Let Z —> Y be the normalization of an affine cuspidal curve over an algebraically closed field k. Let i : Z —> X be a closed immersion over Y with X smooth over Y.<\/p>\n
Question: Does there exist a coherent module F on X, flat over Y, whose support is equal to Z set theoretically?<\/p>\n
Answer: No in characteristic 0 and yes in characteristic p > 0.<\/p>\n
To see that the answer is no in characteristic 0 you show that the map O_Y —> O_Z has an O_Y-linear section if you have F (and of course this isn’t possible for the normalization of the cuspidal curve). Namely, consider the map tau : O_Z —> O_Y which sends an element f of O_Z to the trace over O_Y of multiplication by f’ on F where f’ is any lift of f to O_X. You show that the choice of f’ doesn’t matter by checking at the generic point; the key fact is that the support condition tells us that f’ which vanish on Z give nilpotent operators on F. Finally, this gets us a section as tau(g) = rg for g in O_Y. Here r = rank_Y(F) > 0 which is invertible as we have char 0.<\/p>\n
Remark: I think there doesn’t even exist a coherent F on X, flat over Y, such that the generic point of Z is an associated point of F. Exercise! [Edit on 2\/12\/22: Jason did the exercise and, uh, it isn’t true!]<\/p>\n
To see that the answer is yes in characteristic p > 2, say Y is the spectrum of A = k[a, b]\/(a^3 – b^2). Let X be the spectrum of A[t] and consider the closed subscheme, finite flat over Y, cut out by t^p – a^{(p – 3)\/2}b. The reduction of this subscheme is isomorphic to Z. For p = 2 use t^2 – a. Cheers!<\/p>\n","protected":false},"excerpt":{"rendered":"
Let Z —> Y be the normalization of an affine cuspidal curve over an algebraically closed field k. Let i : Z —> X be a closed immersion over Y with X smooth over Y. Question: Does there exist a … Continue reading