Algebraic Geometry Sample exam solutions 2017 Question 1. Sub-question 1(a). Define an affine algebraic set. (3 marks) An affine algebraic set is a subset V ⊆ An which consists of the common zeros of some finite set of polynomials f1, . , fm with coefficients in k. Alternatively: an affine algebraic set is a set of the form n V = {(x1, . , xn) ∈ A : f1(x1, . , xn) = ··· = fm(x1, . , xn) = 0} for some polynomials f1, . , fm ∈ k[X1,...,Xn]. Sub-question 1(b). Prove that if V is an affine algebraic set, then V is a union of finitely many irreducible closed subsets. You may use Hilbert’s Basis Theorem provided you state it clearly and correctly. (8 marks) Suppose for contradiction that V is an affine algebraic set which cannot be written as a union of finitely many irreducible closed subsets. V is not itself irreducible, so V = V1 ∪ W1 where V1 and W1 are proper closed subsets of V . If each of V1 and W1 is a union of finitely many irreducible closed subsets, then V is also a union of finitely many irreducible closed subsets. So at least one of V1 and W1 is not a union of finitely many irreducible closed subsets. Without loss of generality, we may suppose that V1 is not a union of finitely many irreducible closed subsets. We repeat the argument: V1 is not reducible, so V1 = V2 ∪ W2 where V2 and W2 are proper closed subsets of V . At least one of V2 and W2 is not a union of finitely many irreducible closed subsets; suppose that V2 is not a union of finitely many irreducible closed subsets. Repeating this, we build up a descending chain of closed subsets V ⊃ V1 ⊃ V2 ⊃ · · · where all the inclusions are strict. Hence we get an ascending chain of ideals in k[X1,...,Xn]: I(V ) ⊆ I(V1) ⊆ I(V2) ⊆ · · · . Since V(I(W )) = W for every affine algebraic set W , the strictness of the inclusions between the Vi implies that the inclusions between the ideals I(Vi) are strict. Recall the statement of the Hilbert Basis Theorem. Theorem. Let I1 ⊆ I2 ⊆ I3 ⊆ · · · be an ascending chain of ideals in k[X1,...,Xn]. Then there is some N such that In = IN for every n > N. This contradicts the existence of the strictly ascending chain of ideals I(V ) ⊂ I(V1) ⊂ I(V2) ⊂ · · · . Sub-question 1(c). Find the irreducible components of the affine algebraic set 3 2 2 3 V = {(x, y, z) ∈ A : y = xz, z = y }. For each irreducible component C of V : (i) Write down polynomials which define C as a subset of A3. (ii) Prove that C is isomorphic to A1. (9 marks) Consider (x, y, z) ∈ V . The equation y2 = xz implies that y3 = xyz. Substituting in z2 = y3 gives z2 = xyz and so either z = 0 or z = xy. If z = 0 then the equations for V become y2 = 0, y3 = 0. Hence V contains the line y = z = 0. If z = xy, then we can substitute this into the equations for V to get y2 = x2y and x2y2 = y3. Hence y = 0 or y = x2. If y = 0 then z2 = y3 implies that z = 0, so we are back in the case considered previously. If y = x2, then we have z = x3. We can see that (x, x2, x3) satisfies the equations for V . Hence V = {(x, 0, 0) : x ∈ k} ∪ {(x, x2, x3): x ∈ k}. (i) The component {(x, 0, 0)} is defined by the equations y = z = 0. The component {(x, x2, x3)} is defined by the equations y = x2, z = xy. (ii) The component {(x, 0, 0)} is isomorphic to A1 via the inverse pair of regular maps x 7→ (x, 0, 0) and (x, y, z) 7→ x. The component {(x, x2, x3)} is isomorphic to A1 via the inverse pair of regular maps x 7→ (x, x2, x3) and (x, y, z) 7→ x. Question 2. In this question, you may use any theorems from the course provided you state them clearly and correctly. Let 3 Σ1,1 = {[w : x : y : z] ∈ P : wz = xy}. Sub-question 2(a). Define a regular map between projective algebraic sets. (3 marks) Let V ⊆ Pn and W ⊆ Pm be projective algebraic sets. A regular map ϕ: V → W is a function V → W such that for every point x ∈ V , there exist a Zariski open set U ⊆ V containing x and a sequence of polynomials f0, . , fm ∈ k[X0,...,Xn] such that: (i) f0, . , fm are homogeneous of the same degree; (ii) for every y ∈ U, f0, . , fm are not all zero at y; (iii) for every y = [y0 : ··· : yn] ∈ U, ϕ(y) = [f0(y0, . , yn): ··· : fm(y0, . , yn)]. 1 1 3 Sub-question 2(b). Write down the Segre embedding σ1,1 : P × P ⊆ P . (You should write it down in such a way that the image of σ1,1 is equal to Σ1,1, but you are not required to prove this.) (2 marks) σ1,1([s : t], [u : v]) = [su : sv : tu : tv]. (Note: other permutations of the coordinates are also possible, but only some of them satisfy the equation for Σ1,1.) 1 1 Sub-question 2(c). Write down two regular maps p1 :Σ1,1 → P , p2 :Σ1,1 → P such that (p1, p2) is the inverse of σ1,1. (You are not required to prove that (p1, p2) is inverse to σ1,1.) (2 marks) p1([w : x : y : z]) = [w : y] if (w, y) 6= (0, 0), p1([w : x : y : z]) = [x : z] if (x, z) 6= (0, 0). p2([w : x : y : z]) = [w : x] if (w, x) 6= (0, 0), p2([w : x : y : z]) = [y : z] if (y, z) 6= (0, 0). Observe that these definitions are well-defined everywhere on Σ1,1. (A different permutation of coordinates for σ1,1 in (b) will require a different per- mutation of coordinates here as well.) Sub-question 2(d). Prove that P1 × P1 and P2 are birational but not isomorphic. (7 marks) A1 × A1 is an open subset of P1 × P1, and so A1 × A1 is birational to P1 × P1. Trivially, A1 × A1 is isomorphic, and hence birational, to A2. A2 is an open subset of P2, and so A2 is birational to P2. Composing these birational equivalences, we conclude that P1 × P1 is birational to P2. Now we prove that P1 × P1 is not isomorphic to P2. (There are a variety of ways to prove this, here’s one.) Pick two points a, b ∈ P1. Then {a} × P1 and {b} × P1 are disjoint irreducible closed subsets of P1 × P1, each of dimension 1. We will prove that P2 cannot contain two disjoint irreducible closed subsets of dimension 1. Assume for contradiction that V, W ⊆ P2 are disjoint irreducible closed subsets of dimension 1. Let f ∈ k[X0,X1,X2] be a homogeneous polynomial which vanishes on V . If f is reducible, say f = f1f2, then we can write V as the union of the closed subsets V1 = {x ∈ V : f1(x) = 0} and V2 = {x ∈ V : f2(x) = 0}. Because V is irreducible, either V = V1 or V = V2. Then we can replace f by either f1 or f2, and it will still vanish on V . Hence we can assume that f is irreducible. Let Z denote the zero set of f. Then Z is a hypersurface in P2. By a theorem from lectures, the dimension of a hypersurface in P2 is 2 − 1 = 1. So dim Z = dim V , V is a closed subset of Z and Z is irreducible. By a theorem from lectures, these conditions imply that V = Z. Thus V is a hypersurface in P2. By a theorem from lectures, a hypersurface and a positive-dimensional closed subset of P2 must have non-empty intersection. We conclude that V ∩ W is non-empty, contradicting the assumption that V and W are disjoint.. Thus P2 cannot contain two disjoint irreducible closed subsets of dimension 1, but P1 × P1 does contain such subsets, so they are not isomorphic. Sub-question 2(e). Let n n n X Γ = {([a0 : ··· : an], [b0 : ··· : bn]) ∈ P × P : aibi = 0}. i=0 Use the Segre embedding to prove that (Pn ×Pn)\Γ is isomorphic to an affine algebraic set. (6 marks) Let N = (n + 1)2 − 1 = b2 + 2n. Label the homogeneous coordinates on PN as [(zij)] for 0 ≤ i, j ≤ n. n n n n N The Segre embedding of P × P is a function σn,n : P × P → P which sends N ([a0 : ... : an], [by0 : ... : bn]) to a point in P with coordinates given by zij = aibj. By a theorem from lectures, the image of the Segre embedding is a Zariski closed set N Σn,n ⊆ P . Using the above explicit description of σn,n, we can describe the image of Γ as n X {[(zij)] ∈ Σn,n : zii = 0}. i=0 N In other words, σn,n(Γ) = Σn,n ∩ H where H is the hyperplane in P defined by the equation n X zii = 0. i=0 By definition, the structure of algebraic variety on Pn × Pn is the same as that on n n Σn,n via σn,n. Hence (P × P ) \ Γ is isomorphic to Σn,n \ σn,n(Γ). By the above, this is equal to N Σn,n ∩ (P \ H). N n n Since Σn,n is closed in P , we conclude that (P × P ) \ Γ is isomorphic to a closed subset of PN \ H. Since H is a hyperplane, PN \ H is isomorphic to AN .