Let \(A\) be a ring, let \(S = A[x_0, \ldots, x_r]\), and let \(X = \Proj S\). We have seen that a homogeneous ideal \(I\) in \(S\) defines a closed subscheme of \(X\) (Ex. 3.12), and that conversely every closed subscheme of \(X\) arises in this way (5.16).
For any homogeneous ideal \(I \subseteq S\), define the saturation \(\bar I\) of \(I\) to be the set of \(s \in S\) such that for each \(i = 0, \ldots, r\) there is an \(n\) with \(x_i^n s \in I\). We say \(I\) is saturated if \(I = \bar I\). Show that \(\bar I\) is a homogeneous ideal of \(S\).
Two homogeneous ideals \(I_1\) and \(I_2\) of \(S\) define the same closed subscheme of \(X\) if and only if they have the same saturation.
If \(Y\) is any closed subscheme of \(X\), then the homogeneous ideal \[ J_Y\coloneqq\bigoplus_{d\ge0}\Gamma(X,\mci_Y(d))\subseteq S \] is saturated. Hence it is the largest homogeneous ideal defining the subscheme \(Y\). For \(r\ge1\), this ideal is \(\Gamma_*(\mci_Y)\); the nonnegative truncation also covers \(r=0\).
There is a bijection between saturated ideals of \(S\) and closed subschemes of \(X\).
(Projective dimension zero)
With the all-integer-degree convention for $\Gamma_*$, its use as an ideal of \(S\) in part (c) needs \(r\ge1\). For a field \(k\), take \(S=k[x_0]\), \(X=\PP_k^0\), and \(Y=\varnothing\). Then \(\mci_Y=\OO_X\) and \[ \Gamma_*(\OO_X)\cong k[x_0,x_0^{-1}], \] which has nonzero negative-degree components and is not an ideal of \(k[x_0]\) as a graded submodule. The nonnegative section ideal is \(J_Y=S\), the saturated ideal defining the empty subscheme.
Solution
For \(0\le i\le r\), put \(T_i=S_{x_i}\), \(R_i=(T_i)_0\), and \(U_i=D_+(x_i)=\Spec R_i\). The degree-one element \(x_i\) is a unit in \(T_i\) and trivializes \(\OO_X(1)\) on \(U_i\).
1. Saturation is a homogeneous ideal, and it satisfies\[ \bar I=\bigcap_{i=0}^r(IT_i\cap S),\qquad \bar I T_i=IT_i,\qquad\overline{\bar I}=\bar I. \]
For \(s\in S\), membership of \(s/1\) in \(IT_i\) is equivalent to \(x_i^n s\in I\) for some \(n\ge0\), by the criterion for membership in a localized ideal. This proves the intersection formula and shows that \(\bar I\) is an ideal containing \(I\). If \(s=\sum_d s_d\) is its homogeneous decomposition and \(x_i^n s\in I\), homogeneity of \(I\) gives \(x_i^n s_d\in I\) for every \(d\). Applying this for every \(i\) shows that each \(s_d\) lies in \(\bar I\). Hence \(\bar I\) is homogeneous, proving part (a).
For fixed \(i\), the inclusions \(I\subseteq\bar I\subseteq IT_i\cap S\) give \(\bar I T_i=IT_i\) after localization. The intersection formula applied to \(\bar I\) then gives \(\overline{\bar I}=\bar I\).
2. The closed subscheme defined by \(I\) is determined on \(U_i\) by the ideal \((IT_i)_0\subseteq R_i\).Equality of these degree-zero ideals for two homogeneous ideals is equivalent to equality of their full localized ideals in \(T_i\).
The restriction of \(\Proj(S/I)\) to \(U_i\) has coordinate ring \[ ((S/I)_{x_i})_0\cong R_i/(IT_i)_0. \] Here localization and taking a fixed graded component are exact. Thus \((IT_i)_0\) is its defining ideal on this affine chart.
For every integer \(d\), multiplication by the unit \(x_i^d\) gives \[ (IT_i)_d=x_i^d(IT_i)_0. \] Since the localized ideals are graded, their degree-zero components therefore determine all their components.
3. The equivalence in part (b) holds.
Two homogeneous ideals define the same closed subscheme of \(X\) exactly when their defining ideal sheaves agree on the affine cover \((U_i)\). By step 2, this is equivalent to \(I_1T_i=I_2T_i\) for every \(i\). The intersection formula in step 1 then gives \(\bar I_1=\bar I_2\). Conversely, equality of the saturations gives \[ I_1T_i=\bar I_1T_i=\bar I_2T_i=I_2T_i \] by step 1, so the closed subschemes agree on every chart and hence on \(X\).
4. For part (c), \(J_Y\) is the saturation of every homogeneous ideal defining \(Y\).
4.1. The canonical map \(S_d\to\Gamma(X,\OO_X(d))\) is an isomorphism for \(d\ge0\).For \(r\ge1\), it is an isomorphism for all integers \(d\), with both sides zero when \(d<0\).
On \(U_i\), the sections of \(\OO_X(d)\) are \((T_i)_d\). Every variable is a non-zero-divisor in the polynomial ring over \(A\), so these modules inject into the degree-\(d\) component of the Laurent polynomial ring \[ B=A[x_0^{\pm1},\ldots,x_r^{\pm1}]. \] The sheaf gluing condition identifies global sections with \(\bigcap_i(T_i)_d\) inside \(B_d\). If \(r\ge1\), membership in every \(T_i\) forbids a negative exponent of any variable: for a fixed variable \(x_j\), choose \(i\ne j\), so \(T_i\) allows no negative powers of \(x_j\). Uniqueness of Laurent polynomial coefficients then gives \(\bigcap_i T_i=S\). Taking degree \(d\) proves the assertion for \(r\ge1\).
If \(r=0\), the single chart is all of \(X\), and \((T_0)_d=Ax_0^d\) for every integer \(d\). It agrees with \(S_d\) exactly in the nonnegative degrees, except when \(A=0\), in which case every module is zero.
4.2. If \(I\) defines \(Y\), then \(J_Y=\bar I\).
The existence of such an \(I\) is the homogeneous-ideal description of closed subschemes recalled in the problem [@Har10a, Proposition II.5.16]. The inclusion \(\mci_Y(d)\to\OO_X(d)\) is injective because twisting by an invertible sheaf is exact. Step 1 therefore identifies \(\Gamma(X,\mci_Y(d))\) with a submodule of \(S_d\) for \(d\ge0\). Multiplication by \(S_e\) carries this submodule into \(\Gamma(X,\mci_Y(d+e))\), so their direct sum is a homogeneous ideal.
1.This identifies every nonnegative graded component of \(J_Y\) with that of \(\bar I\), proving equality.
1.3. Q.E.D.
Step 2 and idempotence of saturation prove that \(J_Y\) is saturated. Step 3 shows that it defines \(Y\). Every homogeneous ideal defining \(Y\) is contained in its saturation \(J_Y\), giving maximality. For \(r\ge1\), step 1 and the inclusions \(\mci_Y(d)\subseteq\OO_X(d)\) show that all negative-degree sections vanish, so \(J_Y=\Gamma_*(\mci_Y)\).
5. For part (d), the inverse bijections are\[ \boxed{I\longmapsto\Proj(S/I),\qquad Y\longmapsto J_Y}. \]
If \(I\) is saturated, step 4 gives \(J_{\Proj(S/I)}=\bar I=I\). Conversely, for every closed subscheme \(Y\), step 4 gives a saturated ideal \(J_Y\) defining \(Y\). Thus the displayed maps are inverse.
6. Q.E.D.
Step 1 proves part (a), step 3 proves part (b), step 4 proves part (c) with the nonnegative-degree clarification, and step 5 proves part (d).