“…This can be seen from the discussions at the end of the proof of [19,Proposition 2.1], and immediately after it; the point there is that the Eagon-Northcott complex becomes the desired linear graded free resolution of R/I a (Σ) (see [6,Theorem A2.60]). (5) If no two of the linear forms of Σ are proportional, then from part (4), I n−1 (Σ), so a = n − 1, has linear graded free resolution. (6) If Σ = ℓ , for some ℓ ∈ R 1 , then for any 1 ≤ a ≤ n, we have I a (Σ) = ℓ a , which, as any principal ideal, has linear graded free resolution.…”