Dice Question Streamline Icon: https://streamlinehq.com

Constructive proof of LM(I) = LM(closure(I)) without degree-compatibility of the monomial order

Develop a constructive proof that, for any ideal I in the commutative formal power series ring K[[x1,...,xn]] over a field K and for any monomial order on the commutative monoid of monomials (without assuming compatibility with degree), the set of leading monomials LM(I) equals the set of leading monomials LM(closure(I)), where closure(I) denotes the closure of I in the (x1,...,xn)-adic topology.

Information Square Streamline Icon: https://streamlinehq.com

Background

Proposition 3.1 establishes LM(I) = LM(closure(I)) under the assumption that the chosen monomial order is compatible with degree. The authors then prove that any ideal I is closed in the adic topology (Theorem 3.1), implying LM(I) = LM(closure(I)) for any monomial order as a consequence of I = closure(I).

However, the proof of Proposition 3.1 relies essentially on degree-compatibility to control valuations and ensure a descending sequence of leading monomials, which underpins the constructive elimination and Cauchy convergence arguments. The authors explicitly note they lack a direct constructive argument proving LM(I) = LM(closure(I)) that does not assume degree-compatibility. Achieving such a constructive proof would remove this hypothesis from the elimination procedure and strengthen the rewriting-theoretic applications.

References

In Theorem~\ref{thm:any-ideal-is-closed}, we will show that $I$ is closed, from which we get that $\lm(I)=\lm(I)$ is true for any monomial order. However, we were not able to provide a constructive proof of this fact that works for monomial orders that are not assumed to be compatible with the degree.

Topological closure of formal power series ideals and application to topological rewriting theory (2402.05511 - Chenavier et al., 8 Feb 2024) in Remark following Proposition 3.1, Subsection 3.1 (Closure in the commutative case)