On Gorenstein sequences of socle degrees 4 and 5

Stanley’s nonunimodal Gorenstein $h$-vector is optimal

“…Theorem 3.2. The Gorenstein h-vectors (1, r, h 2 , r, 1) of socle degree 4 and codimension ≤ 17 are precisely the ones with r ≤ h 2 ≤ r+1 2 , together with (1, 13, 12, 13, 1), (1,14,13,14,1), (1,15,14,15,1), (1,16,15,16,1) and (1,17,16,17,1). In particular, nonunimodal Gorenstein h-vectors of socle degree 4 and codimension r exist if and only if r ≥ 13.…”

“…The first example of a nonunimodal Gorenstein h-vector, namely (1,13,12,13,1), which has socle degree 4 (i.e., length 5), was produced by Stanley [22], using the technique of trivial extensions, also introduced in that paper and useful in some of our proofs here. Notice that 4 is the smallest socle degree allowing the existence of a nonunimodal Gorenstein h-vector, because Gorenstein h-vectors are symmetric.…”

“…(It is easy to see that if a nonunimodal Gorenstein h-vector exists in codimension r, then one must also exist in all larger codimensions.) Several incremental results, both characteristic-free or with assumptions on the characteristic (see for instance [1,6,18,22]) finally left open only the existence of (1,12,11,12,1) as a possible Gorenstein h-vector, in order to show that all Gorenstein h-vectors of socle degree 4 and codimension r ≤ 12 must be unimodal. In this paper we settle this problem, by showing that (1,12,11,12,1) cannot be Gorenstein in any characteristic.…”

“…In socle degree 5, the work of several authors culminated in Theorem 3.3 of [1], which showed that all socle degree 5 Gorenstein h-vectors of codimension ≤ 15 are unimodal. The authors of [1] also pointed out that (1,18,16,16 (1,17,16,16,17,1). In this note, as an application of our method, we prove that H (1) and H (2) cannot be Gorenstein, while we see using trivial extensions that H (3) is.…”

On almost revlex ideals with Hilbert function of complete intersections

“…n=1: J [1] = (x 3 1 ); n=2: In order to compute J [2] up to degree d 3 , we add to J ′ = (J [1] ) ≤4 K[x 1 , x 2 ] only the term x 2 1 x 2 2 , and we do not need to explicitly compute the other generators of J [2] . Observe indeed that K[x 1 , x 2 ]/J ′ , with J ′ = (x 3 1 , x 2 1 x 2 2 ), does not have Hilbert function H [2] , because H R/J ′ (t) = H [2] (t) only up to t = d 3 . n=3: Let now J ′ := (x 3 1 ,…”