Let $P$ be a prime ideal in $S=\mathbb{C}[x_1,\ldots , x_n],$ and write $[n] = \{ 1, \ldots , n \}.$ The algebraic matroid of $P$ can be defined according to circuit axioms as follows: $C\subset [n]$ is a circuit if $P \cap \mathbb{C} [x_i \mid i \in C]$ is principal, and we call a generator of this ideal a circuit polynomial. The circuit ideal $P_{\mathcal{C}}\subset S$ is generated by all circuit polynomials.

Question For which $P$ do we have $\sqrt{P_{\mathcal{C}}}=P$?

For context, I include the following facts:

  1. If $P$ is generated by monomials, the answer is trivially always.
  2. If $P$ is generated by binomials, the answer is always, though seemingly less trivial. This follows from results in the article "Binomial Ideals" by Eisenbud and Sturmfels.
  3. If $P$ is homogeneous, the circuit polynomials need not be scheme-theoretic generators for $P$ (even in the binomial case.)
  • $\begingroup$ why is $P\cap\mathbb C[x_i|i\in C]$ principal implies $C$ is a circuit? $\endgroup$
    – Turbo
    Dec 1, 2020 at 21:44
  • $\begingroup$ @1.. that's just the definition of what it means for $C$ to be a circuit. The content of the statement is showing that this really does definite a matroid, essentially because deleting any variable from $C$ yields an algebraically independent set in the fraction field of $S/P.$ You might have a look at Section 3 of this paper for more details. $\endgroup$
    – tim
    Dec 2, 2020 at 4:04


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