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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.

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  • $\begingroup$ Hint: You already know how to define a closed subscheme $V(f)$ on affine schemes, and your projective scheme is covered by affines. $\endgroup$ Commented Mar 15, 2018 at 23:10
  • $\begingroup$ @user45878 Actually he hasn't defined closed subschemes yet in chapter 4...but thank you for the hint anyway... $\endgroup$
    – No One
    Commented Mar 16, 2018 at 1:56
  • $\begingroup$ Hint 2: If $f$ has degree $1$, what is $V(f)$? $\endgroup$ Commented Mar 16, 2018 at 11:53
  • $\begingroup$ @user45878 By the way, where do we need the condition "S∙ is generated in degree 1"? $\endgroup$
    – No One
    Commented Mar 19, 2018 at 16:23
  • $\begingroup$ @Armandoj18eos why consider the case when $f$ has degree 1? $\endgroup$
    – No One
    Commented Mar 19, 2018 at 16:24

2 Answers 2

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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).

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    $\begingroup$ Thank you so much for you endeavor! I would take me a while to digest though... $\endgroup$
    – No One
    Commented Apr 15, 2018 at 4:42
  • $\begingroup$ You are too good! ;) $\endgroup$ Commented Apr 15, 2018 at 9:04
  • $\begingroup$ @Armandoj18eos Is it true that the scheme $V(f)$ is just $\operatorname{Proj}(S_\bullet /(f)$ as NoOne has suggested in his answer? $\endgroup$
    – user5826
    Commented Jul 25, 2021 at 22:25
  • $\begingroup$ 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. $\endgroup$ Commented Jul 27, 2021 at 11:12
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    $\begingroup$ @Sisyphus Your thought is right! I corrected the statement. $\endgroup$ Commented Jul 5, 2022 at 16:45
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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.

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  • $\begingroup$ I am also confused by this exercise. I don't understand what's the point of the condition that $S_{\bullet}$ is generated in degree $1$. $\endgroup$ Commented Jun 15 at 15:59

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