MATH 8253 ALGEBRAIC GEOMETRY WEEK 12 CIHAN_ BAHRAN 3.2.1. Let Y be a Noetherian scheme. Show that any Y -scheme X of finite type is Noetherian. Moreover, if Y is of finite dimension, then so is X. Write f : X ! Y for the structure morphism which is assumed to be of finite type. To show that X is Noetherian, it suffices to show that it is quasi-compact and locally Noetherian. Since Y is quasi-compact, it can be covered by finitely many affine open subsets, say −1 −1 V1;:::;Vn. Then f (Vj)'s cover X. Now f is quasi-compact, so each f (Vj) is quasi- compact and hence their finite union X is quasi-compact. To show that every point in X has a Noetherian neighborhood, let x 2 X. Since Y is locally Noetherian, f(x) is contained in an affine open subset V = Spec B of Y such that B is Noetherian. Let U = Spec A be an affine neighborhood of x such that U ⊆ f −1(V ). Then since f is of finite-type, A is a finitely generated B-algebra, hence A is Noetherian. Moreover, in the above notation, if Y has finite dimension, then dim B < 1 and so by Corollary 2.5.17 the finitely generated B-algebra A has finite dimension. So every x 2 X has a neighborhood with finite dimension. Since X is quasi-compact, X itself has finite dimension. 3.2.2. Show that any open immersion into a localy Noetherian scheme is a mor- phism of finite type. Let f : X ! Y be an open immersion where Y is locally Noetherian. Let V be an affine open subscheme of Y . Then by Exercise 2.3.16 V is Noetherian. Let U = f −1(V ), then clearly the restriction fjU : U ! V is also an open immersion; hence by Proposition 2.3.46(a) U is Noetherian. By definition, U is then a finite union of affine open (hence quasi-compact) subschemes, therefore U = f −1(V ) is quasi-compact. This shows that f is quasi-compact. Let fVigi be an open covering of Y by affine open subsets. Via f, we can identify X as −1 an open subset of Y . For each i, f (Vi) is quasi-compact by above. Identifying X as −1 an open subscheme of Y , f (Vi) = Vi \X is an open subset of the affine scheme Vi and hence by quasi-compactness can be covered by finitely many principal open sets Uij of Vi. We can make the identifications Vi = Spec Ai for some ring Ai and Uij = D(gij) where gij 2 Ai. Observe that OY (Vi) = Ai and OX (Uij) = OY (Uij) = OVi (Uij) = (Ai)gij such that the induced map OY (Vi) !OX (Uij) is the localization Ai ! (Ai)gij . In general, if A is a ring and g 2 A, A[X] surjects on Ag by the A-algebra homo- morphism X 7! 1=g, so the localization A ! Ag makes Ag into a finitely generated A-algebra. Thus OX (Uij) is a finitely generated algebra over OY (Vi) for every j. Hence by Proposition 3.2.2 f is of finite type. 1 MATH 8253 ALGEBRAIC GEOMETRY WEEK 12 2 3.2.3. An immersion of schemes is a morphism which is an open immersion followed by a closed immersion. (a) Let f : X ! Y be an immersion. Show that it can be decomposed into a closed immersion followed by an open immersion. (b) Show that the converse of (a) is true if f is moreover quasi-compact (e.g. if Y is locally Noetherian). Use the scheme-theoretic closure of f (Exercise 2.3.17). (c) Let f : X ! Y , g : Y ! Z be two immersions with g quasi-compact. Show that g ◦ f is an immersion. (a) Y has a closed subscheme C such that there is an open immersion h : X ! C such that when composed with the closed immersion i : C,! Y we get f, that is, f = i ◦ h. Identify X as an open subspace of C. Then Y has an open subset V such that X = C \ V . Since f(X) ⊆ V and V is open, f restricts to a scheme morphism g : X ! V , that is, if we write j : V,! Y for the open immersion, then f = j ◦ g. Since X is identified with the closed subspace C \ V of V via f, the morphism g is a closed immersion of schemes. So j ◦ g is a decomposition of the desired form. (b) LATER (c) Write g = a ◦ b and f = c ◦ d where b; d are open immersions and a; c are closed immersions. Since g is quasi-compact and a is an open immersion b is also quasi- compact. Also c, being an open immersion is quasi-compact. Thus the composite b ◦ c is quasi-compact. Now by (b), b ◦ c = b0 ◦ c0 where b0 is an open immersion and c0 is a closed immersion. Thus g ◦ f = (a ◦ b0) ◦ (c0 ◦ d) ?????????????????????????? 3.2.4. Let X; Y be schemes over a locally Noetherian scheme S, with Y of fi- nite type over S. Let x 2 X. Show that for any morphism of S-schemes fx : Spec OX;x ! Y , there exist an open subset U 3 x of X and a morphism of S-schemes f : U ! Y such that fx = f ◦ ix, where ix : Spec OX;x ! U is the canonical morphism (in other words, the morphism fx extends to an open neigh- borhood of x). We have a commutative diagram of schemes fx Spec OX;x / Y & S: where the diagonal morphism is given by the structure morphism h : X ! S Consider the image of x in S under the diagonal arrow and find an affine open neighborhood V around it. Then h−1(V ) is an open subset in X containing x so it contains an affine neighborhood U of x in X. Then since OX (U) maps to OX;x, Spec OX;x ! S factors through the restriction U ! V . That is, the diagonal arrow maps to an affine open subset of S. Then making the same argument for fx by taking an affine open W in Y which maps to V and shrinking U if necessary, we reduce to the case where all X, Y and S are affine. MATH 8253 ALGEBRAIC GEOMETRY WEEK 12 3 Writing X = Spec A, Y = Spec B and S = Spec C and x = p 2 Spec A, the commu- tative diagram of (now affine) schemes above correspond to the following commutative diagram of rings: Ap o B c O C In addition to this, there is a ring homomorphism from C to A which is compatible with the above picture. So more compactly, both A and B are C-algebras and ' : B ! Ap is a C-algebra homomorphism. We want to find an f 2 Ar p such that ' factors through the canonical map Af ! Ap; because on the scheme level this amounts to extending the map from Spec OX;x to the principal open set D(f). Note that since C came from an affine open subset of a locally Noetherian scheme, we may assume C is Noetherian. And the finite type assumption ensures that B is a finitely generated C-algebra. So there exists n 2 N such that there is a surjective C-algebra homomorphism π : C[T1;:::;Tn] B say with kernel I. Because C is noetherian, so is C[T1;:::;Tn] (Hilbert basis theorem) therefore I is finitely generated. So I = (g1; : : : gr) for some polynomials g1; : : : ; gr 2 C[T1;:::;Tn]. ai Write π(Ti) = bi 2 B and '(bi) = 2 Ap (so ai 2 A and si 2 Ar p) for each si i = 1; : : : ; n. a1 an Fix j 2 f1; : : : ; mg. Now we know that gj(b1; : : : ; bn) = 0 in B, therefore gj( ;::: ) = s1 sn 0 in Ap. Clearing the denominators in the latter equation, we get a polynomial gej 2 A[T1;:::;Tn] such that g (a ; : : : ; a ) ej 1 n = 0 2 A 1 p hence there exists tj 2 Ar p such that tjgej(a1; : : : ; an) = 0 Qn in A (we may take gej = ( i=1 si) gj). Now let n m Y Y f = si · tj 2 Ar p : i=1 j=1 Since f is invertible in Af , every si is invertible in Af . So we can define a C-algebra homomorphism : C[T1;:::;Tn] ! Af −1 Ti 7! si ai : Now since every tj in Af , we have g (a ; : : : ; a ) ej 1 n = 0 2 A 1 f MATH 8253 ALGEBRAIC GEOMETRY WEEK 12 4 so by decomposing back we get −1 −1 gj(s1 a1; : : : ; sn an) = 0 : Therefore the polynomial gj 2 C[T1;:::;Tn] lies in the kernel of . Since this is true for every j = 1; : : : ; m we get I ⊆ ker . Thus uniquely factors through B, that is, we get a C-algebra homomorphism B ! Af which by construction makes ' B / Ap O # Af commute. This is what we wanted. 3.2.5. Let S be a locally Noetherian scheme. Let X; Y be S-schemes of finite type. Let us fix s 2 S. Let ' : X ×S Spec OS;s ! Y ×S Spec OS;s be a morphism of S-schemes. Show that there exist an open set U 3 s and a morphism of S- schemes f : X ×S U ! Y ×S U such that ' is obtained from f by the base change Spec OS;s ! U.