# The vanishing scheme of for a graded ring generated by elements of degree 1 (Vakil 4.5.P)

I am working on the following exercise of Ravi Vakil's Foundations of algebraic geometry.

4.5.P. EXERCISE. If $S_•$ is generated in degree 1, and $f ∈ S_+$ is homogeneous, explain how to define $V(f)$ “in” $\text{Proj} S_•$, the vanishing scheme of $f$. (Warning: f in general isn’t a function on $\text{Proj} S_•$. We will later interpret it as something close:a section of a line bundle, see for example §14.1.2.) Hence define $V(I)$ for any homogeneous ideal I of $S_+$.

I guess as a set $V(f)=\{P\in \operatorname{Proj} S_•: f\in P\}$. But I think this problem shouldn't be this trivial and he probably wants us to construct a scheme structure on it and I don't see how to do it. I guess we need to use the condition "$S_•$ is generated in degree $1$" (i.e. generated by degree $1$ elements as an algebra) and construct a structure sheaf on it ( really ?).

Let me know if you think I interpret it wrongly.

• Hint: You already know how to define a closed subscheme $V(f)$ on affine schemes, and your projective scheme is covered by affines. Mar 15, 2018 at 23:10
• @user45878 Actually he hasn't defined closed subschemes yet in chapter 4...but thank you for the hint anyway... Mar 16, 2018 at 1:56
• Hint 2: If $f$ has degree $1$, what is $V(f)$? Mar 16, 2018 at 11:53
• @user45878 By the way, where do we need the condition "S∙ is generated in degree 1"? Mar 19, 2018 at 16:23
• @Armandoj18eos why consider the case when $f$ has degree 1? Mar 19, 2018 at 16:24

I explain better the construction of "vanishing scheme of a homogeneous element $$f\in S_{+}$$ in $$\operatorname{Proj}S_{\bullet}$$".

The numeration of citations is refered to Vakil's FOAG November 18th 2017 version.

One knows that the sets $$D_{+}(f)$$ (defined by exercises 4.5.E and 4.5.F, where $$f\in S_{+}$$) are open subsets of $$\operatorname{Proj}S_{\bullet}$$, which determine a base for the Zariski topology of $$\operatorname{Proj}S_{\bullet}$$ (exercise 4.5.G). In particular one has $$\begin{equation} \forall f,g\in S_{+},\,D_{+}(fg)=D_{+}(f)\cap D_{+}(g),V_{+}(f)=\operatorname{Proj}S_{\bullet}\setminus D_{+}(f); \end{equation}$$ since by hypothesis, $$S_{\bullet}$$ is a $$\mathbb{Z}_{\geq0}$$-graded ring genetared by $$S_1$$ as $$S_0$$-algebra, one has: $$\begin{gather} \forall g\in S_{+},\,g=\sum_{I\in\left(\mathbb{N}_{\geq0}\right)^n,}\lambda_If_I^{a_I},\, \text{where:}\,n=\deg g,\,a_{i_k}\in\mathbb{N}_{\geq0},\,\lambda_I\in S_0,\,f_{i_k}\in S_1,\\ |a_I|=a_{i_1}+\dots+a_{i_n}=n,f_I^{a_I}=f_{i_1}^{a_1}\cdot\dots f_{i_n}^{a_n},\,\lambda_I=0\,\text{for almost all multi-indexes}\,I, \end{gather}$$ it is a good idea to understand what $$V_{+}(f)$$ is when $$f\in S_1$$!

By exercises 4.5.E.(a), 4.5.F and 4.5.J $$\begin{equation} \forall f\in S_1,\,D_{+}(f)=\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid f\notin\mathfrak{p}\}\cong\operatorname{Spec}((S_{\bullet})_f)_0\leftrightarrow\left\{\mathfrak{p}\in\operatorname{Spec}S_{\bullet}\mid\mathfrak{p}\,\text{is homogeneous,}\,S_{+}\not\subseteq\mathfrak{p},\,f\notin\mathfrak{p}(S_{\bullet})_f\cap((S_{\bullet})_f)_0\right\}, \end{equation}$$ where: the isomorphism $$\cong$$ is in the category $$\mathbf{Sch}$$ of schemes; $$\leftrightarrow$$ indicates a bijection of sets.

Remark 1. Easily one proves that: $$\begin{equation} \forall n\in\mathbb{N}_{\geq2},f\in S_1,\,D_{+}(f)=D_{+}(f^n) \end{equation}$$ as sets, moreover they are the same scheme (see for example Bosch - Algebraic Geometry and Commutative Algebra, Lemma 9.1.7).

Proof of Remark 1.

In other words (cfr. exercises 4.5.L and 4.5.M): $$\begin{equation} \forall f\in S_1,\,\mathcal{O}_{\operatorname{Proj}S_{\bullet}}(D_{+}(f))=((S_{\bullet})_f)_0 \end{equation}$$ and $$\begin{equation} \forall f\in S_1,\,D_{+}(f)=\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid[f]\in\left(\mathcal{O}_{\operatorname{Proj}S_{\bullet},\mathfrak{p}}\right)^{\times}\}; \end{equation}$$ from all this, it turns out that $$\begin{equation} \forall f\in S_1,\,V_{+}(f)=\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid[f]\notin\left(\mathcal{O}_{\operatorname{Proj}S_{\bullet},\mathfrak{p}}\right)^{\times}\}=\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid f\in\mathfrak{p}\} \end{equation}$$ in according to previous definition of $$V_{+}(\cdot)$$.

Because previous reasoning does not depend by degree of $$f$$, one has: $$\begin{equation} \forall f\in S_{+},\,V_{+}(f)=\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid[f]\notin\left(\mathcal{O}_{\operatorname{Proj}S_{\bullet},\mathfrak{p}}\right)^{\times}\}=\left\{\mathfrak{p}\in\operatorname{Proj}S_{\bullet}\mid\mathcal{O}_{\operatorname{Proj}S_{\bullet},\mathfrak{p}\displaystyle/f\mathcal{O}_{\operatorname{Proj}S_{\bullet},\mathfrak{p}}}\neq0\right\}=\operatorname{Supp}\mathcal{O}_{\operatorname{Proj}S_{\bullet}\displaystyle/f\mathcal{O}_{\operatorname{Proj}S_{\bullet}}}! \end{equation}$$ Remark 2.

• Until this point, $$V_{+}(f)$$ is the vanishing set of the homogeneous element $$f$$ of $$S_{\bullet}$$, and it is closed in $$\operatorname{Proj}S_{\bullet}$$.
• This idea comes from exercises 2.7.F and 3.4.I.(a).

Let $$g\in S_{+}$$ and let $$\{f_a\in S_1\}_{a\in A}$$ such that: $$\begin{equation} V_{+}(g)\subseteq\bigcup_{a\in A}D_{+}(f_a); \end{equation}$$ let $$\begin{equation} \forall a\in A,\,\mathcal{O}_{V_{+}(g)|D_{+}(f_a)}=\widetilde{((S_{\bullet})_{f_a})_{0\displaystyle/g(S_{\bullet})_{f_a}\cap((S_{\bullet})_{f_a})_0}} \end{equation}$$ that is $$\mathcal{O}_{V_{+}(g)|D_{+}(f_a)}$$ is the $$\mathcal{O}_{\operatorname{Spec}((S_{\bullet})_{f_a})_0}$$-module associated to $$((S_{\bullet})_{f_a})_0$$-quotient module of base ring over the ideal generated by $$0$$-degree part of $$g$$ in this ring as well (see exercise 4.1.D); by exercise 4.5.K: these sheaves can be glue together (cfr. exercises 2.5.D and 4.4.A) in a sheaf $$\mathcal{O}_{V_{+}(g)}$$ of rings; in other words, $$V_{+}(g)$$ is a scheme!

Moreover, via this construction the following statement holds:

Let $$S_{\bullet}$$ be a $$\mathbb{Z}_{\geq0}$$-graded ring, which is generated as $$S_0$$-algebra by $$S_1$$, let $$f\in S_{+}$$ and let $$V_{+}(f)$$ the vanishing scheme of $$f$$ in $$\operatorname{Proj}S_{\bullet}$$ constructed as showed. For any point $$x\in V_{+}(f)$$ there exists an affine open neighbourhood $$U=\operatorname{Spec}R$$ of $$x$$ in $$\operatorname{Proj}S_{\bullet}$$ such that $$V_{+}(f)\cap U$$ is a closed subscheme of $$U$$; that is, there exists an ideal $$I$$ of $$R$$ such that $$V_{+}(f)\cap U\cong\operatorname{Spec}R_{\displaystyle/I}$$ (affine local property on target of closed subschemes, cfr. definition 7.1.2).

• Thank you so much for you endeavor! I would take me a while to digest though... Apr 15, 2018 at 4:42
• You are too good! ;) Apr 15, 2018 at 9:04
• @Armandoj18eos Is it true that the scheme $V(f)$ is just $\operatorname{Proj}(S_\bullet /(f)$ as NoOne has suggested in his answer? Jul 25, 2021 at 22:25
• Yes, it is; one can prove that $\mathrm{Proj}\left(S_{\bullet}/(f)\right)$ is isomorphic to the closed subscheme $V_{+}(f)$ of $\mathrm{Proj}S_{\bullet}$. However, in the general setting, this proof is different form previous one; for exmaple see Bosch - Algebraic Geometry and Commutative Algebra, section 9.1. Jul 27, 2021 at 11:12
• @Sisyphus Your thought is right! I corrected the statement. Jul 5, 2022 at 16:45

Possible solution

It seems to me we can define $V(f)=\operatorname{Proj} (S_•/(f))$. When $f$ is homogeneous, then $S_•/(f)$ is a graded ring and this definition makes sense.

In general, $V(I)=\operatorname{Proj} (S_•/I)$, when $I$ is a homogeneous ideal of $S_•$.

I didn't use the condition $S_∙$ is generated in degree $1$. Please let me know if I am wrong.