Betti Posets and the Stanley Depth

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Fix a field $K$. Let $S=K[x_1,\ldots,x_n]$ and $S'=K[y_1,\ldots,y_m]$, with their fine multigradings, and let $I\subset S$ and $I'\subset S'$ be nonzero proper monomial ideals. For a multigraded module $M$, a Stanley space is a subspace $uK[Z]\subset M$ generated by a homogeneous $u$ and free as a module over the polynomial ring on a subset $Z$ of the variables. A Stanley decomposition is a finite direct sum $M=\bigoplus_j u_jK[Z_j]$ as a $K$-vector space. Define $\operatorname{sdepth}_S M=\max_{\mathcal D}\min_j|Z_j|$ and $\operatorname{spdim}_S M=n-\operatorname{sdepth}_S M$. The Betti poset $\mathcal B(I)$ consists of the monomial multidegrees with nonzero $\operatorname{Tor}_i^S(S/I,K)$ for some $i\ge1$, ordered by monomial divisibility; equivalently these are the positive homological degrees of the minimal fine-graded free resolution. Use the analogous definitions over $S'$.

Statement recorded in the supplied source.

If $\mathcal B(I)\cong\mathcal B(I')$ as posets, must both equalities hold? \[\operatorname{spdim}_S(S/I)=\operatorname{spdim}_{S'}(S'/I'),\qquad \operatorname{spdim}_S I=\operatorname{spdim}_{S'}I'.\] [S2]

Short English statement

Does the Betti poset determine the Stanley projective dimensions of both a monomial ideal and its quotient ring?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 2900001 (AMR-028-0001). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 L. Katthän, Betti posets and the Stanley depth (2016), Sections 2.1, 2.2 and 2.4; Definition 2.2 and Conjecture 2.4. https://arxiv.org/pdf/1509.08275.