Castelnuovo-mumford Regularity: Examples of Curves and Surfaces

The behaviour of Castelnuovo-Mumford regularity under “geometric” transformations is not well understood. In this paper we are concerned with examples which will shed some light on certain questions concerning this behaviour. One simple question which was open (see e.g. [R]) is: May the regularity increase if we pass to the radical or remove embedded primes? By examples, we show that this happens. As a by-product we are also able to answer some related questions. In particular, we provide examples of licci ideals related to monomial curves in P (resp. in P) such that the regularity of their radical is essentially the square (resp. the cube) of that of the ideal. It is well known that the regularity cannot increase when points (embedded or not) are removed. Hence, in order to construct examples where on removing an embedded component the regularity increases, we have to consider surfaces. More surprisingly, we find an irreducible surface such that, after embedding a line into it, the depth of the coordinate ring increases! Another important concept to understand is the limit of validity of Kodaira type vanishing theorems. The Castelnuovo-Mumford regularity of the canonical module of a reduced curve is 2. An analogous result holds true for higher dimensional varieties with isolated singularities (in characteristic zero), thanks to Kodaira vanishing. As a consequence one can give bounds for CastelnuovoMumford regularity (see [CU]). In [Mum], Mumford proves that for an ample line bundle L on a normal surface S, H(S,OS ⊗ L−1) = 0. He remarks that this is false if S doesn’t satisfy S2 (i.e. S is not Cohen-Macaulay) and asks if the S2 condition is sufficient. The first counter-example was given in [AJ]. In this article, we show that counter-examples satisfying S1 give rise to counterexamples satisfying S2. We then provide monomial surfaces whose canonical module has large Castelnuovo-Mumford regularity, so that this vanishing fails. We give a simple proof to show that if S satisfies R1, then H1(S, ωS ⊗L) = 0