A mathematical preprint reports explicit counterexamples to the idea that the strong finite type, or SFT, property must survive the addition of a polynomial variable. Its main construction starts with a one-dimensional local SFT domain R, but the polynomial ring R[X] is not SFT. That extension contains a prime ideal that is itself not SFT and is an upper to zero: it is nonzero, yet it contracts to the zero ideal of R. The result gives a negative answer to the general question, which the paper says was still listed as open in 2024.
SFT is a condition for controlling ideals with finite information. Under the paper’s definition, an ideal I is SFT when it contains a finitely generated subideal B and there is an integer N at least 1 such that the N-th power of every element of I lies in B. A ring is SFT when this holds for every ideal. The distinction matters here because the starting ring satisfies that condition even though adjoining X creates an ideal that does not.
How the counterexample is built
The construction begins with spaces V_n that organize coefficients into filtration levels. Those spaces are multiplicatively filtered, remain stable under squaring, and obey an escape condition: the polynomial ring F2[u] is not contained in any one fixed level V_E. From these spaces, the authors build a graded ring A, take its positive-degree ideal m, and localize at m to obtain R=A_m with M=mR.
Two estimates then establish the finiteness property of the coefficient ring. Every z in M has its square in cR, while every nonzero a in M has some power of c in aR. These facts lead to the structural conclusion that R is a one-dimensional local SFT domain. Its spectrum—the collection of its prime ideals—is exactly {(0), M}.
To expose what changes in R[X], the paper uses an evaluation ideal Q. Q is prime, contracts to (0) in R, and is nonzero because c(X−u) belongs to it. Those properties make Q an upper to zero, providing a concrete witness for the failure of SFT in the polynomial extension.
The decisive step is a coefficient comparison. If Q were SFT, the resulting relations would force a growing list of powers of u into one fixed filtration level, written in the construction as V_{q+d}. Letting the degree grow would then place all of F2[u] inside V_{q+d}. That contradicts the escape condition, so Q is not SFT and R[X] cannot be SFT.
The pattern crosses characteristics
The phenomenon is not confined to the initial construction. The paper gives, for every prime characteristic, a one-dimensional local SFT domain R_p whose polynomial extension is not SFT; the non-SFT prime can again be chosen as an upper to zero. It also constructs a one-dimensional local SFT domain R_0 in characteristic zero with a non-SFT polynomial extension and a non-SFT upper to zero. For every prime p, it further gives a two-dimensional local SFT domain T_p of mixed characteristic (0,p) whose polynomial extension is not SFT.
The consequences extend to larger polynomial rings. For every displayed counterexample coefficient domain S, the witness prime is not finitely generated, S is non-Noetherian, and S[X_1,…,X_m] is not SFT for every m at least 1. In practical terms, once one of these examples loses SFT after adjoining a variable, adding more polynomial variables does not restore the property.
The result has a defined boundary. It concerns polynomial rings, and the authors make no claim about whether the corresponding formal-power-series rings are SFT or what their dimension is. The supplied document is arXiv:2608.26044v1 in math.AC, dated 26 August 2026.
Paper data and sources
Original title: Polynomial extensions do not preserve the strong finite type property
Authors: Viet-Hoang Tran, Phan Thanh Toan, Thieu N. Vo, Tan M. Nguyen
Journal/Repository: arXiv
Status: Preprint, not yet peer-reviewed
First online: 2026-08-26
DOI: Not available
Original paper · Full text