon$. Then $\overline{s_{1}}=\overline{\lfloor s_{2}b_{2}\rfloor_{m_{2}}}$, $\lfloor b_{2}\rfloor\in\lfloor X^{+}\rfloor$ and $|\widetilde{s_{1}}|+|\widetilde{s_{2}}|\geq 3$. We thus have done by the triviality of short and long intersection composition.

Suppose that $b_{1}\neq\varepsilon$ and $b_{2}\neq\varepsilon$. Thus $s_{1},s_{2}$ are strong. Here we need to consider two cases:

Case 1. $\widetilde{s_{1}}$ and $\widetilde{s_{2}}$ are mutually disjoint. We may assume that $\lfloor b_{1}\rfloor=\lfloor\widetilde{s_{2}}c\rfloor,\lfloor b_{2}\rfloor=\lfloor\widetilde{s_{1}}c\rfloor$, where $\lfloor c\rfloor\in\lfloor X^{*}\rfloor$. This splits into two cases depending on whether $m=1$ or $m=2$. By (4) and Lemmas 11 and 5,

[TABLE]

[TABLE]

Case 2. $\widetilde{s_{1}}$ and $\widetilde{s_{2}}$ have a nonempty intersection. In this case, we need to discuss three sub-cases:

Case 2.1. $|\widetilde{s_{1}}|+|\widetilde{s_{2}}|=2$. Thus $\overline{s_{1}}=\overline{s_{2}}$ and $\lfloor b_{1}\rfloor=\lfloor b_{2}\rfloor$. We have two possibilities depending on whether $m=1$ or $m=2$. By the triviality of equal composition and Lemmas 11 and 5,

[TABLE]

Case 2.2. $|\widetilde{s_{1}}|+|\widetilde{s_{2}}|=3$. Then $m_{1}=m_{2}=1$. We may assume that $|\widetilde{s_{1}}|=2$ and $|\widetilde{s_{2}}|=1$, i.e. $\widetilde{s_{1}}=\lfloor\widetilde{s_{2}}x\rfloor$ and $\lfloor b_{2}\rfloor=\lfloor xb_{1}\rfloor$, where $x\in X$. It follows that $\overline{s_{1}}=\overline{\lfloor s_{2}x\rfloor_{l}}$ for some $l=1,2$. By the triviality of short composition and Lemmas 11 and 5,

[TABLE]

Case 2.3. $|\widetilde{s_{1}}|+|\widetilde{s_{2}}|>3$. Then $m_{1}=m_{2}=1$. If $\widetilde{s_{1}}=\widetilde{s_{2}}$, then $\lfloor b_{1}\rfloor=\lfloor b_{2}\rfloor$. We are done by the triviality of equal composition and equal multiplication composition, and Lemmas 11 and 5. Suppose that $\widetilde{s_{1}}\neq\widetilde{s_{2}}$ and $lcm\{\widetilde{s_{1}},\widetilde{s_{2}}\}=\lfloor w_{1}\rfloor=\lfloor\widetilde{s_{1}}a_{1}\rfloor=\lfloor\widetilde{s_{2}}a_{2}\rfloor$, where $\lfloor a_{1}\rfloor,\lfloor a_{2}\rfloor\in\lfloor X^{*}\rfloor$. Then $\lfloor b_{1}\rfloor=\lfloor a_{1}c\rfloor,\ \lfloor b_{2}\rfloor=\lfloor a_{2}c\rfloor$ for some $\lfloor c\rfloor\in\lfloor X^{*}\rfloor$ and $|w_{1}|<|\widetilde{s_{1}}|+|\widetilde{s_{2}}|$. Since $\widetilde{s_{1}}\neq\widetilde{s_{2}}$, we may assume that $\lfloor a_{2}\rfloor\in\lfloor X^{+}\rfloor$. Then we have the long composition $\lfloor s_{1}a_{1}\rfloor_{1}-\lfloor s_{2}a_{2}\rfloor_{1}$. By Lemmas 11 and 5,

[TABLE]

The proof is complete. ∎

The following theorem is the main result in this article.

Theorem 14**.**

(Composition-Diamond lemma for commutative dialgebras)  Let $S$ be a monic subset of $Di[X]$, $>$ a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ and $Id(S)$ the ideal of $Di[X]$ generated by $S$. Then the following statements are equivalent.

1. (i)

$S$* is a Gröbner–Shirshov basis in $Di[X]$.* 2. (ii)

$0\neq f\in Id(S)\Rightarrow\overline{f}=\overline{\lfloor sb\rfloor_{m}}$* for some normal *$S$​-polynomial $\lfloor sb\rfloor_{m}$. 3. (iii)

The set

[TABLE]

is a $\mathbf{k}$-basis of the commutative dialgebra $Di[X|S]:=Di[X]/Id(S)$.

Proof.

$(i)\Rightarrow(ii)$. Let $0\neq f\in Id(S)$. Then by Lemma 11 $f$ has an expression

[TABLE]

where each $0\neq\alpha_{i}\in\mathbf{k},\ \lfloor b_{i}\rfloor\in\lfloor X^{*}\rfloor,\ s_{i}\in S$. Write $\lfloor w_{i}\rfloor_{m_{i}}=\overline{\lfloor s_{i}b_{i}\rfloor_{m_{i}}}=\lfloor\widetilde{s_{i}}b_{i}\rfloor_{m_{i}},i=1,2,\cdots$. We may assume without loss of generality that

[TABLE]

If $l=1$, then $\overline{f}=\overline{\lfloor s_{1}b_{1}\rfloor_{m_{1}}}$ and the result holds. Suppose that $l\geq 2$. Then

[TABLE]

By Lemma 13, we can rewrite the first two summands of (5) in the form

[TABLE]

where each $\lfloor s^{\prime}_{j}d_{j}\rfloor_{n_{j}}$ is a normal $S$​-polynomial and $\overline{\lfloor s^{\prime}_{j}d_{j}\rfloor_{n_{j}}}<\lfloor w_{1}\rfloor_{m_{1}}$. Thus the result follows from induction on $(\lfloor w_{1}\rfloor_{m_{1}},l)$.

$(ii)\Rightarrow(iii)$. By Lemma 12, the set $Irr(S)$ generates $Di[X|S]$ as a linear space. On the other hand, suppose that $h=\sum_{i}\alpha_{i}\lfloor u_{i}\rfloor_{l_{i}}=0$ in $Di[X|S]$, where each $\alpha_{i}\in\mathbf{k}$, $\lfloor u_{i}\rfloor_{l_{i}}\in{Irr(S)}$. This means that $h\in Id(S)$. Then all $\alpha_{i}$ must be equal to zero. Otherwise, $\overline{h}=\lfloor u_{j}\rfloor_{l_{j}}$ for some $j$ which contradicts $(ii)$.

$(iii)\Rightarrow(i)$. Suppose that $h$ is a composition of elements of $S$. Clearly, $h\in Id(S)$. By Lemma 12,

[TABLE]

where each $\lfloor u_{i}\rfloor_{n_{i}}\in Irr(S),\ \alpha_{i},\beta_{j}\in\mathbf{k},\ \lfloor b_{j}\rfloor\in\lfloor X^{*}\rfloor,\ s_{j}\in S$, $\lfloor u_{i}\rfloor_{n_{i}}\leq\overline{h}$ and $\lfloor\widetilde{s_{j}}b_{j}\rfloor_{m_{j}}\leq\overline{h}$. Then $\sum\alpha_{i}\lfloor u_{i}\rfloor_{n_{i}}\in Id(S)$. By $(iii)$, we have all $\alpha_{i}=0$ and $h$ is trivial modulo $S$. ∎

Buchberger–Shirshov algorithm If a monic subset $S$ of $Di[X]$ is not a Gröbner–Shirshov basis then one can add to $S$ all nontrivial compositions. Continuing this process repeatedly, we finally obtain a Gröbner–Shirshov basis $S^{comp}$ that contains $S$ and generates the same ideal, $Id(S^{comp})=Id(S)$.

Remark 15**.**

In Theorem 14, if $\dashv\ =\ \vdash$, then $Di[X]=\mathbf{k}[X]$ is free commutative algebra generated by $X$ and Theorem 14 is Buchberger Theorem in Buchberger (1965).

A Gröbner–Shirshov basis $S$ in $Di[X]$ is minimal if for any $s\in S$, $\overline{s}\in Irr(S\backslash\{s\})$. A Gröbner–Shirshov basis $S$ in $Di[X]$ is reduced if for any $s\in S$, $supp(s)\subseteq Irr(S\backslash\{s\})$. If $S$ is a Gröbner–Shirshov basis in $Di[X]$, then we also call $S$ a Gröbner–Shirshov basis for $I=Id(S)$.

Lemma 16**.**

Let $I$ be an ideal of $Di[X]$ and $S$ a monic subset of $I$. If, for all nonzero $f\in I$, there exists a normal $S$​-polynomial $\lfloor sb\rfloor_{m}$ such that $\overline{f}=\overline{\lfloor sb\rfloor_{m}}$, then $I=Id(S)$ and $S$ is a Gröbner–Shirshov basis for $I$.

Proof.

Clearly, $Id(S)\subseteq I$. For any nonzero $f\in I$, $\overline{f}=\overline{\lfloor sb\rfloor_{m}}$ for some normal $S$​-polynomial $\lfloor sb\rfloor_{m}$. Thus $f_{1}=f-lc(f)\lfloor sb\rfloor_{m}\in I$ and $\overline{f_{1}}<\overline{f}$. By induction on $\overline{f}$, $f_{1}\in Id(S)$. Hence $f=f_{1}+lc(f)\lfloor sb\rfloor_{m}\in Id(S)$. This shows that $I=Id(S)$. Now the result follows from Theorem 14. ∎

Lemma 17**.**

Let $S$ be a Gröbner–Shirshov basis in $Di[X]$ and $f\in S$. If $\overline{f}\notin Irr(S\backslash\{f\})$, then $S\backslash\{f\}$ is a Gröbner–Shirshov basis for $Id(S)$.

Proof.

Let $S_{1}=S\backslash\{f\}$. By Lemma 16, we only need to show that for any non-zero $h\in Id(S)$ there exists a $g\in S_{1}$ and $\overline{h}=\overline{\lfloor gb\rfloor_{m}}$ for some normal $g$-polynomial $\lfloor gb\rfloor_{m}$. Indeed, $\overline{h}=\overline{\lfloor sa\rfloor_{m}}$ for some normal $S$​-polynomial $\lfloor sa\rfloor_{m}$ by Theorem 14. If $s\neq f$, then we are done. Suppose that $s=f$ and $\overline{f}=\overline{\lfloor s_{1}c\rfloor_{l}}$ for some normal $S\!_{1}$​-polynomial $\lfloor s_{1}c\rfloor_{l}$. When $f$ is not strong or $s_{1}$ is strong, we conclude that

[TABLE]

and we have done. When $f$ is strong and $s_{1}$ is not strong, by Proposition 4, we have

[TABLE]

As $f-s_{1}\in Id(S)$ and $\overline{f-s_{1}}<\overline{f}$, $\lfloor x_{i}x_{j}\rfloor_{1}=\overline{f-s_{1}}=\overline{\lfloor s^{\prime}a^{\prime}\rfloor_{1}}$ some normal $S\!_{1}$​-polynomial $\lfloor s^{\prime}a^{\prime}\rfloor_{1}$ by Theorem 14. It follows that $s^{\prime}\in S\!_{1}$ is strong and

[TABLE]

for some normal $s^{\prime}$-polynomial $\lfloor s^{\prime}a^{\prime}a\rfloor_{m}$. Therefore the result holds. ∎

Lemma 18**.**

Let $S$ be a monic subset of $Di[X]$ and $s\in S$. Then $s$ has an expression $s=s^{\prime}+s^{\prime\prime}$, where $s^{\prime},s^{\prime\prime}\in Di[X],\ supp(s^{\prime})\subset Irr(S\backslash\{s\}),\ s^{\prime\prime}\in Id(S\backslash\{s\})$ and for any $\lfloor a\rfloor_{m}\in supp(s^{\prime})$, $\lfloor a\rfloor_{m}\leq\overline{s}$. Moreover, if $S$ is a minimal Gröbner–Shirshov basis in $Di[X]$, then for any $s\in S$, $\overline{s}=\overline{s^{\prime}}$ and $s^{\prime}$ is strong if $s$ is strong.

Proof.

Analysis similar to that in the proof of Lemma 12 shows the first claim.

If $S$ is a minimal Gröbner–Shirshov basis, then we have $\overline{s}=\overline{s^{\prime}}\in Irr(S\backslash\{s\})$ for any $s\in S$. Recall $r\!_{{}_{s}}:=s-\overline{s}$. It follows that $r\!_{{}_{s}}=r\!_{{}_{s^{\prime}}}+s^{\prime\prime}$ and $\overline{r\!_{{}_{s}}}\geq\overline{r\!_{{}_{s^{\prime}}}}$. Thus $\widetilde{s^{\prime}}=\widetilde{s}>\widetilde{r\!_{{}_{s}}}\geq\widetilde{r\!_{{}_{s^{\prime}}}}$ and the result holds. ∎

Let $S$ be a subset of $Di[X]$ and $\lfloor a\rfloor_{m}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. We set

[TABLE]

Theorem 19**.**

Let $I$ be an ideal of $Di[X]$ and $>$ a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then there exists a unique reduced Gröbner–Shirshov basis for $I$.

Proof.

It is clear that $\{lc(f)^{-1}f\mid 0\neq f\in I\}$ is a Gröbner–Shirshov basis for $I$. Let $S$ be an arbitrary Gröbner–Shirshov basis for $I$. For any $g\in S$, we set

[TABLE]

By Lemma 17, we conclude that $S_{1}$ is a minimal Gröbner–Shirshov basis for $I$.

For any $s\in S_{1}$, by Lemma 18, we have $s=s^{\prime}+s^{\prime\prime}$, where $supp(s^{\prime})\subset Irr(S_{1}\backslash\{s\}),\ s^{\prime\prime}\in Id(S_{1}\backslash\{s\})$. Let

[TABLE]

We claim that $S_{2}$ is a reduced Gröbner–Shirshov basis for $I$. Indeed, it is clear that $S_{2}\subset Id(S_{1})$. For any $f\in Id(S_{1})$, by Theorem 14 and Lemma 18, $\overline{f}=\overline{\lfloor sb\rfloor_{m}}=\overline{\lfloor s^{\prime}b\rfloor_{m}}$ for some normal $S\!_{1}$-polynomial $\lfloor sb\rfloor_{m}$ and normal $S\!_{2}$-polynomial $\lfloor s^{\prime}b\rfloor_{m}$. According to Lemma 16, we have $S_{2}$ is a Gröbner–Shirshov basis for $I$. Suppose there exists $s^{\prime}\in S_{2}$ such that $supp(s^{\prime})\nsubseteq Irr(S_{2}\backslash\{s^{\prime}\})$, i.e. there exists $\lfloor a\rfloor_{m}\in supp(s^{\prime})\subset Irr(S_{1}\backslash\{s\})$ but $\lfloor a\rfloor_{m}\notin Irr(S_{2}\backslash\{s^{\prime}\})$. Then $\lfloor a\rfloor_{m}=\overline{\lfloor t^{\prime}b\rfloor_{m}}$ for some normal $t^{\prime}$-polynomial $\lfloor t^{\prime}b\rfloor_{m}$, where $\lfloor b\rfloor\in\lfloor X^{*}\rfloor,t^{\prime}\in S_{2}\backslash\{s^{\prime}\}$, $t^{\prime}=t-t^{\prime\prime}$, $t\in S_{1}\backslash\{s\}$ and $t^{\prime\prime}\in Id(S_{1}\backslash\{t\})$. By Lemma 18, $\overline{t}=\overline{t^{\prime}}$. We must have $m=1$, $\lfloor b\rfloor\in\lfloor X^{+}\rfloor$, $t^{\prime}$ is strong and $t$ is not strong. Otherwise $\lfloor tb\rfloor_{m}$ is a normal $S_{1}\backslash\{s\}$-polynomial and $\lfloor a\rfloor_{m}=\overline{\lfloor t^{\prime}b\rfloor_{m}}=\overline{\lfloor tb\rfloor_{m}}$, which contradicts our assumption. Then $\overline{t^{\prime}}=\overline{t}=\lfloor c\rfloor_{2}$, where $\lfloor c\rfloor\in\lfloor X^{+}\rfloor,|c|=2$. Recall $r\!_{{}_{t}}:=t-\overline{t}$. Hence $\overline{r\!_{{}_{t}}}=\lfloor c\rfloor_{1}=\overline{t^{\prime\prime}}$. There exist $f\in S_{1}\backslash\{t\}$ and a normal $f$-polynomial $\lfloor fd\rfloor_{1}$ such that $\overline{\lfloor fd\rfloor_{1}}=[c]_{1}$. It follows that $f$ is strong. Since $\lfloor a\rfloor_{m}=\lfloor a\rfloor_{1}\in supp(s^{\prime})$, by Lemma 18, we have $\overline{f}<\overline{\lfloor fdb\rfloor_{1}}=\lfloor cb\rfloor_{1}=\overline{\lfloor t^{\prime}b\rfloor}_{1}=\lfloor a\rfloor_{1}\leq\overline{s}$. Hence $f\neq s$ and $\lfloor a\rfloor_{1}=\overline{\lfloor fdb\rfloor_{1}}$, where $\lfloor fdb\rfloor_{1}$ is a normal $S_{1}\backslash\{s\}$-polynomial. This contradicts the fact that $\lfloor a\rfloor_{1}\in Irr(S_{1}\backslash\{s\})$. We thus get $supp(s^{\prime})\subseteq Irr(S_{2}\backslash\{s^{\prime}\})$ for all $s^{\prime}\in S_{2}$ and our claim holds.

Suppose that $T$ is a reduced Gröbner–Shirshov basis for $I$. Let $\overline{s_{0}}=\min\overline{S_{2}}$ and $\overline{r_{0}}=\min\overline{T}$, where $s_{0}\in S_{2},r_{0}\in T$. By Theorem 14 and Lemma 8, $\overline{s_{0}}=\overline{\lfloor r^{\prime}a^{\prime}\rfloor_{p}}\geq\overline{r^{\prime}}\geq\overline{r_{0}}$ for some $r^{\prime}\in T$ and normal $T$-polynomial $\lfloor r^{\prime}a^{\prime}\rfloor_{p}$. Similarly, $\overline{r_{0}}\geq\overline{s_{0}}$. Then $\overline{r_{0}}=\overline{s_{0}}$. We claim that $r_{0}=s_{0}$. Otherwise, $0\neq r_{0}-s_{0}\in I$. We apply the above argument again, with replace $\overline{s_{0}}$ by $\overline{r_{0}-s_{0}}$, to obtain that $\overline{r_{0}}>\overline{r_{0}-s_{0}}\geq\overline{r^{\prime\prime}}\geq\overline{r_{0}}$ for some $r^{\prime\prime}\in T$, a contradiction. As both $T$ and $S_{2}$ are reduced Gröbner–Shirshov bases, we have $S_{2}^{\overline{s_{0}}}=\{s_{0}\}=\{r_{0}\}=T^{\overline{r_{0}}}$. Given any $\lfloor w\rfloor_{m}\in\overline{S_{2}}\cup\overline{T}$ with $\lfloor w\rfloor_{m}>\overline{r_{0}}$. Assume that $S_{2}^{<\lfloor w\rfloor_{m}}=T^{<\lfloor w\rfloor_{m}}$. To prove $T=S_{2}$, it is sufficient to show that $S_{2}^{\lfloor w\rfloor_{m}}\subseteq T^{\lfloor w\rfloor_{m}}$. For any $s\in S_{2}^{\lfloor w\rfloor_{m}}$, we can see that $\overline{s}=\overline{\lfloor ra^{\prime\prime}\rfloor_{q}}\geq\overline{r}$ for some $r\in T$ and normal $T$-polynomial $\lfloor ra^{\prime\prime}\rfloor_{q}$. Now, we claim that $\lfloor w\rfloor_{m}=\overline{s}=\overline{r}$. Otherwise, $\lfloor w\rfloor_{m}=\overline{s}>\overline{r}$. Then $r\in T^{<\lfloor w\rfloor_{m}}=S_{2}^{<\lfloor w\rfloor_{m}}$ and $r\in S_{2}\backslash\{s\}$. But $\overline{s}=\overline{\lfloor ra^{\prime\prime}\rfloor_{q}}$, which contradicts the fact that $S_{2}$ is a reduced Gröbner–Shirshov basis. We next claim that $s=r\in T^{\lfloor w\rfloor_{m}}$. If $s\neq r$, then $0\neq s-r\in I$. By Theorem 14 and Lemma 8, $\overline{s-r}=\overline{\lfloor r_{1}u\rfloor_{n}}=\overline{\lfloor s_{1}v\rfloor_{n}}$ for some normal $T$-polynomial $\lfloor r_{1}u\rfloor_{n}$ and normal $S_{2}$-polynomial $\lfloor s_{1}v\rfloor_{n}$ with $\overline{r_{1}},\overline{s_{1}}\leq\overline{s-r}<\overline{s}=\overline{r}$, where $r_{1}\in T,s_{1}\in S_{2}$. This means that $s_{1}\in S_{2}\backslash\{s\}$ and $r_{1}\in T\backslash\{r\}$. Noting that $\overline{s-r}\in supp(s)\cup supp(r)$, we may assume that $\overline{s-r}\in supp(s)$. As $S_{2}$ is a reduced Gröbner–Shirshov basis, we have $\overline{s-r}\in Irr(S_{2}\backslash\{s\})$, which contradicts the fact that $\overline{s-r}=\overline{\lfloor s_{1}v\rfloor_{n}}$, where $s_{1}\in S_{2}\backslash\{s\}$. Thus $s=r$. This shows that $S_{2}^{\lfloor w\rfloor_{m}}\subseteq T^{\lfloor w\rfloor_{m}}$. ∎

Remark 20**.**

Theorem 19 together with Buchberger–Shirshov algorithm gives a method to find the unique reduced Gröbner–Shirshov basis $S^{red}$ for the ideal $I=Id(S)$, where $S$ is a monic subset of $Di[X]$. One can find $S^{red}$ by the following steps.

1. (i)

Buchberger–Shirshov algorithm gives a Gröbner–Shirshov basis $S_{0}:=S^{comp}$ for $I$. 2. (ii)

Let

[TABLE]

Then $S_{1}$ is a minimal Gröbner–Shirshov basis for $I$. 3. (iii)

For each $s\in S_{1}$, by the same method as in the proof of Lemma 12, $s$ has an expression $s=s^{\prime}+s^{\prime\prime}$, where $supp(s^{\prime})\subset Irr(S_{1}\backslash\{s\})$, $s^{\prime\prime}\in Id(S_{1}\backslash\{s\})$. Then $S^{red}=\{s^{\prime}\mid s\in S_{1}\}$ is the reduced Gröbner–Shirshov basis for $I$.

4 Polynomial dialgebra has a finite Gröbner–Shirshov basis

In this section, we use the Hilbert’s basis theorem to prove that the polynomial diring $Di_{R}[X]$ is left Noetherian if $R$ is left Noetherian and $X$ is a finite set. Furthermore, we show that each ideal of the polynomial dialgebra $Di[X]$ has a finite Gröbner–Shirshov basis, if $X$ is a finite set.

Throughout this section, $R$ is an associative ring with unit.

Definition 21**.**

A diring is a quaternary $(T,+,\vdash,\dashv)$ such that both $(T,+,\vdash)$ and $(T,+,\dashv)$ are associative rings with the identities (1) in $T$.

Definition 22**.**

Let $(D,\vdash,\dashv)$ be a disemigroup, $R$ an associative ring with unit and $T$ the free left $R$-module with an $R$-basis $D$. Then $(T,+,\vdash,\dashv)$ is a diring with a natural way: for any $f=\sum_{i}r_{i}f_{i},\ g=\sum_{j}r_{j}^{\prime}g_{j}\in T,\ r_{i},r_{j}^{\prime}\in R,\ f_{i},g_{j}\in D$,

[TABLE]

Such a diring is called a disemigroup-diring of $D$ over $R$.

For the free commutative disemigroup $Disgp[X]$ generated by a set $X$, we denote $Di_{R}[X]$ the disemigroup-diring of $Disgp[X]$ over $R$ which is also called the polynomial diring over $R$. In particular, $Di_{\mathbf{k}}[X]\ (=Di[X])$ is the free commutative dialgebra (polynomial dialgebra) generated by $X$ when $\mathbf{k}$ is a field.

A left (right, resp.) ideal $I$ of $Di_{R}[X]$ is an $R$-submodule of $Di_{R}[X]$ such that $f\vdash g,f\dashv g\in I$ $(g\vdash f,g\dashv f\in I$, resp.) for any $f\in Di_{R}[X]$ and $g\in I$. We call $I$ an ideal of $Di_{R}[X]$ if $I$ is a left and right ideal.

Clearly, $Di_{R}[X]$ is a commutative disemigroup-diring if and only if $R$ is a commutative ring.

Let $U$ be a semigroup, $R$ an associative ring and $RU$ the semigroup ring of $U$ over $R$. Note that $RU$ is a free left $R$-module with an $R$-basis $U$.

Thus, $R\lfloor X^{*}\rfloor$ is the ring of polynomials over $R$ in indeterminates $X$. It is known that $R\lfloor X^{*}\rfloor$ is left Noetherian if $R$ is left Noetherian and $X$ is a finite set.

Theorem 23**.**

Let $X$ be a finite set. If $R$ is left Noetherian, then so is $Di_{R}[X]$.

Proof.

Let $S=\{\lfloor a\rfloor_{1}\in\lfloor X^{+}\rfloor_{{}_{1}}\mid|a|\geq 3\}$ and $I=Id(S)$ the ideal of $Di_{R}[X]$ generated by $S$. We first prove that $Di_{R}[X]/I$ is left Noetherian. It suffices to show that any left ideal $H$ of $Di_{R}[X]/I$ has a finite set of generators. Since $X$ is finite, $\{\lfloor a\rfloor_{p}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}\mid|a|\leq 2\}$ is finite, say, $\{\lfloor a\rfloor_{p}\in\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}\mid|a|\leq 2\}=\{\lfloor a_{1}\rfloor_{p_{1}},\dots,\lfloor a_{m}\rfloor_{p_{m}}\}$ with $\lfloor a_{1}\rfloor_{p_{1}}<\dots<\lfloor a_{m}\rfloor_{p_{m}}$, where $<$ is the deg-lex-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then $Di_{R}[X]/I=R\lfloor a_{1}\rfloor_{p_{1}}+R\lfloor a_{2}\rfloor_{p_{2}}+\cdots+R\lfloor a_{m}\rfloor_{p_{m}}$. For $j=1,2,\dots,m$, let $I_{j}$ be the set of elements $\alpha_{j}\in R$ such that there exists an element of the form

[TABLE]

It is easy to check that $I_{j}$ is a left ideal of $R$. Hence each $I_{j}$ has a finite set of generators and we may assume that $I_{j}$ is generated by $\alpha_{j1},\cdots,\alpha_{j{n_{j}}}$, where $n_{j}\in\mathbb{Z}^{+}$. Then for $1\leq i\leq n_{j}$, we have polynomials $f_{ji}=\alpha_{ji}\lfloor a_{j}\rfloor_{p_{j}}+h_{ji}\in H$ where $\overline{h_{ji}}<\lfloor a_{j}\rfloor_{p_{j}}$. Thus we obtain a finite set $T:=\{f_{ji}\in H\mid 1\leq j\leq m,1\leq i\leq n_{j}\}$.

Now we claim that the set $T$ generates $H$. For any nonzero $f\in H$, we prove $f$ belongs to the left ideal of $Di_{R}[X]/I$ generated by $T$, by induction on $\overline{f}$. Assume that $f=\alpha\lfloor a_{k}\rfloor_{p_{k}}+f_{1}$, where $1\leq k\leq m$, $0\neq\alpha\in R$ and $\overline{f_{1}}<\lfloor a_{k}\rfloor_{p_{k}}$. Thus $\alpha\in I_{k}$ and $\alpha=\Sigma_{i}r_{i}\alpha_{ki}$, where $r_{i}\in R$. Clearly, $\Sigma_{i}r_{i}f_{ki}\in H$ and $lt(\Sigma_{i}r_{i}f_{ki})=\alpha\lfloor a_{k}\rfloor_{p_{k}}=lt(f)$. It follows that $f_{1}=f-\Sigma_{i}r_{i}f_{ki}\in H$ and $\overline{f_{1}}<\overline{f}=\lfloor a_{k}\rfloor_{p_{k}}$. If $\lfloor a_{k}\rfloor_{p_{k}}=\min\{\overline{g}\mid 0\neq g\in H\}$, then $f_{1}=0$ and $f=\Sigma_{i}r_{i}f_{ki}$. Hence our assertion holds. Suppose that $\lfloor a_{k}\rfloor_{p_{k}}>\min\{\overline{g}\mid g\in H\}$. By induction on $\overline{f}$, we obtain the assertion.

Now, for any ascending chain of left ideals of $Di_{R}[X]$

[TABLE]

we have an ascending chain of left ideals of $Di_{R}[X]/I$

[TABLE]

Since $Di_{R}[X]/I$ is left Noetherian, it follows that there is a $p\in\mathbb{Z}^{+}$ such that $(L_{p}+I)/I=(L_{p+1}+I)/I=\cdots$. Therefore $L_{p}+I=L_{p+1}+I=\cdots$. On the other hand, note that for any $f\in Di_{R}[X],\ h\in I$, we have $f\vdash h=f\dashv h$, in particular, in $I,\ ``\vdash"=``\dashv"$. Thus, $I$ is also a left ideal of the associative ring $(R\lfloor X^{*}\rfloor,+,\dashv)$. Then for the ascending chain of left ideals of $R\lfloor X^{*}\rfloor$

[TABLE]

since $(R\lfloor X^{*}\rfloor,+,\dashv)$ is left Noetherian, there is an $l\in\mathbb{Z}^{+}$ such that $L_{l}\cap I=L_{l+1}\cap I=\cdots$. Take $n=\max\{p,l\}$. We thus get $L_{n}=L_{n}+(L_{n}\cap I)=L_{n}+(L_{n+1}\cap I)=L_{n+1}\cap(L_{n}+I)=L_{n+1}\cap(L_{n+1}+I)=L_{n+1}=\dots.$ This shows that $Di_{R}[X]$ is left Noetherian. ∎

Corollary 24**.**

The polynomial dialgebra $Di[X]$ is Noetherian, if $X$ is a finite set.

Theorem 25**.**

Let $X$ be a finite set and $I$ an ideal of $Di[X]$. Then for any monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$, $I$ has a finite Gröbner–Shirshov basis.

Proof.

By Theorem 19, $I$ has a minimal Gröbner–Shirshov basis $S$. The leading monomial of a non-strong polynomial $f\in S$ is $\lfloor x_{i}x_{j}\rfloor_{2}$ for some $x_{i},x_{j}\in X$ by Proposition 4. Since $X$ is finite, it follows that the set $\{\lfloor x_{i}x_{j}\rfloor_{2}\mid x_{i},x_{j}\in X\}$ is finite. As $S$ is a minimal Gröbner–Shirshov basis, we conclude that the set $\{s\in S\mid s\ \mbox{is not strong}\}$ is finite. If $S$ is infinite, then there exists an infinite sequence $s_{1},s_{2},\dots$, where each $s_{i}\in S$ is strong and $\overline{s_{i}}\neq\overline{s_{j}}$ if $i\neq j$. For $n=1,2,\dots$, let $I_{n}$ be the ideal of $Di[X]$ generated by $T_{n}=\{\overline{s_{1}},\dots,\overline{s_{n}}\}$. Then $\overline{s_{n+1}}\not\in I_{n}$. Otherwise, $\overline{s_{n+1}}$ has an expression

[TABLE]

where $f_{i},h_{i}\in Di[X],\ \alpha_{i}\in\mathbf{k}$. This implies that either $\overline{s_{n+1}}=\overline{s_{i}}\vdash x$ for some $\overline{s_{i}}\in T_{n}$ with $|\widetilde{s_{i}}|=1$ and $x\in X$, or $\overline{s_{n+1}}=\overline{s_{i}}\dashv\lfloor a\rfloor_{1}$ for some $\overline{s_{i}}\in T_{n}$ and $\lfloor a\rfloor\in\lfloor X^{+}\rfloor$. Because each $s_{i}$ is strong, we can conclude that $s_{n+1}=\overline{s_{i}}\vdash x=\overline{s_{i}\vdash x}$ or $\overline{s_{n+1}}=\overline{s_{i}}\dashv\lfloor a\rfloor_{1}=\overline{s_{i}\dashv\lfloor a\rfloor_{1}}$. It follows that $\overline{s_{n+1}}=\overline{\lfloor s_{i}b\rfloor_{m}}$ for some normal $s_{i}$-polynomial $\lfloor s_{i}b\rfloor_{m}$, which contradicts the fact that $S$ is a minimal Gröbner–Shirshov basis. Then we have an infinite sequence $I_{1}\subset I_{2}\subset\dots$ which contradicts that $Di[X]$ is Noetherian. ∎

Remark 26**.**

Under the conditions in Theorem 25, in $Di[X]$, every minimal Gröbner–Shirshov basis is finite. In particular, every reduced Gröbner–Shirshov basis is finite.

By Theorem 25, we immediately have

Corollary 27**.**

If $X$ is a finite set, then each congruence on the commutative disemigroup $Disgp[X]$ is finitely generated.

Proof.

Suppose that $\rho$ is a congruence on $Disgp[X]$ and $\rho$ is not finitely generated. Then we can obtain a strictly increasing infinite chain of congruences on $Disgp[X]$

[TABLE]

Let $S_{i}=\{\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\mid(\lfloor u\rfloor_{m},\lfloor v\rfloor_{n})\in\rho_{i},\lfloor u\rfloor_{m}>\lfloor v\rfloor_{n}\},i=1,2,\cdots$. It is easy to check that each $S_{i}$ is a Gröbner–Shirshov basis in $Di[X]$ and

[TABLE]

Assume that $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\in S_{i+1}\subseteq Id(S_{i+1})$ and $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\notin S_{i}$. We claim that $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\notin Id(S_{i})$. Otherwise, by Theorem 14, we may assume that

[TABLE]

where each $\alpha_{j}\in\mathbf{k},\ \lfloor u_{j}\rfloor_{m_{j}}-\lfloor v_{j}\rfloor_{n_{j}}\in S_{i}$ and $\lfloor u_{1}\rfloor_{m_{1}}>\cdots>\lfloor u_{l}\rfloor_{m_{l}}$. This implies that $\alpha_{1}=1,\lfloor u\rfloor_{m}=\lfloor u_{1}\rfloor_{m_{1}}$. Now we prove that $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\in S_{i}$ by induction on $\lfloor u\rfloor_{m}$. If $\lfloor u\rfloor_{m}=\min\{\overline{s}\mid s\in Id(S_{i})\}$, then $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}=\lfloor u_{1}\rfloor_{m_{1}}-\lfloor v_{1}\rfloor_{n_{1}}\in S_{i}$. Otherwise, we can assume that $\lfloor v_{1}\rfloor_{n_{1}}-\lfloor v\rfloor_{n}=\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}-(\lfloor u_{1}\rfloor_{m_{1}}-\lfloor v_{1}\rfloor_{n_{1}})\neq 0$ and $\lfloor v_{1}\rfloor_{n_{1}}>\lfloor v\rfloor_{n}$. Then

[TABLE]

By induction on $\lfloor u\rfloor_{m}$, $\lfloor v_{1}\rfloor_{n_{1}}-\lfloor v\rfloor_{n}\in S_{i}$. It follows that $(\lfloor v_{1}\rfloor_{n_{1}},\lfloor v\rfloor_{n})\in\rho_{i}$. Since $(\lfloor u_{1}\rfloor_{m_{1}},\lfloor v_{1}\rfloor_{n_{1}})\in\rho_{i}$, we have $(\lfloor u\rfloor_{m},\lfloor v\rfloor_{n})=(\lfloor u_{1}\rfloor_{m_{1}},\lfloor v\rfloor_{n})\in\rho_{i}$ and $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\in S_{i}$. This contradicts our assumption that $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}\notin S_{i}$. Therefore $Id(S_{i})\subsetneq Id(S_{i+1}),i=1,2,\cdots$, and we can obtain a infinite properly ascending chain of ideals, which is contradicts the fact that $Di[X]$ is Noetherian. ∎

5 Applications

5.1 Normal forms of commutative disemigroups

Recall that $Disgp[X]=\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$ is the free commutative disemigroup generated by a set $X$. Then for an arbitrary commutative disemigroup $D$, $D$ has an expression

[TABLE]

for some set $X$ and $S\subseteq Disgp[X]\times Disgp[X]$, where $\rho(S)$ is the congruence of $Disgp[X]$ generated by $S$.

It is natural to ask how to find normal forms of elements of the commutative disemigroup $D=Disgp[X|S]$?

Let $>$ be a monomial-center ordering on $Disgp[X]$ and $S=\{(\lfloor u_{i}\rfloor_{m_{i}},\lfloor v_{i}\rfloor_{n_{i}})\mid\lfloor u_{i}\rfloor_{m_{i}}>\lfloor v_{i}\rfloor_{n_{i}},\ i\in I\}$. Consider the commutative dialgebra $Di[X|S]$, where $S=\{\lfloor u_{i}\rfloor_{m_{i}}-\lfloor v_{i}\rfloor_{n_{i}}\mid i\in I\}$. By Buchberger–Shirshov algorithm, we have a Gröbner–Shirshov basis $S^{comp}$ in $Di[X]$ and $Id(S^{comp})=Id(S)$. It is clear that each element in $S^{comp}$ is of the form $\lfloor u\rfloor_{m}-\lfloor v\rfloor_{n}$, where $\lfloor u\rfloor_{m}>\lfloor v\rfloor_{n},\ \lfloor u\rfloor_{m},\lfloor v\rfloor_{n}\in Disgp[X]$.

The following theorem is an application of Theorem 14.

Theorem 28**.**

Let $>$ be a monomial-center ordering on $Disgp[X]$ and $D=Disgp[X|S]$, where $S=\{(\lfloor u_{i}\rfloor_{m_{i}},\lfloor v_{i}\rfloor_{n_{i}})\mid\lfloor u_{i}\rfloor_{m_{i}}>\lfloor v_{i}\rfloor_{n_{i}},\ i\in I\}$. Then the set $Irr(S^{comp})$ is a set of normal forms of elements of the commutative disemigroup $D=Disgp[X|S]$, where $S^{comp}$ is a Gröbner–Shirshov basis in $Di[X]$ obtained from $S$ by Buchberger–Shirshov algorithm.

5.2 Word problem for commutative dialgebras

In this subsection, we present algorithms for computing a (reduced) Gröbner–Shirshov basis. By using these algorithms and Composition-Diamond lemma for commutative dialgebras, we prove that for finitely presented commutative dialgebras (disemigroups), the word problem and the problem whether two ideals of $Di[X]$ are identical are solvable.

Let $f\in Di[X]$ be a nonzero polynomial and let $S$ be a subset of $Di[X]$. We say $f$ is reducible by $S$ if there exist a polynomial $g$ in $S$ and a diword $\lfloor u\rfloor_{m}$ in $supp(f)$ such that $\lfloor u\rfloor_{m}=\overline{\lfloor gb\rfloor_{m}}$ for some normal $g$-polynomial $\lfloor gb\rfloor_{m}$. Moreover, if $f$ is reducible by $S$ and $\alpha$ is the coefficient of $\lfloor u\rfloor_{m}$, then $f\rightarrow_{{}_{S}}f-\frac{\alpha}{lc(g)}\lfloor gb\rfloor_{m}$ is called a one-step-reduction by $S$. In particular, we say $f$ is partially reducible by $S$ if there exists a polynomial $g$ in $S$ and $\overline{f}=\overline{\lfloor gb\rfloor_{m}}$ for some normal $g$-polynomial $\lfloor gb\rfloor_{m}$. If an element $r\in Di[X]$ is obtained from $f$ by finitely many one-step-reductions by $S$ and $r$ is not reducible by $S$, then we say that $r$ is a remainder of $f$ modulo $S$.

In what follows, we assume that $X$ is a finite set. The following algorithm is an analogue of the Buchberger’s Algorithm for polynomial algebras in Buchberger (1965).

Algorithm 29**.**

(Algorithm for computing Gröbner–Shirshov bases in $Di[X]$)

$\mathbf{input}:F=\{f_{1},\cdots,f_{m}\}\subseteq Di[X]$.

$\mathbf{output}:$* A Gröbner–Shirshov basis $S$ for $Id(F)$.*

$S:=\{lc(f)^{-1}f\mid 0\neq f\in F\}$**

$\mathcal{G}:=\{s,(s,g)\mid s,g\in S\}$**

$\mathbf{for}$* $u\in\mathcal{G}$ $\mathbf{do}$*

$\mathbf{if}$* there exists a composition of $u$ $\mathbf{then}$*

$h:=$* a remainder of the composition of $u$ modulo $S$*

$\mathbf{if}$* $h\neq 0$ $\mathbf{then}$*

$h:=lc(h)^{-1}h$**

$\mathcal{G}:=\mathcal{G}\cup\{h\}\cup\{(h,s)\mid s\in S\}\cup\{(s,h)\mid s\in S\}$**

$\mathcal{S}:=\mathcal{S}\cup\{h\}$**

$\mathbf{end}$* $\mathbf{if}$*

$\mathbf{end}$* $\mathbf{if}$*

$\mathbf{end}$* $\mathbf{do}$*

$\mathbf{return}$* $S$*

Lemma 30**.**

Let $>$ be a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then Algorithm 29 terminates after finitely many steps and returns a Gröbner–Shirshov basis in $Di[X]$.

Proof.

(Correctness.) If Algorithm 29 terminates after finitely many steps, then its correctness follows clearly from Definition 10.

(Termination.) Suppose to the contrary that Algorithm 29 does not terminate. Then, as the algorithm progress, we construct a set $S_{i}$ strictly larger than $S_{i-1}$ and obtain a strictly increasing infinite sequence

[TABLE]

Each $S_{i+1}$ is obtained from $S_{i}$ by adding $h\in Id(F)$ to $S_{i+1}$, where $h$ is a non-zero remainder of a composition of elements in $S_{i}$ modulo $S_{i}$. It is easy to see that $\overline{h}\in Irr(S_{i})$. By the method similar to that in the proof of Theorem 25, we can obtain a infinite properly ascending chain of ideals, which is contradicts the fact that $Di[X]$ is Noetherian. ∎

By mimicking the method used in Buchberger (1965), we obtain the following algorithm.

Algorithm 31**.**

(Algorithm for computing the reduced Gröbner–Shirshov bases in $Di[X]$)

$\mathbf{input}:$* A Gröbner–Shirshov basis $S=\{g_{1},\cdots,g_{n}\}$ in $Di[X]$.*

$\mathbf{output}:$* The reduced Gröbner–Shirshov basis $RG$ for $Id(S)$.*

$C:=S$**

$\mathbf{for}$* $f\in C$ $\mathbf{do}$*

$S:=S\backslash\{f\}$**

$\mathbf{if}$* $f$ is not partially reducible by $S$ $\mathbf{then}$*

$S:=S\cup\{f\}$**

$\mathbf{end}$* $\mathbf{if}$*

$\mathbf{end}$* $\mathbf{do}$*

$RG:=\emptyset$**

$\mathbf{for}$* $h\in S$ $\mathbf{do}$*

$G:=S\backslash\{h\}$**

$r:=$* a remainder of $h$ modulo $G$*

$\mathbf{if}$* $r\neq 0$ $\mathbf{then}$*

$RG:=RG\cup\{r\}$**

$\mathbf{end}$* $\mathbf{if}$*

$\mathbf{end}$* $\mathbf{do}$*

$\mathbf{return}$* $RG$*

The following lemma follows from Theorem 19 directly.

Lemma 32**.**

Let $>$ be a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then Algorithm 31 terminates after finitely many steps and returns the reduced Gröbner–Shirshov basis in $Di[X]$.

By Theorems 19 and 25 and Algorithm 31, it follows that

Corollary 33**.**

Let $X$ be a finite set, $I$ an ideal of $Di[X]$ and $>$ a monomial-center ordering on $\lfloor X^{+}\rfloor_{{}_{1}}\cup\lfloor XX\rfloor_{{}_{2}}$. Then there exists a unique finite reduced Gröbner–Shirshov basis for I.

Theorem 34**.**

The word problem for any finitely presented commutative dialgebra is algorithmically solvable.

Proof.

Let $D=Di[X|S]$ be a finitely presented commutative dialgebra. By Algorithm 29, we can obtain a Gröbner–Shirshov basis $G$ from $S$. For any $f\in Di[X]$, by Theorem 14, $f\in Id(S)$ if and only if the remainder of $f$ modulo $G$ is equal to zero. ∎

Theorem 35**.**

Let $X$ be a finite set, $S$ and $T$ two finite subsets in $Di[X]$. The problem whether $Id(S)$ and $Id(T)$ are identical is algorithmically solvable.

Proof.

By Algorithms 29 and 31, we can obtain the finite reduced Gröbner–Shirshov bases $G_{1}$ for $Id(S)$ and $G_{2}$ for $Id(T)$ from $S$ and $T$, respectively. By Corollary 33, $Id(S)=Id(T)$ if and only if $G_{1}=G_{2}$. ∎

6 Lifting commutative Gröbner–Shirshov bases to associative ones

Let $Di\langle X\rangle$ be the free dialgebra over a field $\mathbf{k}$ generated by a set $X$. Recall that $X^{+}$ is the free semigroup generated by $X$ without unit. Write

[TABLE]

It is well known that $[X^{+}]_{\omega}$ is a $\mathbf{k}$-basis of $Di\langle X\rangle$.

We first recall from the following result concerning the Gröbner–Shirshov bases theory for dialgebras in Zhang and Chen (2017).

Lemma 36**.**

(Zhang and Chen, 2017, Theorem 4.4)* Let $S$ be a monic subset of $Di\langle X\rangle$, $>$ a monomial-center ordering on $[X^{+}]_{\omega}$ and $Id(S)$ the ideal of $Di\langle X\rangle$ generated by $S$. Then $S$ is a Gröbner–Shirshov basis in $Di\langle X\rangle$ if and only if the set*

[TABLE]

is a $\mathbf{k}$-basis of the quotient dialgebra $Di\langle X|S\rangle:=Di\langle X\rangle/Id(S)$.

Now, we construct a Gröbner–Shirshov basis in $Di\langle X\rangle$ by lifting a Gröbner–Shirshov basis in $Di[X]$ and adding some relations.

Let $X=\{x_{i}\mid i\in I\}$ be a well-ordered set. Note that $\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}\subseteq[X^{+}]_{\omega}$. We consider the natural map $\gamma:[X^{+}]_{\omega}\rightarrow\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$ defined by, for any $[a]_{m}\in[X^{+}]_{\omega}$,

[TABLE]

and define a $\mathbf{k}$-linear map $\delta:Di[X]\rightarrow Di\langle X\rangle$ which is given by

[TABLE]

Let $\succ$ be any monomial-center ordering on $\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$ and $>_{d}$ the deg-lex-center ordering on $[X^{+}]_{\omega}$. For any $[a]_{m},[b]_{n}\in[X^{+}]_{\omega}$, define

[TABLE]

It is clear that $>$ is a monomial-center ordering on $[X^{+}]_{\omega}$. In what follows, a monomial-center ordering $\succ$ on $\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$ and the ordering $>$ on $[X^{+}]_{\omega}$ defined as (6) will be used unless otherwise stated.

For any $\lfloor a\rfloor_{m},\lfloor b\rfloor_{n}\in\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$, if $\lfloor a\rfloor_{m}\succ\lfloor b\rfloor_{n}$, then $\delta(\lfloor a\rfloor_{m})>\delta(\lfloor b\rfloor_{n})$. Thus, for any $f\in Di[X]$, $\delta(\overline{f})=\overline{\delta(f)}$. Let $s\in Di[X]$, $\lfloor sb\rfloor_{m}$ be a normal $s$-polynomial in $Di[X]$. Then $\overline{\delta(\lfloor sb\rfloor_{m})}=\delta(\overline{\lfloor sb\rfloor_{m}})=\delta(\lfloor\widetilde{s}b\rfloor_{m})$.

Let $S$ be a monic subset of $Di[X]$, $\lfloor v\rfloor=\lfloor x_{i_{1}}x_{i_{2}}\cdots x_{i_{r}}\rfloor\in\lfloor X^{+}\rfloor,\ x_{i_{l}}\in X,\ 1\leq l\leq r$. We set

[TABLE]

[TABLE]

For example, let $\lfloor u\rfloor=x_{1}x_{2}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7}x_{7}$ and $\lfloor v\rfloor=x_{2}x_{6}$. Then $\lfloor u\rfloor=x_{1}x_{2}\lfloor vx_{3}x_{4}x_{5}\rfloor x_{7}x_{7}$, where $x_{3}x_{4}x_{5}\in U(\lfloor v\rfloor)$. That is to say, $\lfloor u\rfloor$ can split into three parts by $\lfloor v\rfloor$ and the center part is $\lfloor va\rfloor$ for some $\lfloor a\rfloor\in U(\lfloor v\rfloor)$.

Throughout the following proof, we follow the notation used in Zhang and Chen (2017).

Lemma 37**.**

Let $S$ be a monic subset of $Di[X]$ and $Y=\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$. Then

[TABLE]

Proof.

Let $[w]_{n}\in[X^{+}]_{\omega}$. Then

[TABLE]

∎

Theorem 38**.**

Let $X=\{x_{i}\mid i\in I\}$ be a well-ordered set, $S$ be a Gröbner–Shirshov basis in $Di[X]$, and $W$ consist of the following polynomials in $Di\langle X\rangle$:

[TABLE]

Then with the monomial-center ordering on $[X^{+}]_{\omega}$ defined as (6), the set $G_{S}\bigcup W$ is a Gröbner–Shirshov basis in $Di\langle X\rangle$.

Proof.

Let $T=G_{S}\bigcup W$ and $Y=\lfloor X^{+}\rfloor_{1}\cup\lfloor XX\rfloor_{{}_{2}}$. It is clear that $Irr_{A}(W)=Y$. By Theorem 14, $Irr(S)$ is a $\mathbf{k}$-basis of the commutative dialgebra $Di[X|S]$.

Noting that $Di\langle X|T\rangle$ is a commutative dialgebra, we have a homomorphism

[TABLE]

which is induced by $X\rightarrow Di\langle X|T\rangle,\ x\mapsto x+Id(T)$. Since $\theta(S)=0$, we have a homomorphism

[TABLE]

On the other hand, the following homomorphism

[TABLE]

is induced by $X\rightarrow Di[X|S],\ x\mapsto x+Id(S)$. Consider the natural homomorphism

[TABLE]

Since $\xi(T)=0$, we have a homomorphism $\zeta:\ Di\langle X|T\rangle\rightarrow Di[X|S]$ such that $\zeta\eta=\xi$ and $\zeta$ is exactly the inverse of $\sigma$. This shows that $\sigma$ is an isomorphism. Thus $\delta(Irr(S))=\{\delta(\lfloor a\rfloor_{n})\mid\lfloor a\rfloor_{n}\in Irr(S)\}$ is a $\mathbf{k}$-basis of the dialgebra $Di\langle X|T\rangle$. Now, for any $[w]_{n}\in[X^{+}]_{\omega}$, by Lemma 37, we have

[TABLE]

Hence, $\delta(Irr(S))=Irr_{A}(T)$. It follows that $Irr_{A}(T)$ is a $\mathbf{k}$-basis of the dialgebra $Di\langle X|T\rangle$. By Lemma 36, $T$ is a Gröbner–Shirshov basis in $Di\langle X\rangle$. ∎

Acknowledgement We wish to express our thanks to the referee for helpful suggestions and comments.