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The Stacks project

Lemma 27.13.7. Let R be a ring. Let Z \subset \mathbf{P}^ n_ R be a closed subscheme. Let

I_ d = \mathop{\mathrm{Ker}}\left( R[T_0, \ldots , T_ n]_ d \longrightarrow \Gamma (Z, \mathcal{O}_{\mathbf{P}^ n_ R}(d)|_ Z)\right)

Then I = \bigoplus I_ d \subset R[T_0, \ldots , T_ n] is a graded ideal and Z = \text{Proj}(R[T_0, \ldots , T_ n]/I).

Proof. It is clear that I is a graded ideal. Set Z' = \text{Proj}(R[T_0, \ldots , T_ n]/I). By Lemma 27.11.5 we see that Z' is a closed subscheme of \mathbf{P}^ n_ R. To see the equality Z = Z' it suffices to check on an standard affine open D_{+}(T_ i). By renumbering the homogeneous coordinates we may assume i = 0. Say Z \cap D_{+}(T_0), resp. Z' \cap D_{+}(T_0) is cut out by the ideal J, resp. J' of R[T_1/T_0, \ldots , T_ n/T_0]. Then J' is the ideal generated by the elements F/T_0^{\deg (F)} where F \in I is homogeneous. Suppose the degree of F \in I is d. Since F vanishes as a section of \mathcal{O}_{\mathbf{P}^ n_ R}(d) restricted to Z we see that F/T_0^ d is an element of J. Thus J' \subset J.

Conversely, suppose that f \in J. If f has total degree d in T_1/T_0, \ldots , T_ n/T_0, then we can write f = F/T_0^ d for some F \in R[T_0, \ldots , T_ n]_ d. Pick i \in \{ 1, \ldots , n\} . Then Z \cap D_{+}(T_ i) is cut out by some ideal J_ i \subset R[T_0/T_ i, \ldots , T_ n/T_ i]. Moreover,

J \cdot R\left[ \frac{T_1}{T_0}, \ldots , \frac{T_ n}{T_0}, \frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i} \right] = J_ i \cdot R\left[ \frac{T_1}{T_0}, \ldots , \frac{T_ n}{T_0}, \frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i} \right]

The left hand side is the localization of J with respect to the element T_ i/T_0 and the right hand side is the localization of J_ i with respect to the element T_0/T_ i. It follows that T_0^{d_ i}F/T_ i^{d + d_ i} is an element of J_ i for some d_ i sufficiently large. This proves that T_0^{\max (d_ i)}F is an element of I, because its restriction to each standard affine open D_{+}(T_ i) vanishes on the closed subscheme Z \cap D_{+}(T_ i). Hence f \in J' and we conclude J \subset J' as desired. \square


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