We consider the scheme $P:=\mathrm{Proj}(S)$ of a commutative, (nonnegatively) graded, finitely generated $\mathbb C$-algebra $S$ with no nilpotents.

(Q1) When is $P$ a projective variety?

Since we have a projective map $P \to \mathrm{Spec}(S_0)$, $P$ is projective if $S_0=\mathbb C$. But the opposite direction is too difficult for me. (I have neither a proof nor a counterexample.)

(Q2) Is $P$ always quasi-projective?

The examples which I know are always quasi-projective, so I suspect that $P$ is always quasi-projective, but I can not find any relevant references.

EDIT (An example of non-projective $P$)

Let $G = \lbrace \mathrm{diag}(t,t^{-1},u) \in GL_3(\mathbb C) | t \in \mathbb C^*, u = \pm 1 \rbrace$ act on $\mathbb C^3$ and define a character of $G$ by $\chi(t,t^{-1},u)=tu$. The character defines a $G$-equivariant line budle $L \to \mathbb C^3$. Put $$S_* = \bigoplus_{d\geq0}H^0(\mathbb C^3,L^{\otimes d})^G.$$ We can show that $S_*$ is a finitely generated $\mathbb C$-algebra (e.g. Lemma 14.1.10) and $$S_{2*} = \bigoplus_{d} S_{2d} = \mathbb C[xy,z^2][x^2].$$ Therefore $$\mathrm{Proj}(S_*) = \mathrm{Proj}(S_{2*}) = \mathrm{Proj}(\mathbb C[xy,z^2][x^2]) = \mathrm{Spec}(\mathbb C[xy,z^2]) \times \mathbb P^0 = \mathbb C^2.$$

Note: This example can be found the book linked above (in Chap. 14).


1 Answer 1


This answer your question just partially.

Let $S=\oplus_{n=0}^\infty S_n$ be a finitely generated graded $\mathbb{C}$-algebra and consider the graded $\mathbb{C}$-algebra $S'$ defined by $S_0'=\mathbb{C}$ and $S_n'=S_n$ for $n>0$. Then there exists an isomorphism of $\mathbb{C}$-schemes $\operatorname{Proj}(S)\cong\operatorname{Proj}(S')$ (EGA II, 2.4.8). Therefore if $S'$ is finitely generated $\mathbb{C}$-algebra then $\operatorname{Proj}(S)$ is projective.

However the condition on $S'$ of being finitely generated is not necessary because of the following: For $d>0$, let $S'^{(d)}$ be the graded $\mathbb{C}$-algebra defined by $S'^{(d)}_n=S'_{nd}$ for $n\geq 0$. Then there exists an isomorphism of $\mathbb{C}$-schemes $\operatorname{Proj}(S')\cong\operatorname{Proj}(S'^{(d)})$ (EGA II, 2.4.7).

Then we can conclude that if for some $d>0$ the $\mathbb{C}$-algebra $S'^{(d)}$ is finitely generated then $\operatorname{Proj}(S)$ is a projective variety.

For an example where $\operatorname{Proj}(S)$ is not a projective variety, I would try with $S=\mathbb{C}[X][Y]=\oplus_{n=0}^\infty \mathbb{C}[X]Y^n$. Not that in this case none of the $S'^{(d)}$ are finitely generated. I am not sure if it will work.

  • $\begingroup$ In your example, the global sections are $\mathbb C[X]$, no?, so it cannot be projective. $\endgroup$ Commented Mar 9, 2015 at 7:10
  • $\begingroup$ @Diego Thanks for your answer. I added a possible counterexample to my question. I am confused now. But I should say that the finitely generated property could not be relevant, as you say. (I do not have any reason why we should assume it. It is just in case.) $\endgroup$
    – H. Shindoh
    Commented Mar 10, 2015 at 12:04
  • $\begingroup$ I am also confused. If $S$ is finitely generated with $S_0=\mathbb{C}$, some of the $S^{(d)}$ is generated in degree 1 (take $d$ as the lcm of the degrees of the generators) and therefore $S^{(d)}$ is a quotient of some $\mathbb{C}[X_0,\ldots,X_n]$ and therefore $Proj(S)$ is a closed subscheme of $P_{\mathbb{C}}^n$. $\endgroup$
    – Diego
    Commented Mar 12, 2015 at 3:45

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