Let k be a field, let R_m=k[x_1,…,x_m], and let I(C_n)⊆ R_n be the edge ideal of the cycle graph C_n. For n≥ 4 and r≥ 0, set I_n,r=operatornameInc(N)_n,n+r(I(C_n))⊆ R_n+r. Determine whether R_n+r/I_n,r is Cohen–Macaulay. The conjectured complete classification is R_n+r/I_n,r is Cohen–Macaulay if and only if r≥ lfloor fracn-42rfloor and r≠ n-4.

Cohen–Macaulayness classification for Inc(N)-invariant chains of edge ideals

Let $k$ be a field, let $R_m=k[x_1,\ldots,x_m]$ , and let $I(C_n)\subseteq R_n$ be the edge ideal of the cycle graph $C_n$ . For $n\ge 4$ and $r\ge 0$ , set $I_{n,r}=\operatorname{Inc}(\mathbb{N})_{n,n+r}(I(C_n))\subseteq R_{n+r}$ . Determine whether $R_{n+r}/I_{n,r}$ is Cohen–Macaulay. The conjectured complete classification is $R_{n+r}/I_{n,r}$ is Cohen–Macaulay if and only if $r\ge \left\lfloor\frac{n-4}{2}\right\rfloor$ and $r\ne n-4$ .

References

Primary source

arXiv

Additional references

Cohen--Macaulayness of Inc(N)-Invariant Chains of Edge Ideals — arXiv

algebraic-combinatorics commutative-algebra

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

A September 2026 paper settles several important families but leaves the main cyclic case as a conjecture.

The problem seeks a complete Cohen–Macaulayness classification for chains of edge ideals invariant under order-preserving injections. The reported work resolves several families and reduces the cyclic case to a precise conjectural criterion.

Known results

Exact criteria are proved for line graphs.
Unconditional results cover several complementary graph families.

September 3, 2026 development

A reported paper proves the line-graph criteria and unconditional results for complementary families, while conjecturing the full cyclic classification for $r \ge \lfloor (n-4)/2 \rfloor$ and $r \ne n-4$ . The cyclic classification is supported by computational and structural evidence but remains unverified.

Current status (as of September 2026): Several graph families are claimed classified, but the cyclic case for $r \ge \lfloor (n-4)/2 \rfloor$ and $r \ne n-4$ remains conjectural.

Sources

No solutions have been posted yet.