MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 8 down vote accepted

Serre's proof can indeed be generalized to noetherian rings - or even noetherian schemes - once you have proved that for a projective morphism between (locally) noetherian schemes the higher direct images of coherent sheaves are again coherent. You find all the arguments in EGA III, Section 2.3, in particular Scholie (2.3.3) answers your question affirmatively for noetherian rings.

share|cite|improve this answer
Great, thanks a lot! – Martin Brandenburg Mar 13 '11 at 12:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.