For an element in the integral closure of an ideal $I$ - which power is in $I$?

Question

Consider an ideal $I$ in a ring $R$. If $f \in R$ belongs to the integral closure of $I$, then there is $k_0 \geq 0$ such that $f^k \in I^{k-k_0}$ for all $k \geq k_0$. Are there any known upper bounds on $k_0$? If that is difficult, what would be an upper bound on the minimum $k$ such that $f^k \in I$?

I am working in a setting of classical algebraic geometry, so the ambient ring is commutative, Noetherian etc. In fact $R$ is the local ring at a point of a (possibly singular and non-reduced) curve, $I$ is a principal ideal, and I know $f$ is in the integral closure of $I$ via the "valuative criterion" using Rees valuations (Definition 10.1.1 and Theorem 10.1.6 in Swanson-Huneke, 2023 edition) - so any bound in terms of those would be great.

Answer 1

Not a complete answer to your question, but the number $k_0$ (or $k_0-1$) is called the Briançon-Skoda number of $R$. For analytic planar curves, there is a formula for this number in J. Sznajdman: The Briançon-Skoda number of analytic irreducible planar curves (Annales de l'Institut Fourier, Volume 64 (2014) no. 1, pp. 177-187), given in terms of it's so called Puiseux characteristics.