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

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1 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. 1. Describe the Zariski topology on V (XY ) A 2. Solution: The algebraic set V (XY ) consists of the union of the two coordinate axis Y = 0 and X = 0. The proper closed subsets are given by the whole X-axis, the whole Y -axis and subsets consisting of finitely many points. 2. Assume that char(k) 2, 3. Show that the polynomial Y 2 X 3 X K[X, Y ] is irreducible. Describe the Zariski topology on V (Y 2 X 3 X) A 2. Solution: We consider the polynomial as an element in the ring (K[X])[Y ]. In the ring K[X] the element X is prime and divides X 3 + X, but its square does not. Using Eisenstein s criterion for the irreducibility of a polynomial, we deduce that Y 2 X 3 X is irreducible. We conclude that V (Y 2 X 3 X) A 2 is an irreducible algebraic variety. Since K[X, Y ] has Krull dimension 2 and (Y 2 X 3 X) is a nonzero prime ideal the coordinate ring O(V (Y 2 X 3 X)) = K[X, Y ]/(Y 2 X 3 X) and thus also the variety has dimension 1. Hence the only proper closed subsets are given by finitely many points. 3. Let C 1, C 2 A 2 be two algebraic curves where C 2 is irreducible. Show that either C 1 C 2 is a finite set or C 2 C 1. Solution: Let a, p K[X, Y ] be ideals such that V (a) = C 1 and V (p) = C 2. Since C 2 is irreducible p is a prime ideal. We have V (a) V (p) = V (a + p). If a p we conclude that C 2 C 1. Otherwise, the irreducible subsets of C 1 C 2 are given by V (m), where m are prime ideals which contain a + p p. Because of the dimension of K[X, Y ] we conclude that all such m are maximal ideals and thus V (m) are points. Every algebraic set can be covered by finitely many irreducible subsets, hence C 1 C 2 is the union of finitely many points. 4. Let A A n and B A m be two algebraic sets. Prove that their product set A B A n+m is algebraic, too. Solution: Let a K[X 1,..., X n ] and b K[Y 1,..., Y m ] be ideals such that A = V (a) and B = V (b). Let f 1,..., f r a and g 1,..., g s b be respective generators of the ideals. It is straightforward to see that A B = V (f 1,..., f r, g 1,..., g s ), where we extended tacitly to the ring K[X 1,..., X n, Y 1,..., Y m ]. 1

2 5. Consider the group SL n (C). Note that it is given by the vanishing set of a polynomial and is thus also an algebraic set in A n2 C, i.e. an algebraic group. Let H SL n (C) be a subgroup. Show that the Zariski closure H is still a subgroup of SL n (C). [Hint: observe that the multiplication and inversion maps are morphisms.] Solution: Let i : SL n (C) SL n (C) be the inversion morphism. Since H is a subgroup we have i : H H. But i is a homeomorphism for the Zariski topology and thus maps i : H H. Hence H is closed under inversion. Now consider the multiplication morphism with a fixed element x H given by SL n (C) SL n (C), a xa. This is a homeomorphism, too and hence xh = H for all x H. We conclude that for all x H we have Hx H. By the same argument as before we conclude that Hx H. Hence H is closed under multiplication. It follows that H is a subgroup of SL n (C). 6. Determine the Zariski closure for the following subsets: (a) {(x, sin(x)) x C} A 2 C (b) SL 2 (Z) SL 2 (C) A 4 C [Hint: Consider the subgroups ( ) 1 Z and 0 1 (c) {(a 2 b 2, 2ab, a 2 + b 2 ) a, b Z} A 3 C. Solution:.] Z 1 (a) Denote the subset by A. We claim that A = A 2 C. To prove this we compute the dimension. Since A is not a finite set, we conclude that the dimension of A must be 1 or 2. Assume that it is 1. Then A is a curve. However, for all real values 1 y 1 the intersection with the irreducible curve given by C y := {(x, y) x C} has infinitely many points. We conclude using exercise 3 that C y A for all real 1 y 1, so A contains infinitely many different curves. But A is an algebraic set and thus must be coverable by finitely many irreducible curves which is a contradiction. We conclude that the dimension of A is 2 and since A 2 C is irreducible we have proven the claim. Aliter: Consider the associated ideal I(A) K[X, Y ]. A polynomial f I(A) must in particular vanish on all points (x, sin(x)). Hence for any x R the polynomial f(x, sin(x)) K[X] has infinitely many roots X = 2πZ + x. Thus f(x, sin(x)) = 0. Since this holds for all x R and sin(r) = [ 1, 1] we conclude that f = 0. Hence I(A) = (0) and A = A 2 C. (b) Since Z is infinite and A 1 C is irreducible, we conclude ( ) that ( the closure ) Z = A1 C. 1 Z 1 C We deduce that the closure of the subgroup is and similarly

3 is Z 1 C 1 the closure of. From exercise 5 we know that SL 2 (Z) is a ( ) ( ) 1 C 1 0 group and from the preceding we know that it contains and. 0 1 C 1 Notice that we have ( ) ( ) ( ) ( ) 1 a c 1 + ab c + abc + a = 0 1 b b 1 + bc ( ) x y Hence all matrices in SL 2 (C) of the form with z 0 are contained z w in SL 2 (Z). This is a Zariski dense subset of SL 2 (C), because A 4 C V (Z) is Zariski dense in A 4 C. We conclude that SL 2(Z) = SL 2 (C). (c) Let A be the subset in question. These points are all zeroes of the polynomial X 2 +Y 2 Z 2, they are Pythagorian triples. Hence A V (X 2 +Y 2 Z 2 ). The polynomial X 2 +Y 2 Z 2 is irreducible in K[X, Y, Z], which can be seen by the Eisenstein criterion with Y Z in the ring (K[Y, Z])[X]. Therefore V (X 2 + Y 2 Z 2 ) is irreducible of dimension 2. We only need to show that A has dimension 2. Containing infinitely many points, we know that it does not have dimension 0. Assume that it has dimension 1. For all c Z, the intersection of A with the irreducible curve given by C c := {(c 2 b 2, 2cb, c 2 + b 2 ) b C} has infinitely many points. By exercise 3 we conclude that C c A for all c Z, hence A contains infinitely many different curves, which proves that it cannot have dimension 1. Thus A has dimension 2 and it follows that it is equal to V (X 2 + Y 2 Z 2 ). 7. Let ϕ : A 1 A 2 be defined by t (t 2, t 3 ). Show that ϕ defines a bijective bicontinuous morphism of A 1 onto the curve y 2 = x 3, but that ϕ is not an isomorphism. This shows that not every morphism whose underlying map of topological spaces is a homeomorphism needs to be an isomorphism. Solution: We see that the image of ϕ is contained in V (y 2 x 3 ). It is bijective with inverse ϕ 1 : (a, b) b a for a 0 and ϕ 1 (0, 0) = 0. Closed sets in A 1 are finitely many points and they are mapped via ϕ to finitely many points (i.e. closed sets) in A 2, so ϕ is a closed map. Hence ϕ and ϕ 1 are both continuous. So ϕ is bijective and bicontinuous. However, it is not an isomorphism. To see this let f : A 1 k be the canonical regular function t t. Note that f ϕ 1 : (a, b) b a is not regular at a = 0, hence ϕ 1 is not a morphism. 8. Let Y A 3 be the set Y := {(t, t 2, t 3 ) t K }. Show that Y is an affine variety of dimension 1. Find generators for the ideal I(Y ) and prove that the coordinate ring O(Y ) is isomorphic to a polynomial ring in one variable over K. Solution: The ideal I(Y ) is equal to (v u 2, w u 3 ) K[u, v, w]. We have the equality Y = V (v u 2, w u 3 ). Note that K[u, v, w]/(v u 2, w u 3 ) = 3

4 K[u, u 2, u 3 ] = K[u]. This proves that O(Y ) is isomorphic to a polynomial ring in one variable over K and furthermore that (v u 2, w u 3 ) is a prime ideal of coheight 1, so Y is an irreducible affine variety of dimension The Segre Embedding. Let ψ : P r P s P N be the map defined by sending the ordered pair (a 0,..., a r ) (b 0,..., b s ) to (..., a i b j,... ) in lexicographic order, where N = rs + r + s. Show that ψ is well-defined and injective. It is called the Segre embedding. Prove that the image of ψ is a projective algebraic set in P N. Solution: The map ψ is homogeneous and thus well-defined. Denote for all indices 0 i r, 0 j s by x ij the coordinates of a point in P N. The points in the image of ψ satisfy x ij x kl = x kj x il. Let Y be the projective algebraic set defined by these equations. Let Q := [x ij ] Y be a point. Then at least one coordinate x kl is non-zero. Using that we are in projective space we have Q = [x kl x ij ] = [x il x kj ] = ψ([x 0l : : x rl ], [x k0 : : x ks ]). Which proves that Y = im(ψ), hence the image of ψ is a projective algebraic set. The above also provides a left inverse to ψ which proves that ψ is injective. 10. Consider the surface Q (i.e. variety of dimension 2) in P 3 defined by the equation xy zw. (a) Show that Q is equal to the image of the Segre embedding of P 1 P 1 in P 3, for suitable choice of coordinates. (b) Show that Q contains two families of lines (i.e. linear varieties of dimension 1) {L t }, {M t }, each parametrized by t P 1, with the property that if L t L u, then L t L u = ; if M t M u then M t M u = and for all t, u we have L t M u = one point. (c) Show that Q contains other curves besides these lines and deduce that the Zariski topology on Q is not homeomorphic via the Segre embedding to the product topology on P 1 P 1. Solution: (a) The image of the Segre embedding is given by im(ψ) = { [a 0 b 0 : a 0 b 1 : a 1 b 0 : a 1 b 1 ] [a 0 : a 1 ], [b 0 : b 1 ] P 1 } We see that every point [x : w : z : y] in the image of ψ satisfies the equation xy zw = 0. Conversely suppose that a point [x : w : z : y] P 3 satisfies xy zw = 0. If x 0, then the image of ([x : z], [x : w]) under the Segre embedding is the point [xx : xw : xz : wz] = [x : w : z : y]. Similarly for w 0 take ([w : y], [x : w]), for z 0 take ([x : z], [z : y]) and for y 0 take ([w : y], [z : y]) to prove that every point [x : w : z : y] with xy zw = 0 is always in the image of the Segre embedding. Thus the image of the Segre embedding is equal to V (xy zw) = Q. 4

5 (b) For any t P 1 we define L t to be ψ(p 1 {t}) and for any u P 1 we define M u to be ψ({u} P 1 ). We have shown in exercise 9 that ψ is injective. Thus L t L u = and M t M u = for t u. Furthermore it follows that L t M u = ψ(u, t), which is one point. (c) The surface Q contains the curve C := V (xy zw, w z) = ψ ({ ([x : z], [x : z]) [x : z] P 1 }). By bijectivity of ψ, we know that C L t and C M t are both one point for all t P 1, hence we conclude that C is a different curve. The curve C is closed in P 3, but the set {([x : z], [x : z]) [x : z] P 1 } is not closed in the product topology of P 1 P 1, because it is the diagonal of a non-hausdorff space. 11. Let n, d > 0 be integers. We denote by M 0,..., M N all monomials of degree d in the n + 1 variables x 0,..., x n, where N := ( n+d n ) 1. We define the d- uple embedding as the map ρ d : P n P N sending the point a = (a 0,..., a n ) to the point (M 0 (a),..., M N (a)). Show that the d-uple embedding of P n is an isomorphism onto its image. [Hint: Look at the inverse map.] Solution: For 0 i, j n denote by M ij the monomial x d 1 i x j and denote for a point [b 0 : : b N ] in the image of ρ d by b ij the coordinate corresponding to the monomial M ij. Note that at least one b ii is non-zero. On the chart b ii 0, define the map ϕ i : [b 0 : : b N ] [b i0 : : b in ]. These maps ϕ i glue to an inverse of ρ d on the whole image. Since they are defined only by projecting to certain coordinates, they are regular and thus glue to a morphism. Hence ρ d is an isomorphism onto its image. 5

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