The Proj Construction


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1 The Proj Construction Daniel Murfet May 16, 2006 Contents 1 Basic Properties 1 2 Functorial Properties 2 3 Products 6 4 Linear Morphisms 9 5 Projective Morphisms 9 6 Dimensions of Schemes 11 7 Points of Projective Space 12 1 Basic Properties Definition 1. Let be a ring. graded algebra is an algebra R which is also a graded ring in such a way that if r R d then ar R d for all a. That is, R d is an submodule of R for all d 0. Equivalently a graded algebra is a morphism of graded rings R where we grade by setting 0 =, n = 0 for n > 0. morphism of graded algebras is a morphism of algebras which preserves grade. Equivalently, this is a morphism of graded rings which is also a morphism of modules, or a morphism of graded rings R S making the following diagram commute R S Let S be a graded algebra, and let X = P rojs. Then there is a canonical morphism of rings θ : Γ(X, P rojs) defined by θ(a)(p) = a/1 S (p) Where as usual, a denotes the element a 1 of S. By Ex 2.4 this induces a morphism of schemes f : P rojs Spec. If γ : S gives the algebra structure on S, then an element a is mapped to the maximal ideal of O P rojs,p via Γ(X, P rojs) O P rojs,p if and only if a γ 1 p. We denote this prime ideal by p. Let κ p : f(p) S (p) be defined by κ p (a/s) = a/s. Then f : P rojs Spec f(p) = p f # U (t)(p) = κ p(t(f(p))) 1
2 So the projective space over any graded algebra is a scheme over. In particular there is a canonical morphism of schemes P n Spec. Notice that if e S + is homogenous then the composite of the open immersion Spec(S (e) ) = D + (e) P rojs with f is the morphism of schemes corresponding to the canonical map S (e) defined by a a/1. Proposition 1. If S is a graded algebra then the morphism f : P rojs Spec is separated. In particular P rojs is separated for any graded ring S. Proof. The open sets D + (a) for homogenous a S + are an open cover of P rojs. Therefore the open sets D + (a) D + (b) are an open cover of P rojs P rojs, and the inverse image of this open set under the diagonal P rojs P rojs P rojs is clearly D + (a) D + (b) = D + (ab). Since the property of being a closed immersion is local on the base, we reduce to showing that the morphism D + (ab) D + (a) D + (b) is a closed immersion. Since we know that this gives the canonical morphisms D + (ab) D + (a), D + (ab) D + (b) when composed with the projections, the corresponding morphism of algebras is the morphism α : S (a) S (b) S (ab) induced by the canonical morphisms S (a) S (ab), S (b) S (ab). Since α is clearly surjective, the proof is complete. Proposition 2. If S is a noetherian graded ring then P rojs is a noetherian scheme. Proof. For any homogenous f S + the ring S (f) is noetherian (GRM,Corollary 15). Since S is noetherian the ideal S + admits a finite set of homogenous generators f 1,..., f p and the schemes D + (f i ) = SpecS (fi) cover P rojs. Therefore P rojs is noetherian. Proposition 3. If S is a finitely generated graded algebra, then the scheme P rojs is of finite type over Spec. Proof. The hypothesis imply that S 0 is a finitely generated algebra, and that S is a finitely generated S 0 algebra. Therefore S + is a finitely generated ideal (GRM,Corollary 8). So using the argument of Proposition 2 we reduce to showing that for homogenous f S +, S (f) is a finitely generated algebra. By (GRM,Proposition 14) it suffices to show that S (d) is a finitely generated algebra, which follows from (GRM,Lemma 11). Proposition 4. If S is a graded domain with S + 0 then P rojs is an integral scheme. Proof. By (3.1) it is enough to show that P rojs is reduced and irreducible. By assumption the zero ideal 0 is a homogenous prime ideal, so P rojs has a generic point and is therefore irreducible. The scheme P rojs is covered by open subsets isomorphic to schemes Spec(S (f) ) for nonzero homogenous f S +, and it is clear that S (f) is an integral domain. This shows that P rojs is reduced and therefore integral. Corollary 5. Let be a ring and consider the scheme P n for n 1. Then (i) The morphism P n Spec is separated and of finite type. (ii) If is noetherian then so is P n. (iii) If is an integral domain, then P n is an integral scheme. 2 Functorial Properties Let ϕ : S T be a morphism of graded rings, G(ϕ) the open subset {p P rojt p ϕ(s + )} and induce the following morphism of schemes using Ex 2.14: Φ : G(ϕ) P rojs Φ(p) = ϕ 1 p Φ # V (s)(p) = ϕ (p)(s(ϕ 1 p)) 2
3 Here ϕ (p) : S (ϕ 1 p) T (p) is defined by a/s ϕ(a)/ϕ(s). If ϕ is an isomorphism of graded rings with inverse ψ then clearly G(ϕ) = P rojt, G(ψ) = P rojs and one checks that if Ψ : P rojs P rojt is induced by ψ then ΦΨ = 1 and ΨΦ = 1 so that P rojs = P rojt. More generally, let ϕ : S T and ψ : T Z be morphisms of graded rings. Then G(ψϕ) G(ψ). Let Φ : G(ϕ) P rojs and Ψ : G(ψ) P rojt be induced as above, and let Ω be induced by the composite ψϕ. Then ΨG(ψϕ) G(ϕ) and it is readily checked that the following diagram of schemes commutes: Ω P rojs G(ψϕ) Φ Ψ G(ψϕ) G(ϕ) If ϕ : S T is a morphism of graded algebras for some ring, then Φ : G(ϕ) P rojs is a morphism of schemes over, where G(ϕ), P rojs are schemes in the canonical way. lso if ϕ : S S is the identity, clearly G(ϕ) = P rojs and Φ is the identity. For any graded ring S and homogenous f S + there is an isomorphism of schemes Spec(S (f) ) = D + (f). We claim that this isomorphism is natural with respect to morphisms of graded rings, in the following sense. Lemma 6. Let ϕ : S T be a morphism of graded rings and Φ : G(ϕ) P rojs the induced morphism of schemes. If f S + is homogenous then Φ 1 D + (f) = D + (ϕ(f)) and the following diagram commutes: Spec(T (ϕ(f)) ) Spec(S (f) ) (1) D + (ϕ(f)) D + (f) where the top morphism corresponds to ϕ (f) : S (f) T (ϕ(f)) defined by s/f n ϕ(s)/ϕ(f) n, and the bottom morphism is the restriction of Φ. Proof. By (H, Ex.2.4) it suffices to check that the diagram commutes on global sections of Spec(S (f) ), which follows immediately from the definition of Φ and the explicit form of the vertical isomorphisms given in our Section 2.2 notes. Lemma 7. Let S be a graded ring and f, g S + homogenous elements. diagram commutes Spec(S (f) ) D + (f) Then the following Spec(S (fg) ) D + (fg) where the vertical morphism on the left corresponds to the ring morphism S (f) S (fg) defined by s/f n sg n /(fg) n. Proof. By (H, Ex.2.4) it suffices to check that the diagram commutes on global sections of Spec(S (f) ). But this is easy to check, using the explicit form of the isomorphisms given in our Section 2.2 notes. Let H be a graded ring and suppose f H 0. Then ϕ f (h) = d 0 f d h d defines a morphism of graded rings ϕ f : H H. If f is a unit with inverse g, then ϕ f is an isomorphism with inverse ϕ g. If f is a unit, then it is easy to see that for a homogenous prime ideal p, ϕ 1 f p = p and the induced morphism (ϕ f ) (p) : H (p) H (p) is the identity. Therefore ϕ f induces the identity morphism P rojh P rojh. This simple observation has the following important consequence 3
4 Lemma 8. Let ϕ : S T be a morphism of graded rings and suppose f T 0 is a unit. Let ϕ f : S T be the following morphism of graded rings ϕ f (s) = d 0 f d ϕ(s d ) Then G(ϕ) = G(ϕ f ) and the induced morphisms G(ϕ) P rojs are the same. Proposition 9. Let S be a graded ring and e > 0. The morphism of graded rings ϕ : S [e] S induces an isomorphism of schemes Φ : P rojs P rojs [e] natural in S. Proof. We defined the graded ring S [e] in (GRM,Definition 6). The inclusion ϕ : S [e] S is a morphism of graded rings. It is clear that G(ϕ) = P rojs, so we get a morphism of schemes Φ : P rojs P rojs [e]. If a is a homogenous ideal of S then a S [e] is homogenous ideal of S [e], and it is not hard to see that for a homogenous prime ideal p we have p a if and only if p S [e] a S [e]. In particular this shows that Φ is injective. To see that it is surjective, let q be a homogenous prime ideal of S [e]. Then the ideal q generated by q in S is homogenous and it is not difficult to check that q S [e] = q. We claim that the homogenous ideal p = q is prime. If a S m and b S n are such that ab p then (ab) ke q for some k > 0 and therefore either a ke q or b ke q, which shows that a p or b p. Clearly p S [e] = q and if q S [e] + then p S +, which shows that Φ is surjective. Our arguments in the previous paragraph also imply that Φ(V (a)) = V (a S [e] ) for any homogenous ideal a, so Φ is in fact a homeomorphism. For homogenous f S + we have D + (f) = D + (f e ) and it is clear that for f S [e] + we have Φ(D + (f)) = D + (f). The induced morphism of rings S [e] (f) S (f) is the morphism ϕ (f) which is clearly an isomorphism. This proves that Φ : P rojs P rojs [e] is an isomorphism. This isomorphism is natural in S in the following sense. Suppose ψ : S T is a morphism of graded rings, ψ [e] : S [e] T [e] the induced morphism of graded rings. It is not hard to check that the isomorphism P rojt = P rojt [e] identifies the open subsets G(ψ) and G(ψ [e] ) and that the following diagram commutes G(ψ) Ψ P rojs G(ψ [e] ) Ψ [e] P rojs [e] which completes the proof. Definition 2. Let S be a graded ring and e > 0. Let S e denote the graded ring with the same ring structure as S, but with the inflated grading S e = S S 1 0 so (S e) 0 = S 0, (S e) e = S 1, (S e) 2e = S 2 and so on. It is not hard to see that an ideal a S is homogenous in S iff. it is homogenous in S e, so as topological spaces we have P rojs = P roj(s e). Moreover for a homogenous prime ideal p we have an equality of rings S (p) = (S e) (p), and therefore the sheaves of rings are also the same. Hence we have an equality of schemes P rojs = P roj(s e). Example 1. For a graded ring S and e > 0 it is clear that S (e) e = S [e]. Therefore as schemes there is an equality P rojs (e) = P rojs [e]. Corollary 10. Let S be a graded ring and e > 0. There is a canonical isomorphism of schemes Ψ : P rojs P rojs (e) natural in S. 4
5 Proof. The isomorphism is the equality P rojs (e) = P rojs [e] followed by the isomorphism of Proposition 9. Naturality in S means that for a morphism of graded rings θ : S T with induced morphism of graded rings θ (e) : S (e) T (e) the isomorphism P rojs = P rojs (e) identifies G(θ) and G(θ (e) ) and the following diagram commutes G(θ) Θ P rojs G(θ (e) ) Θ (e) P rojs (e) This follows easily from the naturality of Proposition 9. Proposition 11. Let S be a graded algebra generated by S 1 as an algebra. Let S be the graded algebra defined by S 0 =, S d = S d for d > 0. Then the canonical morphism of graded algebras S S induces an isomorphism of schemes P rojs P rojs natural in S. Proof. We make the direct sum S = d 1 S d into a graded algebra in the canonical way (see our note on Constructing Graded lgebras). Let ϕ : S S be induced by S and all the inclusions S d. This is a morphism of graded algebras. It is clear that G(ϕ) = P rojs, so let Φ : P rojs P rojs be the induced morphism of schemes. It follows from (GRM,Lemma 17) that Φ is injective and from (GRM,Proposition 18), (GRM,Lemma 17) that Φ is surjective. For a homogenous ideal a of S we have Φ(V (a)) = V (ϕ 1 a) (GRM,Lemma 16) so Φ is a homeomorphism. To show that Φ is an isomorphism of schemes it suffices by Lemma 1 to show that for d > 0 and f S d the following ring morphism is bijective ϕ (f) : S (f) S (f) a/f n a/f n This map is clearly surjective. To see that it is injective, suppose a/f n maps to zero in S (f). If n > 0 then a S nd and f k a = 0 for some k 0 implies that a/f n = 0 in S (f). If n = 0 and a then we can write a/1 = af/f and reduce to the case where n > 0. This shows that ϕ (f) is an isomorphism and completes the proof that Φ is an isomorphism of schemes. To prove naturality, let ψ : S T be a morphism of graded algebras, where T is also generated by T 1 as an algebra. There is an induced morphism of graded algebras ψ = 1 d 1 ψ d : S T making the following diagram commute S ψ T S ψ T We claim that the isomorphisms Φ S : P rojs P rojs and Φ T : P rojt P rojt are natural in the sense that Φ T identifies the open subsets G(ψ) and G(ψ ) and the following diagram commutes Ψ G(ψ) P rojs The details are easily checked. G(ψ ) Φ S Ψ P rojs Definition 3. Let ϕ : S T be a morphism of graded rings. We say that ϕ is a quasimonomorphism, quasiepimorphism or quasiisomorphism if it has this property as a morphism of graded Smodules (GRM,Definition 10). 5
6 Corollary 12. Let ϕ : S T be a morphism of graded algebras where S is generated by S 1 as an algebra and T is generated by T 1 as an algebra. If Φ : G(ϕ) P rojs is the induced morphism of schemes then (i) If ϕ is a quasiepimorphism then G(ϕ) = P rojt and Φ is a closed immersion. (ii) If further ϕ is a quasiisomorphism then Φ is an isomorphism. Proof. Using Proposition 11 we can assume that S 0 = and T 0 =. If ϕ is a quasiepimorphism (resp. quasiisomorphism) then for some e > 0 the morphism ϕ (e) : S (e) T (e) is surjective (resp. bijective) so by Corollary 10 we can reduce to proving that if ϕ is surjective then Φ is a closed immersion, and further that if ϕ is an isomorphism then so is Φ. But we proved the former claim in our solution to (H,Ex.3.12) and the latter claim is trivial. Remark 1. Note that if S is a graded algebra and S denotes the graded algebra S 0 =, S d = S d for d > 0 defined in Proposition 11 then the canonical morphism of graded algebras ϕ : S S is trivially a quasiisomorphism. Proposition 13. Let ϕ : S T be a quasiisomorphism of graded algebras with S 0 = and T 0 =. Then there is an integer E > 0 such that for all e E, ϕ (e) : S (e) T (e) is an isomorphism of graded algebras. Proof. Let E > 0 be large enough that ϕ n : S n T n is bijective for all n E. Then it is not hard to see that ϕ (e) : S (e) T (e) is an isomorphism for e E. 3 Products Lemma 14. Let, B be rings, S a graded algebra, and T a graded Balgebra. Let ψ : B be a morphism of rings and ϕ : S T a morphism of graded rings making the following diagram commute: ψ B (2) S ϕ T Then the following diagram commutes, where Ψ = Spec(ψ) and the vertical morphisms are canonical: Φ G(ϕ) P rojs (3) SpecB Ψ Spec Proof. The open sets D + (f) P rojs for homogenous f S + are an open cover, so the open sets D + (ϕ(f)) = Φ 1 D + (f) for homogenous f S + must be an open cover of G(ϕ). So it suffices to show that the two legs of the diagram agree when composed with Spec(T (ϕ(f)) ) G(ϕ) for every homogenous f S +. Using the commutativity of (1), this reduces to checking commutativity of the following diagram of rings, where the vertical morphisms are canonical: T (ϕ(f)) ϕ (f) S (f) (4) B ϕ But this square is trivially commutative, completing the proof. lternatively, use the fact that morphisms into Spec are in bijection with ring morphisms out of and just check the diagram on global sections. 6
7 Proposition 15. Let, B be rings, S a graded algebra, ψ : B a morphism of rings. Then T = S B is a graded Balgebra and P rojt = P rojs SpecB. Proof. Consider, B as graded rings in the canonical way. Then S and B become graded  modules, so the tensor product S B becomes a graded Balgebra (GRM,Section 6) with grading { } (S B) n = s i b i s i S n, b i B i The map ϕ : S T defined by ϕ(s) = s 1 is a morphism of graded rings, which makes the diagram (2) commute. Since T + is generated as a Bmodule by ϕ(s + ) it follows that G(ϕ) = P rojt and we get a morphism of schemes Φ : P rojt P rojs fitting into a commutative diagram P rojt Φ P rojs (5) SpecB Ψ Spec We have to show that this diagram is a product of schemes over Spec. The open sets D + (f) for homogenous f S + cover P rojs and by our earlier notes on the local nature of products, it suffices to show that D + (ϕ(f)) = Φ 1 D + (f) is a product for D + (f) and SpecB over Spec for all homogenous f S +. Using commutativity of (1) we reduce this to showing that the following diagram is a pullback: Spec(T (ϕ(f)) ) Spec(S (f) ) Spec(B) Spec() Which is equivalent to showing that (4) is a pushout of rings. The map S f B T ϕ(f) defined by (s/f n, b) (s b)/ϕ(f) n is welldefined and bilinear, so induces S f B T ϕ(f), which is easily checked to be a morphism of rings. The canonical map S S f is a morphism of  modules, so there is a welldefined morphism T = S B S f B defined by s b s/1 b. This is a morphism of rings mapping multiples of ϕ(f) = f 1 to units, so it induces a morphism of rings T ϕ(f) S f B defined by (s b)/ϕ(f) n s/f n b. Since we have already constructed the inverse, this is an isomorphism of rings T ϕ(f) = Sf B. The ring S f is a Zgraded ring, and thus can be considered as a graded module. The ring S f B is therefore also Zgraded. Putting the canonical Zgrading on T ϕ(f) it is easily checked that T ϕ(f) = Sf B is an isomorphism of Zgraded rings, which restricts to give an isomorphism of the degree zero subrings T (ϕ(f)) = (Sf B) 0. We have to show that this latter ring is isomorphic as a ring to S (f) B. Let α : S f Z B S f B be canonical (GRM,Section 6). The kernel of α is the abelian group P generated by elements (a x) b x (a b) where x is homogenous. The morphism of groups S (f) Z B S f Z B is injective since S (f) is a direct summand of S f and tensor products preserve colimits. Therefore the group S (f) Z B is isomorphic to its image in S f Z B, which is mapped by α onto (S f B) 0. So there is an isomorphism of abelian groups (S (f) Z B)/P = (S f B) 0 where P = P (S (f) Z B). But since S (f) Z B can be identified with the degree zero subring of S f Z B, P is generated as an abelian group by elements (a x) b x (a b) where x is homogenous of degree zero, that is, x S (f). Hence there is an isomorphism of abelian groups S (f) B = (S (f) Z B)/P = (Sf B) 0 s/f n b s/f n b So finally we have an isomorphism of rings T (ϕ(f)) = S(f) B defined by (s b)/ϕ(f) n s/f n b. Using this isomorphism one checks easily that (4) is a pushout, completing the proof. 7
8 Let ψ : B be a morphism of rings (n 0) and ϕ : [x 0,..., x n ] B[x 0,..., x n ] the morphism of graded rings induced by ψ. Then it is not difficult to check that the following diagram is a pushout of rings: ψ B [x 0,..., x n ] ϕ B[x 0,..., x n ] By uniqueness of the pushout, there is an isomorphism of rings α B[x 0,..., x n ] B [x 0,..., x n ] f(x 0,..., x n ) f(1 x 0,..., 1 x n ) f(α)x α0 0 xαn n α f(α) x α0 0 xαn n This is the morphism of Balgebras induced by x i 1 x i. Similarly x i x i 1 induces an isomorphism of rings θ : B[x 0,..., x n ] [x 0,..., x n ] B. If we give the tensor product the canonical graded structure, these are isomorphisms of graded rings. Note that θϕ is just the canonical map of [x 0,..., x n ] into the tensor product. Now let n 1 and T = [x 0,..., x n ] B. Then G(ϕ) = P n B and G(θ) = G(θϕ) = P rojt. Let Ω : P rojt Pn be induced by θϕ. Consider the following diagram: P n B (6) Φ Θ Ω P rojt P n SpecB Ψ Spec ll unmarked morphisms are the canonical ones, and by the previous Proposition the square is a pullback. The top triangle commutes by the results of the previous section. To check commutativity of the left triangle, we need only check the two legs give the same morphism on global sections (morphisms into SpecB are in bijection with ring morphisms out of B). n element b B determines the global section q b/1 of P n B and p (1 b)/1 of P rojt. These clearly correspond under the localisations of θ, so (6) commutes. Hence the outside square is a pullback, and we have proved the following result. Proposition 16. Let ψ : B be a morphism of rings and for n 1 let ϕ : [x 0,..., x n ] B[x 0,..., x n ] be the corresponding morphism of graded rings. Then the following diagram is a pullback: Φ P n B P n SpecB Ψ Spec In other words, P n B = SpecB Spec P n. In particular, for any ring we have Pn = Spec Pn Z. P n B One could probably run all the proofs with n = 0, but since in that case P n Spec and SpecB are isomorphisms, the diagram in the proposition is trivially a pullback. 8
9 4 Linear Morphisms Let be a ring n, m 0 and f 0 (x 0,..., x n ),..., f m (x 0,..., x n ) [x 0,..., x n ] homogenous polynomials of degree 1. Then we define φ : [y 0,..., y m ] [x 0,..., x n ] φ(y i ) = f i (x 0,..., x n ) This is a morphism of graded algebras, hence induces a morphism of schemes Φ : G(φ) P m. Here G(φ) = {p P n p φ(s + )} = {p P n p {f 0,..., f m }} m = D + (f i ) i=0 For example, suppose m 1 and 0 i m and let φ i : [x 0,..., x m ] [x 0,..., x m 1 ] be the morphism of graded algebras determined by the following assignments: x 0 x 0,..., x i 1 x i 1, x i+1 x i,..., x m x m 1 and x i 0. Then clearly G(φ i ) = P m 1 and so we have a morphism of schemes Φ i : P m 1 P m The morphism φ i is surjective, so by Ex 3.12 the morphism Φ i is a closed immersion whose image is the closed set V (Kerφ) = V ((x i )) = P m D +(x i ). So for m 1 there are m + 1 closed immersions of P m 1 in P m. Notice that the following diagram also commutes for any 0 i m: P m 1 Φ i Spec In particular there are two closed immersions Spec = P 0 P1. lso note that for n 1 and 0 i n the open set D + (x i ) of P n is isomorphic to Spec[x 0,..., x n ] (xi) and thus to Spec[x 1,..., x n ]. In fact these are isomorphisms of schemes over Spec. nother example are automorphisms arising from invertible matrices. Fix n 1 and let = (a ij ) be an invertible (n + 1) (n + 1) matrix over a field k. For convenience we use the indices 0 i, j n. Then the map x i a ij x j determines a kalgebra automorphism P m ϕ : k[x 0,..., x n ] k[x 0,..., x n ] which gives rise to an automorphism of schemes over k, Φ = P rojϕ : P n k Pn k. 5 Projective Morphisms Definition 4. Let be a scheme and n 0. projective nspace over is a pullback SpecZ P n Z. This consists of two morphisms Z, Z P n Z making the following diagram a pullback: Z P n Z SpecZ Two projective nspaces over are canonically isomorphic as schemes over and P n Z, and we denote any projective nspace over by P n. morphism X is projective if it factors as a 9
10 closed immersion X P n followed by the projection Pn for some n 1 and projective nspace P n. This definition is independent of the projective nspace chosen, in the sense that if it factors through one projection via a closed immersion, then it factors through all of them by a closed immersion. We refer to this situation by saying that X factors via a closed immersion through projective nspace. For a ring and n 0 the scheme P roj[x 0,..., x n ] is a projective nspace over Spec with the canonical morphisms, so the notation P n is unambiguous. Since the morphism P0 Z SpecZ is an isomorphism, projective 0spaces over correspond to isomorphisms Z, in the sense that if Z, Z P 0 Z is a projective 0space then Z is an isomorphism, and any isomorphism Z can be paired with Z SpecZ P 0 Z to make a projective 0space. For any ring and n 1 there are n + 1 closed immersions P n 1 P n of schemes over Spec. We can extend this to projective space over any scheme: Lemma 17. For any scheme there are n + 1 canonical closed immersions P n 1 P n of schemes over for any n 1. Proof. Let P n 1 and P n be any projective n 1 (resp. n)spaces over. We induce a morphism P n 1 P n into the pullback so that the following diagram commutes: P n Z P n 1 Z SpecZ P n P n 1 Where for 0 i n we let P n 1 Z P n Z be the ith closed immersion. Using the usual pullback argument, we see that P n 1 P n is the pullback of Pn 1 Z P n Z, and since closed immersions are stable under pullback the proof is complete. Lemma 18. If for some n 0 a morphism X factors via a closed immersion through projective nspace, then it factors via a closed immersion through projective mspace for any m n. The lemma implies that we could just as well define a projective morphism X to be a morphism factoring via a closed immersion through a projection P n for some n 0, since if X factors via a closed immersion through projective 0space, then it factors through projective 1space by a closed immersion, and is thus projective. So our final definition is: Definition 5. morphsim X is projective if it factors via a closed immersion through projective nspace over for some n 1 (equivalently, some n 0). morphism X is quasiprojective if it factors via an immersion through projective nspace for some n 1 (equivalently, some n 0). Notice that a morphism X factors through projective 0space over iff. it is a closed immersion. In particular any closed immersion is projective. Lemma 19. Let be a ring and S a graded algebra finitely generated as an algebra by S 1. Then the structural morphism P rojs Spec is projective. Proof. By hypothesis we can find a surjective morphism of graded algebras [x 0,..., x n ] S with n 1, so the induced morphism P rojs P n is a closed immersion of schemes over, as required. 10
11 Let f : X be a morphism of schemes, fix an integer n 1 and pullbacks P n X, Pn. Then the morphisms P n X Pn Z and Pn X X induce a morphism Pn f : Pn X Pn which is the unique morphism of schemes over P n Z fitting into the following commutative diagram P n X P n f P n X f In fact using standard properties of pullbacks it is not hard to see that this is a pullback diagram. Clearly P n 1 = 1 and if we have a morphism g : Z and fix a pullback P n Z then Pn g P n f = Pn gf. If X = Spec, = SpecB and f is induced by a ring morphism ψ : B, and if we choose P n, Pn B as our pullbacks, then Pn f is the morphism induced by the canonical ring morphism [x 0,..., x n ] B[x 0,..., x n ] as in Proposition Dimensions of Schemes We want to translate some results about varieties into results about schemes. To be specific, recall the following (proved in I, 3.4): Proposition 20. Let P n be a projective variety and 0 i n such that i = U i, where U i = n is the open set x i 0. Then i corresponds to an affine variety in n and S( ) (xi) = ( i ), where S( ) = k[x 0,..., x n ]/I( ) and ( i ) = k[y 1,..., y n ]/I( i ). Proof. In outline: let S = k[x 0,..., x n ]. Then k[y 1,..., y n ] = S (xi) and this isomorphism identifies I( i ) with the prime ideal I( )S xi S (xi) (which is denoted I( )S (xi) in the proof of I, 3.4). It follows that ( i ) = k[y 1,..., y n ]/I( i ) = S (xi)/i( )S (xi)) = (S/I( )) (xi) = S( ) (xi) Let k be a field, and for n 1 set S = k[x 0,..., x n ] and consider the the projective space P n k = P rojs. There are n + 1 open affine subsets D +(x i ) = SpecS (xi) = Speck[x 0 /x i,..., x n /x i ] which cover P n k. We know from our divisor notes that the closed irreducible subsets of Pn k are in bijection with the homogenous primes of S other than S +. Let P n k be a closed irreducible subset, and suppose D + (x i ). Then I( ) D + (x i ) and D + (x i ) is the closure of I( ). The prime I( ) corresponds to the prime I( )S (xi) = I( )S xi S (xi) of SpecS (xi). Let i Speck[x 0 /x i,..., x n /x i ] be the closed irreducible subset corresponding to D + (x i ). Then I( i ) is the prime ideal of k[x 0 /x i,..., x n /x i ] corresponding to I( )S (xi). D + (x i ) = = SpecS (xi) Speck[x 0 /x i,..., x n /x i ] 11
12 Let ( i ) denote k[x 0 /x i,..., x n /x i ]/I( i ) and S( ) denote S/I( ). Then immediately we have an isomorphism of kalgebras ( i ) = S (xi)/i( )S (xi). It follows from the next Lemma that S (xi)/i( )S (xi) = (S/I( )) (xi). So finally we have an isomorphism of kalgebras ( i ) = S( ) (xi). Lemma 21. Let be a ring and S a graded algebra. If I is a homogenous ideal of S and x S homogenous, then there is a canonical isomorphism of algebras α : (S/I) (x+i) S (x) /(IS x S (x) ) (f + I)/(x n + I) f/x n + IS x S (x) Proof. It is easy to check the given map is an isomorphism of algebras. We proved in our Chapter 1, Section 3 notes that for a graded domain S and nonzero x S 1 there is an isomorphism of S 0 algebras S x = S(x) [z, z 1 ] (the ring of Laurent polynomials). s in the proof of Ex I, 2.6, this implies that for the case S = k[x 0,..., x n ] we have an isomorphism of kalgebras S( ) xi = S( )(xi)[z, z 1 ] = ( i )[z] z It follows that the quotient field of S( ) is kisomorphic to the quotient field of ( i )[z]. Since both S( ) and ( i )[z] are finitely generated domains over the field k, this implies that dims( ) = dim( i )[z] = dim( i ) + 1. Proposition 22. Let S = k[x 0,..., x n ] for a field k and n 1. If P n k irreducible closed subset then dims( ) = 1 + dim where I( ) is the homogenous prime associated with and S( ) = S/I( ). = P rojs is an Proof. The previous discussion shows that is covered by open subsets homeomorphic to schemes of the form Spec( i ) where dim( i ) = dims( ) 1. The claim now follows from I, Ex1.10 and the fact that for a ring, dim = dim(spec). Since dims( ) = coht.i( ), another way to state the result is dim = coht.i( ) 1. This is in contrast to the affine case, where for an irreducible closed we would have dim = coht.i( ). 7 Points of Projective Space Let S = k[x 0,..., x n ] where k is a field and n 1. If k is algebraically closed, then we know from our Varieties as Schemes notes that the closed points of P n k = P rojs are homeomorphic to the variety P n defined in Chapter 1. Corresponding to a point P = (a 0,..., a n ) with a i 0 is the homogenous prime ideal I(P ) = (a i x 0 a 0 x i,..., a i x n a n x i ). In fact, we can still use this idea of points when k is not algebraically closed. Throughout this section k is an arbitrary field and S = k[x 0,..., x n ] for fixed n 1. Lemma 23. Given a nonzero tuple P = (a 0,..., a n ) k n+1 with say a i 0, I(P ) = (a i x 0 a 0 x i,..., a i x n a n x i ) is a homogenous prime ideal of S of height n 1. Proof. First we show that the ideal I(P ) does not depend on i. That is, if also a j 0 then we claim I(P ) = (a j x 0 a 0 x j,..., a j x n a n x j ). This follows easily from the following formula for any 0 k n a j x k a k x j = a j /a i (a i x k a k x i ) a k /a i (a i x j a j x i ) Since I(P ) is generated by linear polynomials, it is a homogenous prime ideal of height n 1 by our Linear Variety notes. It is clear that if P = λq for some nonzero λ k then I(P ) = I(Q), as one would expect for points of projective space. Since ht.s + = n it is clear that I(P ) P rojs, so we have associated points of P n k with tuples of kn+1. 12
13 Lemma 24. Given a nonzero tuple P = (a 0,..., a n ) k n+1 there is an automorphism Φ : P n k P n k of schemes over k mapping P to (1, 0,..., 0). Proof. Suppose that a i k[x 0,..., x n ] by 0 and define an automorphism of kalgebras ψ : k[x 0,..., x n ] x 0 x i x 1 a i x 0 a 0 x i. x n a i x n a n x i It is not hard to see that this induces a kautomorphism of P n k identifying P and (1, 0,..., 0) (that is, identifying the homogenous prime ideals I(P ) and (x 1,..., x n )). 13
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