math 532 – final problems due: april 25 Do six of the following problems. If you hand in solutions to more than six problems, I will count the best six. The solutions are due by the end of April 25. If you have typed solutions, you can send them to me by e-mail. Otherwise please leave them in my mailslot at the department or under my oο¬ce door. algebraic sets In the following three problems all spaces are assumed to be quasi-projective algebraic sets over an algebraically closed ground ο¬eld. Problem 1. Let π βΆ π → π be a dominant morphism of irreducible quasi-projective algebraic sets. Show that there is a non-empty open subset π of π such that every component of the ο¬ber π −1 (π) has dimension dim π − dim π for all π ∈ π . (Hint: Reduce to π ⊆ πΈuοΏ½ and π ⊂ πΈuοΏ½ aο¬ne. By considering the graph (π, π (π)) ∈ πΈuοΏ½+uοΏ½ , reduce to the case where π βΆ πΈuοΏ½+1 → πΈuοΏ½ is the projection map.) From this one can deduce that for a surjective proper map π βΆ π → π one has dim π −1 (π¦) ≥ dim π − dim π , π¦ ∈ π, with equality on a non-empty open subset of π . You many use this result in your solution to the next problem. Problem 2. Let π be an irreducible quasi-projective algebraic set and π ⊆ π × βuοΏ½ a closed subset, π βΆ π → π the projection to the ο¬rst factor. Assume that all ο¬bres of π are irreducible and have the same dimension. Prove that π is irreducible. (Hint: prove that every component of π is a union of ο¬bres of π . Then study the images of the components.) Problem 3. For given π, π > 0 let π0 , π1 , …, πuοΏ½ (with π = (uοΏ½+uοΏ½ uοΏ½ ) − 1) be all the monomials of degree π in the π + 1 variables π₯0 , …, π₯uοΏ½ . Deο¬ne a map πuοΏ½ βΆ βuοΏ½ → βuοΏ½ by sending a point π = (π0 βΆ β― βΆ πuοΏ½ ) to (π0 (π) βΆ β― βΆ πuοΏ½ (π)). (i) Show that πuοΏ½ is a homeomorphism of βuοΏ½ with an irreducible closed subset of βuοΏ½ . (Hint: Consider the kernel of the homomorphism π βΆ π[π¦0 , … , π¦uοΏ½ ] → π[π₯0 , … , π₯uοΏ½ ] deο¬ned by π¦uοΏ½ β¦ πuοΏ½ .) (ii) Compute the degree of the image of πuοΏ½ . resolution of surface singularities In the following two problems all spaces are quasi-projective algebraic sets over an algebraically closed ο¬eld of characteristic zero. You are allowed to use the result of Exercise 9.22 in Gathmann’s notes. We can sometimes resolve the singularities of a surface by blowing up singular points and taking the strict transform. Suppose π has only one singular point π. Then in the resolution π βΆ π ′ → π we can consider the inverse image π −1 (π), consisting of a ο¬nite union of irreducible curves. The resolution graph of π is constructed as follows. Take one node for each component of π −1 (π) and connect two nodes with an edge if the two components intersect. Problem 4. Let π have the π΄uοΏ½ singularity: π = π(π₯ uοΏ½+1 + π¦2 + π§2 ) ⊆ πΈ3 , where π ≥ 1. Find the resolution of π by a sequence of blowups of points. Find the graph of the resolution. (Hint: After one blowup the inverse image of the singular point consists of two irreducible curves. In the next blowup the strict transforms of these curves do not lie in the chart containing the singularity. However, in another chart, the strict transforms should be disjoint, each meeting one of the two new exceptional curves.) Problem 5. Let π have the π·uοΏ½ singularity: π = π(π₯ uοΏ½−1 + π₯π¦2 + π§2 ) ⊆ πΈ3 , where π ≥ 4. Find the resolution of π when π = 4 by a sequence of blowups of points. Find the graph of the resolution. (Hint: After the ο¬rst blowup the inverse image of 0 consists of one irreducible curve. The remaining singular points lie on this curve. When we blow up a point that is not the origin using our usual method, we ο¬rst have to change coordinates so that it becomes the origin.) schemes In the remaining problems, all spaces are schemes. Problem 6. Show that Spec β€[π₯] consists of the following points: (i) (0); (ii) (π), for π ∈ β€ prime; (iii) principal ideals of the form (π ), where π ∈ β€[π₯] is a polynomial irreducible over β whose coeο¬cients have greatest common divisor 1; and (iv) maximal ideals of the form (π, π ), where π ∈ β€ is a prime and π ∈ β€[π₯] a monic polynomial whose reduction mod π is irreducible. Describe the closure of each non-closed point. (Hint: Consider the map Spec β€[π₯] → Spec β€.) Problem 7. Let (π, πͺuοΏ½ ) be a ringed space and πΊ a group acting on π . Then the quotient π/πΊ can again be given the structure of a ringed space. Let π/πΊ be the quotient space, that means, the set of πΊ-orbits. This set is given the quotient topology where π ⊆ π/πΊ is open if and only if π−1 (π) is open in π . Here π βΆ π → π/πΊ is the quotient map. The sheaf of rings on π/πΊ consists of πΊ-invariant sections of πͺuοΏ½ : πͺuοΏ½/uοΏ½ (π) = πͺuοΏ½ (π−1 (π))uοΏ½ . Note that if π is covered by πΊ-invariant open sets πuοΏ½ , then π/πΊ is covered by the quotients πuοΏ½ /πΊ. Let π be a ο¬eld. (i) Let π = πΈ2uοΏ½ − {(0, 0)}, let πΊ = π ∗ , and let πΊ act by π‘ · (π₯, π¦) = (π‘π₯, π‘π¦), π‘ ∈ π ∗ , (π₯, π¦) ∈ π. Show that π/πΊ = β1uοΏ½ . (Hint: show that π is covered by two πΊ-invariant aο¬nes, isomorphic to πΈ1uοΏ½ × πΊ. Describe the quotients of these charts and how the quotients are glued in π/πΊ.) (ii) Let π and πΊ be as in the previous part, but let the action be π‘ · (π₯, π¦) = (π‘π₯, π‘ −1 π¦), π‘ ∈ π ∗ , (π₯, π¦) ∈ π. Show that π/πΊ is the line πΈ1uοΏ½ with doubled origin. Problem 8. In this problem we consider schemes over a ο¬xed ο¬nite ο¬eld π½uοΏ½ . The zeta function of a scheme π is deο¬ned as πuοΏ½ −uοΏ½uοΏ½ π ), π uοΏ½=1 ∞ π(π, π ) = exp( ∑ where πuοΏ½ = #π(π½uοΏ½uοΏ½ ) is the number of π½uοΏ½uοΏ½ -rational points of π , i.e. the number of morphisms Spec π½uοΏ½uοΏ½ → π over π½uοΏ½ . (i) By explicit computations, show that π(πΈ1π½uοΏ½ , π ) and π(β1π½uοΏ½ , π ) are rational functions in π−uοΏ½ . (Hint: You will need the power series for log(1 − π₯).) (ii) It is known that for an elliptic curve πΈ (i.e., a smooth cubic curve in β2π½uοΏ½ ) π(πΈ, π ) = (1 − πΌπ−uοΏ½ )(1 − π½π−uοΏ½ ) , (1 − π−uοΏ½ )(1 − π1−uοΏ½ ) where πΌ, π½ ∈ β are complex conjugate numbers with absolute value √π. Show that β£π1 − (π + 1)β£ ≤ 2√π. Note: You can ignore any convergence questions.