Lecture 4: Grassmannians, Finite and Affine Morphisms

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1 Algebraic Geometry I Lecture 4 Lecture 4: Grassmaias, Fiite ad Affie Morphisms Remarks o last time 1. Last time, we proved the Noether ormalizatio lemma: If A is a fiitely geerated k-algebra, the, A cotais B = k[x 1,..., x ] (free subrig) such that A is a fiitely geerated B-module. Questio: Whe is A a fiitely geerated B-module? Aswer: If ad oly if A is a Cohe-Macauley rig. I particular, this does t deped o the choice of B (which is very ot uique...) 2. A remark o the homework problem (Problem 3(e) of Problem Set 2): The aswer to the optioal problem: P 2 (F ) = ( q 2 q ) + q. This is a quadric i P 2+1 (F q ). The middle term q also comes up elsewhere ad this geeralizes to the Weil cojectures. Also, the same problem ca be used to compute H (Q C ) (classical topology). This has the same cohomology as projective space for the middle degree. H is 1-dimesioal i degree 2, 4,..., 4 except for H 2, which is 2-dimesioal. The fact that the cohomology H is the same as for CP except for the middle degree geeralizes to the Lefschetz Hyperplae Theorem, which will be covered i O the isomorphism X = P 1 for irreducible degree 2 curves X P 2 : The degree 2 curve C = (XY Z 2 ) i P 2 from last lecture ca be covered by two affie ope pieces: Y Z (a) X = 0: a =, b =, (a = b 2 ) = A 1 = U 1 X X X Z (b) Y = 0: a =, b =, (a = b 2 ) = A 1 = U 2 Y Y Note that U 1 U 2 = A 1 \ {0}. By chagig coordiates, we ca take the degree 2 curve i P 2 to be X 2 + Y 2 = Z 2. Coect poits i a quadric to a fixed poit. I practice, we ca work with the poit (1 : 0 : 1). We idetify the set of all a 1 lies through a give poit with P 1. Takig this to affie coordiates, we sed (a, b). Writig b a = tb + 1, we express a ad b via t. The, we get a bijectio P 1 k X. This map seds poits with ratioal coordiates to poits with ratioal coordiates. Oe applicatio is the classificatio of Pythagorea triples. (Exercise: Work out the details.) Noetheria topological spaces ad irreducible compoets Propositio 1. A Noetheria topological space X is a fiite uio of its compoets (i.e. maximal irreducible subsets). Remark 1. Here, we ca see that the coditio that X is Noetheria ca be a aalogue of compactess. Lemma 1. A Noetheria topological space X is a fiite uio of closed irreducible subsets. Proof. We are doe if X is irreducible. Suppose that X is ot such a fiite uio. Write X = X 1 X 2, where X 1 ad X 2 are proper closed subsets of X. If the claim is false, the oe of either X 1 or X 2 is ot a uio of fiitely may irreducibles. Cotiuig this process, we get a sequece of closed subsets X X 1 X 2, which cotradicts the assumptio that X is Noetheria. Now we begi the proof of the propositio. 1

2 Proof. Write X = X i, where the X i are closed irreducible subsets of X. Without loss of geerality, we i=1 ca assume that oe of the X i are a subset of aother. The, X i is ot a subset of Xj (follows from irreducibility). Otherwise, we would have that X i is a uio of proper closed subsets Xj X i. Sice every irreducible closed subset Z X is a subset of X i for some i, the X i are exactly the compoets (i.e. maximal irreducible closed subsets) of X. Remark 2. A lot of thigs are ot Noetheria i the classical topology (e.g. R ). Corollary 1. A radical ideal i a fiitely geerated rig without ilpotets A is a fiite itersectio of prime ideals. Remark 3. This gives us a correspodece prime ideals of A irreducible subsets of Spec A. Proof. Let I be a radical ideal of A. The, I = I for some closed subset Z Spec A. Sice Z is Noetheria, Z = Z Z i, where the Z i are irreducible compoets of Z. The, I = I Zi. Note that I Zi is prime sice i=1 i=1 Z i is irreducible. Thus, I is a fiite itersectio of prime ideals. Claim: Spec A is irreducible if ad oly if A has o zerodivisors. Corollary 2. A closed subset Z Spec A is irreducible if ad oly if I Z is prime. Now we begi the proof of the claim. Proof. Let f ad g be ozero elemets of A Fu k (Spec A), where Fu k (Spec A) is the set of k-valued fuctios o Spec A. Suppose that Spec A is irreducible. If fg = 0, the Z f Z g = Spec A, where Z f are the zeros of f ad Z g are the zeros of g. If Z f, Z g Spec A, the Spec A is reducible. Thus, we must either have f = 0 or g = 0 ad A has o zerodivisors. Coversely, suppose that Spec A is ot irreducible. Let X = Spec A. The, we ca write X = Z 1 Z 2, where Z 1, Z 2 X are proper closed subsets of X. Sice proper closed subsets correspod to ozero ideals, we ca pick ozero f I Z1 ad ozero g I Z2. The, fg = 0 ad f ad g are zerodivisors of A. A example of a projective variety (Grassmaias) Last time, we started to discuss some properties of projective varieties ad looked at liear subvarieties of P. Here is aother example of a projective variety. Example 1. The Grassmaia Gr(k, ) is the set of liear subspaces of dimesio k i the -dimesioal vector space K := V. For example, Gr(1, ) = P 1. Here, we have the usual topology ad regular fuctios o P 1. I geeral, the topology ad regular fuctios are characterized as follows: Let W be a k-dimesioal subspace of V with complemet U (i.e. V = W U). If T Gr(k, ) is trasversal to U (i.e. T U = {0}), the T is the graph of a uique liear map W U. I other words, we have j=i {T Gr(k, ) : T U = {0}} = Hom k (W, U) = Mat k, k (K) = A k( k), where Mat k, k (K) is the set of k ( k) matrices with etries i K. 2

3 We require that this subset is ope ad that the isomorphism with A k( k) is a isomorphism of varieties. Notatio: PV := P is the projectivizatio of V = k (choose a basis for this). Theorem 1.1. This defies a projective algebraic variety. The embeddig of Gr(k, ) ito projective space is defied by W the lie W V. k k ( ) k Claim: This map realizes Gr(k, ) as a closed subvariety i P V = P( ) k 1. Example 2. Cosider the case = 4 ad k = 2. These are lies i P 3. 2 There is a lemma from liear algebra which gives a basic classificatio of elemets of V. 2 4 Lemma 2. Take ω V. If ω = v 1 v 2, the ω ω V = 0. If dim V = 4, the the coverse holds. 2 Proof. A elemet ω of V ca be thought of as a biliear skew form (2-form) of the 4-dimesioal vector space V. Note that ker ω is of eve dimesio. If dim ker ω = 0, the ω = v 1 v 2 v 3 v 4 for some basis v 1, v 2, v 3, v 4 of V. If dim ker ω = 2 (pullback from 3-dimesioal quotiet), the ω = v 1 v 2 for some v 1, v 2. Fially, ω = 0 if dim ker ω = 4, the the form ω = 0. Thus, Gr(2, 4) is isomorphic to a quadric i P 5 ad Gr(2, 4) = Q(P 5 ), where Q is defied by ω ω = 0. (Exercise: Show this is a isomorphism of varieties.) Usig some liear algebra, we ca show that the quadratic form is ot degeerate. For more details o work above ad o Grassmaias i geeral: See Chapter 6 of Algebraic Geometry (1992) by Joe Harris or p (i 3rd editio) i Sectio ( Closed Subsets of Projective Space ) of Basic Algebraic Geometry 1 by Igor Shafarevich. Fiite ad affie morphisms Defiitio 1. A morphism of algebraic varieties f : X Y is called affie if Y has a ope cover Y = U i where the U i are affie ope pieces such that the f 1 (U i ) X are affie. The affie pieces allow us to use commutative algebra. Note that we have a equivalece of categories {Affie varieties} = {Fiitely geerated k-algebras with o ilpotets}, where the secod category is the opposite category of the first oe. Defiitio 2. The morphism f is fiite if there is a affie ope cover Y = U i such that f 1 (U i ) = Spec A ad U i = Spec B with A a fiitely geerated B-module (see Noether ormalizatio theorem/noether s lemma). This reduces everythig to commutative algebra locally o a lie. Lemma 3. A fiite map satisfies the followig properties: 1. It is closed: f(z) Y is closed for every closed Z X. 2. It has fiite fibers. Corollary 3. If B A ad A is fiitely geerated over B as a B-module ( A is fiite over B ), the Spec A Spec B has fiite oempty fibers. 3

4 Proof. We oly eed to check that the map Spec A Spec B is oto. The image is ot cotaied i Z I for all ozero I B sice B A. Otherwise, we would have a ideal of B which kills A. Sice a fiite map is closed, we have that the map is surjective. Now we begi the proof of the lemma (use similar ideas as last time) (compare with Lemma o p. 19 of Kempf). Proof. Let f : X Y be a fiite map. We ca assume X ad Y are affie (statemet local o lie). Sice the compositio of two fiite maps is fiite, we ca also assume that Z = X. Write X = Spec A ad Y = Spec B ad let I = A B (A). This is a radical ideal sice A has o ilpotets. Sice I is a radical ideal, it correspods to the closed subset Z I of Spec B. The, we have the surjectio X Z I ad f(x) Z I. For x Z I, we have that A/m x A = 0 by Nakayama s lemma. Otherwise, there exists r 1 (mod m x ) such that ra = 0. However, this is ot possible sice r 1 (mod m x ) r / I. It follows from Hilbert s Nullstellesatz that Z I f(x). Sice A is a fiite B-module, A/m x A is a fiite dimesioal ozero k- algebra. This meas that there exists a maximal ideal m x such that Spec A/m x A = Hom(A/m x A, k) is a fiite oempty set (oempty sice quotiet rig ozero). Thus, f has fiite oempty fibers. Example 3. (Examples of affie morphisms) 1. Let Z X be a closed subvariety. The, the map i : Z X is affie ad fiite sice Spec A/I is a closed subset of Spec A (this is a local questio). Ay affie ope coverig of X works. 2. Let Y be ay algebraic variety ad X = Y \Z f, where f k[x]. Cosider the ope embeddig X Y. This map is affie, but usually ot fiite. Locally, it looks like Spec A (f) = A[t]/(1 tf) Spec A. Example 4. The morphism A 2 \ {0} A 2 is ot affie. This is similar to a exercise i the homework (Problem 3 of Problem Set 1). It actually follows from this ad the exactess of localizatio. Let U A 2 be a ope eighborhood of 0 such that U = A 2 \ Z f for some f. Sice k[u] = k[u \ {0}], U \ {0} is ot affie. We also have a short exact sequece 0 k[u \ {0}] k[u 1 ] k[u 2 ] k[u 1 U 2 ], where U = U 1 U 2 (U 1 = (X = 0), U 2 = (Y = 0)). The sequece above is exact because it is obtaied from the correspodig sequece i A 2 by localizatio, which is a exact fuctor. Thus, there is o affie eighborhood of 0 whose complemet is affie. Preview of ext lecture Lemma 4. Let Z 1 Z 2 be irreducible closed subsets of a algebraic variety X. If f : X Y is a fiite morphism, the f(z 1 ) f(z 2 ). Note that f(z 1 ) ad f(z 2 ) are closed by the previous lemma. We also have that the image of a irreducible set is irreducible. This lemma shows that the images are actually distict. We will check this result (see Lemma o p. 19 of Kempf) i the ext lecture. Defiitio 3. The dimesio of a Noetheria topological space is the maximal umber such that there exists a chai X Z Z 1 Z 2 Z 0 of irreducible subsets i X. For example, the dimesio of a poit is equal to 0. Remark 4. The dimesio may ot ecessarily be fiite sice the Noetheria coditio is oly for a give chai. Here are some facts about the dimesio of a Noetheria topological space: dim A = If X = U i, the dim X = max dim U i. i i=1 If f : X Y is a fiite ad surjective morphism, the dim X = dim Y. 4

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