Projective Varieties. Chapter Projective Space and Algebraic Sets


 Scot Lucas
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1 Chapter 1 Projective Varieties 1.1 Projective Space and Algebraic Sets Definition. Consider A n+1 = A n+1 (k). The set of all lines in A n+1 passing through the origin 0 = (0,..., 0) is called the ndimensional projective space and is denoted by P n (k), or simply P n when k is understood. We also have the identification P n = (A n+1 \ {0})/k, where (x 1,..., x n+1 ) (λx 1,..., λx n+1 ) for all λ k, i.e., two points in A n+1 \{0} are equivalent if they are on the same line through the origin. An element of P n is called a point. If P is a point, then any (n+1)tuple (a 1,..., a n+1 ) in the equivalence class P is called a set of homogeneous coordinates for P. Equivalence classes are often denoted by P = [a 1 : : a n+1 ] to distinguish from the affine coordinates. Note that [a 1 : : a n+1 ] = [λa 1 : : λa n+1 ] for all λ k. We defined P n as the collection of all one dimensional subspaces of the vector space A n+1, but P n may also be thought of as n+1 (overlapping) copies of affine nspace. Indeed, we can express any any point [x 1 : : x n1 ] U i in terms of n affine coordinates: [ x1 [x 1 : : x n+1 ] = : : 1 : : x ] n+1. x i x i Thus, U i = A n Example. Consider P 2, where [x : y : z] are the homogeneous coordinates. Then {[ U x = x : y x : z ] } P 2 x 0 x = {[1 : u : v] u, v k} = {(u, v) u, v k} = A 2, 1
2 so (u, v) are the affine coordinates on U x. Similarly, {[ x U y = {[x : y : z] P 2 y 0} = y : 1 : z ] y } P 2 y 0, and {[ x U z = {[x : y : z] P 2 z 0} = z : y ] z : 1 } P 2 y 0. Moreover, A n can be considered as a subspace of P n, where the inclusion is given by identifying A n with U n+1 P n, i.e. the inclusion is the map ϕ : A n P n given by ϕ(u 1,..., u n ) = [u 1 : : u n : 1]. One could also introduce a 1 to any other position, but we will usually use this convention. For each i = 1,..., n + 1, H i = {[x 1 : : x n+1 ] x i = 0} = P n \ U i is a hyperplane, which can be identified with P n 1 by the correspondence [x 1 : : 0 : : x n+1 ] [x 1 : : x i 1 : x i+1 : : x n+1 ]. In particular, H n+1 is often denoted H and is called the hyperplane at infinity, and P n = U n+1 H = A n P n 1, so P n is the union of a copy of A n and H, which can be seen as the set of all directions in A n Examples. (i) P 0 = { } is a single point. (ii) P 1 = A 1 P 0 = A 1 { }, the one point compactification of A 1. (iii) P 2 = A 2 P 1, where the copy of P 1 here is often referred to as the line at infinity, and is denoted by l. Remarks. In P n, any two lines intersect. For example, consider two distinct parallel lines in A 2 : L : au + bv + c = 0 L : au + bv + c = 0, where c c. If one considers A 2 as the subset {[u : v : 1] P 2 u, v k} = {[x : y : z] P 2 x, y k, z 0} = U z of P 2, we can see that the equations can be rewritten as follows by letting u = x/z and v = y/z: ax + by + cz = 0 ax + by + c z = 0. Subtracting the two equations, we get z(c c ) = 0, so z = 0, as we assumed that c c. Hence any solution is of the form [x : y : 0] and lies on the line 2
3 at infinity. Substituing this value of z back into either equation gives that ax + by = 0, so x = bt and y = at for some t k. Therefore, the solutions to the system are {(bt, at, 0) t k } = [ b : a : 0], which is a single point in P 2 on the line at infinity Definition. If f k[x 1,..., x n+1 ], then P = [a 1 : : a n+1 ] P n is a zero of f if f(λa 1,..., λa n+1 ) = 0 for every λ k, in which case we write f(p ) = 0. For any S k[x 1,..., x n+1 ], let V p (S) = {P P n f(p ) = 0 for all f S} be the zero set of S in P n. Moreover, if Y P n is such that Y = V p (S) for some S k[x 1,..., x n+1 ], then we say that Y is a projective algebraic set. Similarly, given Y P n, let be the ideal of Y. I p (Y ) = {f k[x 1,..., x n+1 ] f(p ) = 0 for all P Y } Remark. To avoid confusion, from now on we will use the notation I a and V a for the ideal of a set of points in A n and the zero set in A n of a set of polynomials, respectively Lemma. Let f k[x 1,..., x n+1 ] be such that f = f m + + f d, where each f i is an iform. Then, for any P P n, f(p ) = 0 if and only if f i (P ) = 0 for i = m,..., d. Proof: Suppose f(p ) = 0 for P = [a 1 : : a n+1 ] P n. Then q(λ) = λ m f m (a 1,..., a n+1 ) + + λ d f d (a 1,..., a n+1 ) = 0 for all λ k, and q is a polynomial in λ with coefficients b i = f i (a 1,..., a n+1 ). Since k is infinite and q(λ) = 0, we have that b i = 0 for each i. Therefore, f i (λa 1,..., λa n+1 ) = 0, and f i (P ) = 0 for each i. The converse is clear. Thus, if f = f m + + f d k[x 1,..., x n+1 ], where each f i is an iform, then V p (f) = V p (f m,..., f d ) and if f I p (Y ) for some Y P n, then f i I p (Y ) for each i Proposition. (i) Every algebraic set in P n is the zero set of a finite number of forms. (ii) If Y P n, then I p (Y ) is generated by homogeneous polynomials. Proof: The result is clear from the preceding discussion. 3
4 This motivates the following definition Definition. An ideal I k[x 1,..., x n ] is said to be homogeneous if whenever f I then each homogeneous component of f is in I. By the above, we see that I p (Y ) is homogeneous for any Y P n. Moreover, one proves as in the affine case that I p (Y ) is radical. We then have a correspondence between projective algebraic sets in P n and homogeneous radical ideals in k[x 1,..., x n+1 ]. We will see that this is almost a onetoone correspondence, but in order to make it onetoone we have to exclude and the ideal of {(0,..., 0)}. Let us begin by stating some properties of homogeneous ideals Proposition. Let I and J be ideals in k[x 1,..., x n+1 ]. Then: (i) I is homogeneous if and only if I can be generated by homogeneous polynomials; (ii) if I and J are homogeneous, then I + J, IJ, I J, and I are homogeneous; (iii) I is a homogeneous prime ideal if and only if whenever f, g k[x 1,..., x n ] are forms such that fg I, then f I or g I. Proof: (i) Exercise (ii) Exercise. (iii) The forward direction is clear. Let f, g k[x 1,..., x n+1 ] be such that fg I. Let f = d i=m f i and g = d i=m g j, where each f i is an iform and each g j is a jform. Then fg = f m g m + d+d k>m+m i+j=k f i g i. Since I is homogeneous, f m g m I. Suppose, for now, that f m / I, so that g m I. Then f(g g m ) = (fg fg m ) I. Since the homogeneous component of f(g g m ) of degree m + m + 1 is f m g m +1 and I is homogeneous, we have f m g m +1 I. And since f m / I, this means that g m +1 I so that f(g g m g m +1) = fg fg m fg m +1 I. Continuing this way, we see that if f m / I, then g i I for all i = m,..., d, implying that g I. If both f m and g m are in I, proceed as above with (f f m )(g g m ) Examples. (i) I = x 2 and J = x 2, y are homogeneous in k[x, y]. 4
5 (ii) I = x 2 + x is not homogeneous since x / I Definition. Let θ : A n+1 \ {0} P n be the standard projection, so that θ(x 1,..., x n+1 ) = [x 1 : : x n+1 ]. If Y P n, the affine cone over Y is C(Y ) = θ 1 (Y ) {(0,..., 0)} A n+1. We note the following properties of the affine cone Proposition. (i) If P P n, then C({P }) is the line in A n+1 through the origin determined by P. (ii) C( ) = {(0,..., 0)}. (iii) C(Y 1 Y 2 ) = C(Y 1 ) C(y 2 ). (iv) C(Y 1 ) = C(Y 2 ) if and only if Y 1 = Y 2. (v) If Y P n is nonempty, then I p (Y ) = I a (C(Y )). (vi) If I k[x 1,..., x n+1 ] is a homogeneous ideal such that V p (I), then C(V p (I)) = V a (I). In particular, C(Y ) = V a (I) for some nonempty Y P n if and only if Y = V p (I). One can use affine cones to compute I p and V p or to determine properties of I p and V p Examples (of projective algebraic sets). (i) V p (I) = V p (1) = for any ideal I such that V a (I) = {(0,..., 0)}. (ii) P n = V p (0). (iii) If P = [a : b] P 1, then {P } = V p (bx ay), since C({P }) is the line through (0, 0) and (a, b) in A 2, which is V a (bx ay). In general, if P = [a 1 : : a n+1 ] P n and a i is a nonzero coordinate of P, then {P } = V p (a i x 1 a 1 x i,..., a i x n+1 a n+1 x i ). (iv) If f is a homogeneous polynomial, then Y = V p (f) is called a hypersurface. (v) Let Y = V p (x y, x 2 yz) P 2. Then so C(Y ) = V a (x y, x 2 yz) = V a (x, y) V a (x y, x z) = {(0, 0, s) s k} {(t, t, t) t k}, Y = V p (x, y) V p (x y, x z) = {[0 : 0 : 1]} {[1 : 1 : 1]}. 5
6 Examples (of ideals). (i) I p (P n ) = 0. (ii) I p ( ) = 1. (iii) If P = [a 1 : : a n+1 ] P n with a i 0, then I p ({P }) = I a (C([a 1 : : a n+1 ])) = a i x 1 a 1 x i,..., a i x n+1 a n+1 x i, since C([a 1 : : a n+1 ]) = V a (a i x 1 a 1 x i,..., a i x n+1 a n+1 x i ). Remark. In the projective case, x 1,..., x n+1 is a homogeneous radical ideal other than 1 whose zero set in P n is empty. We must therefore remove and x 1,..., x n+1 from the onetoone correspondence between projective algebraic sets and radical homogeneous ideals Proposition. The union of two projective algebraic sets is a projective algebraic set. The intersection of any family of projective algebraic sets is a projective algebraic set. Moreover, and P n are projective algebraic sets. Therefore, the projective algebraic subsets of P n are the closed sets of a topology on P n Definition. The Zariski topology on P n is the topology whose open sets are the complements of projective algebraic sets Examples. (i) For each i, the hyperplane H i = V p (x i ) is a closed set and its complement U i = P n \ H i is an open set in the Zariski topology. Therefore, {U i } n+1 i=1 is an open cover of P n. (ii) We have seen that A n can be identified with the open set U n+1 in P n. For any affine variety X A n, we define the projective closure of X in P n to be the smallest projective algebraic set containing X. For example, Y = V p (y 2 z x 3 ) P 2 is the projective closure of X = V a (y 2 x 3 ) A 2 since V p (y 2 z x 3 ) = V a (y 2 x 3 ) {[0 : 1 : 0]}, i.e., Y is the onepoint compactification of X in P Definition. A nonempty closed subset of P n is irreducible if it cannot be expressed as the union of two proper closed subsets. A projective (algebraic) variety is an irreducible algebraic set in P n equipped with the induced Zariski topology. As in the affine case, we have the following result. 6
7 Proposition. Let Y P n be a projective algebraic set. Then Y is irreducible if and only if I p (Y ) is prime. Proof: Let f, g k[x 1,..., x n+1 ] be forms such that fg I p (Y ). Then V p (f) and V p (g) are projective algebraic sets and Y = (Y V p (f)) (Y V p (g)), so by the irreducibility of Y, Y = Y V p (f) or Y = Y V p (g), implying that f I p (Y ) or g I p (Y ). The reverse direction is as in the affine case Proposition. Let Y be a subset of P n. Then: (i) Y is a projective algebraic set if and only if C(Y ) is an affine algebraic set. (ii) Y is an irreducible projective algebraic set if and only if C(Y ) is an irreducible affine algebraic set. (iii) If Y is algebraic, then it is the union of a finite number of irreducible projective algebraic sets. Proof: (i) Y is algebraic if and only if I p (Y ) = I a (C(Y )) is radical, which happens if and only if C(Y ) is algebraic. (ii) Y is an irreducible algebraic set if and only if I p (Y ) = I a (C(Y )) is prime, which happens if and only if C(Y ) an irreducible algebraic set. (iii) If Y is algebraic, then so is C(Y ), which is the union of a finite number of irreducible affine algebraic sets. If C(Y ) = W 1 W n with each W i irreducible, then Y = W 1 W n with W i = V p (I a ( W i )) irreducible for all i. Remark. One defines the (irredundant) decomposition of a projective algebraic set Y as in the affine case. This decomposition is unique up to a permutation of its irreducible components since W 1 W m is the decomposition of Y if and only if C(W 1 ) C(W n ) is the decomposition of C(Y ) (exercise) Examples. (i) If f is a form in k[x 1,..., x n+1 ], then V p (f) is irreducible C(V p (f)) = V a (f) is irreducible f is irreducible. 7
8 (ii) Let Y = V p (x 2 + y 2 + 2yz) P 2, and let f = x 2 + y 2 + 2yz ( (x ) 2 ( y ) 2 ( y ) ) = z z z z ( x = z 2 g z, y ), z where g(u, v) = u 2 + v 2 + 2v is irreducible. Thus f is irreducible and Y is irreducible. (iii) Is Y = V p (xz 3 + y 2 z 2 x 3 z x 2 y 2 ) irreducible? This time g(u, v) = u + v 2 u 3 u 2 v 2 = (u + v 2 )(1 u 2 ) = (u + v 2 )(1 u)(1 + u), so f = xz 3 +y 2 x 2 x 3 z x 2 y 2 = z 4 g(x/z, y/z) = (xz +y 2 )(z x)(z +x), and Y = V p (f) = V p (xz + y 2 ) V p (z x) V p (z + x), which is the irreducible decomposition of Y since xz +y 2, z x, and z +x are irreducible Theorem (Projective Nullstellensatz). Let I k[t 1,..., t n+1 ] be a homogeneous ideal. Then: (i) V p (I) = if and only if there exists N N such that I contains every form of degree at least N; (ii) if V p (I) then I p (V p (I)) = I. Proof: (i) The following statements are equivalent: V p (I) = V a (I) = or {(0,..., 0)} V a (I) {(0,..., 0)} x 1,..., x n+1 = I a ({(0,..., 0)}) I a (V a (I)) = I x mi i I for some m i N, for all i any form of degree at least N is contained in I for some N max{m 1,..., m n+1 }. (ii) I p (V p (I)) = I a (C(V p (I)) = I a (V a (I)) = I, by the affine Nullstellensatz. As a consequence of the projective Nullstellensatz, we have the following onetoone correspondences: 8
9 (nonempty algebraic sets in P n ) (varieties in P n ) ( proper homogeneous radical ideals I x 1,..., x n+1 homogeneous prime ideals I x 1,..., x n+1 ). The empty set is usually thought of as corresponding to x 1,..., x n Regular and Rational Functions Definition. Let I k[x 1,..., x n+1 ] be a homogeneous ideal. A residue class in k[x 1,..., x n+1 ]/I is said to be an mform if it contains an mform. In particular, 0forms are constants Proposition. Let I k[x 1,..., x n+1 ] be a homogeneous ideal. Every f k[x 1,..., x n+1 ]/I may be expressed uniquely as f = f f d, where d = deg f and each f i is an iform. Proof: Existence is clear, so we need only show uniqueness. Suppose that f = f m + + f d = g m + + g d, where each f i is an iform and each g j is a jform. Then (f i g i ) = 0, i where we set f i = 0 for i < m and i > d and g i = 0 for i < m and i > d. Thus (f i g i ) I, so by the homogeneity of I, f i g i I for all i, so i f m + + f d = i f i = i g i = g m + + g d Definition. Let Y P n be a projective variety, so that I p (Y ) = I a (C(Y )) is prime and homogeneous. Then Γ H (Y ) = k[x 1,..., x n+1 ]/ I p (Y ) = Γ(C(Y )) is an integral domain, called the homogeneous coordinate ring. 9
10 Note that, unlike the case of affine coordinate rings, elements of Γ H (Y ) cannot be considered functions unless they are constant. Indeed, f Γ H (Y ) defines a function on Y if and only if f(λx 1,..., λx n+1 ) = f(x 1,..., x n+1 ) for all λ k. But if f = f f d is the decomposition of f into forms, then this happens if and only if f(x 1,..., x n+1 ) = f(λx 1,..., λx n+1 ) = f 0 (x 1,..., x n+1 ) + λf 1 (x 1,..., x n+1 ) + + λ d f d (x 1,..., x n+1 ), for all λ k, which can only happen if f is constant on Y, so that f is constant in Γ H (Y ) Definition. Let Y P n be a projective variety. The field of fractions of Γ H (Y ) is denoted by k H (Y ), and is called the homogeneous function field. Note that k H (Y ) = k(c(y )). But the only elements of k H (Y ) that define functions on Y are of the form f/g with f, g Γ H (Y ) forms of the same degree and g 0. This is because if f = f m + + f d and g = g m + + g d, then f(λx 1,..., λx n+1 ) g(λx 1,..., λx n+1 ) = λm f m (x 1,..., x n+1 ) + + λ d f d (x 1,..., x n+1 ) λ m g m (x 1,..., x n+1 ) + + λ d g d (x 1,..., x n+1 ) = f(x 1,..., x n+1 ) g(x 1,..., x n+1 ) for all λ k if and only if m = m and f = f m, g = g m on Y, so that f = f m and g = g m are forms of the same degree. We then define k(y ) = { } f g f, g Γ H(Y ) are forms of the same degree, which is the function field of Y, whose elements are called rational functions on Y. We have k k(y ) k H (Y ) = k(c(y )), but Γ H (Y ) k(y ) in general Definition. If p Y and z k(y ), we say that z is regular at p (or defined at p) if there exist forms f, g Γ H (Y ) of the same degree such that g(p) 0 and z = f/g, in which case z(p) = f(p)/g(p) is the value of f at p. The set of points where z is not defined is called its pole set Proposition. Let Y P n be a projective variety. Then the pole set of any rational function on Y is an algebraic subset of Y. Proof: The pole set of z k(y ) is the intersection of the algebraic sets V p (g) Y, taken over all forms g for which there is a form f such that z = f/g. Hence it is algebraic. 10
11 1.2.7 Definition. Let Y be a projective variety, and let p be a point in Y. Then O p (Y ) = {z k(y ) z is regular at p} k(y ) is the local ring of Y at p, is the maximal ideal of Y at p, and is the ring of regular functions on Y. M p (Y ) = {z O p (Y ) z(p) = 0} O(Y ) = O p (Y ) p Y As in the affine case, O p (Y ) is a local ring and M p (Y ) is its unique maximal ideal. However, in contrast to the affine case, O(Y ) is not isomorphic to Γ H (Y ) Proposition. Let Y be a projective variety. Then O(Y ) = k. Proof: O(Y ) O(C(Y )) = Γ(C(Y )) = Γ H (Y ) and the only functions in Γ H (Y ) are the constants. Nonetheless, one proves as in the affine case that if two rational functions on Y are equal on an open set U Y, then they are equal on Y. In particular, one has the following Proposition. Let Y P n be a projective variety. Then k(y ) k(y U i ) are isomorphic as kalgebras for all i, where U i is the affine open subset of P n given by x i 0. Moreover, if p Y U i, then as kalgebras. O p (Y ) O p (Y U i ) Proof: Define Φ : k(y ) k(y U i ) by ( ) f Φ = f(x 1,..., 1,..., x n+1 ) g g(x 1,..., 1,..., x n+1 ), and Ψ : k(y U i ) k(y ) by ( ) a Ψ = x i d a(x 1 /x i,..., x n+1 /x i ) b x d i b(x 1 /x i,..., x n+1 /x i ), where d = max{deg a, deg b}. It is then easy to check that these are kalgebra homomorphisms that are mutual inverses. The second statement is a direct consequence of the first. 11
12 From the above proposition, we see that the local properties of a projective variety Y can be completely described in terms of its affine pieces Y U i. This is for instance the case with dimension and smoothness Definition. Let Y P n be a variety. The dimension of Y is defined as dim Y := tr. deg k (k(y )). By the above, dimension is a local property as dim Y = dim(y U i ) for all i. In particular, dim P n = dim A n = n. Moreover, projective varieties of dimensions 1, 2, 3, are called curves, surfaces, 3folds etc... In particular, if Y is the zero set in P n set of a single irreducible form, then Y has dimension n 1 and is called a hypersurface. Remark. One can also show that dim Y = dim C(Y ) 1 (exercise). Furthermore, one defines smoothness as follows Definition. Let Y P n be an rdimensional projective variety and p Y. The kvector space T p (Y ) := ( M p (Y )/(M p (Y )) 2) is called the Zariski tangent space of Y at p, and Y is said to be smooth at p if and only if dim k T p (Y ) = r. Otherwise, p is called singular. Moreover, Y is called smooth if it is smooth at every point. Remark. Clearly, Y is smooth at p if and only if Y U i is smooth at p since O p (Y ) O p (Y U i ) for any affine open subset U i containing p. To check the smoothness of Y at p one then just has to compute the rank of the Jacobian of the polynomials giving Y U i A n at p. However, it is in practice not necessary to consider the restriction of Y to Y U i since one can show, as in the affine case, the following. If Y = V p (f 1,..., f s ) P n for some forms f 1,..., f s k[x 1,..., x n ], then T p (Y ) = ker(jac(f 1,..., f s )(p)), so that Y is smooth at p if and only if Jac(f 1,..., f s )(p) has rank n dim Y. The proof is left to the reader as an exercise Example. Consider the projective plane curve Y = V p (f) P 2 where f = axy + bxz + cyz k[x, y, z]. Then Y is smooth if and only if a, b, c 0. Indeed, Jac(f) = (ay + bz, ax + cz, bx + cy). If one of the a, b, c is zero, say a, then Jac(f) has rank 0 at [x : y : z] = [c, b, 0] Y, and Y has singular points. But if a, b, c 0, then Jac(f) has rank 0 if and only if (x, y, z) = (0, 0, 0), which does not correspond to a point on Y, implying that Y is smooth. 12
13 Finally, as in the affine case, we have: Proposition. A projective curve Y is smooth at p if and only if O p (Y ) is a DVR. Proof: Y is smooth at p if and only if M p (Y )/(M p (Y )) 2 is a 1dimensional kvector space, which happens if and only if M p (Y ) is principal. 1.3 Regular and Rational Maps Definition. Let X P n and Y P m be projective varieties. A map ϕ : X Y is called rational if it can be written as ϕ(x 1 : : x n+1 ) = [F 1 (x 1,..., x n+1 ) : : F m+1 (x 1,..., x n+1 )] for some forms F 1,..., F m+1 k[x 1,..., x n+1 ] of the same degree. Moreover, ϕ is said to be regular, or defined, at p X if it can be represented by forms F 1,..., F m+1 that do not vanish simultaneously at p. If ϕ is not defined at p, then p is said to be a pole of ϕ. If ϕ is regular at every point in X, then it is called a regular map. A rational map ϕ : X Y that has a rational inverse ϕ 1 : Y X is called a birational equivalence, in which case we say that X and Y are birational and write X Y. If ϕ and ϕ 1 are both regular, then ϕ is called a isomorphism, in which case we say that X and Y are isomorphic and write X = Y. Remarks. (i) One can also define rational maps ϕ : X Y as maps that can be written as ϕ(x 1 : : x n+1 ) = [h 1 (x 1,..., x n+1 ) : : h m+1 (x 1,..., x n+1 )] for some rational functions h 1,..., h m+1 k(y ). But each h i = f i /g i with f i and g i forms of the same degree in k[x 1,..., x n+1 ]. By clearing denominators, we can write ϕ(x 1 : : x n+1 ) = [F 1 (x 1,..., x n+1 ) : : F m+1 (x 1,..., x n+1 )], where F i := g 1 g i 1 f i g i+1 g m+1 are forms of the same degree. The two definitions of rational map are therefore equivalent. (ii) As in the affine case, we can define the pullback of a rational map, and we have that X Y if and only if k(x) and k(y ) are isomorphic as kalgebras. Consequently, dimension is preserved under birational equivalences. (iii) As in the affine case, smoothness is preserved under isomorphisms Examples. (i) Any rational function h : X k can be considered as a rational map ϕ from X to P 1. If h = f/g, set ϕ(x 1 : : x n+1 ) = [f(x 1,..., x n+1 ) : g(x 1,..., x n+1 )]. 13
14 (ii) Any invertible matrix A GL(n + 1, k) defines an isomorphism T : P n P n, x Ax, called a projective coordinate change, since matrix multiplication commutes with scalar multiplication in A n+1 and therefore descends to the quotient P n = (A n+1 \{0})/k. (iii) Any hyperplane H = V p (a 1 x a n+1 x n+1 ) in P n is isomorphic to P n 1 since it can be mapped isomorphically onto the hyperplane at infinity H = P n 1 under an appropriate projective coordinate change. (iv) Let Y = V p (xz y 2 ) P 2, and define ϕ : P 1 Y by ϕ(u : v) = [u 2 : uv : v 2 ]. Since u and v can not be simultaneously zero, u 2, uv, and v 2 can not be simultaneously zero, so we see that ϕ is regular at every point. Also, ϕ has a regular inverse defined by { ϕ 1 [x : y] if x 0, (x : y : z) = [y : z] if z 0. We only need to verify that ϕ 1 is welldefined. Note that if x, z 0 on Y, then y 0, in which case [x : y] = [xz : yz] = [y 2 : yz] = [y : z]. So ϕ 1 is welldefined on Y, showing that Y is isomorphic to P 1. More generally, one can show that if Y = V p (f) P 2 is the zero set of an irreducible 2form f k[x, y, z], then Y is isomorphic to P 1 (exercise), implying it is a smooth curve. (v) Let Y = V p (y 2 z x 3 ) P 2, and define ϕ : P 1 Y by ϕ(u : v) = [u 2 v : u 3 : v 3 ]. As in the previous example, it is easy to see that ϕ is regular at every point. However, it can not have a regular inverse, as then restricting to U z would imply that V a (y 2 x 3 ) is isomorphic A 1, which we know is not true. Nonetheless, ϕ does have a rational inverse ϕ 1 : Y P 1 given by ϕ 1 (x : y : z) = [y : x], so ϕ is a birational equivalence, showing that Y is birational to P 1. This is to be expected as Y U z = V a (y 2 x 3 ) and Y = V a (y 2 x 3 ) {[0 : 1 : 0]}, and we have seen that V a (y 2 x 3 ) is birational to A 1. The above examples motivate the following definition Definition. Let Y be a projective variety. We say that Y is rational if it is birational to P n for some n. 14
15 We will show the existence of nonrational varieties in the next chapter. We end this chapter with a few facts about projective curves Proposition. Let C be a projective curve in P n, and let ϕ : C P m be a rational map. Then ϕ is regular at every smooth point of C. Proof: Let p be a smooth point of C. Let F 1,..., F m+1 k[x 1,..., x n+1 ] be forms such that ϕ(x 1 : : x n+1 ) = [F 1 (x 1,..., x n+1 ) : : F m+1 (x 1,..., x n+1 )]. Let t O p (C) be a local parameter. Then each F i can be written in the form F i = t ki u i for some k i Z and unit u i O p (C). After a possible change of coordinates in P m, we may assume that k 1 k 2 k m+1. Then ϕ(x 1 : : x n+1 ) = [F 1 (x 1,..., x n+1 ) : : F m+1 (x 1,..., x n+1 )] = [t k1 u 1 (x 1,..., x n+1 ) : : t km+1 u m+1 (x 1,..., x n+1 )] = [u 1 : t k2 k1 u 2 : : t km+1 k1 u m+1 ]. The first component is nonzero at p since u 1 is a unit in O p (C), and each of the components is regular since k 1 k 2 k m+1 and each u i is a unit in O p (C). Therefore, ϕ is regular at p Corollary. Let C and C be smooth projective curves and let ϕ : C C be a birational equivalence. Then ϕ is an isomorphism. Proof: Since C and C are smooth, the preceding proposition implies that both ϕ and ϕ 1 are regular everywhere. Therefore, ϕ is an isomorphism Corollary. A smooth rational projective curve is isomorphic to P 1. We have seen above that projective plane curves Y = V p (f) P 2 given by 1forms or irreducible 2forms f k[x, y, z], which are called lines or irreducible conics, respectively, are isomorphic to P 1 and therefore rational. We will see in the next chapter that smooth plane cubics, which are projective plane curves given by 3forms, cannot be rational because they cannot be isomorphic to P 1 ; this will be done using divisors. Smooth plane cubics are the simplest examples of projective varieties that are not rational. 15
16
17 Chapter 2 Projective Plane Curves 2.1 Projective Plane Curves Definition. A projective plane curve is an equivalence class of nonconstant forms in k[x, y, z], where f g if and only if f = αg for some α k. The degree of a curve is defined to be the degree of the defining form. Curves of degrees 1, 2, 3, and 4 are called lines, conics, cubics, and quartics, respectively. If f is irreducible, then the projective curve given by f is the projective variety V p (f) in P 2. Local properties of a curve C = V p (f) are given by restricting C to the affine open sets U x, U y, and U z : C U x = V a (f(1, y, z)) C U y = V a (f(x, 1, z)) C U z = V a (f(x, y, 1)). For example, if p = [x 0 : y 0 : 1] C U z, then C is smooth at p if and only if C U z is smooth at (x 0, y 0 ) and the multiplicity of C at p is defined to be m p (f) = m (x0,y 0)(f(x, y, 1)) so that p is singular if and only if m p (f) 2. Intersection multiplicity is similarly defined as the usual affine intersection multiplicity on any affine open set containing the points. Using Proposition 1.2.9, it is easy to check that each of these definitions is independent of the affine open set chosen. Remark. How does one find intersection points of projective plane curves? If C = V p (f) and D = V p (g), to find C D solve the two systems f(x, y, z) = 0 g(x, y, z) = 0 z = 0, 17
18 and f(x, y, z) = 0 g(x, y, z) = 0 z = 1, to find the points on the line at infinity and the points on U z respectively, and discard (0, 0, 0). For example, if C = V p (f), where f = x 2 y 2 + xz, and D = V p (g), where g = x + y, the points at infinity are given by solving the system x 2 y 2 + xz = 0 x + y = 0 z = 0, which has the solutions y = x and z = 0, which represent the single point [ 1 : 1 : 0] in P 2. The points on U z are given by solving the system x 2 y 2 + xz = 0 x + y = 0 z = 1, which has the solutions y = x = 0 and z = 1, which corresponds to the point [0 : 0 : 1] P 2. Therefore, there are only two points of intersection, [1 : 1 : 0] and [0 : 0 : 1]. 2.2 Bézout s Theorem This section will be devoted to the proof of the following theorem and some of its corollaries Theorem (Bézout). Let C = V p (f) and D = V p (g) be projective plane curves that do not have a common component. Then, if C has degree m and D has degree n, C and D intersect in mn points counting multiplicity. Suppose that C and D are given by the forms f and g respectively, so that deg(f) = m and deg(g) = n. Then, since C and D do not have a common component, f and g can not have a common factor; moreovoer, C and D intersect in a finite set of points. We may therefore assume, after an appropriate projective change of coordinates, that none of the intersection points lie on the line at infinity. Indeed, since C and D intersect in a finite set of points, we can find a line in P 2 that does not contain any of the intersection points. This line is then given by a 1form ax + by + cz k[x, y, z]. At least one of the constants a, b, c must 18
19 be nonzero. After a simple projective coordinate change, we may assume that a 0. The matrix M = a b c is then invertible and corresponds to the projective coordinate change [x : y : z] [z : y : ax + by + cz] = [u : v : w]. This transformation then takes the line ax + by + cz = 0 to the line at infinity, i.e. the line w = 0. We therefore assume that C and D do not intersect on the line at infinity given by z = 0, so that f and g do not have any common zeros on z = 0. This implies, in particular, that z does not divide f or g. Indeed, suppose that z divides f. Then f = zf for some form f k[x, y, z]. Consider the restriction of g to oints of the form [x : 1 : 0]. Then g(x, 1, 0) must have at least one zero since k is algebraically closed. If [x 0 : 1 : 0] is such a zero, then g(x 0, 1, 0) = 0 and f(x 0, 1, 0) = f(x 0, 1, 0), since f = zf, contradicting the fact that f and g do not have any common zeros at infinity. Therefore, our assumption that z divides f is false. A symmetric argument establishes that z also does not divide g. These facts are crucial for the proof of Bézout s Theorem. Now, since C and D do not intersect at infinity, we have that C D U Z. We therefore only have to prove that C and D intersect in mn points in U z, counting multiplicity. Also, recall that I(p, C D) = I(p, V a (f(x, y, 1)) V a (g(x, y, 1))) where p C D p C D = dim k (k[x, y]/ f(x, y, 1), g(x, y, 1) ) = dim k (Γ ) Γ = k[x, y]/ f(x, y, 1), g(x, y, 1). We thus have to prove that dim k (Γ ) = mn. We will do this by showing that Γ = Γd, where Γ d is the kvector space of dforms in Γ = k[x, y, z]/ f, g, whenever d m + n, and then we will show that dim k (Γ d ) = mn. Before we continue, we will fix some more notation. Let R = k[x, y, z], and let R d be the kvector space of all dforms in R. Remarks. (i) Let f, g d be the set of all dforms in f, g. Then Γ d = Rd / f, g d. Indeed, if h 1, h 2 Γ d, one can choose h 1, h 2 R d. So, if h 1 = h 2, then (h 1 h 2 ) R d f, g = f, g d. 19
20 (ii) If F is a polynomial of degree d in k[x, y], then z d F (x/z, y/z) is a dform in R. (iii) If F is a dform in R, then F = z d F (x/z, y/z, 1) Proposition. With the above notation, (i) dim k (R d ) = (d + 1)(d + 2)/2, (ii) dim k (Γ d ) = mn whenever d m + n. Proof: (i) There are (d + 1)(d + 2)/2 monomials x r y s z t of degree d = r + s + t in R. Moreover, any dform in R is a linear combination of monomials of degree d. (ii) Since Γ d = Rd / f, g d, it ise nough to show that dim k ( f, g d ) = dim k (R d ) mn. Define ϕ : R d m R d n f, g d by ϕ(a, b) = af + bg. Then ker(ϕ) = {(a, b) R d m R d n af + bg = 0}. If af + bg = 0, then af = bg, so since f and g do not have common factors, this implies that a = gc and b = fc for some c R d m n. Define ψ : R d m n ker(ϕ) by ψ(c) = gcf + ( f c)g. Clearly, ψ is klinear and injective, and the above argument establishes that it is surjective. Hence ker(ϕ) = R d m n, so that Therefore, f, g d = (Rd m R d n )/R d m n. dim k ( f, g d ) = dim k (R d m ) + dim k (R d n ) dim k (R d m n ) (d + 1)(d + 2) = mn 2 = dim k (R d ) mn, as desired Proposition. With the above notation, (i) the map ϕ : Γ d Γ given by ϕ(h) = h(x, y, 1) is a welldefined injective klinear map that is surjective for all d m + n, (ii) dim k (Γ ) = mn. 20
21 Proof: (i) For any r N, define α : Γ Γ by α(h) = z r h. We claim that α is injective. It is enough to show the case when r = 1, i.e. that if zh = 0, then h = 0. Note that since z does not divide either f or g, f(x, y, 0) and g(x, y, 0) are both not identically zero. Moreover, since f and g do not have a common factor, f(x, y, 0) and g(x, y, 0) are relatively prime. We will use these facts in the proof. Suppose that zh = 0. Then there exist a, b R such that zh = af + bg. Substituting z = 0, we have that so 0 = a(x, y, 0)f(x, y, 0) + b(x, y, 0)g(x, y, 0), a(x, y, 0)f(x, y, 0) = b(x, y, 0)g(x, y, 0), and since f(x, y, 0) and g(x, y, 0) are relatively prime, there exists a c k[x, y] such that Let Then and a(x, y, 0) = cf(x, y, 0) and b(x, y, 0) = cg(x, y, 0). a 1 = a + cg and b 1 = b cf. a 1 f + b 1 g = af + bg = zh, a 1 (x, y, 0) = b 1 (x, y, 0) = 0. Hence there exist a, b R such that implying that a 1 = za and b 1 = zb, zh = a 1 f + b 1 g = z(a f + b g), so h = a f + b g and h = 0, proving the injectivity of α. Consider the induced map α : Γ d Γ d+r. Then α is injective, but it is also surjective if d m + n, since by part (ii) Proposition Γ t is a kvector space of dimension mn for all t m + n. Define a klinear map ϕ : Γ d Γ by ϕ(h) = h(x, y, 1). We must first check that ϕ is welldefined. Indeed, if h 1 = h 2 in Γ d, then h 1 = h 2 = af + bg, so that h(x, y, 1) = h 2 (x, y, 1) + a(x, y, 1)f(x, y, 1) + b(x, y, 1)g(x, y, 1), implying that h 1 (x, y, 1) = h 2 (x, y, 1) in Γ. We will now show that ϕ is a bijection. Suppose h Γ d, where h is a dform, and h(x, y, 1) = 0. Then h(x, y, 1) = a(x, y)f(x, y, 1) + b(x, y)g(x, y, 1), 21
22 so for sufficiently large t d, z t m a(x/z, y/z) and z t n b(x/z, y/z) are both forms and z t d h = z t h(x/z, y/z, 1) = af + bg, so z t d h = 0 in Γ t, and h = 0, because the map α constructed early was injective. Fix Q Γ. Let s = deg(q) and let t = max s, d. Then q = z t Q(x/z, y/z) is a tform, so q Γ t = Γd, since t d m + n. Hence q = z t d h for some h Γ d. Therefore, Q(x, y) = q(x, y, 1) = h(x, y, 1) = ϕ(h), showing that ϕ is surjective. (ii) By (i), Γ d and Γ are isomorphic as kvector spaces whenever d m + n. By part (ii) of Proposition 2.2.2, dim k (Γ d ) = mn, so dim k (Γ ) = mn. Therefore, by the discussion preceding these propositions, we have established Bézout s Theorem. There are a number of immediate corollaries Corollary. Let C = V p (f) and D = V p (g) be projective plane curves, of degrees m and n respectively. If C and D have no common component, then m q (f) m q (g) I(q, C D) = mn. q C D q C D Corollary. Let C = V p (f) and D = V p (g) be projective plane curves, of degrees m and n respectively. If C and D intersect in mn distinct points, then these points are smooth points on C and D. Proof: In this case, the previous corollary implies that m q (f) = m q (g) = 1, which is equivalent to q being a smooth point on both C and D Corollary. Let C = V p (f) and D = V p (g) be projective plane curves, of degrees m and n respectively. if C and D intersect in more than mn points, counting multiplicity, then they have a common component. We continue with some less obvious applications of Bézout s Theorem Proposition. Any smooth projective plane curve is irreducible. Proof: Suppose instead that the smooth projective plane curve C = V p (f) is reducible, i.e. that f = ab for some forms a, b k[x, y, z]. By Bézout s Theorem, V p (a) V p (b). Fix q V p (a) V p (b). Then, since f = ab, m q (f) = m q (ab) = m q (a) + m q (b) 2, showing that C is not smooth at q. 22
23 Remark. The preceding proposition is certainly not true for affine plane curves, since an affine plane curve can be the disjoint union of smooth curves, e.g. C = V a (x) V a (x 1) A 2. There is a converse to the above proposition for degree 2 curves Proposition. Let C be an irreducible projective plane curve of degree 2. Then C is smooth. Proof: Suppose instead that C is singular at p. Then if C = V p (f), m p (f) 2. Let q be another point on C and let L be the line joining p and q. Let h be a 1form such that L = V p (h). By Bézout s Theorem, p and q are the only two points of intersection of L and C. If L is not a component of C, then by Corollary 2.2.4, 2 = deg(l) deg(c) = deg(h) deg(f) = I(p, L C) + I(q, L C) m p (L) m p (C) + m q (L) m p (C) = 3, which is absurd, so L must be a component of C, showing that C is reducible, a contradiction to our original assumption that C is singular at p. Therefore, C is smooth. Remark. In particular, this shows that any singular curve of degree 2 in P 2 is the union of lines that intersect at the singular point, e.g. C = V p (x 2 y 2 ) = V p (x y) V p (x + y). 23
24
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