# Coherent sheaves on projective space over a general ring

Let $K$ be a field and $n \geq 0$. Serre proved that $\text{Qcoh}_f(\mathbb{P}^n_K)$ is equivalent to the localization of $\text{grMod}_f(K[x_0,...,x_n])$, in which the inclusions $M_{\geq a} \to M$ become inverted. Here the index $f$ means "finitely presented".

What happens if we replace $K$ by an arbitrary ring? If it fails in general, what happens for nice rings, for example finitely generated reduced algebras over a field, or just $\mathbb{Z}$? If it even fails in that case, are the categories equivalent after passing to the ind-categories?

Let me elaborate a bit the question: Let $R$ be a ring. Define the following category $C$. Objects are finitely presented graded $R[x_0,...,x_n]$-modules (it's OK for me to restrict to $R$ noetherian, so that in this case these are just the finitely generated graded modules). Alternatively and perhaps a little bit more naturally in this context, we could take graded modules $M$ as objects such that there is some $a$ such that $M_{\geq a}$ is finitelys presented. A morphism in $C$ is an equivalence class of homomorphisms of graded modules. Here two homomorphisms $f,g : M \to N$ are equivalent iff they are equals in all large degrees, i.e. if there is some integer $a$ such that $f_{\geq a} = g_{\geq a}$. Then we have a functor $C \to \text{Qcoh}_f(\mathbb{P}^n_R), M \mapsto \widetilde{M}$. If $R$ is a field, then in Serre's FAC (Faisceaux Algebriques Coherents, III.3., Prop. 5 and Prop. 6) states that this functor is an equivalence of categories. Perhaps we can just use the same proof to generalize it to other rings, too?

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I've already asked a variant of this question some weeks ago, but I had to delete it because some of my assumptions were wrong. – Martin Brandenburg Mar 11 '11 at 14:25
I was wondering about this myself the other day. – B. Bischof Mar 11 '11 at 14:42
Is R commutative or not? If R is noncommutative, what is the definition of n-projective space over R – Shizhuo Zhang Mar 11 '11 at 15:56
Rings are assumed to be commutative here. – Martin Brandenburg Mar 13 '11 at 12:19