# Quasi-projective A-schemes are locally of finite type over A, The Rising Sea, Ex.5.3.D

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?

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

## 1 Answer

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 $$e_i for all $$i$$. But this means that $$a$$ is exactly one of the elements in our generating set, and we are finished. $$\blacksquare$$

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