First we give some definitions from Section 3 of the paper Monomials, Binomials, and Riemann-Roch by Manjunath and Sturmfels and then we restate a claim from that paper offered without proof. Finally we provide an example that seems to contradict that claim. The question is

Question: Is the example given below a counterexample to the claim? And if not, why not?


Fix an Artinian monomial ideal $I$ of a polynomial ring $K[\mathbf{x}] = K[x_1, \dots, x_n]$. A monomial $\mathbf{x}^{\mathbf{b}}$ is a socle monomial of $I$ if $\mathbf{x}^{\mathbf{b}} \notin I$ and $x_i\mathbf{x}^{\mathbf{b}} \in I$ for all $i$. Let $\mathrm{MonSoc}(I)$ be the set of all socle monomials of $I$.

Def: $I$ is reflection invariant if there is a canonical monomial $\mathbf{x}^{\mathbf{K}}$ such that the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$.

Following these definitions the authors note the following.

The Claim

Claim: $I$ is reflection invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$ if and only if the monomial ideal generated by $\mathrm{MonSoc}(I)$ equals the Alexander dual $I^{[\mathbf{K} + \mathbf{e}]}$ where $\mathbf{e} = (1,1,\dots, 1)$.

The (Counter?) Example

Let $I = \langle a^4,~ab^2,~b^3,~a^3c,~abc,~c^3 \rangle \subset K[a,b,c]$ and let $\mathbf{K} = (3,2,2)$. Then

$$\mathrm{MonSoc}(I) = \left\{a^{3}b,~a^{2}c^{2},~b^{2}c^{2}\right\}.$$

By (the constructive proof of) Proposition 5.2 in this paper the ideal $J = \langle a^4, a^2b, b^3, ac, b^2c, c^3 \rangle$ is the unique Artinian ideal with

$$\mathrm{MonSoc}(J) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}} \mid \mathbf{x}^{\mathbf{b}} \in \mathrm{MonSoc}(I)\right\}.$$

Moreover, the same algorithm can be used to show that $I$ is the unique Artinian ideal with

$$\mathrm{MonSoc}(I) = \left\{\mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}} \mid \mathbf{x}^{\mathbf{c}} \in \mathrm{MonSoc}(J)\right\}.$$

In particular, the map that sends a monomial $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{b}}$ is an involution on $\mathrm{MonSoc}(I)$, so $I$ is reflection-invariant with canonical monomial $\mathbf{x}^{\mathbf{K}}$. We now get a contradiction to the claim above by computing the Alexander dual (in Macaulay2, for example) and noting that the minimal generators of $I^{[(4,3,3)]}$ are $\{a^4bc,~a^2b^3c,~ab^2c^3\} \neq \mathrm{MonSoc}(I)$.

Again, the question is

Question: Is the example just given a counterexample to the above claim? And if not, why not?

  • $\begingroup$ Note that the generators of Alexander dual $I^{[4,3,3]}$ are precisely the monomials $\mathbf{x}^{\mathbf{K} + \mathbf{1}}/\mathbf{x}^{\mathbf{b}}$ for $\mathbf{x}^{\mathbf{b}} \in \mathrm{MonSoc}(I)$. This is similar to Proposition 1 on page 5 of this paper where the facet ideal of a matroid is expressed as the Stanley-Reisner ideal of the Alexander dual of the (independence complex) of the dual matroid. $\endgroup$
    – Aaron Dall
    Oct 21, 2018 at 15:14

1 Answer 1


Note: The following is the result of an email exchange with one of the coauthors of the paper cited in the OP.

In the definition of a reflection-invariant monomial ideal the requirement that the map $\phi: \mathbf{x}^{\mathbf{c}} \mapsto \mathbf{x}^{\mathbf{K}}/\mathbf{x}^{\mathbf{c}}$ be an involution on $\mathrm{MonSoc}(I)$ is not the usual definition of involution, i.e., that the map is its own inverse. Instead, the map $\phi$ is required to be both an involution on and a permutation of $\mathrm{MonSoc}(I)$. This additional requirement is not mentioned in the paper, though upon closer reading, it is used in a number of the proofs.

As for the example in the OP, the map $\phi$ is clearly not a permutation of $\mathrm{MonSoc}(I)$. Moreover, one can show that $I$ is not reflection-invariant for any choice of $\mathbf{K}$. So the ideal $I$ is not reflection-invariant. Also, with this new restriction on permissible involutions it is easy to prove the note following the definition of reflection invariance in the paper in question.

Edit: Thanks to @benblumsmith's comment the heart of the matter is now clear. The map that sends $\mathbf{x}^{\mathbf{b}} \mapsto \mathbf{x}^{\mathbf{K}-\mathbf{b}}$ has domain and codomain equal to $\mathrm{MonSoc}(I)$ (as opposed to the set of all monomials in the quotient ring $K[\mathbf{x}]/\left\langle x_1^{K_1+1}, \dots, x_n^{K_n+1}\right\rangle$). So it's clear that the example in the OP is not a counterexample.

  • 1
    $\begingroup$ I do not understand why the definition of involution as "a map that is its own inverse" does not already imply that the map is a permutation of $\operatorname{MonSoc}(I)$. Surely in order to be its own inverse a map must have identical domain and codomain? Otherwise, the inverse does not even have the same domain, so how could it be the same map? $\endgroup$ Oct 24, 2018 at 2:31
  • $\begingroup$ @benblumsmith: Your comment clarifies my confusion even further. The domain and codomain of the map $\phi$ aren't given. If one assumes that both are equal to $\mathrm{MonSoc}(I)$ (which is what the authors intended and the right thing to do), then the map must be a permutation. I, however, took the domain and codomain to be the set of all monomials and the statement "an involution of $\mathrm{MonSoc}(I)$" to mean $\phi (\phi (\mathrm{MonSoc}(I))) = \mathrm{MonSoc}(I)$. Then $\phi$ need not take socle monomials to socle monomials. But this (my) interpretation is not the intended one. $\endgroup$
    – Aaron Dall
    Oct 24, 2018 at 6:30
  • $\begingroup$ I see. (I think you must have meant you took the domain and codomain as the set of all Laurent monomials? B/c $\phi$ does not send monomials to monomials in general.) My sense of usage is that "$\phi$ is an involution of/on $X$" always implies that $\phi:X\to X$. In the present case, $\phi$ is an involution on the set $L$ of all Laurent monomials; but this means it's automatic that $\phi^2(X)=X$ (and $\phi^2|_X = 1_X$) for any subset $X$ of $L$. Thus "$\phi$ is an involution on (specific) $X$" isn't actually a statement about $X$ unless you also take it to mean that $\phi$ maps $X$ to $X$. $\endgroup$ Oct 25, 2018 at 12:26
  • $\begingroup$ Exactly @benblumsmith. If you want to write up your points as a new answer I'll happily accept/upvote it. Otherwise, I'll write it up and reference your helpful comments. $\endgroup$
    – Aaron Dall
    Oct 25, 2018 at 13:07

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