Lecture 7: Product of Varieties, Separatedness

Lecture  18.725 Algebraic Geometry I Lecture 7: Product of Varieties, Separatedness Here are some additions to last time. Recall that if R(X) ∼ = R(Y ), then there are open subsets U ⊆ X, V ⊆ Y which are isomorphic. To see this, replace X, Y with U, V such that we have morphisms f : U → V ∼ R(Y ). Then f g : V 0 → V 0 and g : V 0 → U (where V 0 ⊆ V ) which are induced by the isomorphism R(X) = is the identity (induced by R(Y ) → R(X) → R(Y ) which is the identity). Then g : V 0 → f −1 (V 0 ), and set U 0 = f −1 (V 0 ). Then gf : U 0 → U 0 is the identity for similar reasons. Hence U 0 ' V 0 . In the proof of a lemma from last time (that the set of unramified points is open), we used that if SpecA → Spec(C = B[t]/P ) → SpecB (where everything has dimension n), then C ⊆ A; that is, C → A is an injection. If not, then the kernel is nontrivial, and consequently Spec(image) has dimension less than n, and hence dim SpecA < n. Products Let C be any category and X, Y ∈ Ob(C). Then X × Y is an object Z ∈ Ob(C) together with maps πX : Z → X, πY : Z → Y such that for any other T ∈ Ob(C), there is an isomorphism ∼ Hom(T, Z) − → Hom(T, X) × Hom(T, Y ) given by f 7→ (πX ◦ f, πY ◦ f ). Equivalently, X × Y is the object corresponding to the functor T 7→ Hom(T, X) × Hom(T, Y ), if it exists. Yoneda’s lemma implies that if it exists, then it is unique up to unique isomorphism. ∼ Similarly, the coproduct X q Y is defined such that Hom(X q Y, T ) − → Hom(X, T ) × Hom(Y, T ) . Example 1. Let C be the category of commutative k-algebras. Then the product is the usual direct product, or direct sum. The coproduct of A, B would be A ⊗k B. We have an equivalence of categories {affine algebraic varieties} = {finitely generated commutative nilpotent-free k-algebras}op, where the op means the opposite category; the objects are the same, but the arrows are reversed. Thus, product of affine algebraic varieties corresponds to the tensor product of their global sections. Exercise 1. Describe the product and coproduct in the category of not necessarily commutative k-algebras. Lemma 1. If A, B are nilpotent-free k-algebras, so is A ⊗k B. Proof. We Xcheck that A ⊗k B injects into Homk−alg (SpecA × SpecB). For contradiction, take a nonzero element ai ⊗ bi ∈ A ⊗k B in the kernel. Without loss of generality, the ai are linearly independent, as well as the bi . Find x ∈ SpecA such that for some i, ai (x) 6= 0. Restricting to {x} × SpecB, we get a contradiction to linear independence of the bi . Therefore, we can identify A ⊗k B with a subspace of Homk−alg (SpecA × SpecB), which clearly contains no nilpotents. Therefore, SpecA ⊗k B makes sense, and Hom(X, SpecA ⊗k B) = Hom(A ⊗k B, k[X]) ' Hom(A, k[X]) × Hom(B, k[X]) = Hom(X, SpecA) × Hom(X, SpecB) implies that SpecA × SpecB = SpecA ⊗k B. Remark 1. Caution: The topology on the product of spaces with functions is not the product topology. Suppose X, Y are algebraic varieties, or spaces with functions. We define a basis of open sets on X × Y to be those X subsets of the form U ⊆ V1 × V2 , where V1 ⊆ X, V2 ⊆ Y are open and U is the complement to zeroes(f = fi gi ) where fi are regular on V1 , gi are regular on V2 . Another construction can be given as follows: suppose that X and Y can be written as X = ∪Ui , Y = ∪Vj for Ui = SpecAi and Vj = SpecBj . Then X × Y will be ∪Spec(Ai ⊗ Bj ), glued properly. Theorem 1.1. dim(X × Y ) = dim(X) + dim(Y ) Proof. The computation is local, so assume X, Y are affine of dimension n, m respectively. Then there are finite onto maps X  An , Y  Am , so their product is a finite onto map X × Y → An+m , which implies that X × Y is of dimension n + m. Lemma 2. Suppose that for i ∈ {1, 2}, Xi is a closed subvariety of Yi . Then X1 × X2 is a closed subvariety of Y1 × Y2 . 1 Lecture  18.725 Algebraic Geometry I Proof. Work locally to reduce to the case when Y1 , Y2 are affine. The corresponding algebraic statement to check is that the tensor product of two surjective maps is still surjective; this is true. Proposition 1. The product of projective varieties is projective. Proof. By the previous lemma, it suffices to check that Pn × Pm is projective. To do so, use the Segre embedding into Pnm+n+m . Geometrically, the Segre embedding takes (x, y) ∈ Pn × Pm , considers the duals of x, y given by lines Lx ⊆ k n+1 = V , Ly ⊆ k m+1 = W , takes the line Lx ⊗ Ly ⊆ V × W = k (n+1)(m+1) , and identifies that with its dual, which is a point in Pnm+n+m . More concretely, it takes ((x0 : · · · : xn ), (y0 : · · · : ym )) 7→ (· · · : xi yj : · · · ). If the coordinate are given by zij such that the xi yj belongs to the zij coordinate, then the image of the Segre embedding is cut out by zij zkl − zkj zil . Separatedness Example 2. Here is a non-quasiprojective variety: the line with a double point. It is given by A1 × {0, 1}/ ((x, 0) ∼ (x, 1)unlessx = 0). Definition 1. An algebraic variety is separated if its diagonal ∆X is a closed subvariety in X × X. In general, the diagonal is always a locally closed subvariety. Furthermore, affine varieties are separated because if X = SpecA, then the multiplication map A ⊗ A  A is surjective. Therefore, if X is an algebraic variety such that X = ∪Ui where the Ui are affine, then ∆X ∩ (Ui × Ui ) is closed in Ui . Lemma 3. A locally closed subvariety in a separated variety is separated. Proof. Suppose X is separated and Z ⊆ X is a subvariety. Then Z × Z ⊆ X × X is a subvariety, and ∆Z = ∆X × (Z × Z). Lemma 4. Pn is separated. Proof. Write Pn = ∪Ani . Then Ani × Anj ⊇ ∆ ∩ (Ani × Anj ). When i = j, we are reduced to the affine case. When i 6= j, say i = 0 and j = 1, we take coordinates x1 , · · · , xn and y0 , y2 , · · · , yn and see that being on the diagonal is the closed condition xa yb = xb ya . Corollary 1. A quasiprojective variety is separated. The line with a doubled origin is not separated. To see this, denote this algebraic variety by X, and note that we have a natural map X → A1 . Then X 2 → A2 , and over 0 we have {0ij }i,j∈{1,2} . The closure of diagonal contains all four points, while only two points 011 and 022 belong to the diagonal. In particular, X cannot be quasiprojective as it is not separated. Remark 2. Often (including Hartshorne), an “abstract variety” is taken to be separated and irreducible. Definition 2. Let f : X → Y be a morphism. Then Γf , called the graph of f , is the image of id × f in X ×Y. Note that Γf is a subvariety isomorphic to X, and Γid is the diagonal. Furthermore, Γf is always locally closed. If Y is separated, then Γ is a closed subvariety. Corollary 2. If X is irreducible and Y is separated and f, g : X → Y agree on a nonempty open set, then f = g. Proof. Suppose f, g agree on a nonempty open set U ⊆ X. Then Γf |U = Γg |U , and taking closures gives that Γf = Γf |U = Γg |U = Γg . Therefore, f = g. Corollary 3. Suppose X is irreducible, Y is separated, U is a nonempty open subset of X, and f : U → Y is a morphism. Then there is a maximal open subset V of X to which f extends. 2