Example 31.31.2. Let $A$ be a graded ring. Let $X = \text{Proj}(A)$ and $S = \mathop{\mathrm{Spec}}(A_0)$. Given a graded ideal $I \subset A$ we obtain a closed subscheme $V_+(I) = \text{Proj}(A/I) \to X$ by Constructions, Lemma 27.11.3. Translating the result of Lemma 31.31.1 we see that if $X$ is quasi-compact, then any closed subscheme $Z$ is of the form $V_+(I(Z))$ where the graded ideal $I(Z) \subset A$ is given by the rule

$I(Z) = \mathop{\mathrm{Ker}}(A \longrightarrow \bigoplus \nolimits _{n \geq 0} \Gamma (Z, \mathcal{O}_ Z(n)))$

Then we can ask the following two natural questions:

1. Which ideals $I$ are of the form $I(Z)$?

2. Can we describe the operation $I \mapsto I(V_+(I))$?

We will answer this when $A$ is Noetherian.

First, assume that $A$ is generated by $A_1$ over $A_0$. In this case, for any ideal $I \subset A$ the kernel of the map $A/I \to \bigoplus \Gamma (\text{Proj}(A/I), \mathcal{O})$ is the set of torsion elements of $A/I$, see Cohomology of Schemes, Proposition 30.14.4. Hence we conclude that

$I(V_+(I)) = \{ x \in A \mid A_ n x \subset I\text{ for some }n \geq 0\}$

The ideal on the right is sometimes called the saturation of $I$. This answers (2) and the answer to (1) is that an ideal is of the form $I(Z)$ if and only if it is saturated, i.e., equal to its own saturation.

If $A$ is a general Noetherian graded ring, then we use Cohomology of Schemes, Proposition 30.15.3. Thus we see that for $d$ equal to the lcm of the degrees of generators of $A$ over $A_0$ we get

$I(V_+(I)) = \{ x \in A \mid (Ax)_{nd} \subset I\text{ for all }n \gg 0\}$

This can be different from the saturation of $I$ if $d \not= 1$. For example, suppose that $A = \mathbf{Q}[x, y]$ with $\deg (x) = 2$ and $\deg (y) = 3$. Then $d = 6$. Let $I = (y^2)$. Then we see $y \in I(V_+(I))$ because for any homogeneous $f \in A$ such that $6 | \deg (fy)$ we have $y | f$, hence $fy \in I$. It follows that $I(V_+(I)) = (y)$ but $x^ n y \not\in I$ for all $n$ hence $I(V_+(I))$ is not equal to the saturation.

Comment #5832 by Anton Mellit on

In the case when $A$ is not generated by $A_1$, how does Proposition 0BXD apply?

By the way, would it make sense to include a general statement (for a not necessarily Noetherian ring)? Something like Maybe there is a better one.

Comment #5848 by on

Thanks! It was just the wrong reference. I haven't checked your formula for the non-Noetherian case and consequently I haven't added it to the Stacks project. I have fixed the reference in this commit.

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