Past exam of the introduction to schemes course of the University of Oxford 2017 3 c Solution 2026-10-25
Fix a coherent sheaf and an integer . Set . Repeatedly apply a to construct short exact sequences of sheaveswhere each is a coherent sheaf and a locally free sheaf. Each kernel sheaf is a coherent sheaf because is a noetherian scheme. Thus the construction can be repeated as many times as needed; it does not require the kernel sheaves themselves to be locally free sheaves.
For every integer , the long exact sequence in sheaf cohomology containsBoth outer groups vanish by the hypothesis, since and exceed . Hence is an isomorphism. Iterating this dimension shifting in sheaf cohomology gives
Choose with , where is the uniform bound from b. The coherent sheaf is a quasi-coherent sheaf, so b makes the last group zero. ThereforeThe argument uses a finite initial segment of a locally free resolution. There is no need to assume that every coherent sheaf has a finite locally free resolution: the uniform bound from b is what terminates the argument.