Lecture Notes for the Algebraic Geometry course held by Rahul Pandharipande

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1 Lecture Notes for the Algebraic Geometry course held by Rahul Pandharipande Endrit Fejzullahu, Nikolas Kuhn, Vlad Margarint, Nicolas Müller, Samuel Stark, Lazar Todorovic July 28, 2014 Contents 0 References 1 1 Affine varieties 1 2 Morphisms of affine varieties 2 3 Projective varieties and morphisms Morphisms of affine algebraic varieties Projective varieties and morphisms II 8 5 Veronese embedding Linear maps and Linear Hypersurfaces in P n Quadratic hypersurfaces Elliptic functions and cubic curves 12 7 Intersections of lines with curves 14 8 Products of varieties and the Segre embedding Four lines in P Intersections of quadrics The Grassmannian and the incidence correspondence The incidence correspondence Irreducibility Images of quasi-projective varieties under algebraic maps Varieties defined by polynomials of equal degrees Images of projective varieties under algebraic maps 21 I

2 15 Bezout s Theorem The resultant Pythagorean triples The Riemann-Hurwitz formula Points in projective space Rational functions Tangent Spaces I Tangent Space II Blow-up Dimension I Dimension II Sheaves Schemes I 33 0 References Good references for the topics here: 1. Harris, Algebraic Geometry, Springer. 2. Hartshorne, Algebraic Geometry, Springer. Chapter I, but also beginning of Chapter II for schemes. 3. Shafarevich, Basic algebraic geometry I, II. 4. Atiyah, Macdonald Commutative Algebra (for basic commutative algebra). 1 Affine varieties In this course we mainly consider algebraic varieties and schemes. It is worth noting that several definitions related to algebraic varieties are formally similar to those involving C - manifolds. (In algebraic geometry the local analysis of algebraic varieties is by commutative algebra, whereas in differential geometry the local analysis of C -manifolds is by calculus.) In the following we take the complex field C to be the underlying field (one could also consider any algebraically closed field instead; the discussion would be identical). By affine n-space over C we mean C n. For a subset X of C[x 1,..., x n ] we denote by V (X) := {x C n f(x) = 0 for all f X} the common zero locus of the elements of X. Notice that V (X) = V ( X ). A subset of C n of the form V (X) for some subset X of C[x 1,..., x n ] is said to be an affine algebraic variety. By Hilbert s basis theorem C[x 1,..., x n ] is Noetherian, so that every affine algebraic variety is in fact the common zero locus of finitely many polynomials. For a subset S of C n the ideal of S is the set of polynomials vanishing on S, I(S) := {f C[x 1,..., x n ] f(x) = 0 for all x S}. 1

3 Both I and V are inclusion-reversing, and are connected by Hilbert s Nullstellensatz: for every ideal I of C[x 1,..., x n ] we have I(V (I)) = I. Hence I and V induce a bijective inclusionreversing correspondence between affine algebraic varieties and radical ideals of C[x 1,..., x n ]. The Zariski topology on C n is defined by letting the closed subsets of C n be the affine algebraic varieties. The Zariski topology on a variety V in C n is the induced (subspace) topology; a more intrinsic definition relies on the following notion: a subvariety W of V is a variety W in C n such that W V. Then a subset U of V is open in the Zariski topology of V if and only if V \ U is a subvariety of V. Let us now consider the sets {U f } f C[x1,...,x n] defined by U f := {x V f(x) 0}. Remark 1.1 {U f } f C[x1,...,x n] is a basis of the Zariski topology: let U V be an open subset. By definition, U c = V (I) for some ideal I of C[x 1,..., x n ]. Since C[x 1,..., x n ] is noetherian, I is finitely generated, i.e. I = f 1,..., f k for some polynomials. Therefore we have U = V (I) c = V ( f 1,..., f k ) c = ( k i=1v ( f i )) c = k i=1u fi, so every open subset is a union of just finitely many basic open subsets. It follows from this also that C n is quasi-compact. Example 1.2 As every nonzero polynomial f C[x] has at most finitely many zeros, the Zariski topology on the affine line C 1 is exactly the cofinite topology. Notice that in this topology every injection f : C C is continuous, in contrast to the standard (euclidean) topology. 2 Morphisms of affine varieties Let V C n an affine algebraic variety, U V an open subset. Definition 2.1 f : U C is an algebraic (regular) function if for each p C there exist an open subset W p U containing p and g p, h p C[x 1,..., x n ] such that 1. 0 g p (W p ). 2. f Wp = hp g p. Theorem 2.2 A map f : C n C is algebraic if and only if f is a polynomial function on C n. Proof. The if part is trivial. For the only if part, let f : C n C be an algebraic map. Then for every p C there is an open neighborhood W p of p in C n and polynomial functions h p and g p such that f Wp = hp g p and g p never vanishes on W p. Inside W p we can find a basic open set p U rp W p, where U rp = {ρ C n r p (ρ) 0}. It follows that for all points p C we have a r p C[x 1,..., x n ] such that f Urp = hp g p and q C n : g p (q) = 0 r p (q) = 0. With Hilbert s Nullstellensatz it follows that r p (g p ) and therefore rp k = α p g p for some α p C[x 1,..., x n ]. On U rp we have that r p, g p, α p are never 0. On U rp we have f Urp = h pα p g p α p = h pα p r k p 2

4 Define ĥp = h p α p and ˆr p = r k p. Then U rp = U r k p, and for every p C n there is r p with r p (p) 0 and f Urp = hp r p (removing the hats again). It follows that finitely many U rp suffice to cover C n. Take p 1,..., p m such that U rp1,..., U rpm cover C m and write r i instead of r pi. We have (because of the cover) (r 1,..., r m ) = (1) and therefore m s 1,..., s m C[x 1,..., x n ] : s i r i = 1 We have f Uri i=1 = h i r i and U ri U rj = U ri r j. On U ri r j we have h i r i = h j r j, h i r j = h j r i, h i r j h j r i = 0 on U ri r j r i }{{} 0 r j }{{} 0 We have to prove f C[x 1,..., x n ]. Let (h i r j h j r i ) = 0 everywhere on C n ( ) h i r j h j r i = 0 C[x 1,..., x n ] F = m h k s k k=1 F is a polynomial. We want to show F = f everywhere. m m r i F = r i h k s k = r k h i s k = h i because of ( ). And on U ri k=1 we have k=1 f Uri = h i r i, F Uri = h i r i m k=1 s k r k = h i, If V C n is an affine algebraic variety and f : V C is an algebraic function, f is restriction of a polynomial (see exercises). If f 1, f 2 : V C are algebraic and the same function f 1 f 2 I(V ). We denote by Γ(V ) the ring of algebraic functions V C. Proposition 2.3 Γ(V ) = C[x 1,..., x n ] I(V ) Proof. By f C[x 1,..., x n ] f V : V C we have C[x 1,..., x n ] Γ(V ) I(V ) So, to prove the proposition, we need to show that this is surjective. Go through the proof of Theorem 2.2 step by step. It is the same, except for one step. Again, we have U ri U rj = U ri r j. On U ri r j we have h i r i = h j r j. Let ĥi = h i r i, ˆr i = ri 2. And we have h i r j r i h j = 0 on U ri r j. And (see exercise 1.3 on the exercise sheets) r i r j (h i r j r i h j ) = 0 everywhere on V 3

5 Example 2.4 V = (x 2 y 3 ) C 2. We look at f : V C { x (x, y) (0, 0) y (x, y) 0 (x, y) = (0, 0) Let U = V \ {(0, 0)}. f U = x. y is never 0 in U. y Claim: f is not algebraic on V. Proof. Suppose it were. Then there is a basic open set U r, r C[x, y], with (0, 0) U r, and there are h, g C[x, y] such that (x, y) U r \ {(0, 0)} : x y = h g and g is never 0 on U r. We have gx hy = 0 on U r and r(gx hy) = 0 on V. We have x 2 y 3 r(gx hy) because (x 2 y 3 ) is a radical ideal. x 2 y 3 can t divide r since r(0, 0) 0. Therefore x 2 y 3 gx hy Because g(0, 0) 0, g has a nonzero constant term. Use the ring homomorphism Then x 2 y 3 0, but C[x, y] C[t] x t 3 y t 2 g(t 3, t 2 )t 3 h(t 3, t 2 )t 2 0 because the t 3 term (which exists, because g has a constant term) cannot cancel. Therefore x 2 y 3 gx hy and f is not algebraic. Let V, W C n affine algebraic varieties. Let U V C n, Û W Cm be Zariski-open subsets of the affine algebraic varieties. Definition 2.5 Φ : U Û is algebraic (regular) if 1. Φ is continuous (with regard to the Z-top.) 2. S open Û and f : S C algebraic the composition is algebraic. Φ 1 (S) Φ S f C Let Φ be an algebraic map. Then there is a correspondence C[y 1,..., y n ] I(W ) = Γ(W ) Γ(V ) = C[x 1,..., x n ] I(V ) f f Φ 4

6 3 Projective varieties and morphisms Definition 3.1 We define projective n-space over C as P n := (C n+1 \ {0})/, where the equivalence relation is defined by x y iff there is λ C with y = λx. Remark P n is not a vector space; it is the set of one-dimensional subspace of C n+1 (lines through the origin). 2. If x P n is represented by (x 0,..., x n ) C n+1, then (x 0,..., x n ) is a set of homogenous coordinates for x and denoted [x 0,..., x n ] (unique only up to multiplication by a scalar), for any λ C we have [λx 0,..., λx n ] = [x 0,..., x n ]. It is evident that polynomials f C[x 0,..., x n ] do not induce well-defined functions on complex projective n-space; there are however certain polynomials which allow meaningful discussion of their zero sets, also called projective algebraic varieties. In addition, we can define the Zariski topology for projective varieties. The closed sets of this topology are defined to be the projective algebraic varieties. First, we need the notion of an homogeneous ideal of the ring C[z 0,..., z n ]. The motivation for this lies in the fact that our objects can only be described by homogeneous polynomials, and homogeneous ideals allow us to decompose any element into its homogeneous parts: Definition 3.3 An ideal I C[z 0,..., z n ] is homogeneous if for any f I all the homogeneous components of f are again elements of I. Example 3.4 Consider f = f 0 + f where f i are the homogeneous components of degree i. Take for example: f = z 0 + 3z 2 1. In particular, f has a set of homogeneous generators. We have the following results concerning the homogeneous ideal: Lemma 3.5 Let I C[z 0,..., z n ] be homogeneous, then I has finitely many homogeneous generators. Proof. Let us consider I C[z 0, z 1, z 2,..., z n ]. Using Hilbert s basis theorem, we have that I is finitely generated in C[z 0, z 1, z 2,...z n ] (say I = (g 1, g 2, g 3,..., g n )). Taking the homogeneous components of the generators we obtain the conclusion (I = (g 00, g 01,..., g 10, g 11,... )). Lemma 3.6 Consider an ideal I C[z 0,..., z n ], which is generated by homogeneous elements. Then I is a homogeneous ideal. Proof. Let (F 1, F 2, F 3,... ) be the homogeneous generators of I. It follows that f I can be expressed f = n i=1 g if i, where F i are homogeneous polynomials of degree d i and g i are some polynomials (not necessarily homogeneous). The degree d part of f is it equal to i G if i where G i represents the degree d d i part of g i. Lemma 3.7 If I is a homogeneous ideal I C[z 0,..., z n ] then I is homogeneous. 5

7 Proof. Consider f I, then k N such that f k I. It follows that f = f l + f d (where f d is the maximal degree term of the polynomial, so f k = (f l + f d ) k can be written as f k = f k d + other terms. Then, f k f k d I and we can conclude (by induction). Given the previous results we can consider W P n be an projective algebraic variety: the zero-locus of the homogeneous polynomials F 1, F 2, F 3,..., F l. Then it follows that the ideal I = (F 1, F 2, F 3,..., F l ) is a homogeneous ideal. Theorem 3.8 (Projective Nullstellensatz) Let I be an ideal in C[z 0,..., z n ] such that its zero locus V (I) in P n is non-empty. Then I(V (I)) = I. Proof. As in the affine Nullstellensatz only I(V (I)) I requires proof; thus let f I(V (I)). If V (I) P n is defined by the homogeneous polynomials F 1, F 2,..., F k, then these polynomials define an associated affine variety ˆV (I) in C n+1. (As it is defined by homogenous polynomials, ˆV (I) is a cone in the sense that x ˆV (I) implies λx ˆV (I) for all λ C.) It is clear that f I(V (I)) vanishes on ˆV (I) \ {0}, for if x ˆV (I) \ {0} then [x] V (I). Now if 0 ˆV (I) (which always holds, unless one of the F i is constant and non-zero) we use that there is [x] V (I), giving f(0) = 0. Thus f vanishes on the whole of ˆV (I); hence by the affine Nullstellensatz f I( ˆV (I)) = I. Remark 3.9 The condition V (I) is necessary, as can be seen by considering the maximal ideal I := (z 0,..., z n ) with ˆV (I) = {0} in C n+1 and hence V (I) = in P n. We have I(V (I)) = (1), which is distinct from I = I. We will now introduce the notion of algebraic functions and maps. Note that they are also known as regular functions and maps, respectively. Let W P n be a projective algebraic variety and U an open subset of W. Definition 3.10 A function f : U C is called an algebraic function if for every p U there is a neighborhood W of p in U and homogeneous polynomials of the same degree h, g C[z 0,..., z n ] with g 0 on W and f W = h/g. Secondly, algebraic maps are defined as follows: Definition Suppose U is an open subset of either an affine algebraic variety or a projective algebraic variety. In this case, we call U a quasi-projective variety. 2. Let U, W be quasi-projective varieties. A map f : U W is algebraic if: i) f is continuous w.r.t. the Zariski topology. ii) f respects precomposition with algebraic functions, i.e. Y W open, g : Y C algebraic : g f : f 1 (Y ) C is an algebraic function. One easily checks that any open or closed subset of a quasi-projective variety is again a quasi-projective variety. Remark 3.12 The reader should note the following basic facts about algebraic maps. The proofs are straightforward from the definitions and left as exercises. 6

8 1. If V is a quasi-projective variety, f 1,..., f n Γ(V ) elements of the ring of algebraic functions on V, then is an algebraic morphism. V C n (or more generally: any affine algebraic variety) ξ (f 1 (ξ),..., f n (ξ)) 2. If V P n is a quasi-projective variety, and if F 0,..., F m C[Z 0,..., Z n ] are homogeneous polynomials of the same degree, which moreover do not have a common zero on V, then defines an algebraic morphism. V P m ξ (F 0 (ξ),..., F m (ξ)) 3. If Φ : V W is an algebraic map between quasi-projective varieties, Ṽ V a subvariety, then the restriction Φ Ṽ : Ṽ W is algebraic. If W W is a subvariety that contains the image of Φ, then the map Φ : V W obtained by restricting the range of Φ is algebraic. 3.1 Morphisms of affine algebraic varieties The study of the morphisms between affine algebraic varieties can be done by understanding their ring of functions. Given two affine algebraic varieties V C n and W C m with their corresponding ring of functions Γ(V ) and Γ(W ) we have the following correspondence: If Φ : V W is an algebraic morphism, then there exists a homomorphism of C-algebras (i.e. takes constants into constants): Φ : Γ(W ) Γ(V ) γ γ Φ Conversely, if there is a homomorphism of the C-algebras f : Γ(W ) Γ(V ), then this gives a unique algebraic morphism Φ : V W such that Φ = f. Let Φ : V C m ξ (f(y 1 )(ξ),..., f(y m )(ξ)) where y 1,..., y m are the coordinate functions on W. We need to show that Im(Φ) W. Let g I(W ) C[Y 1,..., Y m ]. Then g Φ = g(f(y 1 ),..., f(y m )) }{{} = f(g(y 1,..., y m )) = f(0) = 0 }{{} f homomorphism =0 This implies Im(Φ) W. Φ is an algebraic morphism by remark (see also Lemma 3.6 in Hartshorne). Let γ Γ(W ). Then Φ (γ)(ξ) = γ(f(y 1 )(ξ),..., f(y m )(ξ)) }{{} = f(γ(y 1,..., y m ))(ξ) f homomorphism and we have Φ = f. In conclusion, Φ and Φ encode the same data. 7

9 Example 3.13 Let Φ : C C 2 t (t 2, t 3 ) Then Im(Φ) = V (x 3 y 2 ). This is an affine algebraic variety. Let Then Φ : C 2 C 2 (x, y) (x, xy) Im(Φ) C = {(0, ρ) φ 0} and Im(Φ) is not a quasi-projective variety. It follows that the image of an algebraic map need not be a quasi-projective variety. 4 Projective varieties and morphisms II Recall that we defined a quasi-projective variety as an open subset of either a projective or an affine variety. This makes sense, as we will see that any open subset of an affine variety is isomorphic to, i.e. can be viewed as an open subset of a projective variety. Let n 1. For 0 i n, we have basic open subsets U i := U Zi = { [ξ 0, ξ 1,..., ξ n ] P n ξ i 0} of projective space, and maps Φ i,ψ i defined by: and Φ i : C n U i (x 1,..., x n ) [x 1,..., x i 1, 1, x i+1,..., x n ] Ψ i : U i C n ( ξ0 [ξ 0, ξ 1,..., ξ n ],..., ξ i 1, ξ i+1,..., ξ ) n ξ i ξ i ξ i ξ i which is well defined on U i. Note that the U i cover all of projective space. Claim 4.1 Φ i and Ψ i are isomorphisms between varieties, which are inverse to each other. Proof. It is easy to check that the composition of Ψ i with Φ i in any order is the identity map. So one only needs to show that both are algebraic maps. For the sake simplicity, we consider only the case when i is zero and drop the index 0 when referring to Ψ 0, Φ To prove continuity, we verify that the preimages of closed sets remain closed under both maps. Let first W U 0 be a closed subset defined by homogeneous polynomials F 1,..., F k C[Z 0,..., Z n ]. Then Φ 1 (W ) is the common zero set of the set of polynomials (check this!), thus closed in C n. (F i (1, X 1,..., X n )) i=1,...,k C[X 0,..., X n ] Conversely, if V C n is defined by (f i ) i=1,...,k C[X 1,..., X n ], then we can construct homogeneous polynomials (F i ) i=1,...,k in n+1 variables, by F i [Z 0,..., Z n ] := f i ( Z 1 Z 0,..., Zn Z 0 )Z deg f 0. Check that these define the preimage Ψ 1 (V ) in U 0 as their set of common zeros. 8

10 2. Let f : U C be any function, where U is open in U 0. Let W := Φ 1 (U). We want to show that f Φ : W C is regular if and only if f is. Note that f Φ(x 1,..., x n ) = f([1, x 1,..., x n ]) and f([ξ 0,..., ξ n ]) = f Φ( ξ 1 ξ 0,..., ξn ξ 0 ). Assume f is regular. Let x = (x 1,..., x n ) W, ξ := Φ(x). By assumption, there is an open neighborhood Ũ of ξ, such that f Ũ = H G Ũ, where H and G C[Z 0,..., Z n ] are homogeneous polynomials of the same degree, and G doesn t vanish on Ũ. Then f Φ equals H(1,X 1,...,X n) G(1,X 1,...,X n) on the preimage of Ũ, which is an open neighborhood of x. Noting 1 that G(1, X 1,..., X n ) is never zero for any point in Φ (Ũ), this means that we have found a neighborhood of x where f Φ is equal to a rational function. This proves that Φ is algebraic. The other direction works similarly: Assume f Ψ 1 is algebraic. Let ξ U, x = Ψ(ξ). By assumption, there is an open neighborhood W of x in W on which f Ψ 1 equals a rational function h g, and g does not vanish on W. Now Ũ := Φ( W ) = Ψ 1 ( W ) is an open neighborhood of ξ in W, and we have by the above remark: f Ũ = h( Z 1 Z 0,..., Zn Z 0 ) g( Z 1 Z 0,..., Zn Z 0 ) = h( Z 1 Z 0,..., Zn Z 0 )Z0 D, g( Z 1 Z 0,..., Zn Z 0 )Z0 D which holds for any positive integer D. In particular if D max{deg h, deg g}, then both the denominator and the numerator are homogeneous polynomials of the same degree in Z 0, Z 1,..., Z n. As ξ was arbitrary, this proves that Ψ is algebraic. From what we have proven we directly obtain Lemma 4.2 Any projective variety can be covered by finitely many affine varieties. Using our newfound knowledge, we now want to compute the algebraic functions on CP n. Lemma 4.3 Any algebraic function f : CP n C is constant. Proof. Let f : CP n C be algebraic. For 1 i 1, the restriction of f to U i gives an algebraic function C n C (1) (x 1,..., x n ) f(x 1,..., x i 1, 1, x i+1,..., x n ), (2) but we already know that such a function must equal a polynomial P i C[X 1,..., X n ]. Let ( Q i (Z 0,..., Z n ) := Zi d Z0 P i,..., Z i 1, Z i+1,..., Z ) n, Z i Z i Z i Z i where d is a positive integer that is strictly bigger than the maximal degree of P i. Q i is homogeneous of degree d and vanishes outside U i. By construction, f Ui = Q i(z 0, Z 1,..., Z n ), Zi d 9

11 and in particular on the overlap regions: f U0 U i = Q 0(Z 0,..., Z n ) Z d 0 = Q i(z 0,..., Z n ). Zi d So the equation Z d i Q 0 (Z 0,..., Z n ) = Z d 0Q i (Z 0,..., Z n ) is fulfilled at every point in C n+1, where the zero th and i th coordinate are not equal to zero. However, at all other points, we see that both sides of the equations are zero. This means, the two polynomials take the same values on all of C n, which implies they are the same. By unique factorization we get, for 0 i n, that Zi d divides Q i, and both have degree d, which means the quotient Q i Zi d =: c i, has to be a constant, and by the above equality, c i = c 0 for any i. As the U i cover CP n, f is everywhere constant and equal to c 0. Corollary 4.4 CP n is not isomorphic to any affine algebraic variety. Proof. An isomorphism between varieties induces a isomorphism of C-algebras between their rings of functions. The only affine varieties whose ring of functions equals C are the points. But CP n is not isomorphic to a point. 5 Veronese embedding The discussion so far has been about quasi-projective varieties and their regular functions f : X C as well as their regular maps p : X Y. We are interested to know what the image of a regular map of a projective variety is. Suppose we have a map P n A, where A is an affine variety. What are the possibilities for this map? The claim is that this map must be constant. To see this, assume that for two points ξ 1 ξ 2 P n we have f(ξ 1 ) f(ξ 2 ). This means f(ξ 1 ) and f(ξ 2 ) differ in at least one coordinate x i. Define a map g : A C by g(x) = x i. Then the composition gf : P C consists of a regular non-constant map, which we have already proved does not exist. Now consider regular maps P n P m. They are given by [z 0,..., z n ] [F 0,..., F m ] where all F i are homogeneous polynomials in C[z 0,..., z n ] of the same degree and moreover have no common zeros. Definition 5.1 The Veronese embedding of degree d N is the map ν d : P n P (d+n n ) 1 [z 0,..., z n ] [(z i 0 0 z in n ) 0 ik d, n k=0 i k=d]. 10

12 Example 5.2 The map P 1 P 2 is the Veronese of degree 2 mapping from P 1. Example 5.3 The Veronese with n = 1, d = 3 is also called twisted cubic. [x, y] [x 2, xy, y 2 ] P 1 P 3 [x, y] [x 3, x 2 y, xy 2, y 3 ] Let X i0,...,x n be the variable in P (n+d n ) 1 that corresponds to the monomial z i 0 0 zn in in the Veronese map. Then, we have on the image of the Veronese map, for the multi-indices I = i 0,..., i n, J = j 0,..., j n, K = k 0,..., k n, L = l 0,..., l n such that I + J = L + K: X I X J X K X L = 0 (3) Proposition 5.4 The image of the Veronese is defined by the quadratic equations (3). Corollary 5.5 The image of each Veronese map is a projective algebraic variety. 5.1 Linear maps and Linear Hypersurfaces in P n Definition 5.6 A linear hypersurface in P n is defined by one equation: 0 = n a k z k. k=0 Remark 5.7 When we deal with one hypersurface, we can assume a 0 = 0, by a change of coordinates. Definition 5.8 The projective linear group is PGL n+1 (C) = GL n+1 (C)/C. Remark 5.9 PGL n+1 (C) is the group of automorphisms of P n. Because homogeneous coordinates are unique only up to a factor λ C, matrices defining a linear map P n P n also define the same map if they differ only by some factor λ C. 5.2 Quadratic hypersurfaces Definition 5.10 A quadratic hypersurface or quadric in projective space P n is the zero locus of some F C[z 0,..., z n ] of the form F = n n n a k zk 2 + a k,l z k z l. k=0 k=0 l=k+1 11

13 Remark 5.11 A quadric F = n a k zk 2 + k=0 n n k=0 l=k+1 2a k,l z k z l can also be written using a symmetric matrix M: F = ( z 0 a 0 a 0,1 a 0,n ) a 0,1 a 1 a 1,n z 0 z n z a 0,n a n 1,n a n n }{{} =:M Definition 5.12 The rank of a quadric is the rank of the symmetric matrix M associated with it. Proposition 5.13 A quadric is determined, up to PGL n+1 (C) by its rank. Proof. M be the symmetric matrix of a quadric. Then, there is a P PGL n+1 (C) such that 1... (P T ) 1 MP 1 1 = where the number of 1 s is equal to the rank of M. Let Q = {x x T Mx = 0} be the quadric associated with M. Then P Q = {P x x T Mx = 0} = {x x T (P T ) 1 MP 1 x = 0} = x x T x = 0 0 Something about the genus of cubic and quadratic surfaces is missing here. 6 Elliptic functions and cubic curves There is an intimate connection between cubic equations and elliptic functions (doubly periodic meromorphic functions), in particular the Weierstrass -function. Consider a meromorphic function f on C which is doubly-periodic with periods 1 and i; let Λ denote the lattice in C generated by 1 and i (which is, as a set, simply Z[i]), defining also Λ := Λ {0}. (One could consider instead of i any element of the upper half plane.) It follows that we also have 12

14 f(z) = f(z+λ) for all λ Λ. It is evident that f is determined by its values on the fundamental parallelogram Γ := [0, 1) [0, i). If f is an entire function, then f is bounded on the compact set Γ by continuity, hence on all of C so that f must be constant by Liouville s theorem. If f has only a simple pole ξ in the interior of Γ, then the residue theorem gives (taking Γ to be positively oriented) 1 f(z)dz = res ξ f. 2πi Γ The integral vanishes as opposite sides of Γ cancel one another by periodicity; hence res ξ f = 0 meaning that ξ is a removable singularity. Thus the argument above applies and shows f to be constant. Therefore, if f is to be nonconstant it must have at least two poles in Γ. We now define the elliptic function, which has double poles at the points of Λ. It is defined by the infinite series (z) := 1 z + ( 1 2 (z ω) 1 ), 2 ω 2 ω Λ which converges compactly for all z / Λ and hence defines a meromorphic function on C. As we may differentiate termwise, the derivative is (z) = 2 z 2 3 (z ω) = (z ω). 3 ω Λ ω Λ As the sum is invariant under z z + λ (λ Λ), we have (z) = (z + λ) for all λ Λ, hence is an elliptic function. From this it follows that (z) (z + λ) is constant and in fact vanishes identically as ( λ/2) (λ/2) = 0 (λ/2 / Λ is not a pole), since is even. Thus is indeed doubly-periodic with periods 1 and i. If f is an elliptic function with periods 1 and i (not vanishing on Γ and without poles on Γ), then the integral 1 f (z) 2πi Γ f(z) dz vanishes by an argument as in the first paragraph. By the argument principle this integral is equal to the number of zeros of f minus the number of poles of f (both counted with multiplicity). Thus has equally many poles as it has zeros. The Laurent expansion of about the origin is (z) = 1 z 2 + ω Λ 1 ω 2 ( k=0 ) 2 z k 1 w k ω 2 = 1 z 2 + vanishes for n odd. Thus this can also be written as n=1 c n z n where c n = (n + 1) ω Λ 1 ω n+2 (z) = 1 z + (2k 1)G 2 k z 2k 2 where G k := 1 ω, 2k ω Λ k=2 is the Eisenstein series of weight 2k (at i), and we find (z) 2 = 4 z 6 24G 2 z 2 80G = 4 z G 2 z G ; defining g 2 := 60G 2 and g 3 := 140G 3 we see that the differential equation 2 = 4 3 g 2 g 3 13

15 holds, since (z) 2 4 (z) G 2 (z) = 140G , the right hand side being an elliptic function without poles (and is hence constant). Thus the points of the form ( (z), (z)) satisfy the cubic equation y 2 = 4x 3 g 2 x g 3. In fact if we define φ : C P 2 by φ(z) = [1, (z), (z)] we get an induced bijection between the cubic curve defined by y 2 = 4x 3 g 2 x g 3 (embedded into P 2 ) and the commutative group C/Λ. The latter is homeomorphic to a torus, as can be seen by considering the construction of the torus from its fundamental polygon; notice that the geometric structure of a torus is intrinsically doubly periodic. The cubic curve is in fact nonsingular (the roots of the polynomial 4x 3 g 2 x g 3 are distinct are in fact given by (1/2), (i/2), (1/2 + i/2)), hence an elliptic curve and as such has a group law (turning the bijection into a group isomorphism). Ahlfors complex analysis, chap. 7, was recommended during the lecture. It was also suggested to read the entire book. 7 Intersections of lines with curves Definition 7.1 Let α = (α 0,..., α n ), β = (β 0,..., β n ) C n be linearly independent vectors. A set S P n of the form is called a line. S := {[α 0 s + β 0 t, α 1 s + β 1 t,..., α n s + β n t] [s, t] P 1 } Let F C[X, Y, Z] be homogeneous of degree d. We want to find the intersection of the zeros of F with the a line given by (α 0, β 0, γ 0 ), (α 1, β 1, γ 1 ). This is given by F (α 0 s + α 1 t, β 0 s + β 1 t, γ 0 s + γ 1 t) = 0 }{{} = ˆF (s,t) Either ˆF = 0 or ˆF = (λ 1 0s + λ 1 1t)(λ 2 0s + λ 2 1t) (λ d 0s + λ d 1t) 8 Products of varieties and the Segre embedding We would like to investigate the products of algebraic varieties. In the affine case, this is a very simple matter. For f 1,..., f r C[x 1,..., x n ], g 1,..., g s C[y 1,..., y m ], let A = V ( f 1,..., f r ) C n and B = V ( g 1,..., g s ) C m be affine varieties. Then A B is a subset of C n C m, which in turn is isomorphic to C n+m. We can now consider the f i and g i as elements of C[x 1,..., x n, y 1,..., y m ]. It is clear that A B = V ( f 1,..., f r, g 1,..., g s ). In the projective case, the above approach does not work, as P n P m is not isomorphic to P n+m. However, we can define an embedding from P n P m to a projective space. Definition 8.1 The map σ : P n P m P (n+1)(m+1) 1 [x 0, x 1,..., x n ] [y 0, y 1,..., y m ] [x 0 y 0, x 0 y 1,..., x i y j,... x n y m ], where 0 i n and 0 j m, is called the Segre embedding. We give P n P m the structure of a projective variety by identifying it with its image. 14

16 We will prove that our definitions make sense: Claim 8.2 σ is injective. Proof. Let c = [c 0,0, c 0,1,..., c i,j,... c n,m ] be an element of Im(σ). Let (a, b) P n P m such that σ(a, b) = c. Without loss of generality, a 0 = b 0 = c 0,0 = 1. This enforces b j = c 0,j for all 0 j m and a i = c i,0, which uniquely determines a and b. Therefore, the Segre embedding is bijective onto its image. Claim 8.3 The image of σ is a variety of P (n+1)(m+1) 1. Proof. We denote the complex polynomials in (n+1)(m+1) variables by C[z 0,0, z 0,1,..., z i,j,... z n,m ]. For i k and j l, let us define P i,j,k,l := z i,j z k,l z k,j z i,l and I, the ideal generated by the P i,j,k,l. It is clear that σ(p n P m ) V (I). For the other inclusion, take a c V (I). Without loss of generality, c 0,0 = 1. For any k, l 0, the polynomial P 0,0,k,l gives us c k,l = c k,0 c 0,l. By taking a 0 = b 0 = 1, a k = c k,0 and b l = c 0,l, we get (a, b) such that σ(a, b) = c. Remark By definition, the Segre embedding is algebraic. This, together with Claim 8.2, implies that the Segre embedding is an isomorphism onto its image. 2. It is also possible to define an algebraic structure on P n P m by using bihomogeneous polynomials. One can check that this is equivalent to our definition. 3. We can also define the product of two quasi-projective varieties X and Y. It is not very hard to check that σ(x Y ) again is a quasi-projective variety. 4. If X and Y are quasi-projective, it is useful to know that their product X Y is a categorical product, i.e. the projections on the first resp. second coordinate are algebraic maps, and if we have algebraic maps then the product map is also algebraic. f : Z X g : Z Y (f, g) : Z X Y z (f(z), g(z)) Example 8.5 Given four skew lines L 1, L 2, L 3, L 4 in P 3, how many lines intersect all of them? Consider the vector space V of homogeneous polynomials of degree two in the variables x, y, z, w. This is a 10 dimensional space and its projectivization parametrizes quadrics in P 3. Pick three generic points on each of the lines L 1, L 2, L 3. The condition for a quadric Q P(V ) to contain a point is a linear condition on V. As the lines are chosen general and there are three points on each line, there will be a 1-dimensional subspace of V of homogeneous polynomials vanishing at all the points and therefore a unique quadric Q that contains the chosen points (Check that the conditions are independent!). The restriction of the polynomial that defines Q, to the lines L i, i = 1, 2, 3 is a polynomial of degree 2 that vanishes at three distinct points, hence must be zero. This shows that Q contains the three lines L 1, L 2, L 3. As the lines chosen are generic, Q is a smooth quadric and after a change of coordinates we may assume it is given by V (xw yz), which is isomorphic to P 1 P 1 via the Segre map. The lines L 1, L 2, L 3 are then given P 1 P i for P i P 1, i = 1, 2, 3 some points (This follows as they are distinct and of degree 1). Any line that meets the three lines L 1, L 2, L 3 must meet the quadric Q in three distinct points and with the same argument as above is contained in Q. It must hence given by M R = R P 1 for R P 1 a point. 15

17 Consider now the fourth general line L 4. L 4 meets the quadric Q in precisely 2 points (R 1, S 1 ), (R 2, S 2 ) P 1 P 1. This shows that there are precisely two lines (namely M R1 and M R2 ) incident to all four lines L 1,..., L Four lines in P 3 Given four skew (i.e. pairwise non-intersecting) lines L 1,..., L 4 in P 3, how many lines intersect all of them? Lemma 8.6 Let L 1, L 2, L 3 be skew lines in P 3 and let K 1, K 2, K 3 be skew lines in P 3. Then, there is a M PGL 4 (C) such that ML i = K i for i = 1, 2, 3. Proof. The ring of polynomials of P 3 is C[X, Y, Z, W ]. Without loss of generality we may assume that K 1 = Z(X, Y ), K 2 = Z(Z, W ), K 3 = Z(X Z, Y W ). Because each line can be defined by two points that are on the line, we can find M PGL 4 (C) such that M L 1 = K 1, M L 2 = K 2, because M can be defined by mapping four points in general linear position to four points in general linear position. Now, consider M L 3. This cannot lie in X = 0 or W = 0, because then M L 3 and M L 2 would not be skew or M L 3 and M L 1 would not be skew. Therefore M L 3 intersects X = 0 and W = 0 in one point each. Let [0 : a : b : c] be the point where M L 3 intersects X = 0 and let [d : e : f : 0] be the point where M L 3 intersects W = 0. K 3 is defined by the points [0 : 1 : 1 : 1] and [1 : 1 : 1 : 0]. Now, we need some N PGL 4 (C) that leaves K 1, K 2 fixed and maps [0 : a : b : c] to [0 : 1 : 1 : 1] and [d : e : f : 0] to [1 : 1 : 1 : 0]. N can be written as a block matrix [ ] A B N = C D where A, B, C, D are 2 2 matrices. To fix K 1 we need that N[0 : 0 : z 2 : z 3 ] = [0 : 0 : z 2 : z 3]. This implies B = 0. Fixing K 3, similarly, implies C = 0. To map [0 : a : b : c] to [0 : 1 : 1 : 1] and [d : e : f : 0] to [1 : 1 : 1 : 0] we can set Set M = NM. N = 1/d /a /f /c. Lemma 8.7 Let L 1, L 2, L 3 be skew lines in P 3. Then, the family of lines which meets all three lines sweeps out a quadric in P 3. Proof. By the previous lemma we can assume that L 1 = Z(X, Y ), L 2 = Z(Z, W ), L 3 = Z(X Z, Y W ) Let Q = V (XW Y Z). Then Q contains L 1, L 2, L 3. Also Q is the image of the Segre map from P 1 P 1. We have that L 1 = σ({[0 : 1]} P 1 ), L 2 = σ({[1 : 0]} P 1 ), L 3 = σ({[1 : 1]} P 1 ) 16

18 For any P P 1 the line σ(p 1 P ) intersects L 1, L 2, L 3 and so Q = σ(p 1 P 1 ) = P P 1 σ(p 1 {P }) and therefore the lines which intersect all of L 1, L 2, L 3 sweep out at least Q. Let L 4 be a line that intersects all of L 1, L 2, L 3. Let P 1 L 4 L 1, P 2 L 4 L 2, P 3 L 4 L 3. Then XW Y Z is zero on P 1, P 2, P 3, and therefore it is zero on all of L 4 as it has degree 2. 9 Intersections of quadrics Let Q 1, Q 2 P 3 be quadrics. We want to determine Q 1 Q 2. We use the Segre to cut out the first quadric: S : P 1 P 1 P 3 [x 0, x 1 ] [y 0, y 1 ] [x 0 y 0, x 0 y 1, x 1 y 0, x 1 y 1 ] =: [z 0, z 1, z 2, z 3 ] The image is the zeros of z 0 z 3 z 1 z 2. Step 1 Choose coordinates, such that Q 1 = }{{} Segre P 1 P 1 Step 2 Now consider S 1 (Q 2 ) P 1 P 1. Q 2 P 3 is defined by the equation c 1 z c 2 z c 3 z c 4 z c 5 z 0 z c 10 z 2 z 3 =: F (z 0, z 1, z 2, z 3 ) What is S 1 (Q 2 )? Let G(x 0, x 1, y 0, y 1 ) = F (x 0 y 0, x 0 y 1, x 1 y 0, x 1 y 1 ) Let p P 1 P 1. Then p S 1 (Q 2 ) G(p) = 0. But being zero for G doesn t even make sense if it is (only) homogeneous. It needs to be bihomogeneous (i.e. homogeneous in each set of variables separately). Definition 9.1 A polynomial p(x 1,..., x n, y 1,..., y m ) is bihomogeneous of bidegree (a, b) if it is homogeneous of degree a in the x s and homogeneous of degree b in the y s. Example 9.2 x 0 x 1 y 3 0 is bihomogeneous with bidegree (2, 3). If G is bihomogeneous of bidegree (a, b), then whether G(p) = 0 is well defined for p P 1 P 1. Let p = [x 0, x 1 ] [y 0, y 1 ] = [λx 0, λx 1 ] [µy 0, µy 1 ] Then G(p) is defined up to a factor of λ a µ b. Proposition 9.3 Let G be bihomogeneous. Z(G) P 1 P 1 is an algebraic subvariety. Proof. Trivial. 17

19 We have that Q 2 = F (z 0, z 1, z 2, z 3 ) is homogenenous of degree 2. Then we have G(x 0, y 1, y 0, y 1 ) = F (x 0 y 0, x 0 y 1, x 1 y 0, x 0 y 1 ) is bihomogeneous of degree (2, 2). We have projection maps π 1, π 2 : P 1 P 1 P 1 for the x and the y coordinates respectively. Look at π 2 (Z(G)) P 1. What are the fibers of Z(G)? For [ξ 0, ξ 1 ] P 1, G(x 0, x 1, ξ 0, ξ 1 ) has either one or two roots as a polynomial in [x 0, x 1 ] P 1. Can G(x 0, x 1, y 0, y 1 ) be zero for all [x 0, x 1 ]? We have G(x 0, x 1, y 0, y 1 ) = x 2 0P 1 (ξ 0, ξ 1 ) + x 0 x 1 P 2 (ξ 0, ξ 1 ) + x 2 1P 3 (ξ 0, ξ 1 ) where P 1, P 2, P 3 C[y 0, y 1 ], homogeneous of degree 2. This can be zero for all [x 0, x 1 ] if and only if [ξ 0, ξ 1 ] is a common root of P 1, P 2, P 3. How many points of π 2 (Z(G)) are double (ramification)? Look at m G : P 1 y P 2 [y 0, y 1 ] [P 1 (y 0, y 1 ), P 2 (y 0, y 1 ), P 3 (y 0, y 1 )] Consider the quadratic equation P 1 x P 2 x 0 x 1 + P 3 x 2 1 in P 2. We have a double zero if the discriminant P 2 2 4P 1 P 3 is zero. this has 4 solutions in general, because it is an equation of degree 4 in P 1. Something about the Euler characteristic is missing here! 10 The Grassmannian and the incidence correspondence Quadrics in different dimensions In P 1 we just have 2 points. In P 2 we have a conic, and P 1 can be embedded as this conic with the Veronese. In P 3, P 1 P 1 can be embedded as the quadric by the Segre. We skip P 4 for now. In P 5 Grassmann and Plücker thought about it and we will consider it now. You can think about P n as Define the Grassmannian P n = {L L C n+1 is a 1-dim subspace} Gr(r, n) = {S S C n is a r-dimensional subspace} Can we express Gr(r, n) as a quotient? Let M be the space of r n matrices, and let U M be the subset consisting of elements with rank r. We have U/Gl r = Gr(r, n) (by the row space of the matrices). A matrix has rank < r if every r r minor has det = 0. Let γ {1,..., n} be a set with γ = r. d γ : M C d γ (ξ) the determinant of the γ-minor of ξ We have U C = {ξ M γ {1,..., n} with γ = r : d γ (ξ) = 0} M 18

20 This is Zariski closed. How can we think of Gr(r, n) as an algebraic projective variety? Let P : Gr(r, n) P (n r) 1 C n S [z γ,... ]. Here, z γ is the γ-minor of a matrix ξ S that has the vectors of a basis of S as columns. The basis is not unique, so ξ S is determined only up to Gl r. For ξ S = g ˆξ S, g Gl r we have (d γ (ξ s )),... ) = (d γ (g ˆξ S ),... ) = det(g)(d γ (ˆξ S ),... ) so [z γ,... ] is fixed up to a scalar! This fits perfectly with the homogeneous coordinates and the map is well-defined. It is called the Plücker embedding P. Example 10.1 Given an element ξ S as ( 1 0 a ) b 0 1 c d we see that the Gr(2, 4) is, in fact, 4-dimensional. We want to compute the Plücker P. It is P(S) = [z 12, z 13, z 14, z 23, z 24, z 34 ] = [1, c, d, a, b, ad bc] The image satisfies the first Plücker equation z 12 z 34 + z 14 z 23 z 13 z 24 = 0 This is a rank 6 quadric in P 5 = P (4 2) 1. We have a bijection of a full-rank quadric in P 5 via the Plücker to Gr(2, 4) The incidence correspondence Fix r, n N. Definition 10.2 The set is called the incidence correspondence. I = {(S, p) S Gr(r, n), p P(S) P n 1 } Theorem 10.3 I Gr(r, n) P n 1 is an algebraic subvariety. Proof. As Gr(r, n) is an projective algebraic variety by the Plücker embedding, it is covered by affine opens U J where J {1,..., n} with J = r. U J corresponds to the r-dimensional subspaces of C n where the determinant of the J-th minor is nonzero. The elements of U J are therefore given (after reordering) by matrices of the form 1 x 1,1... x 1,n r x r,1... x r,n r 19

21 Therefore U I = C r(n r). To show that I is Z-closed it suffices to show that I U J P n 1 is Z-closed for all possible indices J. It is sufficient to check this for J = {1,..., r}; for all other indices, the procedure is the same. We have the variables x 11, x 12,..., x 1,n r,..., x r,n r in C r(n r) and homogeneous coordinates [z 1,..., z n ]. We want to find equations F 1 (x ij, z k ),..., (where the x are honest coordinates, and the z are homogeneous, projective coordinates) that cut out I J C r(n r) P n 1. We get equations F 1 (x, z) = z r+1 x 11 z 1 x 21 z 2 x r,1 z r F 2 (x, z) = z r+2 x 12 z 1 x 22 z 2 x r,2 z r. F n r (x, z) = z n x 1,n r z 1 x r,n r z r In conclusion I Gr(r, n) P n 1 is an algebraic subvariety. Proposition 10.4 Let V P 3 be a projective algebraic variety. Then the set of lines in P 3 which meet V form an algebraic subvariety of Gr(2, 4). Proof. Let π 1 : I Gr(2, 4), π 2 : I P 3 be the projection maps. Then π 1 π2 1 (V ) is the set of lines in P 3 which intersect V. π2 1 (V ) is closed. π 1 π2 1 (V ) is closed because the image of a projective variety under an algebraic map is closed (see theorem 14.3). Example 10.5 Example defining the Catalan numbers.. 11 Irreducibility Basically page 5 in Hartshorne. 12 Images of quasi-projective varieties under algebraic maps Definition 12.1 Let X be a subset of a topological space. X is called locally closed if X = U V for U open and V closed. A constructible set is a finite union of locally closed subsets. Theorem 12.2 Let X, Y be quasi-projective varieties. Let f : X Y be algebraic. Then Im(f) is constructible. 13 Varieties defined by polynomials of equal degrees Let V P n be a projective algebraic variety. V = Z(F 1,..., F k ), where F i C[z 0,..., z n ] is homogeneous of degree d i. Let d = max{d 1,..., d n }. We want to find polynomials G i all of degree d such that V = Z(G 1,..., G m ). Construct them as follows: Let G i,j := z d d j i F j 20

22 where 0 i n, 1 j k. Obviously V Z({G i,j : 0 i n, 1 j k}) =: V. Let p V. We know that z d d j 0 F j (p) = 0 z d d j n F j (p) = 0, and it follows that F j (p) = 0. V = Z(G 1,1,..., G n,k ).. Theorem 13.1 Let V P n be a projective algebraic variety, then we can find an embedding of V in P N such that the image is defined by quadratic equations. Proof. Let V P n be a projective variety. Assume that V is cut out by F 1,..., F k all of degree d. Let ν d : P n P N [z 0,..., z n ] [z0, d }{{}..., zn] d all other monomials be the d-veronese map. The image of P n is cut out by quadratic equations. V P n is cut out by {quadrics that cut out ν d (P n ), linear terms corresponding to F 1,..., F k }. 14 Images of projective varieties under algebraic maps The following will be useful: Fact 14.1 If G C[Z 0,..., Z n ] is a homogeneous polynomial of degree d 1, then U G = {ξ P n G(ξ) 0} is an affine open, i.e. isomorphic to a closed subset of affine space. Proof. We identify U G with its image U under the d th Veronese embedding ν : P n P N. Let V = Im(ν) Then G gives rise to a linear polynomial L C[Z 0,..., Z N ] s.t. U = U L V. So U is a closed subset of U L which, as L is linear, is isomorphic to C N. Fact 14.2 Any quasi-projective variety Y can be covered by finitely many affine open sets. This means that we can write Y = N i=1u i, where for each i, U i is an open subset of Y isomorphic to a closed subset of the affine space C m. Proof. There are n 1, F 1,..., F k, G 1,... G l C[Z 0,..., Z n ] homogeneous polynomials, s.t. Y P n is the set of points p, s.t. p is a zero of every F i but not of every G j. If we restrict our attention to the open set U Gi, we see that Y U Gi = V (F 1,..., F k ) U Gi. So it is clearly an open set of Y and at the same time a closed subset of U Gi which can be viewed as a closed subset of affine space by the previous fact. Theorem 14.3 Let f : X Y be an algebraic morphism between varieties, where X is projective and Y quasi-projective. Then (A) Im(f) is Zariski-closed in Y. (B) for Z X Zariski-closed, f(z) is Zariski-closed in Y. 21

23 Remark 14.4 Note that it suffices to prove either (A) or (B), as the other part is then easily implied. Proof. We will prove Part (A) of the theorem. More precisely, we will prove first that the graph of f (defined below) is Zariski-closed in X Y, and second that the projection Π Y : X Y Y is a closed map, whenever X is projective, Y quasi-projective, which can be reduced to the proof of the special case X = P n, Y = C m. The first part works for any algebraic morphismf : X Y between varieties. Define the graph of f, Γ(f) := {(x, y) X Y f(x) = y} We want to show Γ(f) X Y is Zariski closed. (4) As Y is a quasi-projective variety, we can view it as a subset Y P n for some n. Let g : X P n be the composition of f and the embedding i : Y P n. Now it is clear that (4) will follow, if we can show instead Γ(g) X P n is Zariski closed. (5) This can further be simplified. Consider the map (g, id) : X P n P n P n (x, y) (g(x), y) This is algebraic by universal property of the product. Moreover, Γ(g) is just the preimage of the diagonal = {(x, x) P n P n x P n } under this map. Thus to prove (5) and thus finish the proof of (4), we have to prove that is Zariski-closed. By taking a look at the Segre embedding P n P n P N ([x 0,..., x n ], [y 0,..., y n ]) [x i y j ] 1 i,j n we see that the as a subset of P N is defined by the equations Z ij = Z ji 0 i < j n together with the equations for the Segre embedding. The second part takes a bit more work. We will start with reformulating the problem several times. More precisely, we will show how any of the following statements implies the one above it, and then prove the last: Claim 14.5 The projection Π Y : X Y Y is a closed map w.r.t. the Zariski topology. (6) (i) (6) holds when X is a projective, Y a quasi-projective variety. (ii) (6) holds when X = P n, Y a quasi-projective variety. (iii) (6) holds for X = P n, Y = U, where U is a closed subset of affine space. (iv) (6) holds for X = P n, Y = C m 22

24 (ii) (i): As X is projective, X P n is a closed subset for some n. But then X Y is closed inside P n Y. Now given V X Y closed, it is also closed in P n Y. So by (ii) its image under projection is closed in Y. (iii) (ii): Cover Y with finitely many affine opens U 1,..., U N. Now note that any subset of Y is closed if and only if for i = 1,..., n its intersection with U i is closed in U i. Also note that Π 1 Y (U i) = X U i. Now let V X Y closed, W Y its image under projection. Then V X U i is closed in X U i. By (iii), its image under Π Y, which is just W U i is closed in U i. This holds for any choice of i, so W is closed in Y. (iv) (iii): trivial What is left is the proof of (iv). Proof. Let Z P n C m be a closed subset. Z is the zero locus of polynomials F 1,... F k C[Z 0,..., Z n, Y 1,..., Y m ], where F i is homogeneous, when viewed as a polynomial in Z 1,... Z n, say of degree d i. We Want to find polynomial equations that describe Π Y (Z) in C m. It is easily seen that Π Y (Z) = {(ξ 1,..., ξ m ) C n F 1 (Z 0,..., Z n, ξ 1,..., ξ m ). F k (Z 0,..., Z n, ξ 1,..., ξ m ) have a common zero in P n } Let us take a step back, and reconsider the general question of when a set of homogeneous polynomials G 1,..., G l in the variables Z 0,... Z n has a common zero in projective space and when the opposite holds. Of the following statements any one is in an obvious way (by definition or by some basic result) equivalent to the next one. 1. G 1,..., G l have no common zero in P n. 2. The common zero locus of G 1,... G l in C n+1 is empty or equal to (0,..., 0). 3. { (1) (G 1,..., G l ) = (Z 0,..., Z n ) 4. (G 1,... G l ) (Z 0,..., Z n ) 5. There is a d 1 such that (G 1,..., G l ) contains all monomials in Z 0,..., Z n of degree d. Now if we apply the equivalence of 1. and 5. to our situation we immediately get Π Y (Z) = {ξ C m d 1 : (F 1 (Z, ξ),..., F k (Z, ξ)) does not contain all monomials of degree d.} = {ξ C m (F 1 (Z, ξ),..., F k (Z, ξ)) does not contain all monomials of degree d.} =: d=1 W d. d=1 As any intersection of closed sets is closed, we can consider one d at a time. For fixed d, all monomials of degree d are contained in the ideal (F 1 (Z, ξ),..., F k (Z, ξ)) if and only if they are contained in the subset of homogeneous elements of degree d, denoted by (F 1 (Z, ξ),..., F k (Z, ξ)) d, which is a vector space, and in particular a subspace of the space of homogeneous polynomials 23

25 of degree d, which is given by the monomial basis. So (F 1 (Z, ξ),..., F k (Z, ξ)) does not contain all monomials of degree d, if and only if (F 1 (Z, ξ),..., F k (Z, ξ)) d C[Z 0,..., Z n ] d. (F 1 (Z, ξ),..., F k (Z, ξ)) d is generated as a vector space by S := {F i (Z, ξ)z α 0 i n : d i d, α = d d i } So the question is if these generate a space of dimension strictly lower than the number of monomials of degree d. By linear algebra this is the case iff for the matrix M where the columns are the coordinates of elements of S w.r.t. the monomial basis, every size-d minor has zero determinant. Note that this gives a finite set of polynomial equations in the entries of M, and this gives us polynomials in ξ 1,..., ξ m, as every entry of M is a polynomial expression of those. W d is the zero locus of the resulting polynomials, thus closed in C m. 15 Bezout s Theorem Last time we talked about Bézout s theorem, which in some sense is the analogue of the Fundamental Theorem of Algebra. In particular, given H 1 and H 2 in P 2 where H 1 is of degree d 1 and H 2 of degree d 2, then H 1 and H 2 intersect d 1 d 2 times, where points are counted with multiplicity. One problem with this statement is that H 1 and H 2 can intersect in infinitely many points, say for example H 1 = H 2 = (x 0 ). But at least we can restrict to the simple case: if H 1 H 2 has finitely many points, then H 1 H 2 = d 1 d 2, with multiplicities counted. It may be the case that, at least locally, H 1 and H 2 intersect in two points, but after some parametrization they intersect in only one point. We want our definition of intersection to be independent of the parametrization, so we must somehow put some weight in each point to count for multiplicities. The way we calculate the weight of the point p i is by considering the local ring at p i and then taking the quotient by the equations of H 1 and H 2. The dimension of this quotient is equal to m pi and we get that H 1 H 2 = m pi. p i H 1 H The resultant Let p, q C[x, y]: p = α 0 x n + α 1 x n 1 y + + α n y n q = β 0 x m + β 1 x m 1 y + + β m y m We want to find a polynomial R such that R(α 0,..., α n, β 0,..., β m ) = 0 p, q have a common root. Let V n be the vector space of homogeneous polymomials of degree n. We have (x n, x n 1 y,..., y n ) as a basis and dim V n = n + 1. Here, denote P n = P(V n ). Let f : P 1 P n 1 P m 1 P n P m [ˆl, ˆp, ˆq] [ˆlˆp, ˆlˆq] 24