Let $I \subseteq K[x_1,...,x_n]$ be a graded ideal. Under what conditions is $\text{in}_{<}(I)$ a reduced Grobner basis?
Here $\text{in}_{<}(I)$ is the ideal:
\begin{align*} \text{in}_{<}(I) := \langle \text{in}(f) : f \in I \rangle \end{align*}
I suspect that if $\text{in}_{<}(I)$ is a complete intersection, it should be enough, but I'm having trouble proving it. My intuition is that since $\mathrm{ht}(\text{in}_{<}(I)) = \mu(I)$ where $\mu(I)$ is the minimal amount of generators, then all generators of $\text{in}_{<}(I)$ are irreducible by the other generators in the basis, but I don't think it guarantees that the leading coefficients are all equal to $1$.