A projective $A$-scheme is a $\operatorname{Proj} S_{\bullet}$ where $S_{\bullet}$ is a finitely generated graded ring over $A=S_0$.

A quasi-projective $A$-scheme is an open quasicompact subscheme of a projective $A$-scheme.

An scheme over $A$ is a scheme where all its rings of sections are $A$-algebras and restriction maps are maps of $A$-algebras.

A scheme over A is called locally of finite type over $A$ if it can be covered by $\operatorname{Spec} A_i$ where the $A_i$’s are finitely generated $A$-algebras.

Now, Exercise 5.3.D of Vakil notes asks us to prove that quasi-projective A-schemes are locally of finite type over A.

Since the $D(f)\cong \operatorname{Spec} ((S_{\bullet})_f)_0)$ cover, I think that proving $((S_{\bullet})_f)_0)$ is a finitely generated $A$-algebra is enough, but I haven’t been capable to do it. Any help?

  • $\begingroup$ Where have your attempts broken down? You're on the right track. $\endgroup$
    – KReiser
    Feb 27, 2022 at 18:23
  • $\begingroup$ I cannot prove that those rings are finitely generated. $\endgroup$
    – Potitov06
    Feb 27, 2022 at 23:24

1 Answer 1


Claim: If $S_\bullet$ is a finitely-generated $S_0$-algebra, then for any nonzero homogeneous $f$ of positive degree $d$, the ring $(S_f)_0$ is finitely generated over $S_0$.

Proof: Let $g_i$ be a finite set of homogeneous generators for $S_\bullet$ as an $S_0$-algebra with $\deg g_i=d_i>0$. Any element in $(S_f)_0$ can be written as $p/f^n$, where $p$ is homogeneous of degree $nd$ for some $p$ and some $n$. Such a $p$ may be written as an $S_0$-linear combination of monomials of the form $\prod g_i^{e_i}$ with $\sum e_id_i=dn$. Therefore to show that $(S_f)_0$ is finitely generated over $S_0$, it suffices to show that there are finitely many elements of the form $\frac{\prod g_i^{e_i}}{f^n}$ which can be multiplied to produce any product of the form $\frac{\prod g_i^{e_i}}{f^n}$.

I claim that the elements $\frac{g_i^d}{f^{d_i}}$ and $\frac{\prod g_i^{e_i}}{f^n}$ where $\sum d_ie_i=nd$ and $0\leq e_i< d$ suffice, and they are finite in number because there are finitely many $g_i$ and $d$ is finite. To show this, let $a=\frac{\prod g_i^{e_i}}{f^n}$ be arbitrary with $\sum e_id_i=dn$. If there is some $i_0$ such that $e_{i_0}\geq d$, then we may write $$a=\frac{g_{i_0}^d}{f^{e_{i_0}}} \cdot \frac{\prod g_i^{e_i'}}{f^{n-e_{i_0}}}$$ and it suffices to show that $\frac{\prod g_i^{e_i'}}{f^{n-e_{i_0}}}$, the second element of the right-hand side, can be written as a product of the elements mentioned above. By repeated applications of this process, we may assume that $e_i<d$ for all $i$. But this means that $a$ is exactly one of the elements in our generating set, and we are finished. $\blacksquare$

  • $\begingroup$ (+1) FYI I made some small corrections to the last sentence. Feel free to correct me if I've messed anything up. $\endgroup$ Feb 28, 2022 at 1:55
  • $\begingroup$ @RaviFernando thanks for the correction. $\endgroup$
    – KReiser
    Feb 28, 2022 at 2:37
  • $\begingroup$ thank you for the answer! $\endgroup$
    – Potitov06
    Mar 1, 2022 at 15:09

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