# Projective Varieties. Chapter Projective Space and Algebraic Sets

Size: px
Start display at page:

Transcription

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 n-dimensional 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 n-space. 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 i-form. 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 i-form, 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 one-to-one correspondence, but in order to make it one-to-one 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 i-form and each g j is a j-form. 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 non-empty, 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 non-empty 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 non-zero 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 one-to-one 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 one-point compactification of X in P Definition. A non-empty 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 one-to-one correspondences: 8

9 (non-empty 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 m-form if it contains an m-form. In particular, 0-forms 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 i-form. 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 i-form and each g j is a j-form. 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 k-algebras 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 k-algebras. 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 k-algebra 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, 3-folds 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 r-dimensional projective variety and p Y. The k-vector 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 1-dimensional k-vector 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 k-algebras. 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 well-defined. 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 well-defined 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 2-form 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 non-rational 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 non-zero 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 1-forms or irreducible 2-forms 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 3-forms, 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 1-form ax + by + cz k[x, y, z]. At least one of the constants a, b, c must 18

19 be non-zero. 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 k-vector space of d-forms 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 k-vector space of all d-forms in R. Remarks. (i) Let f, g d be the set of all d-forms 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 d-form in R. (iii) If F is a d-form 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 d-form 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 k-linear 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 well-defined injective k-linear 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 k-vector space of dimension mn for all t m + n. Define a k-linear map ϕ : Γ d Γ by ϕ(h) = h(x, y, 1). We must first check that ϕ is well-defined. 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 d-form, 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 t-form, 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 k-vector 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 1-form 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

### Pure Math 764, Winter 2014

Compact course notes Pure Math 764, Winter 2014 Introduction to Algebraic Geometry Lecturer: R. Moraru transcribed by: J. Lazovskis University of Waterloo April 20, 2014 Contents 1 Basic geometric objects

### Math 203A - Solution Set 3

Math 03A - Solution Set 3 Problem 1 Which of the following algebraic sets are isomorphic: (i) A 1 (ii) Z(xy) A (iii) Z(x + y ) A (iv) Z(x y 5 ) A (v) Z(y x, z x 3 ) A Answer: We claim that (i) and (v)

### 12. Hilbert Polynomials and Bézout s Theorem

12. Hilbert Polynomials and Bézout s Theorem 95 12. Hilbert Polynomials and Bézout s Theorem After our study of smooth cubic surfaces in the last chapter, let us now come back to the general theory of

### div(f ) = D and deg(d) = deg(f ) = d i deg(f i ) (compare this with the definitions for smooth curves). Let:

Algebraic Curves/Fall 015 Aaron Bertram 4. Projective Plane Curves are hypersurfaces in the plane CP. When nonsingular, they are Riemann surfaces, but we will also consider plane curves with singularities.

### 2. Intersection Multiplicities

2. Intersection Multiplicities 11 2. Intersection Multiplicities Let us start our study of curves by introducing the concept of intersection multiplicity, which will be central throughout these notes.

### MATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties

MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently,

### 3. The Sheaf of Regular Functions

24 Andreas Gathmann 3. The Sheaf of Regular Functions After having defined affine varieties, our next goal must be to say what kind of maps between them we want to consider as morphisms, i. e. as nice

### 10. Smooth Varieties. 82 Andreas Gathmann

82 Andreas Gathmann 10. Smooth Varieties Let a be a point on a variety X. In the last chapter we have introduced the tangent cone C a X as a way to study X locally around a (see Construction 9.20). It

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### Algebraic Varieties. Chapter Algebraic Varieties

Chapter 12 Algebraic Varieties 12.1 Algebraic Varieties Let K be a field, n 1 a natural number, and let f 1,..., f m K[X 1,..., X n ] be polynomials with coefficients in K. Then V = {(a 1,..., a n ) :

### Math 203A - Solution Set 1

Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in

### DIVISORS ON NONSINGULAR CURVES

DIVISORS ON NONSINGULAR CURVES BRIAN OSSERMAN We now begin a closer study of the behavior of projective nonsingular curves, and morphisms between them, as well as to projective space. To this end, we introduce

### Math 418 Algebraic Geometry Notes

Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = {P R

### Math 203A - Solution Set 1

Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in

### where m is the maximal ideal of O X,p. Note that m/m 2 is a vector space. Suppose that we are given a morphism

8. Smoothness and the Zariski tangent space We want to give an algebraic notion of the tangent space. In differential geometry, tangent vectors are equivalence classes of maps of intervals in R into the

### This is a closed subset of X Y, by Proposition 6.5(b), since it is equal to the inverse image of the diagonal under the regular map:

Math 6130 Notes. Fall 2002. 7. Basic Maps. Recall from 3 that a regular map of affine varieties is the same as a homomorphism of coordinate rings (going the other way). Here, we look at how algebraic properties

### INTERSECTION NUMBER OF PLANE CURVES

INTERSECTION NUMBER OF PLANE CURVES MARGARET E. NICHOLS 1. Introduction One of the most familiar objects in algebraic geometry is the plane curve. A plane curve is the vanishing set of a polynomial in

### Resolution of Singularities in Algebraic Varieties

Resolution of Singularities in Algebraic Varieties Emma Whitten Summer 28 Introduction Recall that algebraic geometry is the study of objects which are or locally resemble solution sets of polynomial equations.

### Yuriy Drozd. Intriduction to Algebraic Geometry. Kaiserslautern 1998/99

Yuriy Drozd Intriduction to Algebraic Geometry Kaiserslautern 1998/99 CHAPTER 1 Affine Varieties 1.1. Ideals and varieties. Hilbert s Basis Theorem Let K be an algebraically closed field. We denote by

### Local properties of plane algebraic curves

Chapter 7 Local properties of plane algebraic curves Throughout this chapter let K be an algebraically closed field of characteristic zero, and as usual let A (K) be embedded into P (K) by identifying

### CHAPTER 1. AFFINE ALGEBRAIC VARIETIES

CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the

### LECTURE 5, FRIDAY

LECTURE 5, FRIDAY 20.02.04 FRANZ LEMMERMEYER Before we start with the arithmetic of elliptic curves, let us talk a little bit about multiplicities, tangents, and singular points. 1. Tangents How do we

### Algebraic Geometry (Math 6130)

Algebraic Geometry (Math 6130) Utah/Fall 2016. 2. Projective Varieties. Classically, projective space was obtained by adding points at infinity to n. Here we start with projective space and remove a hyperplane,

### ALGEBRAIC GROUPS. Disclaimer: There are millions of errors in these notes!

ALGEBRAIC GROUPS Disclaimer: There are millions of errors in these notes! 1. Some algebraic geometry The subject of algebraic groups depends on the interaction between algebraic geometry and group theory.

### Algebraic Geometry. Instructor: Stephen Diaz & Typist: Caleb McWhorter. Spring 2015

Algebraic Geometry Instructor: Stephen Diaz & Typist: Caleb McWhorter Spring 2015 Contents 1 Varieties 2 1.1 Affine Varieties....................................... 2 1.50 Projective Varieties.....................................

### Institutionen för matematik, KTH.

Institutionen för matematik, KTH. Contents 7 Affine Varieties 1 7.1 The polynomial ring....................... 1 7.2 Hypersurfaces........................... 1 7.3 Ideals...............................

### Projective Spaces. Chapter The Projective Line

Chapter 3 Projective Spaces 3.1 The Projective Line Suppose you want to describe the lines through the origin O = (0, 0) in the Euclidean plane R 2. The first thing you might think of is to write down

### Math 203A - Solution Set 1

Math 203A - Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

### Math Midterm Solutions

Math 145 - Midterm Solutions Problem 1. (10 points.) Let n 2, and let S = {a 1,..., a n } be a finite set with n elements in A 1. (i) Show that the quasi-affine set A 1 \ S is isomorphic to an affine set.

### MATH 8253 ALGEBRAIC GEOMETRY WEEK 12

MATH 8253 ALGEBRAIC GEOMETRY WEEK 2 CİHAN BAHRAN 3.2.. Let Y be a Noetherian scheme. Show that any Y -scheme X of finite type is Noetherian. Moreover, if Y is of finite dimension, then so is X. Write f

### Algebraic Geometry. Contents. Diane Maclagan Notes by Florian Bouyer. Copyright (C) Bouyer 2011.

Algebraic Geometry Diane Maclagan Notes by Florian Bouyer Contents Copyright (C) Bouyer 2011. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation

### Elliptic Curves and Public Key Cryptography (3rd VDS Summer School) Discussion/Problem Session I

Elliptic Curves and Public Key Cryptography (3rd VDS Summer School) Discussion/Problem Session I You are expected to at least read through this document before Wednesday s discussion session. Hopefully,

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton. Timothy J. Ford April 4, 2016

MAS 6396 Algebraic Curves Spring Semester 2016 Notes based on Algebraic Curves by Fulton Timothy J. Ford April 4, 2016 FLORIDA ATLANTIC UNIVERSITY, BOCA RATON, FLORIDA 33431 E-mail address: ford@fau.edu

### HARTSHORNE EXERCISES

HARTSHORNE EXERCISES J. WARNER Hartshorne, Exercise I.5.6. Blowing Up Curve Singularities (a) Let Y be the cusp x 3 = y 2 + x 4 + y 4 or the node xy = x 6 + y 6. Show that the curve Ỹ obtained by blowing

### CHEVALLEY S THEOREM AND COMPLETE VARIETIES

CHEVALLEY S THEOREM AND COMPLETE VARIETIES BRIAN OSSERMAN In this note, we introduce the concept which plays the role of compactness for varieties completeness. We prove that completeness can be characterized

### k k would be reducible. But the zero locus of f in A n+1

Math 145. Bezout s Theorem Let be an algebraically closed field. The purpose of this handout is to prove Bezout s Theorem and some related facts of general interest in projective geometry that arise along

### Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #18 11/07/2013 As usual, all the rings we consider are commutative rings with an identity element. 18.1 Regular local rings Consider a local

### ALGEBRAIC GEOMETRY CAUCHER BIRKAR

ALGEBRAIC GEOMETRY CAUCHER BIRKAR Contents 1. Introduction 1 2. Affine varieties 3 Exercises 10 3. Quasi-projective varieties. 12 Exercises 20 4. Dimension 21 5. Exercises 24 References 25 1. Introduction

### Basic facts and definitions

Synopsis Thursday, September 27 Basic facts and definitions We have one one hand ideals I in the polynomial ring k[x 1,... x n ] and subsets V of k n. There is a natural correspondence. I V (I) = {(k 1,

### FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS.

FILTERED RINGS AND MODULES. GRADINGS AND COMPLETIONS. Let A be a ring, for simplicity assumed commutative. A filtering, or filtration, of an A module M means a descending sequence of submodules M = M 0

### Exercises for algebraic curves

Exercises for algebraic curves Christophe Ritzenthaler February 18, 2019 1 Exercise Lecture 1 1.1 Exercise Show that V = {(x, y) C 2 s.t. y = sin x} is not an algebraic set. Solutions. Let us assume that

### Introduction to Algebraic Geometry. Jilong Tong

Introduction to Algebraic Geometry Jilong Tong December 6, 2012 2 Contents 1 Algebraic sets and morphisms 11 1.1 Affine algebraic sets.................................. 11 1.1.1 Some definitions................................

### Rings and groups. Ya. Sysak

Rings and groups. Ya. Sysak 1 Noetherian rings Let R be a ring. A (right) R -module M is called noetherian if it satisfies the maximum condition for its submodules. In other words, if M 1... M i M i+1...

### Algebraic Geometry. Andreas Gathmann. Class Notes TU Kaiserslautern 2014

Algebraic Geometry Andreas Gathmann Class Notes TU Kaiserslautern 2014 Contents 0. Introduction......................... 3 1. Affine Varieties........................ 9 2. The Zariski Topology......................

### Math 145. Codimension

Math 145. Codimension 1. Main result and some interesting examples In class we have seen that the dimension theory of an affine variety (irreducible!) is linked to the structure of the function field in

### Summer Algebraic Geometry Seminar

Summer Algebraic Geometry Seminar Lectures by Bart Snapp About This Document These lectures are based on Chapters 1 and 2 of An Invitation to Algebraic Geometry by Karen Smith et al. 1 Affine Varieties

### The most important result in this section is undoubtedly the following theorem.

28 COMMUTATIVE ALGEBRA 6.4. Examples of Noetherian rings. So far the only rings we can easily prove are Noetherian are principal ideal domains, like Z and k[x], or finite. Our goal now is to develop theorems

### 10. Noether Normalization and Hilbert s Nullstellensatz

10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.

### INTRODUCTION TO ALGEBRAIC GEOMETRY. Throughout these notes all rings will be commutative with identity. k will be an algebraically

INTRODUCTION TO ALGEBRAIC GEOMETRY STEVEN DALE CUTKOSKY Throughout these notes all rings will be commutative with identity. k will be an algebraically closed field. 1. Preliminaries on Ring Homomorphisms

### Chapter 1. Affine algebraic geometry. 1.1 The Zariski topology on A n

Chapter 1 Affine algebraic geometry We shall restrict our attention to affine algebraic geometry, meaning that the algebraic varieties we consider are precisely the closed subvarieties of affine n- space

### Introduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #15 10/29/2013 As usual, k is a perfect field and k is a fixed algebraic closure of k. Recall that an affine (resp. projective) variety is an

### be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore

### (1) is an invertible sheaf on X, which is generated by the global sections

7. Linear systems First a word about the base scheme. We would lie to wor in enough generality to cover the general case. On the other hand, it taes some wor to state properly the general results if one

### Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #23 11/26/2013 As usual, a curve is a smooth projective (geometrically irreducible) variety of dimension one and k is a perfect field. 23.1

### COMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY

COMPLEX VARIETIES AND THE ANALYTIC TOPOLOGY BRIAN OSSERMAN Classical algebraic geometers studied algebraic varieties over the complex numbers. In this setting, they didn t have to worry about the Zariski

### ALGEBRAIC GEOMETRY (NMAG401) Contents. 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30

ALGEBRAIC GEOMETRY (NMAG401) JAN ŠŤOVÍČEK Contents 1. Affine varieties 1 2. Polynomial and rational maps 9 3. Hilbert s Nullstellensatz and consequences 23 References 30 1. Affine varieties The basic objects

### π X : X Y X and π Y : X Y Y

Math 6130 Notes. Fall 2002. 6. Hausdorffness and Compactness. We would like to be able to say that all quasi-projective varieties are Hausdorff and that projective varieties are the only compact varieties.

### The Group Structure of Elliptic Curves Defined over Finite Fields

The Group Structure of Elliptic Curves Defined over Finite Fields A Senior Project submitted to The Division of Science, Mathematics, and Computing of Bard College by Andrija Peruničić Annandale-on-Hudson,

### 14. Rational maps It is often the case that we are given a variety X and a morphism defined on an open subset U of X. As open sets in the Zariski

14. Rational maps It is often the case that we are given a variety X and a morphism defined on an open subset U of X. As open sets in the Zariski topology are very large, it is natural to view this as

### NONSINGULAR CURVES BRIAN OSSERMAN

NONSINGULAR CURVES BRIAN OSSERMAN The primary goal of this note is to prove that every abstract nonsingular curve can be realized as an open subset of a (unique) nonsingular projective curve. Note that

### MATH 631: ALGEBRAIC GEOMETRY: HOMEWORK 1 SOLUTIONS

MATH 63: ALGEBRAIC GEOMETRY: HOMEWORK SOLUTIONS Problem. (a.) The (t + ) (t + ) minors m (A),..., m k (A) of an n m matrix A are polynomials in the entries of A, and m i (A) = 0 for all i =,..., k if and

### Algebraic Geometry I Lectures 14 and 15

Algebraic Geometry I Lectures 14 and 15 October 22, 2008 Recall from the last lecture the following correspondences {points on an affine variety Y } {maximal ideals of A(Y )} SpecA A P Z(a) maximal ideal

### THE ENVELOPE OF LINES MEETING A FIXED LINE AND TANGENT TO TWO SPHERES

6 September 2004 THE ENVELOPE OF LINES MEETING A FIXED LINE AND TANGENT TO TWO SPHERES Abstract. We study the set of lines that meet a fixed line and are tangent to two spheres and classify the configurations

### A course in. Algebraic Geometry. Taught by Prof. Xinwen Zhu. Fall 2011

A course in Algebraic Geometry Taught by Prof. Xinwen Zhu Fall 2011 1 Contents 1. September 1 3 2. September 6 6 3. September 8 11 4. September 20 16 5. September 22 21 6. September 27 25 7. September

### 4. Images of Varieties Given a morphism f : X Y of quasi-projective varieties, a basic question might be to ask what is the image of a closed subset

4. Images of Varieties Given a morphism f : X Y of quasi-projective varieties, a basic question might be to ask what is the image of a closed subset Z X. Replacing X by Z we might as well assume that Z

### Elliptic Curves Spring 2017 Lecture #5 02/22/2017

18.783 Elliptic Curves Spring 017 Lecture #5 0//017 5 Isogenies In almost every branch of mathematics, when considering a category of mathematical objects with a particular structure, the maps between

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE ABOUT VARIETIES AND REGULAR FUNCTIONS.

ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS. ANDREW SALCH. More about some claims from the last lecture. Perhaps you have noticed by now that the Zariski topology

### Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013

18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #17 11/05/2013 Throughout this lecture k denotes an algebraically closed field. 17.1 Tangent spaces and hypersurfaces For any polynomial f k[x

### Algebraic Varieties. Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra

Algebraic Varieties Notes by Mateusz Micha lek for the lecture on April 17, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra Algebraic varieties represent solutions of a system of polynomial

### Plane Algebraic Curves

Plane Algebraic Curves Andreas Gathmann Class Notes TU Kaiserslautern 2018 Contents 0. Introduction......................... 3 1. Affine Curves......................... 6 2. Intersection Multiplicities.....................

### The Topology and Algebraic Functions on Affine Algebraic Sets Over an Arbitrary Field

Georgia State University ScholarWorks @ Georgia State University Mathematics Theses Department of Mathematics and Statistics Fall 11-15-2012 The Topology and Algebraic Functions on Affine Algebraic Sets

### Draft: January 2, 2017 INTRODUCTION TO ALGEBRAIC GEOMETRY

Draft: January 2, 2017 INTRODUCTION TO ALGEBRAIC GEOMETRY MATTHEW EMERTON Contents 1. Introduction and overview 1 2. Affine space 2 3. Affine plane curves 3 4. Projective space 7 5. Projective plane curves

### Math 40510, Algebraic Geometry

Math 40510, Algebraic Geometry Problem Set 1, due February 10, 2016 1. Let k = Z p, the field with p elements, where p is a prime. Find a polynomial f k[x, y] that vanishes at every point of k 2. [Hint:

### LECTURE Affine Space & the Zariski Topology. It is easy to check that Z(S)=Z((S)) with (S) denoting the ideal generated by elements of S.

LECTURE 10 1. Affine Space & the Zariski Topology Definition 1.1. Let k a field. Take S a set of polynomials in k[t 1,..., T n ]. Then Z(S) ={x k n f(x) =0, f S}. It is easy to check that Z(S)=Z((S)) with

### A MODEL-THEORETIC PROOF OF HILBERT S NULLSTELLENSATZ

A MODEL-THEORETIC PROOF OF HILBERT S NULLSTELLENSATZ NICOLAS FORD Abstract. The goal of this paper is to present a proof of the Nullstellensatz using tools from a branch of logic called model theory. In

### 12. Linear systems Theorem Let X be a scheme over a ring A. (1) If φ: X P n A is an A-morphism then L = φ O P n

12. Linear systems Theorem 12.1. Let X be a scheme over a ring A. (1) If φ: X P n A is an A-morphism then L = φ O P n A (1) is an invertible sheaf on X, which is generated by the global sections s 0, s

### The set of points at which a polynomial map is not proper

ANNALES POLONICI MATHEMATICI LVIII3 (1993) The set of points at which a polynomial map is not proper by Zbigniew Jelonek (Kraków) Abstract We describe the set of points over which a dominant polynomial

### ABSTRACT NONSINGULAR CURVES

ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine n-space the collection A n k of points P = a 1, a,..., a

### Lecture 6. s S} is a ring.

Lecture 6 1 Localization Definition 1.1. Let A be a ring. A set S A is called multiplicative if x, y S implies xy S. We will assume that 1 S and 0 / S. (If 1 / S, then one can use Ŝ = {1} S instead of

### HILBERT FUNCTIONS. 1. Introduction

HILBERT FUCTIOS JORDA SCHETTLER 1. Introduction A Hilbert function (so far as we will discuss) is a map from the nonnegative integers to themselves which records the lengths of composition series of each

### 8. Prime Factorization and Primary Decompositions

70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings

### BEZOUT S THEOREM CHRISTIAN KLEVDAL

BEZOUT S THEOREM CHRISTIAN KLEVDAL A weaker version of Bézout s theorem states that if C, D are projective plane curves of degrees c and d that intersect transversally, then C D = cd. The goal of this

### FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 43 RAVI VAKIL CONTENTS 1. Facts we ll soon know about curves 1 1. FACTS WE LL SOON KNOW ABOUT CURVES We almost know enough to say a lot of interesting things about

### David Eklund. May 12, 2017

KTH Stockholm May 12, 2017 1 / 44 Isolated roots of polynomial systems Let f 1,..., f n C[x 0,..., x n ] be homogeneous and consider the subscheme X P n defined by the ideal (f 1,..., f n ). 2 / 44 Isolated

### 9. Birational Maps and Blowing Up

72 Andreas Gathmann 9. Birational Maps and Blowing Up In the course of this class we have already seen many examples of varieties that are almost the same in the sense that they contain isomorphic dense

### SPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS

SPACES OF RATIONAL CURVES IN COMPLETE INTERSECTIONS ROYA BEHESHTI AND N. MOHAN KUMAR Abstract. We prove that the space of smooth rational curves of degree e in a general complete intersection of multidegree

### IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.

IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. IAN KIMING 1. Non-singular points and tangents. Suppose that k is a field and that F (x 1,..., x n ) is a homogeneous polynomial in n variables

### D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski. Solutions Sheet 1. Classical Varieties

D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski Solutions Sheet 1 Classical Varieties Let K be an algebraically closed field. All algebraic sets below are defined over K, unless specified otherwise.

### Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed.

Reid 5.2. Describe the irreducible components of V (J) for J = (y 2 x 4, x 2 2x 3 x 2 y + 2xy + y 2 y) in k[x, y, z]. Here k is algebraically closed. Answer: Note that the first generator factors as (y

### TROPICAL SCHEME THEORY

TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),

### TWO IDEAS FROM INTERSECTION THEORY

TWO IDEAS FROM INTERSECTION THEORY ALEX PETROV This is an expository paper based on Serre s Local Algebra (denoted throughout by [Ser]). The goal is to describe simple cases of two powerful ideas in intersection

### B 1 = {B(x, r) x = (x 1, x 2 ) H, 0 < r < x 2 }. (a) Show that B = B 1 B 2 is a basis for a topology on X.

Math 6342/7350: Topology and Geometry Sample Preliminary Exam Questions 1. For each of the following topological spaces X i, determine whether X i and X i X i are homeomorphic. (a) X 1 = [0, 1] (b) X 2

### ARCS IN FINITE PROJECTIVE SPACES. Basic objects and definitions

ARCS IN FINITE PROJECTIVE SPACES SIMEON BALL Abstract. These notes are an outline of a course on arcs given at the Finite Geometry Summer School, University of Sussex, June 26-30, 2017. Let K denote an

### V (f) :={[x] 2 P n ( ) f(x) =0}. If (x) ( x) thenf( x) =

20 KIYOSHI IGUSA BRANDEIS UNIVERSITY 2. Projective varieties For any field F, the standard definition of projective space P n (F ) is that it is the set of one dimensional F -vector subspaces of F n+.

### Polynomials, Ideals, and Gröbner Bases

Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields

### Algebraic Geometry. 1 4 April David Philipson. 1.1 Basic Definitions. 1.2 Noetherian Rings

Algebraic Geometry David Philipson 1 4 April 2008 Notes for this day courtesy of Yakov Shlapentokh-Rothman. 1.1 Basic Definitions Throughout this course, we let k be an algebraically closed field with