# Set-theoretic generation by circuit polynomials

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.)
• why is $P\cap\mathbb C[x_i|i\in C]$ principal implies $C$ is a circuit? Dec 1, 2020 at 21:44
• @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.
– tim
Dec 2, 2020 at 4:04