$t$-Unique Reductions for M\'esz\'aros's Subdivision Algebra

Fix a commutative ring $\mathbf{k}$, two elements $\beta,\alpha\in\mathbf{k}$ and a positive integer $n$. Let $\mathcal{X}$ be the polynomial ring over $\mathbf{k}$ in the $n(n-1)/2$ indeterminates $x_{i,j}$ for all $1\leq i<j\leq n$. Consider the ideal $\mathcal{J}$ of $\mathcal{X}$ generated by all polynomials of the form $x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ for $1\leq i<j<k\leq n$. The quotient algebra $\mathcal{X}/\mathcal{J}$ (at least for a certain choice of $\mathbf{k}$, $\beta$ and $\alpha$) has been introduced by Karola M\'esz\'aros as a commutative analogue of Anatol Kirillov's quasi-classical Yang-Baxter algebra. A monomial in $\mathcal{X}$ is said to be pathless if it has no divisors of the form $x_{i,j}x_{j,k}$ with $1\leq i<j<k\leq n$. The residue classes of these pathless monomials span the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$, but (in general) are $\mathbf{k}$-linearly dependent. Recently, the study of Grothendieck polynomials has led Laura Escobar and Karola M\'esz\'aros to defining a $\mathbf{k}$-algebra homomorphism $D$ from $\mathcal{X}$ into the polynomial ring $\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]$ that sends each $x_{i,j}$ to $t_{i}$. We show the following fact (generalizing a conjecture of M\'esz\'aros): If $p\in\mathcal{X}$, and if $q\in\mathcal{X}$ is a $\mathbf{k}$-linear combination of pathless monomials satisfying $p\equiv q\operatorname{mod}\mathcal{J}$, then $D(q)$ does not depend on $q$ (as long as $\beta$, $\alpha$ and $p$ are fixed). Thus, reducing a $p\in\mathcal{X}$ modulo $\mathcal{J}$ may lead to different results depending on the choices made in the reduction process, but all of them become identical once $D$ is applied. We also find an actual basis of the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$, using what we call forkless monomials.

March 26, 2018 x i,j x j,k | m (sometimes, there are several choices). Now, replace this monomial m by x i,k x i,j + x j,k + β + α x i,j x j,k m in p.
Thus, each move modifies the polynomial, replacing a monomial by a sum of four monomials (or fewer, if β or α is 0). The game ends when no more moves are possible (i.e., no monomial m appearing in your polynomial is divisible by x i,j x j,k for any 1 ≤ i < j < k ≤ n).
According to a conjecture of Mészáros, the result of this substitution is indeed independent of the choices made during the game (as long as p is fixed).
Why would one play a game like this? The interest in the reduction rule m → x i,k x i,j + x j,k + β x i,j x j,k m (this is a particular case of our above rule, when α is set to 0) originates in Karola Mészáros's study [Meszar09] of the abelianization of Anatol Kirillov's quasi-classical Yang-Baxter algebra (see, e.g., [Kirill16] for a recent survey of the latter and its many variants). To define this abelianization 1 , we let β be an indeterminate (unlike in Example 0.1, where it was an element of Q). Furthermore, fix a positive integer n. The abelianization of the (n-th) quasiclassical Yang-Baxter algebra is the commutative Q [β]-algebra S (A n ) with generators x i,j for all 1 ≤ i < j ≤ n and relations x i,j x j,k = x i,k x i,j + x j,k + β for all 1 ≤ i < j < k ≤ n.
A natural question is to find an explicit basis of S (A n ) (as a Q-vector space, or, if possible, as a Q [β]-module). One might try constructing such a basis using a reduction algorithm (or "straightening law") that takes any element of S (A n ) (written as any polynomial in the generators x i,j ) and rewrites it in a "normal form". The most obvious way one could try to construct such a reduction algorithm is by repeatedly rewriting products of the form x i,j x j,k (with 1 ≤ i < j < k ≤ n) as x i,k x i,j + x j,k + β , until this is no longer possible. This is precisely the game that we played in Example 0.1 (with the only difference that β is now an indeterminate, not a number). Unfortunately, the result of the game turns out to depend on the choices made while playing it; consequently, the "normal form" it constructs is not literally a normal form, and instead of a basis of S (A n ) we only obtain a spanning set. 2 Nevertheless, the result of the game is not meaningless. The idea to substitute t i for x i,j (in the result, not in the original polynomial!) seems to have appeared in work of Postnikov, Stanley and Mészáros; some concrete formulas (for specific 1 The notations used in this Introduction are meant to be provisional. In the rest of this paper, we shall work with different notations (and in a more general setting), which will be introduced in Section 1. 2 Surprisingly, a similar reduction algorithm does work for the (non-abelianized) quasi-classical Yang-Baxter algebra itself. This is one of Mészáros's results ( [Meszar09,Theorem 30]).
values of the initial polynomial and specific values of β) appear in [Stanle15, Exercise A22] (resulting in Catalan and Narayana numbers). Recent work on Grothendieck polynomials by Laura Escobar and Karola Mészáros [EscMes15,§5] has again brought up the notion of substituting t i for x i,j in the polynomial obtained at the end of the game. This has led Mészáros to the conjecture that, after this substitution, the resulting polynomial no longer depends on the choices made during the game. She has proven this conjecture for a certain class of polynomials (those corresponding to "noncrossing trees").
The main purpose of this paper is to establish Mészáros's conjecture in the general case. We shall, in fact, work in greater generality than all previously published sources. First, instead of the relation x i,j x j,k = x i,k x i,j + x j,k + β , we shall consider the "deformed" relation x i,j x j,k = x i,k x i,j + x j,k + β + α; the idea of this deformation again goes back to the work of Anatol Kirillov (see, e.g., [Kirill16, Definition 5.1 (1)] for a noncommutative variant of the quotient ring X /J , which he calls the "associative quasi-classical Yang-Baxter algebra of weight (α, β)"). Instead of requiring β to be either a rational number (as in Example 0.1) or an indeterminate over Q (as in the definition of S (A n )), we shall let β be any element of the ground ring, which in turn will be an arbitrary commutative ring k. Rather than working in an algebra like S (A n ), we shall work in the polynomial ring X = k x i,j | 1 ≤ i < j ≤ n , and study the ideal J generated by all elements of the form Instead of focussing on the reduction algorithm, we shall generally study polynomials in X that are congruent to each other modulo the ideal J . A monomial in X will be called "pathless" if it is not divisible by any monomial of the form x i,j x j,k with i < j < k. A polynomial in X will be called "pathless" if all monomials appearing in it are pathless. Thus, "pathless" polynomials are precisely the polynomials p ∈ X for which the game in Example 0.1 would end immediately if started at p.
Our main result (Theorem 1.7) will show that if p ∈ X is a polynomial, and if q ∈ X is a pathless polynomial congruent to p modulo J , then the image of q under the substitution x i,j → t i does not depend on q (but only on α, β and p). This, in particular, yields Mészáros's conjecture; but it is a stronger result, because it does not require that q is obtained from p by playing the game from Example 0.1 (all we ask for is that q be pathless and congruent to p modulo J ), and of course because of the more general setting.
After the proof of Theorem 1.7, we shall rewrite the definition of J (and of X ) in a more symmetric form (Section 2.11). Then, we shall also answer the (easier) question of finding a basis for the quotient ring X /J (Proposition 3.4). This basis will be obtained using an explicit Gröbner basis of the ideal J .
We shall close with further considerations and open questions, some of which relate the quotient ring X /J with the Orlik-Terao algebra of the braid arrangement.
A recent preprint by Mészáros and St. Dizier [MesDiz17] proves a fact [MesDiz17, Theorem A] which, translated into our language, confirms the conjecture stated in Example 0.1 at least in the case when α = 0 and the game is started with a monomial p. This might provide a different route to some of our results. (The arguments in [MesDiz17] are of combinatorial nature, involving flows on graphs, and so is the language used in [MesDiz17]; in particular, monomials are encoded by graphs.)

Acknowledgments
The SageMath computer algebra system [SageMath] was of great service during the development of the results below. Conversations with Nick Early have led me to the ideas in Subsection 4.3, and Victor Reiner has helped me concretize them.

Remark on previous versions
In a previous version (arXiv:1704.00839v2) of this preprint, a weaker version of the main result was proven (which corresponds to the case α = 0 in our notations). The proof used a somewhat different construction (involving formal power series instead of Laurent series, and a different map A).

Definitions and results
Let us now start from scratch, and set the stage for the main result. Let k be a commutative ring. (We fix k throughout this paper.) Fix two elements β and α of k. The word "monomial" shall always mean an element of a free abelian monoid (written multiplicatively). For example, the monomials in two indeterminates x and y are the elements of the form x i y j with (i, j) ∈ N 2 . Thus, monomials do not include coefficients (and are not bound to a specific base ring). Definition 1.2. Fix a positive integer n. Let X be the polynomial ring This is a polynomial ring in n (n − 1) /2 indeterminates x i,j over k.
We shall use the notation M for the set of all monomials in these indeterminates x i,j . Notice that M is an abelian monoid under multiplication. Definition 1.3. A monomial m ∈ M is said to be pathless if there exists no triple (i, j, k) ∈ [n] 3 satisfying i < j < k and x i,j x j,k | m (as monomials).
A polynomial p ∈ X is said to be pathless if it is a k-linear combination of pathless monomials. Definition 1.4. Let J be the ideal of X generated by all elements of the form The following fact is easy to check: Proposition 1.5. Let p ∈ X . Then, there exists a pathless polynomial q ∈ X such that p ≡ q mod J .
In general, this q is not unique. 3 We shall roughly outline a proof of Proposition 1.5 now; a more detailed writeup of this proof can be found in Section 2.10 below.
Proof of Proposition 1.5 (sketched). The weight of a monomial ∏ If we have a monomial m ∈ M that is not pathless, then we can find a triple (i, j, k) ∈ [n] 3 satisfying i < j < k and x i,j x j,k | m; then, we can replace m by a polynomial m = m · x i,k x i,j + x j,k + β + α x i,j x j,k , which is congruent to m modulo J but has the property that all monomials appearing in it have a smaller weight than m. This gives rise to a recursive algorithm 4 for reducing a polynomial modulo the ideal J . The procedure will necessarily terminate (although its result might depend on the order of operation); the polynomial resulting at its end will be pathless.
In [EscMes15, §5 and §7], Escobar and Mészáros (motivated by computations of Grothendieck polynomials) consider the result of substituting t i for each variable x i,j in a polynomial f ∈ X . In our language, this leads to the following definition: Definition 1.6. Let T ′ be the polynomial ring k [t 1 , t 2 , . . . , t n−1 ]. We define a k-algebra homomorphism D : X → T ′ by The goal of this paper is to prove the following fact, which (in a less general setting) was conjectured by Karola Mészáros in a 2015 talk at MIT: Theorem 1.7. Let p ∈ X . Consider any pathless polynomial q ∈ X such that p ≡ q mod J . Then, D (q) does not depend on the choice of q (but merely on the choice of α, β and p).
It is not generally true that D (q) = D (p); thus, Theorem 1.7 does not follow from a simple "invariant".

Preliminaries
The proof of Theorem 1.7 will occupy most of this paper. It proceeds in several steps. First, we shall define four k-algebras Q, T ′ [[w]], T and T [[w]] (with T ′ being a subalgebra of T ) and three k-linear maps A, B and E (with A and E being k-algebra homomorphisms) forming a diagram O O (where the vertical arrow is a canonical injection) that is not commutative. We shall eventually show that: • (Proposition 2.5 below) the homomorphism A annihilates the ideal J , • (Proposition 2.10 below) the homomorphism E is injective, and These three facts will allow us to prove Theorem 1.7. Indeed, the first and the third will imply that each pathless polynomial in J is annihilated by E • D; because of the second, this will show that it is also annihilated by D; and from here, Theorem 1.7 will easily follow.

The algebra Q of Laurent series
Let us begin by defining the notion of (formal) Laurent series in n indeterminates r 1 , r 2 , . . . , r n . This is somewhat slippery terrain, and it is easy to accidentally get a non-working definition (e.g., a notion of "Laurent series" not closed under multiplication, or not allowing multiplication at all), but there are also several different working definitions (see, e.g., [MonKau13] for a systematic treatment revealing many degrees of freedom). The definition we shall give here has been tailored to make our constructions work.
We begin by defining a k-module Q ± of "two-sided infinite formal power series over k"; this is not going to be a ring: Definition 2.1. Consider n distinct symbols r 1 , r 2 , . . . , r n . Let R denote the free abelian group on these n symbols, written multiplicatively. (That is, R is the free Z-module on n generators r 1 , r 2 , . . . , r n , but with the addition renamed as multiplication.) The elements of R thus have the form r a 1 1 r a 2 2 · · · r a n n for (a 1 , a 2 , . . . , a n ) ∈ Z n ; we shall refer to such elements as Laurent monomials in the symbols r 1 , r 2 , . . . , r n .
Informally, we let Q ± denote the k-module of all "infinite k-linear combinations" of Laurent monomials. Formally speaking, we define Q ± as the direct product ∏ r∈R k of copies of k indexed by Laurent monomials. We want to write each element (λ r ) r∈R ∈ ∏ r∈R k of this direct product as the formal k-linear combination ∑ r∈R λ r r; in order for this to work, we make several further conventions: First, we identify each Laurent monomial s ∈ R with the element (δ s,r ) r∈R of Q ± (where δ s,r is the Kronecker delta). Second, we equip the k-module Q ± with a topology: namely, the product topology, defined by recalling that it is a direct product ∏ r∈R k of copies of k (each of which is equipped with the discrete topology). Having made these conventions, we can easily verify that each element (λ r ) r∈R of ∏ r∈R k = Q ± is indeed identical with the infinite sum ∑ r∈R λ r r (which makes sense because of the topology on Q ± ). As usual, if f = (λ r ) r∈R is an element of Q ± , then λ r (for a given r ∈ R) will be called the coefficient of r in f and denoted by [r] f . As we know, the Laurent monomials in R have the form r a 1 1 r a 2 2 · · · r a n n for (a 1 , a 2 , . . . , a n ) ∈ Z n ; thus, sums of the form ∑ r∈R λ r r can also be rewritten in the form ∑ (a 1 ,a 2 ,...,a n )∈Z n λ a 1 ,a 2 ,...,a n r a 1 1 r a 2 2 · · · r a n n ; this is the usual way in which elements of Q ± are written. For example, for n = 1, an element of Q ± will have the form ∑ a∈Z λ a r a 1 for some family (λ a ) a∈Z of elements of k. Already in this simple situation, we see that Q ± is not a ring (or, at least, the usual recipe for multiplying power series does not work in Q ± ): Multiplying ∑ which is not a convergent sum in any reasonable topology (it contains each Laurent monomial infinitely many times). We shall define Laurent series as a subring of Q ± :

Definition 2.2. (a)
If d is an integer and r ∈ R is a Laurent monomial, then we say that r lives above d if and only if r = r a 1 1 r a 2 2 · · · r a n n for some (a 1 , a 2 , . . . , a n ) ∈ {d, d + 1, d + 2, . . .} n .
(b) If d is an integer and f is an element of Q ± , then we say that f is supported above d if and only if every (a 1 , a 2 , . . . , a n ) ∈ Z n \ {d, d + 1, d + 2, . . .} n satisfies r a 1 1 r a 2 2 · · · r a n n f = 0. In other words, f is supported above d if and only if f is an infinite k-linear combination of Laurent monomials that live above d.
(c) An element f ∈ Q ± is said to be a Laurent series if and only if there exists some d ∈ Z such that f is supported above d.
(e) A multiplication can be defined on k ((r 1 , r 2 , . . . , r n )) by extending the multiplication in the group R (in such a way that the resulting map is bilinear and continuous). Explicitly, this means that if f = ∑ r∈R λ r r and g = ∑ r∈R µ r r are two Laurent series, then their product f g is defined as the Laurent series The inner sum ∑ (u,v)∈R 2 ; uv=r λ u µ v here is well-defined, because all but finitely many of its addends are zero. (In fact, if f is supported above d, and if g is supported above e, then (for each given r ∈ R) there are only finitely many pairs (u, v) ∈ R 2 such that uv = r and u lives above d and v lives above e; but these are the only pairs that can contribute nonzero addends to the sum ∑ Thus, k ((r 1 , r 2 , . . . , r n )) becomes a k-algebra with unity 1 = r 0 1 r 0 2 · · · r 0 n . We denote this k-algebra by Q. Note that Q is a topological k-algebra; its topology is inherited from Q ± .
(f) An element f ∈ Q ± is said to be a formal power series if and only if f is supported above 0.
We now define certain Laurent monomials q 1 , q 2 , . . . , q n that we shall often use: , we define a Laurent monomial q i in the indeterminates r 1 , r 2 , . . . , r n by q i = r i r i+1 · · · r n . Notice that this q i is an actual monomial, not only a Laurent monomial.
Notice that each Laurent monomial in R belongs to Q. Each of the elements q 1 , q 2 , . . . , q n of Q is a Laurent monomial, and thus has an inverse (in R and thus also in Q). Hence, it makes sense to speak of quotients such as q i /q j for Thus, for any i ∈ [n] and j ∈ [n] satisfying i < j, the difference 1 − q i /q j = 1 − r i r i+1 · · · r j−1 is a formal power series in k [[r 1 , r 2 , . . . , r n ]] having constant term 1; it is therefore invertible in k [[r 1 , r 2 , . . . , r n ]].
It is easy to see that q a 1 1 q a 2 2 · · · q a n n = r a 1 1 r a 1 +a 2 2 r a 1 +a 2 +a 3 3 · · · r a 1 +a 2 +···+a n n (2) for all (a 1 , a 2 , . . . , a n ) ∈ Z n . Also, Thus, each Laurent monomial r ∈ R can be written uniquely in the form q a 1 1 q a 2 2 · · · q a n n with (a 1 , a 2 , . . . , a n ) ∈ Z n . Thus, q a 1 1 q a 2 2 · · · q a n n (a 1 ,a 2 ,...,a n )∈Z n is a topological basis 5 of the k-module Q ± . 5 The notion of a "topological basis" that we are using here has nothing to do with the concept of a basis of a topology (also known as "base"). Instead, it is merely an analogue of the concept of a basis of a k-module. It is defined as follows: A topological basis of a topological k-module M means a family (m s ) s∈S ∈ M S with the following two properties:

The algebra homomorphism
Notice that this is well-defined, since all denominators appearing here are invertible (indeed, q j is an invertible Laurent monomial in R, and 1 − q i /q j is an invertible formal power series in k [[r 1 , r 2 , . . . , r n ]]).
Proof of Proposition 2.5. The ideal J of X is generated by all elements of the form For example, r b 1 1 r b 2 2 · · · r b n n (b 1 ,b 2 ,...,b n) ∈N n is a topological basis of the topological k-module k [[r 1 , r 2 , . . . , r n ]], because each power series can be uniquely represented as an infinite klinear combination of all the monomials.
But A is a k-algebra homomorphism. Thus, as an unpleasant but straightforward computation reveals. This proves Proposition 2.5.

The algebras T and T [[w]] of power series
Definition 2.6. (a) Let T be the topological k-algebra k [[t 1 , t 2 , . . . , t n ]]. This is the ring of formal power series in the n indeterminates t 1 , t 2 , . . . , t n over k.
The topology on T shall be the usual one (i.e., the one defined similarly to the one on Q ± ).
(b) We shall regard the canonical injections ]. These are the k-algebras of formal power series in a (new) indeterminate w over T and over T ′ , respectively. We endow the k-algebra T [[w]] with a topology defined as the product topology, where T [[w]] is identified with a direct product of infinitely many copies of T (each of which is equipped with the topology we previously defined).

The continuous k-linear map B : Q → T [[w]]
We have T = k [[t 1 , t 2 , . . . , t n ]]. Thus, T [[w]] can be regarded as the ring of formal power series in the n + 1 indeterminates t 1 , t 2 , . . . , t n , w over k. (Strictly speaking, this should say that there is a canonical topological k-algebra isomorphism from T [[w]] to the latter ring). Let us now show a simple lemma: Lemma 2.7. Let m be a monomial in the indeterminates t 1 , t 2 , . . . , t n , w (with nonnegative exponents). Then, there exist only finitely many (a 1 , a 2 , . . . , a n ) Hence, S n is also a finite set. We want to prove that there exist only finitely many (a 1 , a 2 , . . . , a n ) ∈ Z n satisfying (4). We shall show that each such (a 1 , a 2 , . . . , a n ) belongs to the set S n .
Let j ∈ [n]. We want to prove that a j ∈ S. We know that (4) holds. Thus  This is an equality between two monomials in the indeterminates t 1 , t 2 , . . . , t n , w.
Comparing exponents on both sides of this equality, we find that and Now, we are in one of the following three cases: Case 1: We have a j < 0. Case 2: We have a j = 0. Case 3: We have a j > 0.
Let us first consider Case 1. In this case, we have a j < 0. Thus, we can split off the addend for i = j from the sum ∑ In other words, a j ≥ −c. Combining this with a j < 0, we find Thus, a j ∈ S is proven in Case 1.
Let us now consider Case 2. In this case, we have a j = 0. Hence, Thus, a j ∈ S is proven in Case 2. Let us finally consider Case 3. In this case, we have a j > 0. Applying (5) to Thus, a j ∈ S is proven in Case 3.
We have now proven a j ∈ S in all three Cases 1, 2 and 3. Hence, a j ∈ S always holds.
Forget that we have fixed j. We thus have shown that a j ∈ S for each j ∈ [n]. In other words, (a 1 , a 2 , . . . , a n ) belongs to S n . Now, forget that we have fixed (a 1 , a 2 , . . . , a n ). We thus have shown that each (a 1 , a 2 , . . . , a n ) ∈ Z n satisfying (4) belongs to S n . Since S n is a finite set, this shows that there exist only finitely many (a 1 , a 2 , . . . , a n ) ∈ Z n satisfying (4). This proves Lemma 2.7.

March 26, 2018
Definition 2.8. We define a continuous k-linear map B : for each (a 1 , a 2 , . . . , a n ) ∈ Z n . This is well-defined, since q a 1 1 q a 2 2 · · · q a n n (a 1 ,a 2 ,...,a n )∈Z n is a topological basis of Q ± , and because of Lemma 2.7 (which guarantees convergence when the map B is applied to an infinite k-linear combination of Laurent monomials).
The k-linear map B : can be restricted to the k-submodule Q of Q ± . We denote this restriction by B as well. In the following, we shall only be concerned with this restriction.
Of course, B is (in general) not a k-algebra homomorphism.
2.6. The k-algebra monomorphism E : This is well-defined (by the universal property of the polynomial ring T ′ = k [t 1 , t 2 , . . . , t n−1 ]), because for each i ∈ [n − 1], the power series 1 − t i w is invertible in T ′ [[w]] (indeed, its constant term is 1).
Proposition 2.10. The homomorphism E is injective.
Proof of Proposition 2.10. Let F : T ′ [[w]] → T ′ be the T ′ -algebra homomorphism that sends each formal power series f ∈ T ′ [[w]] (regarded as a formal power series in the single indeterminate w over T ′ ) to its constant term f (0) ∈ T ′ . Thus, F is a k-algebra homomorphism, and it sends w to 0 while sending each element of T ′ to itself.
Notice that the map G • F • E : T ′ → T ′ is a k-algebra homomorphism (since it is the composition of the three k-algebra homomorphisms E, F, G).

March 26, 2018
For each i ∈ [n − 1], we have and thus Hence, the two k-algebra homomorphisms G • F • E : T ′ → T ′ and id : T ′ → T ′ agree on the generating set {t 1 , t 2 , . . . , t n−1 } of the k-algebra T ′ . Thus, these two homomorphisms must be identical. In other words, G • F • E = id. Hence, the map E has a left inverse, and thus is injective. This proves Proposition 2.10.
Thus, we have defined the following spaces and maps between them: O O (but this is not a commutative diagram). It is worth reminding ourselves that A, D and E are k-algebra homomorphisms, but B (in general) is not.

Pathless monomials and subsets S of [n − 1]
Next, we want to study the action of the compositions B • A and E • D on pathless monomials. We first introduce some more notations: Definition 2.11. Let S be a subset of [n − 1]. (a) Let P S be the set of all pairs (i, j) ∈ S × ([n] \ S) satisfying i < j.

(b)
A monomial m ∈ M is said to be S-friendly if it is a product of some of the indeterminates x i,j with (i, j) ∈ P S . In other words, a monomial m ∈ M is S-friendly if and only if every indeterminate x i,j that appears in m satisfies i ∈ S and j / ∈ S. We let M S denote the set of all S-friendly monomials. (c) We let X S denote the polynomial ring k x i,j | (i, j) ∈ P S . This is clearly a subring of X . The k-module X S has a basis consisting of all S-friendly monomials m ∈ M.
(d) An n-tuple (a 1 , a 2 , . . . , a n ) ∈ Z n is said to be S-adequate if and only if it satisfies (a i ≥ 0 for all i ∈ S) and (a i ≤ 0 for all i ∈ [n] \ S). We let Q S denote the subset of Q consisting of all infinite k-linear combinations of the Laurent monomials q a 1 1 q a 2 2 · · · q a n n for S-adequate n-tuples (a 1 , a 2 , . . . , a n ) ∈ Z n (as long as these combinations belong to Q). It is easy to see that Q S is a topological k-subalgebra of Q (since the entrywise sum of two S-adequate n-tuples is S-adequate again).
(At this point, it is helpful to recall once again that the q 1 , q 2 , . . . , q n are not indeterminates, but rather monomials defined by q i = r i r i+1 · · · r n . But their products q a 1 1 q a 2 2 · · · q a n n are Laurent monomials. Explicitly, they can be rewritten as products of the r 1 , r 2 , . . . , r n using (2). Thus, it is easy to see that the elements of Q S are the infinite k-linear combinations of the Laurent monomials r (f) We define a k-algebra homomorphism A S : X S → Q S by Notice that this is well-defined 6 . (g) We define a continuous k-linear map B S : Q S → T S [[w]] by setting B S q a 1 1 q a 2 2 · · · q a n n = ∏ i∈S t a i i for each S-adequate (a 1 , a 2 , . . . , a n ) ∈ Z n . This is well-defined, as we will see below (in Proposition 2.12 (b)).
. This is a k-subalgebra of T ′ . Hence, the ring T ′ S [[w]] (that is, the ring of formal power series in the (single) variable w over T ′ S ) is a k-subalgebra of the similarly-defined ring March 26, 2018 (i) We define a k-algebra homomorphism D S : X S → T ′ S by This is well-defined, since each (i, j) ∈ P S satisfies i ∈ S. (j) We define a k-algebra homomorphism E S : This is well-defined (by the universal property of the polynomial ring T ′ S ), because for each i ∈ S, the power series 1 − t i w is invertible in T ′ S [[w]] (indeed, its constant term is 1).
Proposition 2.12. Let S be a subset of [n − 1].
(a) We have B q a 1 1 q a 2 2 · · · q a n n = ∏ i∈S t a i i for each S-adequate n-tuple (a 1 , a 2 , . . . , a n ) ∈ Z n . (b) The map B S (defined in Definition 2.11 (g)) is well-defined.
We must prove (7). The n-tuple (a 1 , a 2 , . . . , a n ) is S-adequate. Thus, (a i ≥ 0 for all i ∈ S) and (a i ≤ 0 for all i ∈ [n] \ S). In particular, (a i ≤ 0 for all i ∈ [n] \ S). Hence, each i ∈ [n] satisfying a i > 0 must belong to S (because otherwise, i would belong For each k ≥ 0, the Laurent monomial q i /q j k belongs to Q S , because the n-tuple (0, 0, . . . , 0, k, 0, 0, . . . , 0, −k, 0, 0, . . . , 0) (where the k stands in the i-th position, and the −k stands in the j-th position) is S-adequate. Hence, ∑ k≥0 q i /q j k ∈ Q S . In view of ∑ k≥0 q i /q j k = 1 1 − q i /q j , this rewrites as 1 1 − q i /q j ∈ Q S . So we know that both Laurent series q i + β + α/q j and 1 1 − q i /q j belong to Q S . Therefore, so does their product proves that A S is well-defined (via the universal property of the polynomial ring X S ).
to [n] \ S, and therefore would have to satisfy a i ≤ 0, which would contradict a i > 0). On the other hand, (a i ≥ 0 for all i ∈ S). Hence, each i ∈ [n] satisfying a i < 0 must belong to [n] \ S (because otherwise, i would belong to S, and therefore would have to satisfy a i ≥ 0, which would contradict a i < 0). Now, the definition of the map B yields B q a 1 1 q a 2 2 · · · q a n n =    Comparing this with (since each i∈[n]\S satisfies either a i =0 or a i <0 (since a i ≤0 for all i∈[n]\S)) we obtain B q a 1 1 q a 2 2 · · · q a n n = ∏  1 q a 2 2 · · · q a n n = ∏ i∈S t a i i ∏ i∈[n]\S w −a i for each S-adequate (a 1 , a 2 , . . . , a n ) ∈ Z n    .
The uniqueness of such a map is clear (because the elements of Q S are infinite k-linear combinations of the Laurent monomials q a 1 1 q a 2 2 · · · q a n n for S-adequate ntuples (a 1 , a 2 , . . . , a n ) ∈ Z n ; but the formula (8) uniquely determines the value of B S on such a k-linear combination). Thus, it remains to prove its existence.
For each f ∈ Q S , we have B ( f ) ∈ T S [[w]] 7 . Hence, we can define a map This map B S is a restriction of the map B; hence, it is a continuous k-linear map (since B is a continuous k-linear map). Furthermore, it satisfies B S q a 1 1 q a 2 2 · · · q a n n = B q a 1 1 q a 2 2 · · · q a n n by the definition of 7 Proof. Let f ∈ Q S . We must show that ]. Since the map B is k-linear and continuous, we can WLOG assume that f is a Laurent monomial of the form q a 1 1 q a 2 2 · · · q a n n for some S-adequate n-tuple (a 1 , a 2 , . . . , a n ) ∈ Z n (because f is always an infinite k-linear combination of such Laurent monomials). Assume this. Consider this (a 1 , a 2 , . . . , a n ) ∈ Z n . Thus, f = q a 1 1 q a 2 2 · · · q a n n . Applying the map B to both sides of this equality, we obtain This is precisely what we wanted to show.

Reductions for the subdivision algebra
March 26, 2018 (where the vertical arrows are the obvious inclusion maps) are commutative.
Proof of Proposition 2.13. The commutativity of the left square of (9) is obvious 8 . So is the commutativity of each of the two squares of (10) 9 . It thus remains to prove the commutativity of the right square of (9). In other words, we must show that B S (p) = B (p) for each p ∈ Q S . So fix p ∈ Q S . Since both maps B S and B are continuous and k-linear, we can WLOG assume that p is a Laurent monomial of the form q a 1 1 q a 2 2 · · · q a n n for an S-adequate n-tuple (a 1 , a 2 , . . . , a n ) ∈ Z n (since the elements of Q S are infinite k-linear combinations of Laurent monomials of this form). Assume this, and fix this (a 1 , a 2 , . . . , a n ).
From p = q a 1 1 q a 2 2 · · · q a n n , we obtain 2.12 (a)). Comparing this with B S (p) = B S q a 1 1 q a 2 2 · · · q a n n since p = q a 1 1 q a 2 2 · · · q a n n = ∏ i∈S t a i i we obtain B S (p) = B (p). This proves the commutativity of the right square of (9). The proof of Proposition 2.13 is thus complete. 8 "Obvious" in the following sense: You want to prove that a diagram of the form is commutative, where A 1 , A 2 , A 3 , A 4 are four k-algebras and f 1 , f 2 , f 3 , f 4 are four k-algebra homomorphisms. (In our concrete case, A 1 = X S , A 2 = Q S , A 3 = X , A 4 = Q, f 1 = A S and f 4 = A, whereas f 2 and f 3 are the inclusion maps X S → X and Q S → Q.) In order to prove this commutativity, it suffices to show that it holds on a generating set of the k-algebra A 1 . In other words, it suffices to pick some generating set G of the k-algebra A 1 and show that all g ∈ G satisfy ( f 3 • f 1 ) (g) = ( f 4 • f 2 ) (g). (In our concrete case, it is most reasonable to pick G = x i,j | (i, j) ∈ P S . The proof then becomes completely clear.) 9 for similar reasons Proposition 2.14. Let S be a subset of [n − 1]. Then, B S : Q S → T S [[w]] is a continuous k-algebra homomorphism.
Proof of Proposition 2.14. We merely need to show that B S is a k-algebra homomorphism. To this purpose, by linearity, we only need to prove that B S (1) = 1 and B S (mn) = B S (m) B S (n) for any two Laurent monomials m and n of the form q a 1 1 q a 2 2 · · · q a n n for S-adequate n-tuples (a 1 , a 2 , . . . , a n ) ∈ Z n (since the elements of Q S are infinite k-linear combinations of Laurent monomials of this form). This is easy and left to the reader.
Proof of Proposition 2.15. From (i, j) ∈ P S , we obtain i ∈ S and j ∈ [n] \ S. Thus, the definition of B S reveals that B S (q i ) = t i and B S q −1 j = w. Proposition 2.14 shows that B S : Applying the map B S to both sides of this equality, we find since B S is a k-algebra homomorphism, and thus respects sums, products and fractions (as long as the denominators of the fractions are invertible) Comparing this with The map (E • D) | X S is a k-algebra homomorphism (since D and E are kalgebra homomorphisms), and the map B S • A S is a k-algebra homomorphism (since both B S and A S are k-algebra homomorphisms 11 ). Hence, we are trying to prove that two k-algebra homomorphisms are equal (namely, the homomorphisms (E • D) | X S and B S • A S ). It is clearly enough to prove this on the generating family x i,j (i,j)∈P S of the k-algebra X S . In other words, it is enough to So let us fix some (i, j) ∈ P S . Proposition 2.13 shows that the diagram (10) is commutative. Thus, (E • D) | X S = E S • D S (provided that we regard E S • D S as a 10 Proof. We need to show that every indeterminate x i,j that appears in m satisfies i ∈ S and j / ∈ S. Indeed, assume the contrary. Thus, some indeterminate x i,j that appears in m does not satisfy i ∈ S and j / ∈ S. Fix such an indeterminate x i,j , and denote it by x u,v . Thus, x u,v is an indeterminate that appears in m but does not satisfy u ∈ S and v / ∈ S. Therefore, we have either u / ∈ S or v ∈ S (or both). We have 1 ≤ u < v ≤ n (since the indeterminate x u,v exists) and thus u ∈ [n − 1]. The definition of b u yields b u = n ∑ j=u+1 a u,j . But v ≥ u + 1 (since u < v). Hence, a u,v is an addend of the sum n ∑ j=u+1 a u,j . Hence, n ∑ j=u+1 a u,j ≥ a u,v . But a u,v > 0 (since the indeterminate x u,v appears in m). Hence, b u = n ∑ j=u+1 a u,j ≥ a u,v > 0. Therefore, u ∈ S (by the definition of S).
Hence, u / ∈ S cannot hold. Therefore, v ∈ S (since we know that we have either u / ∈ S or v ∈ S). In other words, v ∈ [n − 1] and b v > 0 (by the definition of S). But the definition of We have v < w (since w ∈ {v + 1, v + 2, . . . , n}), hence u < v < w. Thus, (u, v) = (v, w). Moreover, the indeterminate x v,w appears in m (since a v,w > 0). Thus, both indeterminates x u,v and x v,w appear in m. Hence, But the monomial m is pathless. In other words, there exists no triple (i, j, k) ∈ [n] 3 satisfying i < j < k and x i,j x j,k | m. This contradicts the fact that (u, v, w) is such a triple (since u < v < w and x u,v x v,w | m). This contradiction completes our proof. 11 Here we are using Proposition 2.14.

Reductions for the subdivision algebra
This completes our proof of (E • D) | X S = B S • A S . Now, from m ∈ X S , we   B is k-linear). Thus, E (D (p)) = (E • D) (p) = 0. Since the klinear map E is injective (by Proposition 2.10), we thus conclude that D (p) = 0. This proves Lemma 2.18.
We are now ready to prove Theorem 1.7: Proof of Theorem 1.7. We need to prove that D (q) does not depend on the choice of q. In other words, we need to prove that if f and g are two pathless polynomials q ∈ X such that p ≡ q mod J , then D ( f ) = D (g).
So let f and g be two pathless polynomials q ∈ X such that p ≡ q mod J . Thus, p ≡ f mod J and p ≡ g mod J . Hence, f ≡ p ≡ g mod J , so that f − g ∈ J . Also, the polynomial f − g ∈ X is pathless (since it is the difference of the two pathless polynomials f and g). Thus, Lemma 2.18 (applied to f − g instead of p) shows that a k-algebra homomorphism). In other words, D ( f ) = D (g). This proves Theorem 1.7.
2.10. Appendix: Detailed proof of Proposition 1.5 Let us now pay a debt and explain the proof of Proposition 1.5 in full detail.
We begin with a few definitions: Definition 2.19. Let X pathless denote the k-submodule of X spanned by all pathless monomials m ∈ M. Thus, X pathless is the set of all pathless polynomials f ∈ X .
Lemma 2.22. Let m ∈ M be a monomial. Then, m ∈ X pathless + J .
Proof of Lemma 2.22. We shall prove Lemma 2.22 by strong induction over weight m. 12 Thus, we fix any N ∈ N, and we assume (as the induction hypothesis) that Lemma 2.22 holds in the case when weight m < N. We then must show that Lemma 2.22 holds in the case when weight m = N.
We have assumed that Lemma 2.22 holds in the case when weight m < N. In other words, if m ∈ M is a monomial such that weight m < N, then m ∈ X pathless + J .
Now, fix a monomial m ∈ M such that weight m = N. We shall show that m ∈ X pathless + J .
If m is pathless, then this is obvious 13 . Hence, for the rest of this proof, we WLOG assume that m is not pathless. In other words, there exists a triple (i, j, k) ∈ [n] 3 satisfying i < j < k and x i,j x j,k | m (as monomials). Consider such a triple (i, j, k).
We have x i,j x j,k | m (as monomials). In other words, there exists a monomial n ∈ M such that m = x i,j x j,k n. Consider this n.
We have x i,j x j,k − x i,k x i,j + x j,k + β − α ∈ J (since x i,j x j,k − x i,k x i,j + x j,k + β − α is one of the designated generators of the ideal J ). Thus, 12 This is legitimate, since Lemma 2.21 (c) shows that weight m ∈ N in the situation of Lemma 2.22. 13 Proof. Assume that m is pathless. We must then show that m ∈ X pathless + J .
But recall that the k-module X pathless is spanned by the pathless monomials. Thus, m ∈ X pathless (since m is a pathless monomial). Hence, m ∈ X pathless ⊆ X pathless + J , qed.
In other words, m ∈ x i,k x i,j n + x i,k x j,k n + βx i,k n + αn + J . (13) We shall now analyze the four monomials x i,k x i,j n, x i,k x j,k n, x i,k n and n on the right hand side of (13): • We have weight x i,k x i,j n < N 14 . Hence, (12) (applied to x i,k x i,j n instead of m) shows that x i,k x i,j n ∈ X pathless + J .
• We have weight x i,k x j,k n < N 15 . Hence, (12) (applied to x i,k x j,k n instead 14 Proof. We have m = x i,j x j,k n = x j,k x i,j n and thus weight m =x j,k x i,j n = weight x j,k x i,j n = weight x j,k =n−k+j (by Lemma 2.21 (a), applied to j and k instead of i and j) + weight x i,j n by Lemma 2.21 (b), applied to p = x j,k and q = x i,j n = n − k + j + weight x i,j n . But Lemma 2.21 (b) (applied to p = x i,k and q = x i,j n) shows that 2.21 (a), applied to k instead of j) + weight x j,k n by Lemma 2.21 (b), applied to p = x i,j and q = x j,k n = n − j + i + weight x j,k n . But Lemma 2.21 (b) (applied to p = x i,k and q = x j,k n) shows that 2.21 (a), applied to k instead of j) of m) shows that x i,k x j,k n ∈ X pathless + J .
• We have weight (x i,k n) < N 16 . Hence, (12) (applied to x i,k n instead of m) shows that x i,k n ∈ X pathless + J .
But Lemma 2.21 (b) (applied to p = x i,k and q = n) yields weight (x i,k n) = weight (x i,k ) =n−k+i (by Lemma 2.21 (a), applied to k instead of j) qed. 17 Proof. Lemma 2.21 (c) (applied to m = x j,k ) yields weight x j,k ∈ N. Thus, weight x j,k ≥ 0.
Also, Lemma 2.21 (a) yields weight 2.21 (b), applied to p=x j,k and q=n) by Lemma 2.21 (b), applied to p = x i,j and q = x j,k n > 0 + weight x j,k ≥0 + weight n ≥ 0 + 0 + weight n = weight n.
Hence, weight n < weight m = N, qed. that n ∈ X pathless + J . a k-module). Now, let us forget that we fixed m. We thus have shown that if m ∈ M is a monomial such that weight m = N, then m ∈ X pathless + J . In other words, Lemma 2.22 holds in the case when weight m = N. This completes the induction step. Thus, Lemma 2.22 is proven. Now, we can prove Proposition 1.5:

Now, (13) becomes
Proof of Proposition 1.5. Lemma 2.22 shows that m ∈ X pathless + J for each m ∈ M. In other words, M ⊆ X pathless + J (where we consider M as being embedded into the polynomial ring X ).
For any subset W of X , we let span W denote the k-submodule of X spanned by W. The set M spans the k-module X (since each polynomial f ∈ X is a k-linear combination of monomials m ∈ M). In other words, X = span M. But from M ⊆ X pathless + J , we obtain span M ⊆ span X pathless + J = X pathless + J (since X pathless + J is a k-submodule of X ). Hence, X = span M ⊆ X pathless + J . Now, p ∈ X ⊆ X pathless + J . In other words, there exist u ∈ X pathless and v ∈ J such that p = u + v. Consider these u and v.
We have p = u + v ∈J ∈ u + J . In other words, p ≡ u mod J . But X pathless is the set of all pathless polynomials in X . Thus, u is a pathless polynomial (since u ∈ X pathless ). Hence, there exists a pathless polynomial q ∈ X such that p ≡ q mod J (namely, q = u). This proves Proposition 1.5.

Appendix: A symmetric description of J
In this section, let us give a different description of J that reveals a symmetry inherent in the setting. First, we introduce auxiliary polynomials. So far, we have only been considering indeterminates x i,j corresponding to pairs (i, j) ∈ [n] 2 satisfying i < j. We shall now also define x i,j for pairs (i, j) ∈ [n] 2 satisfying i > j; but these x i,j will not be new indeterminates, but rather will be polynomials in X : Definition 2.23. (a) Let (i, j) ∈ [n] 2 be a pair satisfying i > j. Then, we define an element x i,j ∈ X by x i,j = −β − x j,i . Thus, an element x i,j ∈ X is defined for any pair (i, j) of two distinct elements of [n] 2 .
(b) For any three distinct elements i, j, k of [n], we define a polynomial J i,j,k ∈ X by Proposition 2.24. The ideal J of X is generated by all polynomials J i,j,k for i, j, k being three distinct elements of [n].
Proof of Proposition 2.24. If u p p∈P is any family of elements of X , then u p p∈P shall mean the ideal of X generated by this family u p p∈P . Thus, we need to prove that J = J i,j,k i,j,k are three distinct elements of [n] .
We know (from the definition of J ) that But for each (i, j, k) ∈ [n] 3 satisfying i < j < k, we have 18 . Hence, (14) rewrites as follows: On the other hand, the definition of J i,j,k shows that J i,j,k is symmetric in its three arguments i, j, k; in other words, we have J i,j,k = J i,k,j = J j,i,k = J j,k,i = J k,i,j = J k,j,i 18 Proof of (15): Let (i, j, k) ∈ [n] 3 be such that i < j < k. Then, the definition of J i,j,k yields This proves (15). i = j (because of (17)). So let us fix some (i, j) ∈ [n] 2 such that i = j. We must prove that φ y i,j + y j,i + β = 0. This statement is clearly symmetric in i and j; thus, we WLOG assume that i ≤ j. Hence, i < j (since i = j). The definition of φ yields φ y i,j = x i,j and φ y j,i = x j,i = −β − x i,j (by the definition of x j,i , since j > i). Now, φ is a k-algebra homomorphism. Thus, This completes our proof. 20 Proof of (20): Let (i, j) ∈ [n] 2 be such that i = j. We must prove that ζ x i,j = π y i,j .
If i < j, then this follows immediately from (19). Thus, we WLOG assume that we don't have i < j. Hence, i ≥ j, so that i > j (since i = j). In other words, j < i. Thus, (19) (applied to (j, i) instead of (i, j)) shows that ζ x j,i = π y j,i .
But the definition of x i,j yields x i,j = −β − x j,i (since i > j). Applying the map ζ to both 3. Forkless polynomials and a basis of X /J

Statements
We have thus answered one of the major questions about the ideal J ; but we have begged perhaps the most obvious one: Can we find a basis of the k-module X /J ? This turns out to be much simpler than the above; the key is to use a different strategy. Instead of reducing polynomials to pathless polynomials, we shall reduce them to forkless polynomials, defined as follows: Definition 3.1. A monomial m ∈ M is said to be forkless if there exists no triple (i, j, k) ∈ [n] 3 satisfying i < j < k and x i,j x i,k | m (as monomials). A polynomial p ∈ X is said to be forkless if it is a k-linear combination of forkless monomials.
The following characterization of forkless polynomials is rather obvious: and Now, we claim the following: Theorem 3.3. Let p ∈ X . Then, there exists a unique forkless polynomial q ∈ X such that p ≡ q mod J .
Proposition 3.4. The projections of the forkless monomials m ∈ M onto the quotient ring X /J form a basis of the k-module X /J .

A reminder on Gröbner bases
Theorem 3.3 and Proposition 3.4 can be proven using the theory of Gröbner bases. See, e.g., [BecWei98] for an introduction. Let us outline the argument. We shall use the following concepts: Definition 3.5. Let Ξ be a set of indeterminates. Let X Ξ be the polynomial ring k [ξ | ξ ∈ Ξ] over k in these indeterminates. Let M Ξ be the set of all monomials in these indeterminates (i.e., the free abelian monoid on the set Ξ).
(For example, if Ξ = x i,j | (i, j) ∈ [n] 2 satisfying i < j , then X Ξ = X and M Ξ = M.) (a) A term order on M Ξ is a total order on the set M Ξ that satisfies the following conditions: • Each m ∈ M Ξ satisfies 1 ≤ m (where 1 is the trivial monomial in M Ξ ).
• If m, u and v are three elements of M Ξ satisfying u ≤ v, then mu ≤ mv.
(b) If we are given a total order on the set Ξ, then we canonically obtain a term order on M Ξ defined as follows: For two monomials m = ∏ ξ∈Ξ ξ m ξ and n = ∏ ξ∈Ξ ξ n ξ in M Ξ , we set m ≤ n if and only if either m = n or the largest ξ ∈ Ξ for which m ξ and n ξ differ satisfies m ξ < n ξ . This term order is called the inverse lexicographical order on the set M Ξ determined by the given total order on Ξ.
(c) Two monomials m = ∏ ξ∈Ξ ξ m ξ and n = ∏ ξ∈Ξ ξ n ξ in M Ξ are said to be nondisjoint if there exists some ξ ∈ Ξ satisfying m ξ > 0 and n ξ > 0. Otherwise, m and n are said to be disjoint. From now on, let us assume that some term order on M Ξ has been chosen. The next definitions will all rely on this term order.
(d) If f ∈ X Ξ is a nonzero polynomial, then the head term of f denotes the largest m ∈ M Ξ such that the coefficient of m in f is nonzero. This head term will be denoted by HT ( f ). Furthermore, if f ∈ X Ξ is a nonzero polynomial, then the head coefficient of f is defined to be the coefficient of HT ( f ) in f ; this coefficient will be denoted by HC ( f ).
(e) A nonzero polynomial f ∈ X Ξ is said to be monic if its head coefficient HC ( f ) is 1. (g) If g 1 and g 2 are two monic polynomials in X Ξ , then the S-polynomial of g 1 and g 2 is defined to be the polynomial s 1 g 1 − s 2 g 2 , where s 1 and s 2 are the unique two monomials satisfying s 1 HT (g 1 ) = s 2 HT (g 2 ) = lcm (HT (g 1 ) , HT (g 2 )). This S-polynomial is denoted by spol (g 1 , g 2 ).
From now on, let G be a subset of X Ξ that consists of monic polynomials. if there exists some p ∈ G and some monomials t ∈ M Ξ and s ∈ M Ξ with the following properties: • The coefficient of t in f is = 0.
• If a is the coefficient of t in f , then g = f − a · s · p. (j) We say that a monomial m ∈ M Ξ is G-reduced if it is not divisible by the head term of any element of G.
We say that a polynomial q ∈ X Ξ is G-reduced if q is a k-linear combination of G-reduced monomials.
(k) Let I be an ideal of X Ξ . The set G is said to be a Gröbner basis of the ideal I if and only if the set G generates I and has the following two equivalent properties: • For each p ∈ X Ξ , there is a unique G-reduced q ∈ X Ξ such that p * −→ G q.
• For each p ∈ I, we have p * −→ G 0.
The definition we just gave is modelled after the definitions in [BecWei98, Chapter 5]; however, there are several minor differences: • We use the word "monomial" in the same meaning as [BecWei98, Chapter 5] use the word "term" (but not in the same meaning as [BecWei98, Chapter 5] use the word "monomial").
• We allow k to be a commutative ring, whereas [BecWei98, Chapter 5] require k to be a field. This leads to some complications in the theory of Gröbner bases; in particular, not every ideal has a Gröbner basis anymore. However, everything we are going to use about Gröbner bases in this paper is still true in our general setting.
• We require the elements of the Gröbner basis G to be monic, whereas [BecWei98, Chapter 5] merely assume them to be nonzero polynomials.
In this way, we are sacrificing some of the generality of [BecWei98, Chapter 5] (a sacrifice necessary to ensure that things don't go wrong when k is not a field). However, this is not a major loss of generality, since in the situation of [BecWei98, Chapter 5] the difference between monic polynomials and arbitrary nonzero polynomials is not particularly large (we can scale any nonzero polynomial by a constant scalar to obtain a monic polynomial, and so we can assume the polynomials to be monic in most of the proofs).
The following fact is useful even if almost trivial: Lemma 3.6. Let Ξ, X Ξ and M Ξ be as in Definition 3.5. Let G be a subset of X Ξ that consists of monic polynomials. Let S be a finite set. For each s ∈ S, let g s be an element of G, and let s s ∈ M Ξ and a s ∈ k be arbitrary. Assume that the monomials s s HT (g s ) for all s ∈ S are distinct. Then, ∑ s∈S a s s s g s * −→ G 0.
The following fact (known as "Buchberger's first criterion") somewhat simplifies dealing with S-polynomials: Proposition 3.8. Let Ξ, X Ξ and M Ξ be as in Definition 3.5. Let G be a subset of X Ξ that consists of monic polynomials. Let g 1 and g 2 be two elements of the set G such that the head terms of g 1 and g 2 are disjoint. Then, spol (g 1 , g 2 ) * −→ We can combine Proposition 3.7 with Proposition 3.8 to obtain the following fact: Proposition 3.9. Let Ξ, X Ξ and M Ξ be as in Definition 3.5. Let I be an ideal of X Ξ . Let G be a subset of X Ξ that consists of monic polynomials. Assume that the set G generates I. Then, G is a Gröbner basis of I if and only if it has the following property: • If g 1 and g 2 are two elements of the set G such that the head terms of g 1 and g 2 are non-disjoint, then spol (g 1 , g 2 ) * −→ G 0.
Proposition 3.9 follows trivially from Proposition 3.7 after recalling Proposition 3.8. Explicitly, Proposition 3.9 appears (at least in the case when k is a field) in [BecWei98, Theorem 5.68, (iii) ⇐⇒ (i)] and [Graaf16, Conclusion after the proof of Lemma 1.1.38].
We shall also use the following simple fact, known as the "Macaulay-Buchberger basis theorem": Proposition 3.10. Let Ξ, X Ξ and M Ξ be as in Definition 3.5. Let I be an ideal of X Ξ . Let G be a Gröbner basis of I. The projections of the G-reduced monomials onto the quotient ring X Ξ /I form a basis of the k-module X Ξ /I. 24 Different sources state slightly different versions of Proposition 3.7. For example, some texts require spol (g 1 , g 2 ) * −→ G 0 not for any two elements g 1 and g 2 of G, but only for any two distinct elements g 1 and g 2 of G. However, this makes no difference, because if g 1 and g 2 are equal, then spol (g 1 , g 2 ) = 0 * −→ G 0. Similarly, other texts require g 1 < g 2 (with respect to some chosen total order on G); this also does not change much, since spol (g 1 , g 2 ) = − spol (g 2 , g 1 ). 25 That said, the proof of [Monass02, Lemme (in the section "Améliorations de l'algorithme")] is incorrect.

Further questions
Let us finally indicate some further directions of research not mentioned so far.

The kernel of A
Question 4.1. (a) Is J the kernel of the map A : X → Q from Definition 2.4? (b) Consider the polynomial ring k [ q 1 , q 2 , . . . , q n ] in n indeterminates q 1 , q 2 , . . . , q n over k. Let Q rat denote the localization of this polynomial ring at the multiplicative subset generated by all differences of the form q i − q j (for 1 ≤ i < j ≤ n). Then, the morphism A : X → Q factors through a k-algebra homomorphism A : X → Q rat which sends each x i,j to − q i + β + α/ q j 1 − q i / q j = − q i q j + β q j + α q j − q i ∈ Q rat . Is J the kernel of this latter homomorphism A ?
Parts (a) and (b) of Question 4.1 are equivalent, since the canonical k-algebra homomorphism Q rat → Q is injective. This question is interesting partly because a positive answer to part (b) would provide a realization of X /J as a subalgebra of a localized polynomial ring in (only) n indeterminates. This subalgebra would probably not be the whole Q rat . (Perhaps it can be shown -by some kind of multidimensional residues -that A maps the forkless monomials in X to linearly independent elements of Q. Such a proof would then immediately yield positive answers to parts (a) and (b) of Question 4.1 as well as an alternative proof of Theorem 3.3.) An approach to Question 4.1 (b) might begin with finding a basis of the kmodule Q rat . It turns out that such a basis is rather easy to construct: Proposition 4.2. In Q rat , consider the family of all elements of the form where each g i has either the form 1 q i − q j m for some j ∈ {i + 1, i + 2, . . . , n} and m > 0 or the form q k i for some k ∈ N. This family is a basis of the k-module Q rat .
Notice that this family is similar to the forkless monomials in Proposition 3.4, but it is "larger" (if we would allow the g i to have the form q k i only for k = 0, then we would obtain a restricted family that would be in an obvious bijection with the forkless monomials).
Proposition 4.2 is closely related to results by Horiuchi and Terao ([HorTer03] and [Terao02]); indeed, if k is a field, then Q rat can be regarded as the ring of regular functions on the complement of the braid arrangement in k n , and such functions are what they have studied (although usually not the whole Q rat ). Notice however that they worked only over fields k of characteristic 0. Question 4.6. (a) Can an arbitrary Orlik-Terao algebra be deformed by two parameters β and α, generalizing our X /J ? (b) Our basis of forkless monomials for X /J can be regarded as an "nbc basis" in the sense of [CorEti01] (except that our monomials are not required to be squarefree). Indeed, if we totally order the monomials x i,j in such a way that x i,j > x u,v whenever i < u, then the broken circuits of the graphical matroid of K n are precisely the sets of the form {{i, j} , {i, k}} for i < j < k; but these correspond to the precise monomials x i,j x i,k that a forkless monomial cannot be divisible by. Does this extend to the arbitrary Orlik-Terao algebras?