## 1. Introduction

Let

P be a finite partially ordered set (poset for short) and

$\mathcal{I}\left(P\right)$ the distributive lattice of the poset ideals of

$P.$ A subset

$\alpha $ of

P is a poset ideal of

P if it satisfies the following condition: for every

$x\in \alpha $ and

$y\in P$, if

$y\le x$, then

$y\in \alpha .$ By a famous theorem of Birkhoff [

1], for every finite distributive lattice

L, there exists a unique subposet

P of

L such that

$L\cong \mathcal{I}\left(P\right).$ The order polytope

$\mathcal{O}\left(P\right)$ and the chain polytope

$\mathcal{C}\left(P\right)$ were introduced in [

2]. In [

3], it was shown that the toric ring

$K\left[\mathcal{O}\right(P\left)\right]$ over a field

K is an algebra with straightening laws (ASL in brief) on the distributive lattice

$\mathcal{I}\left(P\right)$ over the field

$K.$ In [

4], it was shown that the ring

$K\left[\mathcal{C}\right(P\left)\right]$ associated with the chain polytope shares the same property.

Let

$S=K[{x}_{1},\dots ,{x}_{n},t]$ be the polynomial ring over a field

K and

${\left\{{w}_{\alpha}\right\}}_{\alpha \in \mathcal{I}\left(P\right)}$ be an arbitrary set of monomials in

${x}_{1},\dots ,{x}_{n}$ indexed by

$\mathcal{I}\left(P\right)$. Let

$K\left[\mathsf{\Omega}\right]\subset S$ be the toric ring generated over

K by the set of monomials

$\mathsf{\Omega}={\left\{{\omega}_{\alpha}\right\}}_{\alpha \in \mathcal{I}\left(P\right)}$ where

${\omega}_{\alpha}={w}_{\alpha}t$ for all

$\alpha \in \mathcal{I}\left(P\right).$ Clearly,

$K\left[\mathsf{\Omega}\right]$ is a graded algebra if we set

$deg\left({\omega}_{\alpha}\right)=1$ for all

$\alpha \in \mathcal{I}\left(P\right).$ Let

$\phi :\mathcal{I}\left(P\right)\to K\left[\mathsf{\Omega}\right]$ be the injective map defined by

$\phi \left(\alpha \right)={\omega}_{\alpha}$ for all

$\alpha \in \mathcal{I}\left(P\right).$ Assume that

$K\left[\mathsf{\Omega}\right]$ is an ASL on

$\mathcal{I}\left(P\right)$ over

$K.$ According to [

4],

$K\left[\mathsf{\Omega}\right]$ is a

compatible ASL if each of its straightening relations is of the form

$\phi \left(\alpha \right)\phi \left({\alpha}^{\prime}\right)=\phi \left(\beta \right)\phi \left({\beta}^{\prime}\right)$ with

$\beta \subseteq \alpha \cap {\alpha}^{\prime}$ and

${\beta}^{\prime}\supseteq \alpha \cup {\alpha}^{\prime},$ where

$\alpha ,{\alpha}^{\prime}$ are incomparable elements in

$\mathcal{I}\left(P\right).$ If

$K\left[\mathsf{\Omega}\right]$ and

$K\left[{\mathsf{\Omega}}^{\prime}\right]$ are compatible ASL on

$\mathcal{I}\left(P\right)$ over

$K,$ we identify them if they have the same straightening relations. In this case, we write

$K\left[\mathsf{\Omega}\right]\equiv K\left[{\mathsf{\Omega}}^{\prime}\right].$In ([

4], Question 5.1), Hibi and Li asked the following questions:

- (a)
Given a finite poset P, find all possible compatible algebras with straightening laws on $\mathcal{I}\left(P\right)$ over $K.$

- (b)
For which posets P, does there exist a unique compatible ASL on $\mathcal{I}\left(P\right)$ over $K?$

In this note, we give a complete answer to question (b). Namely, we prove the following:

**Theorem** **1.** Let P be a finite poset. Then, the following statements are equivalent:

- (i)
There exists a unique compatible ASL on$\mathcal{I}\left(P\right)$over$K.$

- (ii)
$K\left[\mathcal{O}\left(P\right)\right]\equiv K\left[\mathcal{C}\left(P\right)\right]\equiv K\left[\mathcal{C}\left({P}^{*}\right)\right],$where${P}^{*}$denotes the dual poset of$P.$

- (iii)
Each connected component of P is a chain, that is, P is a direct sum of chains.

An answer to question (a) seems to be quite difficult. In ([

4], Example 5.2), it was observed that, if one considers the poset

$P=\{a,b,c,d,\mathbf{e}\}$ with

$a<c<\mathbf{e}$ and

$b<c<d,$ then there exist nine compatible ASL structures on

$\mathcal{I}\left(P\right)$ over

$K,$ while if one considers

$P=\{a,b,c,d\}$ with

$a<c$,

$b<c,b<d$, then there are three compatible ASL structures on

$\mathcal{I}\left(P\right)$ over

$K,$ namely,

$K\left[\mathcal{O}\right(P\left)\right],$ $K\left[\mathcal{C}\right(P\left)\right]$, and

$K\left[\mathcal{C}\left({P}^{*}\right)\right].$## 2. Order Polytopes, Chain Polytopes, and Their Associated Toric Rings

Let

$P=\{{p}_{1},\dots ,{p}_{n}\}$ be a finite poset. For the basic terminology regarding posets used in this paper, we refer to [

1] and ([

5], Chapter 3). The order polytope

$\mathcal{O}\left(P\right)$ is defined as

In ([

2], Corollary 1.3), it was shown that the vertices of

$\mathcal{O}\left(P\right)$ are

${\sum}_{{p}_{i}\in \alpha}{\mathbf{e}}_{i},\alpha \in \mathcal{I}\left(P\right).$ Here,

${\mathbf{e}}_{i}$ denotes the unit coordinate vector in

${\mathbb{R}}^{n}.$ If

$\alpha =\varnothing ,$ then the corresponding vertex in

$\mathcal{O}\left(P\right)$ is the origin of

${\mathbb{R}}^{n}.$The chain polytope

$\mathcal{C}\left(P\right)$ is defined as

In ([

2], Theorem 2.2), it was proved that the vertices of

$\mathcal{C}\left(P\right)$ are

${\sum}_{{p}_{i}\in A}{\mathbf{e}}_{i}$, where

A is an antichain in

$P.$ Recall that an antichain in

P is a subset of

P such that any two distinct elements in the subset are incomparable. Since every poset ideal is uniquely determined by its antichain of maximal elements, it follows that

$\mathcal{O}\left(P\right)$ and

$\mathcal{C}\left(P\right)$ have the same number of vertices. However, as it was observed in [

2],

$\mathcal{O}\left(P\right)$ and

$\mathcal{C}\left(P\right)$ need not have the same number of

i-dimensional faces for

$i>0.$ Therefore, in general, they are not combinatorial equivalent. Combinatorially, equivalence of order and chain polytopes are studied in [

6].

#### The Toric Rings $K\left[\mathcal{O}\right(P\left)\right]$ and $K\left[\mathcal{C}\right(P\left)\right]$

To each subset $W\subset P$, we attach the squarefree monomial ${u}_{W}\in K[{x}_{1},\dots ,{x}_{n}],$ ${u}_{W}={\prod}_{{p}_{i}\in W}{x}_{i}.$ If $W=\varnothing ,$ then ${u}_{W}=1.$ The toric ring $K\left[\mathcal{O}\right(P\left)\right]$, known as the Hibi ring associated with the distributive lattice $\mathcal{I}\left(P\right),$ is generated over K by all the monomials ${u}_{\alpha}t\in S$, where $\alpha \in \mathcal{I}\left(P\right).$ The toric ring $K\left[\mathcal{C}\right(P\left)\right]$ is generated by all the monomials ${u}_{A}t$ where A is an antichain in $P.$ In addition, as we have already mentioned in the Introduction, both rings are algebras with straightening laws on $\mathcal{I}\left(P\right)$ over $K.$

We recall the definition of an ASL as it was introduced in [

7]. For a quick introduction to this topic, we refer to [

7] and ([

8], Chapter XIII). Algebras with straightening laws turned out to be useful tools in studying determinantal rings. Let

K be a field,

$R={\u2a01}_{i\ge 0}{R}_{i}$ with

${R}_{0}=K$ be a graded

K-algebra,

H a finite poset, and

$\phi :H\to R$ an injective map which maps each

$\alpha \in H$ to a homogeneous element

$\phi \left(\alpha \right)\in R$ with

$deg\phi \left(\alpha \right)\ge 1$. A

standard monomial in

R is a product

$\phi \left({\alpha}_{1}\right)\phi \left({\alpha}_{2}\right)\cdots \phi \left({\alpha}_{k}\right)$ where

${\alpha}_{1}\le {\alpha}_{2}\le \cdots \le {\alpha}_{k}$ in

$H.$**Definition** **1.** The K-algebra R is called an algebra with straightening laws on H over K if the following conditions hold:

- (1)
The set of standard monomials is a K–basis of$R;$

- (2)
If$\alpha ,\beta \in H$are incomparable and if$\phi \left(\alpha \right)\phi \left(\beta \right)=\sum {c}_{i}\phi \left({\gamma}_{i1}\right)\dots \phi \left({\gamma}_{i{k}_{i}}\right),$where${c}_{i}\in K\backslash \left\{0\right\}$and${\gamma}_{i1}\le \dots \le {\gamma}_{i{k}_{i}},$is the unique expression of$\phi \left(\alpha \right)\phi \left(\beta \right)$as a linear combination of standard monomials, then${\gamma}_{i1}\le \alpha ,\beta $for all$i.$

The above relations $\phi \left(\alpha \right)\phi \left(\beta \right)=\sum {c}_{i}\phi \left({\gamma}_{i1}\right)\dots \phi \left({\gamma}_{i{k}_{i}}\right)$ are called the straightening relations of R and they generate all the relations of of $R.$

Let us go back to the toric rings $K\left[\mathcal{O}\right(P\left)\right]$ and $K\left[\mathcal{C}\right(P\left)\right].$

One considers

$\phi :\mathcal{I}\left(P\right)\to K\left[\mathcal{O}\right(P\left)\right]$ defined by

$\phi \left(\alpha \right)={u}_{\alpha}t$ for every

$\alpha \in \mathcal{I}\left(P\right).$ As it was proved by Hibi in [

3],

$K\left[\mathcal{O}\right(P\left)\right]$ is an ASL on

$\mathcal{I}\left(P\right)$ over

K with the straightening relations

$\phi \left(\alpha \right)\phi \left(\beta \right)=\phi (\alpha \cap \beta )\phi (\alpha \cup \beta ),$ where

$\alpha ,\beta $ are incomparable elements in

$\mathcal{I}\left(P\right).$On the other hand, one defines

$\psi :\mathcal{I}\left(P\right)\to K\left[\mathcal{C}\right(P\left)\right]$ by setting

$\psi \left(\alpha \right)={u}_{max\alpha}t$ for all

$\alpha \in \mathcal{I}\left(P\right)$ where

$max\alpha $ denotes the set of the maximal elements in

$\alpha .$ Note that, for every

$\alpha \in \mathcal{I}\left(P\right),$ $max\alpha $ is an antichain in

P and each antichain

$A\subset P$ determines a unique ideal

$\alpha \in \mathcal{I}\left(P\right),$ namely, the poset ideal generated by

$A.$ Therefore,

$\psi $ is an injective well defined map and by ([

4], Theorem 3.1), the ring

$K\left[\mathcal{C}\right(P\left)\right]$ is an ASL on

$\mathcal{I}\left(P\right)$ over

K with the straightening relations

where

$\alpha \ast \beta $ is the poset ideal of

P generated by

$max(\alpha \cap \beta )\cap (max\alpha \cup max\beta ).$We observe that one may also consider

$K\left[\mathcal{C}\left({P}^{*}\right)\right]$ as an ASL on

$\mathcal{I}\left(P\right)$, where

${P}^{*}$ is the dual poset of

$P.$ We may define

$\delta :\mathcal{I}\left(P\right)\to K\left[C\left({P}^{*}\right)\right]$ by

$\delta \left(\alpha \right)={u}_{min\overline{\alpha}}t$ for

$\alpha \in \mathcal{I}\left(P\right),$ where

$min\overline{\alpha}$ is the set of minimal elements in

$\overline{\alpha}$ and

$\overline{\alpha}$ is the filter

$P\backslash \alpha $ of

$P.$ We recall that a

filter $\gamma $ in

P (or

dual order ideal) is a subset of

P with the property that for every

$p\in \gamma $ and every

$q\in P$ with

$q\ge p,$ we have

$q\in \gamma .$ Thus, a filter in

P is simply a poset ideal in the dual poset

${P}^{*}.$ The ring

$K\left[C\left({P}^{*}\right)\right]$ is an ASL on

$\mathcal{I}\left(P\right)$ over

K as well with the straightening relations

for incomparable elements

$\alpha ,\beta \in \mathcal{I}\left(P\right),$ where

$\alpha \circ \beta $ is the poset ideal of

P which is the complement in

P of the filter generated by

$min(\overline{\alpha}\cap \overline{\beta})\cap (min\overline{\alpha}\cup min\overline{\beta}).$ Let us also observe that all the algebras

$K\left[\mathcal{O}\right(P\left)\right],K\left[\mathcal{C}\right(P\left)\right]$, and

$K\left[\mathcal{C}\left({P}^{*}\right)\right]$ are compatible algebras with straightening laws.

## 3. Proof of Theorem 1

We clearly have (i) ⇒ (ii). Let us now prove (ii) ⇒ (iii). By hypothesis, the straightening relations of

$K\left[\mathcal{O}\right(P\left)\right],K\left[\mathcal{C}\right(P\left)\right],$ and

$K\left[\mathcal{C}\left({P}^{*}\right)\right]$ coincide. Therefore, we must have

for all

$\alpha ,\beta $ incomparable elements in

$\mathcal{I}\left(P\right).$ From the second equality in (

1), it follows that

$\overline{\alpha}\cap \overline{\beta}$ is the filter of

P generated by

$min(\overline{\alpha}\cap \overline{\beta})\cap (min\overline{\alpha}\cup min\overline{\beta}).$ Assume that there exists two incomparable elements

$p,{p}^{\prime}\in P$ such that there exists

$q\in P$ with

$q>p$ and

$q>{p}^{\prime}.$ Consider

$\overline{\alpha}$ the filter generated by

p and

$\overline{\beta}$ the filter generated by

${p}^{\prime}.$ Then,

$min(\overline{\alpha}\cap \overline{\beta})\cap (min\overline{\alpha}\cup min\overline{\beta})=\varnothing $, but obviously,

$\overline{\alpha}\cap \overline{\beta}\ne \varnothing .$ This shows that, for any two incomparable elements

$p,{p}^{\prime}\in P,$ there is no upper bound for

p and

${p}^{\prime}.$Similarly, by using the first equality in Equation (

1), we derive that, for any two incomparable elements

$p,{p}^{\prime}\in P,$ there is no lower bound for

p and

${p}^{\prime}$. This shows that every connected component of the poset

P is a chain.

Finally, we prove (iii) ⇒ (i). Let P be a poset such that all its connected components are chains and assume that the cardinality of P is equal to $n.$ Let ${\left\{{\omega}_{\alpha}\right\}}_{\alpha \in \mathcal{I}\left(P\right)}$ be the generators of $K\left[\mathsf{\Omega}\right]\subset S$ and assume that the straightening relations of $K\left[\mathsf{\Omega}\right]$ are $\phi \left(\alpha \right)\phi \left({\alpha}^{\prime}\right)=\phi \left(\beta \right)\phi \left({\beta}^{\prime}\right)$ where $\beta \subseteq \alpha \cap {\alpha}^{\prime},$ ${\beta}^{\prime}\supseteq \alpha \cup {\alpha}^{\prime},$ and $\alpha ,{\alpha}^{\prime}$ are incomparable elements in $\mathcal{I}\left(P\right).$ We have to show that, for all $\alpha ,{\alpha}^{\prime}$ incomparable elements in $\mathcal{I}\left(P\right),$ we have $\beta =\alpha \cap {\alpha}^{\prime}$ and ${\beta}^{\prime}=\alpha \cup {\alpha}^{\prime}.$

We proceed by induction on

Let us recall that, if $\gamma \in \mathcal{I}\left(P\right),$ then $rank\gamma $ denotes the rank of the subposet of $\mathcal{I}\left(P\right)$ consisting of all elements $\delta \in \mathcal{I}\left(P\right)$ with $\delta \subseteq \gamma .$

If

$k=0,$ that is,

$rank(\alpha \cup {\alpha}^{\prime})-rank(\alpha \cap {\alpha}^{\prime})=n$, then

$\alpha \cup {\alpha}^{\prime}=P$ and

$\alpha \cap {\alpha}^{\prime}=\varnothing ,$ thus

$\beta =\alpha \cap {\alpha}^{\prime}$ and

${\beta}^{\prime}=\alpha \cup {\alpha}^{\prime}.$ Assume that the desired conclusion is true for

$rank(\alpha \cup {\alpha}^{\prime})-rank(\alpha \cap {\alpha}^{\prime})=n-k$ with

$k\ge 0.$ Let us choose now

$\alpha ,{\alpha}^{\prime}$ incomparable in

$\mathcal{I}\left(P\right)$ such that

$rank(\alpha \cup {\alpha}^{\prime})-rank(\alpha \cap {\alpha}^{\prime})=n-k-1$ and assume that we have a straightening relation

$\phi \left(\alpha \right)\phi \left({\alpha}^{\prime}\right)=\phi \left(\beta \right)\phi \left({\beta}^{\prime}\right)$ with

$\beta \u228a\alpha \cap {\alpha}^{\prime}$ or

${\beta}^{\prime}\u228b\alpha \cup {\alpha}^{\prime}.$ By duality, we may reduce to considering

${\beta}^{\prime}\u228b\alpha \cup {\alpha}^{\prime}.$ In other words, in

$K\left[\mathsf{\Omega}\right],$ we have

As

P is a direct sum of chains, we may find

$p\in max(\alpha \cup {\alpha}^{\prime})$ and

$q\in {\beta}^{\prime}\backslash (\alpha \cup {\alpha}^{\prime})$ such that

q covers

p in

P, that is,

$q>p$ and there is no other element

${q}^{\prime}$ in

P with

$q>{q}^{\prime}>p.$ Without loss of generality, we may assume that

$p\in {\alpha}^{\prime}.$ Let

${\alpha}_{1}$ be the poset ideal of

P generated by

${\alpha}^{\prime}\cup \left\{q\right\}.$ As all the connected components of

P are chains, we have

${\alpha}_{1}={\alpha}^{\prime}\cup \left\{q\right\}$ since there are no other elements in

P which are smaller than

q except those that are on the same chain as

p and

q, which are in

${\alpha}^{\prime}.$ Moreover, by the choice of

$q,$ we have

By the inductive hypothesis, it follows that

$\phi \left(\alpha \right)\phi \left({\alpha}_{1}\right)=\phi (\alpha \cap {\alpha}_{1})\phi (\alpha \cup {\alpha}_{1}),$ or, equivalently, in

$K\left[\mathsf{\Omega}\right]$ we have the equality

${\omega}_{\alpha}{\omega}_{{\alpha}_{1}}={\omega}_{\alpha \cap {\alpha}_{1}}{\omega}_{\alpha \cup {\alpha}_{1}}.$ Thus, we have obtained the following equalities in

$K\left[\mathsf{\Omega}\right]:$This implies that the monomials in Equation (

2) are distinct standard monomials in

$K\left[\mathsf{\Omega}\right],$ which is in contradiction to the condition that the standard monomials form a

K-basis in

$K\left[\mathsf{\Omega}\right].$ Therefore, our proof is completed.