Proof of Theorem 3. 1.1 The total number of subsets of [n] having fewer than n1/4(log n)2 elements is 2o(n1/3). Therefore we can focus on B3-sets of size n1/4(log n)2 ≤ t < n1/3. In particular, by Theorem 3.2.1(i), |Zn| ≤ 2 o(n1/3) n1/3 Σ + t=n1/4(log n)2 t3
Proof of Theorem 3. 2.1
(i ) estimate the number of B3-sets which are not bounded in the sense of Definition 3.3.9 and account for at most n possible extensions of each t—s of those sets;
(ii ) for each bounded B3-set S, we estimate the number of extensions of S to a B3-set of cardinality t. To establish (ii ) we will describe a graph-based approach for bounding the number of extensions of an arbitrary B3-set S. This approach will shape Defi- nition 3.3.9 below. If two distinct elements x, y ∈ [n] \ S satisfy 2 x + a + a = y + b + b , for some {a , a }, {b , b } ∈ S , (3.10) then S ∪ {x, y} is clearly not a B3-set. This simple observation motivates our next definition.
3.3.1. The collision graph CGS has vertex set [n]\S and edges given by all pairs of distinct elements x, y ∈ [n] \ S satisfying (3.10). By construction, any set of elements of [n] \ S which extends S to a larger B3-set must induce an independent set in CGS. The following lemma provides an upper bound on the number of independent sets of graphs that have many edges in each sufficiently large vertex subset (see (3.12)). The proof will be given in Section 3.3.1.
Proof of Theorem 3. 3.1
3.1 At the outset we note that, by its very definition, η has non-zero component at each simple component of Aea. As xea and yea are both assumed to be invertible in Aea, the product xya · η also has non-zero component at each simple component of Aea and consequently we have that ⟨xya · η⟩ ∼= Aea. In the sequel, we denote xya · η by η˜. To prove the claimed short exact sequence, firstly we observe that there is a natural identification \a H1(C) ea ∗ −→~ e · \a H1(C) ∗ ^ = ea · (H1(C))∗ (3.22) where the arrow follows from Proposition 2.5.2(iii). By definition ^a (H1(C))∗ · e (η˜) = ^a (H1(C))∗ · (η˜) = I(η˜) = xya · I(η), the assignment Φ ∈ ea · Va (H1(C))∗ to Φ(η˜) thus gives an isomorphism \a H1(C) ea ∗ → I(η˜). (3.23) Similarly, we also have an isomorphism T given by sending Φ ∈ ⟨η˜⟩∗ to Φ(η˜). ⟨η˜⟩∗ −→~ Aea (3.24) Let Q = ( a H1(C))ea /⟨η˜⟩∗. Since ⟨η˜⟩ ∼= Aea, we have that Q is a finite module. By the finiteness of Q, we have that HomA(Q, A) = 0 and Ext1 (Q, A) = QV. On the other hand, since A is Gorenstein, Ext1 (⟨η˜⟩, A) = Ext1 (Aea, A) = 0. Consequently, if we apply the functor HomA(−, A) to the tautological exact sequence 0 → ⟨η˜⟩ → (\a H1(C))ea → Q → 0, then we obtain another short exact sequence Combining with the isomorphisms (3.23) and (3.24), ~=y ~=y 0 −→ ((Ta H1(C))ea )∗ −→ ⟨η˜⟩∗ −→ QV −→ 0 0 −→ xya · I(η) −→ Aea −→ Aea xya · I(η) −→ 0 By the definitions of the isomorphisms (3.23) and (3.24), the square in the above diagram is commutative and hence one has that QV =
a 1 ea T V( H (C)) ⟨η˜⟩ ea =∼ xya · I(η) By Theorem 3.2.1 (i), we have that xya · I(η) ⊂ Fita (H2(C))ea , the proof is finished by combining (3.25) and the tautological exact sequence 0 A Fita (H2(C))ea Aea Aea → a → 0 xya · I(η) xya · I(η) FitA(H2(C))ea together with an easy observation as stated in the following lemma.
Proof of Theorem 3. 1.5. For a given value of g, the family given by Proposition 3.2.5 and Lemma 3.4.5 comprises a positive proportion of all hyperelliptic curves with a rational Weierstrass point, since the latter is defined by finitely many congruence conditions. By Corollary 3.2.3, at least 25% of the curves in this family have rank r ≤
1; this is still a positive proportion of all the curves. Furthermore, under Assumption 3.3.7, a positive proportion of these curves satisfy condition (†). Let C be such a curve. Any triple of conjugate cubic points on C that reduce to F3-points will have to lie in D∞ × D∞ × D∞. We may choose P1 = P2 = P3 = ∞ and apply Lemma 3.5.2 to conclude there are at most 26 ordered triples (Q1, Q2, Q3) of conjugate cubic points in this residue class. But since Q1 = Q2 = Q3, each unordered triple is overcounted by a factor of 6, so there are at most |26/6∫ = 4 unordered triples of such cubic points. √ √ √ If the minimal Weierstrass model of C is y2 = f (x), note that (for some choice of square root) the pair of quadratic points (i, ± f (i)) reduces to (i, ±α) for each i = 0, 1, 2. Any triple of conjugate cubic points on C where two of the points reduce to (F9 \ F3)-points will have to lie in D(i,α) × D(i,−α) × D∞ for some i. In D(i,α) × D(i,−α) × D∞, we may choose P1 = (i, f (i)), P2 = (i, − f (i)), and P3 = ∞ and apply Lemma 3.5.3 with the values n(ΛC, (i, α)), n(ΛC, (i, −α)), and n(ΛC, ∞) to count ordered triples. The last case is when a triple of conjugate cubic points on C reduces to (necessarily distinct) (F27 \ F3)-points. In this setting, Lemma 3.5.4 asserts that there are no triples in this residue class away from their centers. Since C is hyperelliptic and has good reduction at 3, any unordered triple of conjugate cubic points (over F3) will F3 have to lie over an unordered triple of cubic points of P1 , of which there are only ((33 + 1) − (3 + 1))/3 = 8. Using Theorem 2.1.1, we get the worst bound on the number of 0-dimensional components in all residue classes by assuming that for some i = 0, 1, 2, n(ΛC, (i, α)) = n(ΛC, (i, −α)) = 1, and all the others will be 0. To conclude, there are at most unordered triples of conjugate cubic points.
Proof of Theorem 3. 1.5 In this section we prove Proposition 3.1.4 and Theorem 3.1.5, the weaker version of our main result. We start by introducing some notation and classical results that we shall need. Recall that, for all non-negative integers n ≥ k, we define Y :=
i=1 —
Remark 3.2.1. For all non-negative integers n ≥ k, we bound n ≤ 2kqk(n—k). This holds as n is the product of k terms, and each term is bounded by 2qn—k.
Proof of Theorem 3. 2.5.1 Our proof relies on the backward induction. It is trivial that equation (3.43)- (3.46) hold for t = T as in chapter 3.1.4. Assuming equation (3.39) and equa- tions (3.43)-(3.46) hold for t ≥ k + ∆t, we now examine the case for t = k. Let u = (uk, u∗k+∆t, ..., u∗T ), from definition (3.37), (3.38) and the Tower Property, the utility function can be written as J(k, Xk, ψk; u) = E [Xu] — γ V ar [Xx] k,Xk,ψk T = Ek,X ,ψ hE 2 k,Xk,ψk T uk Xu∗ i k k k+∆t,Xk+∆t
Proof of Theorem 3
