Lecture 4: Grassmannians, Finite and Affine Morphisms

Lecture 4 18.725 Algebraic Geometry I Lecture 4: Grassmannians, Finite and Affine Morphisms Remarks on last time 1. Last time, we proved the Noether normalization lemma: If A is a finitely generated k-algebra, then, A contains B ∼ = k[x1 , . . . , xn ] (free subring) such that A is a finitely generated B-module. Question: When is A a finitely generated B-module? Answer: If and only if A is a Cohen-Macauley ring. In particular, this doesn’t depend on the choice of B (which is very not unique...) 2. A remark on the homework problem (Problem 3(e) of Problem Set 2): The answer to the optional problem: |P2n (Fq )| = (1 + . . . + q 2n ) + q n . This is a quadric in P2n+1 (Fq ). The “middle” term q n also comes up elsewhere and this generalizes to the Weil conjectures. Also, the same problem can be used to compute H ∗ (QC ) (classical topology). This has the same cohomology as projective space for the middle degree. H ∗ is 1-dimensional in degree 2, 4, . . . , 4n except for H 2n , which is 2-dimensional. The fact that the cohomology H ∗ is the same as for CPn except for the middle degree generalizes to the Lefschetz Hyperplane Theorem, which will be covered in 18.726. 3. On the isomorphism X ∼ = P1 for irreducible degree 2 curves X ⊂ P2 : The degree 2 curve C = (XY − Z 2 ) in P2 from last lecture can be covered by two affine open pieces: Y Z , b = , (a = b2 ) ∼ = A1 = U1 X X X Z (b) Y 6= 0: a0 = , b0 = , (a0 = b02 ) ∼ = A1 = U2 Y Y (a) X 6= 0: a = Note that U1 ∩ U2 ∼ = A1 \ {0}. By changing coordinates, we can take the degree 2 curve in P2 to be X 2 + Y 2 = Z 2 . Connect points in a quadric to a fixed point. In practice, we can work with the point (1 : 0 : 1). We identify the set of all a−1 lines through a given point with P1 . Taking this to affine coordinates, we send (a, b) 7→ . Writing b 1 a = tb + 1, we express a and b via t. Then, we get a bijection Pk ←→ X. This map sends points with rational coordinates to points with rational coordinates. One application is the classification of Pythagorean triples. (Exercise: Work out the details.) Noetherian topological spaces and irreducible components Proposition 1. A Noetherian topological space X is a finite union of its components (i.e. maximal irreducible subsets). Remark 1. Here, we can see that the condition that X is Noetherian can be an analogue of compactness. Lemma 1. A Noetherian topological space X is a finite union of closed irreducible subsets. Proof. We are done if X is irreducible. Suppose that X is not such a finite union. Write X = X1 ∪X2 , where X1 and X2 are proper closed subsets of X. If the claim is false, then one of either X1 or X2 is not a union of finitely many irreducibles. Continuing this process, we get a sequence of closed subsets X ) X1 ) X2 ) · · · , which contradicts the assumption that X is Noetherian. Now we begin the proof of the proposition. 1 Lecture  18.725 Algebraic Geometry I Proof. Write X = n [ Xi , where the Xi are closed irreducible subsets of X. Without loss of generality, we [ can assume that none of the Xi are a subset of another. Then, Xi is not a subset of Xj (follows from i=1 j 6=i irreducibility). Otherwise, we would have that Xi is a union of proper closed subsets Xj ∩ Xi . Since every irreducible closed subset Z ⊂ X is a subset of Xi for some i, the Xi are exactly the components (i.e. maximal irreducible closed subsets) of X. Remark 2. A lot of things are not Noetherian in the classical topology (e.g. Rn ). Corollary 1. A radical ideal in a finitely generated ring without nilpotents A is a finite intersection of prime ideals. Remark 3. This gives us a correspondence prime ideals of A ←→ irreducible subsets of Spec A. Proof. Let I be a radical ideal of A. Then, I = IZ for some closed subset Z ⊂ Spec A. Since Z is Noetherian, n n \ [ IZi . Note that IZi is prime since Zi , where the Zi are irreducible components of Z. Then, I = Z= i=1 i=1 Zi is irreducible. Thus, I is a finite intersection of prime ideals. Claim: Spec A is irreducible if and only if A has no zerodivisors. Corollary 2. A closed subset Z ⊂ Spec A is irreducible if and only if IZ is prime. Now we begin the proof of the claim. Proof. Let f and g be nonzero elements of A ⊂ Funk (Spec A), where Funk (Spec A) is the set of k-valued functions on Spec A. Suppose that Spec A is irreducible. If f g = 0, then Zf ∪ Zg = Spec A, where Zf are the zeros of f and Zg are the zeros of g. If Zf , Zg ( Spec A, then Spec A is reducible. Thus, we must either have f = 0 or g = 0 and A has no zerodivisors. Conversely, suppose that Spec A is not irreducible. Let X = Spec A. Then, we can write X = Z1 ∪ Z2 , where Z1 , Z2 ( X are proper closed subsets of X. Since proper closed subsets correspond to nonzero ideals, we can pick nonzero f ∈ IZ1 and nonzero g ∈ IZ2 . Then, f g = 0 and f and g are zerodivisors of A. An example of a projective variety (Grassmannians) Last time, we started to discuss some properties of projective varieties and looked at linear subvarieties of Pn . Here is another example of a projective variety. Example 1. The Grassmannian Gr(k, n) is the set of linear subspaces of dimension k in the n-dimensional vector space K n := V . For example, Gr(1, n) = Pn−1 . Here, we have the “usual” topology and regular functions on Pn−1 . In general, the topology and regular functions are characterized as follows: Let W be a k-dimensional subspace of V with complement U (i.e. V = W ⊕U ). If T ∈ Gr(k, n) is transversal to U (i.e. T ∩ U = {0}), then T is the graph of a unique linear map W −→ U . In other words, we have {T ∈ Gr(k, n) : T ∩ U = {0}} = Homk (W, U ) ∼ = M atk,n−k (K) ∼ Ak(n−k) , = where M atk,n−k (K) is the set of k × (n − k) matrices with entries in K. 2 Lecture  18.725 Algebraic Geometry I We require that this subset is open and that the isomorphism with Ak(n−k) is an isomorphism of varieties. Notation: PV := Pn is the projectivization of V = k n (choose a basis for this). Theorem 1.1. This defines a projective algebraic variety. The embedding of Gr(k, n) into projective space k k ^ ^ is defined by W 7→ the line W ⊂ V. Claim: This map realizes Gr(k, n) as a closed subvariety in P k ^ ! V n = P(k )−1 . Example 2. Consider the case n = 4 and k = 2. These are lines in P3 . There is a lemma from linear algebra which gives a basic classification of elements of Lemma 2. Take ω ∈ 2 ^ V . If ω = v1 ∧ v2 , then ω ∧ ω ∈ 4 ^ 2 ^ V. V = 0. If dim V = 4, then the converse holds. 2 ^ Proof. An element ω of V can be thought of as a bilinear skew form (2-form) of the 4-dimensional vector space V ∗ . Note that ker ω is of even dimension. If dim ker ω = 0, then ω = v1 ∧ v2 ∧ v3 ∧ v4 for some basis hv1 , v2 , v3 , v4 i of V . If dim ker ω = 2 (pullback from 3-dimensional quotient), then ω = v1 ∧ v2 for some v1 , v2 . Finally, ω = 0 if dim ker ω = 4, then the form ω = 0. Thus, Gr(2, 4) is isomorphic to a quadric in P5 and Gr(2, 4) ∼ = Q(P5 ), where Q is defined by ω ∧ ω = 0. (Exercise: Show this is an isomorphism of varieties.) Using some linear algebra, we can show that the quadratic form is not degenerate. For more details on work above and on Grassmannians in general: See Chapter 6 of Algebraic Geometry (1992) by Joe Harris or p. 42 – 44 (in 3rd edition) in Section 1.4.1 (“Closed Subsets of Projective Space”) of Basic Algebraic Geometry 1 by Igor Shafarevich. Finite and affine morphisms Definition 1. A morphism of algebraic varieties f : X −→ Y is called affine if Y has an open cover [ Y = Ui where the Ui are affine open pieces such that the f −1 (Ui ) ⊂ X are affine. The affine pieces allow us to use commutative algebra. Note that we have an equivalence of categories {Affine varieties} ∼ = {Finitely generated k-algebras with no nilpotents}, where the second category is the opposite category of the first one. [ Definition 2. The morphism f is finite if there is an affine open cover Y = Ui such that f −1 (Ui ) = Spec A and Ui = Spec B with A a finitely generated B-module (see Noether normalization theorem/Noether’s lemma). This reduces everything to commutative algebra locally on a line. Lemma 3. A finite map satisfies the following properties: 1. It is closed: f (Z) ⊂ Y is closed for every closed Z ⊂ X. 2. It has finite fibers. Corollary 3. If B ⊂ A and A is finitely generated over B as a B-module (“A is finite over B”), then Spec A −→ Spec B has finite nonempty fibers. 3 Lecture  18.725 Algebraic Geometry I Proof. We only need to check that the map Spec A −→ Spec B is onto. The image is not contained in ZI for all nonzero I ⊂ B since B ⊂ A. Otherwise, we would have an ideal of B which kills A. Since a finite map is closed, we have that the map is surjective. Now we begin the proof of the lemma (use similar ideas as last time) (compare with Lemma 2.4.3 on p. 19 of Kempf). Proof. Let f : X −→ Y be a finite map. We can assume X and Y are affine (statement local on line). Since the composition of two finite maps is finite, we can also assume that Z = X. Write X = Spec A and Y = Spec B and let I = AnnB (A). This is a radical ideal since A has no nilpotents. Since I is a radical ideal, it corresponds to the closed subset ZI of Spec B. Then, we have the surjection X  ZI and f (X) ⊂ ZI . For x ∈ ZI , we have that A/mx A 6= 0 by Nakayama’s lemma. Otherwise, there exists r ≡ 1 (mod mx ) such that rA = 0. However, this is not possible since r ≡ 1 (mod mx ) ⇒ r ∈ / I. It follows from Hilbert’s Nullstellensatz that ZI ⊂ f (X). Since A is a finite B-module, A/mx A is a finite dimensional nonzero kalgebra. This means that there exists a maximal ideal mx such that Spec A/mx A = Hom(A/mx A, k) is a finite nonempty set (nonempty since quotient ring nonzero). Thus, f has finite nonempty fibers. Example 3. (Examples of affine morphisms) 1. Let Z ⊂ X be a closed subvariety. Then, the map i : Z ,→ X is affine and finite since Spec A/I is a closed subset of Spec A (this is a local question). Any affine open covering of X works. 2. Let Y be any algebraic variety and X = Y \Zf , where f ∈ k[X]. Consider the open embedding X ,→ Y . This map is affine, but usually not finite. Locally, it looks like Spec A(f ) = A[t]/(1 − tf ) −→ Spec A. Example 4. The morphism A2 \ {0} −→ A2 is not affine. This is similar to an exercise in the homework (Problem 3 of Problem Set 1). It actually follows from this and the exactness of localization. Let U ⊂ A2 be an open neighborhood of 0 such that U = A2 \ Zf for some f . Since k[U ] = k[U \ {0}], U \ {0} is not affine. We also have a short exact sequence 0 −→ k[U \ {0}] −→ k[U1 ] ⊕ k[U2 ] −→ k[U1 ∩ U2 ], where U = U1 ∪ U2 (U1 = (X 6= 0), U2 = (Y 6= 0)). The sequence above is exact because it is obtained from the corresponding sequence in A2 by localization, which is an exact functor. Thus, there is no affine neighborhood of 0 whose complement is affine. Preview of next lecture Lemma 4. Let Z1 ( Z2 be irreducible closed subsets of an algebraic variety X. If f : X −→ Y is a finite morphism, then f (Z1 ) ( f (Z2 ). Note that f (Z1 ) and f (Z2 ) are closed by the previous lemma. We also have that the image of an irreducible set is irreducible. This lemma shows that the images are actually distinct. We will check this result (see Lemma 2.4.4 on p. 19 of Kempf) in the next lecture. Definition 3. The dimension of a Noetherian topological space is the maximal number such that there exists a chain X ⊃ Zn ) Zn−1 ) Zn−2 ) · · · ) Z0 of irreducible subsets in X. For example, the dimension of a point is equal to 0. Remark 4. The dimension may not necessarily be finite since the Noetherian condition is only for a given chain. Here are some facts about the dimension of a Noetherian topological space: • dim An = n n [ • If X = Ui , then dim X = max dim Ui . i=1 i • If f : X −→ Y is a finite and surjective morphism, then dim X = dim Y . 4