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I have seen it stated that Proj of any graded ring $A$, finitely generated as an $A_0$-algebra, is isomorphic to Proj of a graded ring $B$ such that $B_0 = A_0$ and $B$ is generated as a $B_0$-algebra by $B_1$.

Could someone either supply a reference for or a sketch a proof of this statement?

Note: An obvious approach to this question is to make $B$ a Veronese subring of $A$. However, when I try this approach, I end up getting a terrible combinatorics problem that I do not know how to approach.

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    $\begingroup$ EGA II, 2.1.6(v) $\endgroup$
    – BCnrd
    Commented Apr 2, 2010 at 6:12
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    $\begingroup$ Red book(old print), 403p. $\endgroup$
    – Moon
    Commented Apr 2, 2010 at 10:15
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    $\begingroup$ It's also handy to note (cf EGA2, Corollary 2.1.5) that a graded ring S is Noetherian if and only if S_0 is Noetherian and S is finitely generated over S_0. Thus the fact you refer to implies that if S is Noetherian then Proj S is projective over Spec S_0. $\endgroup$
    – Chuck Hague
    Commented Apr 7, 2010 at 19:23

1 Answer 1

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Bourbaki Commutative Algebra Chapter 3:

Let $A$ be a non-negatively graded algebra. Assume that $A$ is finitely generated over $A_0$. There exists $e \geq 1$ such that $A^{me} = A_0[A^{me}_1]$ for any $m \geq 1$.

Here $A^{e} = \oplus_{n \in\mathbb Z} A^{e}_n$, where $A^{e}_n := A_{ne}$.

The desired result follows because replacing $A$ with $A^{me}$ does not change Proj.

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  • $\begingroup$ More precisely, this is Chapter III, 1.3 Proposition 3. $\endgroup$
    – Charles Staats
    Commented Apr 2, 2010 at 13:15
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    $\begingroup$ Now we know how to solve that combinatorics problem ! :) $\endgroup$
    – Nico Bellic
    Commented Jan 12, 2012 at 4:03

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