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
Then we can ask the following two natural questions:
Which ideals I are of the form I(Z)?
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
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
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.
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