Morphisms with Only Mild Singular Fibers and Bertini Theorems over Finite Fields

Transcription

1 Morphisms with Only Mild Singular Fibers and Bertini Theorems over Finite Fields Ziquan Yang A Thesis Submitted to the Department of Mathematics for Honors Duke University, NC,

2 Abstract We compute the asymptotic density of non-singular hypersurfaces of bidegree (3, d) in P n P 1 with only mild singular fibers over P 1 over a finite field F q for n = 1, 2. When n = 1 and char F q > 2, this asymptotic density is ζ P 1(2) 2. When n = 2 and char F q > 3, it is bounded below and above by ζ P 1(2) 1 ζ P 1(3/2) 1 and ζ P 1(2) 2. 2

3 Acknowledgments I am deeply indebted to my research advisor, Prof. Chad Schoen, for suggesting these problems to me and for his tireless patience and guidance throughout this research project. I would like to thank Profs. Justin Curry, Richard Hain, and Mike Lipnowski for their helpful comments on this research, and Prof. William Pardon for volunteering his time to be on my thesis committee. I am glad to have the opportunity to express my deep gratitude to Prof. David Kraines, who generously funded me twice with PRUV Fellowship. I also thank Dean s Summer Research Fellowship for funding my research from late June to August Finally, thank you to my parents, all professors who have taught me math, and my wonderful friends, who have made my undergraduate career a very enjoyable experience. Special thanks go to my departmental advisor Prof. Hubert Bray and my first research advisor Prof. Robert Calderbank, without whose encouragement and advice I might not even be doing mathematics now. 3

4 Contents 1 Introduction 5 2 Simply Ramified Curves in P 1 P Introduction A Density Result over Finite Fields Points of Low and Medium Degree Points of High Degree Elliptic Surfaces in P 2 P Introduction Plane Cubics Classification Counting Plane Cubics of Different Types A Density Result over Finite Fields Points of Low and Medium Degree Points of High Degree Proof of the Main Result

5 1 Introduction Let X be a non-singular closed subvariety of P N of dimension m over a ground field κ. The classical Bertini theorem states that if κ is algebraically closed, then for almost all hyperplanes H, H X is non-singular. More precisely, the set of all hyperplanes with this property forms an open dense subset of the complete linear system O P N (1), considered as a projective space. The classical Bertini theorem is known to fail for finite fields, as Poonen demonstrated in [4]. However, he used the closed point sieve to show that the asymptotic density of degree d hypersurfaces that intersect X transversely as d is ζ X (m+1) 1. More precisely, if S d H0 (P n, O(d)), the asymptotic density of S, denoted by µ(s), is defined by µ(s) := lim d #S H 0 (P n, O(d)) #H 0 (P n, O(d)) if it exists, and Poonen proved the following theorem: Theorem 1.1. Let X be a smooth quasi-projective subscheme of P n of dimension m 0 over F q. Define P := {f d 0 H 0 (P n, O(d)) : {f = 0} X is smooth of dimension m 1} Then µ(p) = ζ X (m + 1) 1. Poonen employed an analogue of the sieve method in number theory, which he called the closed point sieve, in the proof of the theorem. We may explain the main idea of his proof here. The condition that {f = 0} X is singular at a closed point Q X amounts to m + 1 linear conditions on the coefficients of f, since if we dehomogenize f in an affine chart of P n that contains Q to obtain a regular function in the neighborhood of Q, then all partial derivatives of this function have to vanish. These linear conditions are over the residue field of Q. When d is large enough, f has more than d + 1 coefficients, the density of f such that f = 0 intersects X transversely at Q is (1 q (m+1)deg Q ). In fact, let P e denote the set d {f H0 (P n, O(d)) : {f = 0} X is smooth at all Q with deg Q e}, it is not hard to prove that µ(p e ) = (1 q (m+1)deg Q ) Q X:deg Q e We may guess that µ(p) = lim d µ(p e ) = ζ X (m + 1) 1. Indeed, essentially what we need to do is to justify a change of the order of two limits # e=1 P e H 0 (P n, O(d)) lim d #H 0 (P n, O(d)) = lim e lim d #P e H 0 (P n, O(d)) #H 0 (P n, O(d)) 5

6 Therefore, the only thing left is to bound the tail terms, and indeed this is the most technical part of the proof. Poonen used a very clever decoupling idea to control the error that arises from ignoring the conditions from points of high degree. In fact, Poonen proved a more general version of the theorem, with which he demonstrated many new results. He used it to construct counterexamples to the original hyperplane Bertini theorem when the base field is finite. He could also construct examples of space-filling and space-avoiding varieties. For example, he proved the following: Theorem 1.2. Let X be a smooth, projective, geometrically integral variety of dimension m 1 over F q, and let E be a finite extension of F q. Then there exists a smooth, projective, geometrically integral curve Y X such that Y (E) = X(E). Poonen s theorem 1.1 has been generalized by Erman and Wood to semiample divisors in [3]. Theorem 1.3. Let X be a smooth projective variety over F q, with a very ample divisor A and a globally generated divisor E. Let π be the map given by the complete linear series on E. π : X E P M There exists an n 0, depending only on dim X and char F q, such that for n n 0, the probability of smoothness for a random D na + de as d is given by the product of local probabilities taken over the fibers of π: Prob(D is smooth) = P P M Prob(D is smooth at all points of π 1 (P )). The product on the right converges, is zero only if some factor is zero, and is always non-zero for n sufficiently large. By probability, they mean an asymptotic probability, which is defined in the same way that Poonen defined the asymptotic density. This thesis studies non-singular hypersurfaces of bidegree (3, d) in P 1 P 1 and P 2 P 1 over a finite field F q. Via the second projection to P 1, these hypersurfaces can viewed as P 1 -families of three collinear points and plane cubics respectively. Due to topological constraints, these families will inevitably contain some singular fibers. We may ask for the density of those families whose fibers have at worst the simplest type of singularity. We borrow techniques from [3] to show that we can express this density as an infinite product taken 6

7 over all closed points in P 1. However, we need to make some nontrivial adjustments to the sieve method to bound the error terms in our case. In particular, in the study of hypersurfaces of bidegree (3, d) in P 2 P 1, we employ the basic idea of linearization. In [1], Poonen proposed a principle that If an existence result about polynomials or n-tuples of polynomials over an infinite field can be proved by dimension counting, then a corresponding result over finite fields can be proved by the closed point sieve. Both our results conform to this principle. The author hopes that these results will be of some use to inspire the formulation of some meta-theorems that make this principle precise. 2 Simply Ramified Curves in P 1 P Introduction Among other things, Erman and Wood showed that for fixed n 3, over a finite field κ = F q, the asymptotic probability for a randomly chosen curve in P 1 P 1 of bidegree (n, d) to be non-singular is ζ P 1 Fq (2) 1 ζ P 1 Fq (3) 1 as d [[3], Theorem 9.5]. In this section, we attempt to study ramifications of these curves with respect to the second projection π : P 1 P 1 P 1. When n 2, d > 1, a non-singular curve C of bidegree (n, d) has genus nd n d+1 > 0 [[5], Chapter III Exercise 5.6 (c)], and hence by Riemann-Hurwitz theorem, π : C P 1 is necessarily ramified. In other words, if we view the curve C as a P 1 -family of three collinear points via π, then C necessarily contains singular fibers. Nonetheless we can show that if the ground field κ is algebraically closed, then for a sufficiently general choice of C, the ramification of π : C P 1 is simple. Moreover, the set of the curves with this property forms a dense subset of O P 1 P1(n, d), viewed as a projective space. Then we borrow techniques from papers [4] and [3] to show that when κ = F q with p = char F q > 2 and n = 3, the asymptotic probability for a randomly chosen curve in the divisor class O P 1 P 1(3, d) to be simply ramified with respect to π is ζ P 1 (2) 2 Fq as d. Notation Let X be P 1 P 1 over a ground field κ. We label its coordinates as ((s 0, s 1 ), (t 0, t 1 )). π : X P 1 is the second projection. If W P 1 is a finite subscheme, the fiber π 1 (W ) X is denoted by X W. Let R n,d κ[s 0, s 1, t 0, t 1 ] be the set of bihomogeneous polynomials that are of degree n in s 0, s 1 and d in t 0, t 1. R n,d can be naturally identified with H 0 (P 1 P 1, O(n, d)). For a section f R n,d, the curve in X cut out by f is denoted by H f. The fiber of H f over P P 1 is denoted by (H f ) P. If Q X is a closed point and C X is a curve 7

8 that is non-singular at Q, then e Q (C) denotes the ramification degree of C with respect to π. When κ = F q is finite, for a subset P d R n,d, we define Prob(f P) := lim d Prob(f d P) = lim d #P R n,d #R n,d where f and f d are randomly chosen from d R n,d and R n,d respectively. In general, if Y is a scheme and Q is a closed point in Y, we use Q (2) to denote the first-order infinitesimal neighborhood of Q in Y. The residue field of Q is denoted by κ(q). 2.2 A Density Result over Finite Fields Theorem 2.1. Suppose p = charf q > 2. Let D d R 3,d be the subset of sections f such that H f is simply ramified with respect to π. Then Prob(f D) = ζ P 1 Fq (2) 2 Furthermore, the conditional probability for a randomly chosen non-singular curve of bidegree (3, d) to be simply ramified is ζ P 1 Fq (3)/ζ P 1 Fq (2) as d. For convenience, we say f R n,d is good at Q X, if H f is non-singular at Q and e Q (H f ) 2. If Q H f, then by default f is good at Q. Otherwise we say that f is bad at Q. For a fixed N, we define P low = d 0{f R 3,d : f is good at all Q X,deg π(q) < } Q med = d 0{f R 3,d : f is bad at some Q, deg π(q) [, d/p ]} Q high = d 0{f R 3,d : f is bad at some Q, deg π(q) > d/p} Points of Low and Medium Degree We first do some local analysis on a fiber. Let P P 1 = Proj F q [t 0, t 1 ] be a closed point. Without loss of generality, we assume t 1 0 at P and P lies in A 1 = Spec F q [t] where t = t 0 /t 1. Let m = (r(t)) F q [t] be the maximal ideal corresponding to P, so that κ(p ) = F q [t]/m, P (2) = Spec F q [t]/m 2 and X P = P 1 F q[t]/m X. Denote the reduction map F q[t]/m 2 F q [t]/m by g g. We extend it to a map H 0 (P 1 F q[t]/m 2, O(3)) H0 (P 1 F q[t]/m, O(3)) by applying F q [t]/m 2 F q [t]/m to coefficients of cubic polynomials. Let ϕ P : R 3,d H 0 (P 1 F q[t]/m 2, O(3)) 8

9 be the restriction map. H f is non-singular at a point Q X P if and only if ϕ P (f) does not vanish at Q (2) P 1 F q[t]/m. If H 2 f is non-singular at Q, then e Q (H f ) < 3 if and only if ϕ P (f) H 0 (P 1 F[t]/m, O(3)) does not have a zero of multiplicity 3 at Q. Therefore we can determine if f is good at all points in (H f ) P by looking at the image of f in H 0 (P 1 F q[t]/m, O(3)). This observation 2 yields an independence result across fibers. More precisely, we have Lemma 2.2. If {P 1, P 2,, P s } is a set of finitely many closed points in P 1, then for f randomly chosen from d R 3,d, Prob(f is good at all Q X Pi ) = In particular, we have Prob(f P low ) = s Prob(f is good at all points Q (H f ) Pi ) i=1 deg (P )< Prob(f is good at all points Q (H f ) P ) since there are only finitely many points with degree <. Proof. Let W := P (2) i exact sequence and δ := deg W = s i=1 2deg P i. Then we have an 0 O X (3, d δ) O X (3, d) O X (3, d) XW 0 which induces an exact sequence H 0 (X, O X (3, d)) H 0 (X W, O X (3, d) XW ) H 1 (X, O X (3, d δ)) That H 1 (X, O X (3, d δ)) = 0 when d δ follows from the isomorphism H 1 (X, O(a, b)) = H 1 (X, O X ( 2 a, 2 b)) given by Serre duality and the fact that H 1 (X, O X (a, b)) = 0 when a, b < 0 [[5], Chapter III Exercise 5.6 (a)]. Therefore the restriction map R 3,d = H 0 (P 1 P 1, O(3, d)) H 0 (X W, O X (3, d) XW ) = s i=1 H 0 (X (2) P, O(3)) i is surjective for d i 2deg P i. This surjectivity, together with our observation that whether f is good at all Q (H f ) Pi is completely determined by ϕ Pi (f) H 0 (P 1, O(3)), implies that when d δ, we have P (2) i #R 3,d P low #R 3,d = s #ϕ Pi ({f R 3,d : f is good at all Q X Pi }) #H 0 (X (2) P, O(3)) i i=1 The lemma follows by taking d on both sides. 9

10 Before proceeding we make the following convention: A pair (F 1, F 2 ) H 0 (P 1 F q[t]/m, O(3))2 is said to be bad if it falls in one of the following 3 types: 1. F 1 has a root of multiplicity 2 at a point where F 2 also vanishes. 2. F 1 has a root of multiplicity 3 at a point where F 2 does not vanish. 3. F 1 0. Otherwise, the pair is said to be good. Let : F q [t]/m F q [t]/m 2 be a F q - linear map that is a section to the previously defined reduction map F q [t]/m 2 F q [t]/m. We extend it to a map H 0 (P 1 F q[t]/m, O(3)) H 0 (P 1 2 F q[t]/m, O(3)) in the same way we extended the reduction map. Then we have the following: Lemma 2.3. The F q -linear map H 0 (P 1 F q[t]/m, O(3)) 2 H 0 (P 1 F q[t]/m 2, O(3)) given by (F 1, F 2 ) F 1 + r(t)f 2 is a bijection. Moreover, f R 3,d is good at all points Q π 1 (P ) if and only if ϕ P (f) corresponds to a good pair. Proof. The map F (F, (F F )/r(t)) is an inverse to the map given in the lemma. Let (F 1, F 2 ) be the pair corresponding to ϕ P (f). ϕ P (f) vanishes on Q (2), i.e. H f is singular at Q, if and only if Q is a double root of F 1 and a root to F 2. Similarly, if H f is non-singular at Q, then e Q (H f ) 3 if and only if Q is root to F 1 of multiplicity 3. Lemma 2.4. Let e := deg P. The density of good pairs in H 0 (P 1 κ(p ), O(3))2 is (1 q 2e ) 2 Proof. We first count type 1 pairs. Let Q P 1 F = q[t]/m P1 F q e be the point that is a double root to F 1 and a root to F 2. Note that deg Q = 1. F 1 is fixed up to rescaling by F q after we choose a third root, which can be any point of e degree 1. Therefore there are (q e + 1)(q e 1) choices for F 1, where (q e + 1) is the number of all possible choices for the third root. The probability that F 2 vanishes at Q as well is q e, so we have q 4e q e = q 3e choices for F 2. Since we have (q e + 1) choices for Q, there are q 3e (q e + 1) 2 (q e 1) type 1 pairs. To count type 2 pairs, let Q P 1 F q e be the triple root to F 1. Again there are q e + 1 choices. At each Q, there are (q e 1) homogeneous polynomials of degree 3 which have Q as a triple root, since the leading coefficient will uniquely determine such a polynomial. We have q 4e q 3e choices for F 2. In total there are (q e + 1)(q e 1)(q 4e q 3e ) pairs of type 2. 10

11 Finally when F 1 = 0, F 2 can be anything, so we have q 4e type 3 pairs. Therefore the density of good pairs is 1 q 8e (q 3e (q e + 1) 2 (q e 1) + (q e + 1)(q e 1)(q 4e q 3e ) + q 4e ) = (1 q 2e ) 2 Lemma 2.5. Proof. By definition, lim Prob(f Qmed e 0 ) = 0 Prob(f Q med ) = lim Prob(f d Q med e d 0 ) for f d randomly chosen from R 3,d. It suffices to show that Prob(f d Q med ) is universally bounded by O(q ), where the implied constant is independent of d or. Let P be a point of degree e d/p on P 1. By the proof of Lemma 2.2, the restriction map R 3,d H 0 (X P (2), O(3)) is surjective since p > 2 and e < d/p < d/2. Lemma 2.3 and Lemma 2.4 hence imply that probability that f d is bad at some point in the fiber X P is 1 (1 q 2e ) 2 < 2q 2e. d/p Prob(f d Q med ) (number of points of degree e in P 1 )(2q 2e ) e= d/p 2 (q e + 1)q 2e q = O( 1 q ) = 1 O(q ) e= Points of High Degree Lemma 2.6. Let W A M A 1 be a closed subscheme. Let t be the coordinate for A 1. Denote by A 0,d = F q [t] d the set of polynomials that are of degree d in t. Let π : A M A 1 A 1 be the second projection. Consider the restriction map ϕ W : A 0,d H 0 (W, O W ) If W contains a closed point Q such that deg π(q) j, then #{Im ϕ W } q min(d+1,j). Proof. If dim π(w ) = 1, then t, and hence any nonzero polynomial in t, does not vanish identically on W. Therefore ϕ W is injective, and #{Im ϕ W } = q d+1. Now suppose dim π(w ) = 0. Note that ϕ W factors as A 0,d = F q [t] d ψ W H 0 (π(w ), O π(w ) ) π# H 0 (W, O W ) 11

12 Since the pull back map π # is injective, it suffices to bound #{Im ψ W }. Let P = π(q). Suppose P = Spec F q [t]/(r(t)) A 1. Note that deg r(t) j since by assumption deg P j. The composition F q [t] d H 0 (π(w ), O π(w ) ) κ(p ) is nothing but the natural map F q [t] d F q [t]/r(t). Therefore #{Im (F q [t] d κ(p ))} = q min(d+1,j) #{Im ψ W } We use the following lemma to bound the probability that a randomly chosen f R 3,d is bad at some point Q when π(q) has high degree (> d/p ). Therefore the following lemma is to be applied with j = d/p. Lemma 2.7. Let j > 2 be an integer and U X be an open subscheme. For a randomly chosen f R 3,d, the probability that there exists a point Q U with deg π(q) j such that H f is non-singular at Q but e Q (H f ) 3 or H f is singular at Q is at most O(d 2 q min( d/p +1,j) ) Proof. Among all U, the lemma is clearly strongest when U = X. Since P 1 P 1 can be covered by 4 affine charts A 1 A 1, we reduce to proving the lemma for all A 1 A 1. Without loss of generality, we may assume s 1 0, t 1 0 on U = A 1 A 1 and work with coordinates s = s 0 /s 1, t = t 0 /t 1. Let A n,d F q [s, t] = H 0 (U, O U ) be the polynomials that are of degree n in s and d in t. Clearly by dehomogenizing sections in R 3,d, we have a natural identification R 3,d = A 3,d. Accordingly, we replace H f by H f A 1 A 1. We call a closed point Q A 1 A 1 admissible if deg π(q) j and a subscheme W A 1 A 1 admissible if it contains an admissible point. By (W ) ad we denote the union of admissible irreducible components of (W ) red. We first deal with the probability that for a randomly chosen f A 3,d, e Q (H f ) 3 at some admissible Q A 1 A 1, which happens only if f(q) = f s (Q) = f ss (Q) = 0 (2.8) Define subschemes W 2 = {f ss = 0}, W 1 = W 2 {f s = 0} and W 0 = W 1 {f = 0} in A 1 A 1. We reduce the problem to bounding the probability that for a randomly chosen f A 3,d, W 0 contains an admissible point. Following Poonen s idea [[4], Proof of Lemma 2.6], we write f in such a way so that the first and second order partial derivatives with respect to s are 12

13 largely independent. If f 0 A 3,d and g 1, g 2, h A 0, d/p are selected uniformly and independently at random, then the distribution of f = f 0 + g p 1s 2 + g p 2s + h p is uniform over A 3,d. Direct computation shows that f s = f 0,s + 2g p 1s + g p 2 f ss = f 0,ss + 2g p 1 Note that W 2 depends only on the choice of f 0, g 1 and W 1 only on f 0, g 1, g 2. Let E denote the event that a. dim (W 1 ) ad = 0 b. f does not vanish identically on any irreducible component of (W 1 ) ad. Clearly if E holds for f, then W 0 does not contain any admissible point. Therefore it suffices to show that for a randomly chosen f A 3,d, Prob(E) = 1 O(d 2 q min( d/p +1,j) ) as d. Now we bound Prob(E) in three steps: Step 1: Conditioned on a choice of f 0, the probability that dim W 2 = 2 is at most q ( d/p +1), since dim W 2 = 2 if and only if g p 1 = f 0,ss /2, for which there is at most one choice of g 1. Step 2: Conditioned on a choice of f 0 and g 1 such that dim W 2 = 1, the probability that dim (W 1 ) ad = 1 is at most O(dq min( d/p +1,j) ). Let V 1,, V l be all the irreducible component of (W 2 ) red. Viewing P 1 P 1 as a subscheme of P 3 via the Segre embedding, we may apply Bézout s theorem to obtain that l = O(d). dim (W 1 ) ad = 1 if and only if f s vanishes identically on V i for some i. We need to bound the set G bad i = {g 2 A 0, d/p : f 0,s + 2g p 1s + g p 2 vanishes identically on V i } If g, g G bad i, then g g vanishes identically on V i. Therefore G bad i is a coset of ker ϕ i, where ϕ i is the F q -linear restriction map A 0, d/p H 0 (V i, O Vi ). Now we apply Lemma 2.6 to obtain that #{Im ϕ i } q min( d/p +1,j). This means the probability that f s vanishes identically on V i is Prob(g 2 G bad i ) = #Gbad i q min( d/p +1,j) #A 0, d/p 13

14 Since there are l = O(d) many V i s, the probability that f s vanishes identically on any of them is bounded by O(dq min( d/p +1,j) ) as claimed. Step 3: Conditioned on a choice of f 0, g 1 and g 2 such that dim (W 1 ) ad = 0, the probability that (W 0 ) ad is at most O(d 2 q min( d/p +1,j) ). Let Q 1, Q 2,, Q l be all irreducible components of (W 1 ) ad. Since W 1 is cut out by f s and f ss, again by considering the Segre embedding P 1 P 1 P 3, we obtain from Bézout s theorem l = O(d 2 ). Since π(q i ) j by assumption, we may apply Lemma 2.6 again obtaining that for each Q i, Im (A 0, d/p H 0 (Q i, O Qi )) q min( d/p +1,j). This implies that the probability that f vanishes at Q i is bounded above by q min( d/p +1,j). Since there are at most O(d 2 ) of Q i s, we obtain the desired conclusion of this step. Finally Step 1 and 2 combine to give that Prob(dim (W 1 ) ad = 0) (1 q ( d/p +1) )(1 q min( d/p +1,j) ) And Step 3 gives = 1 O(dq min( d/p +1,j) ) Prob(E) Prob(dim (W 1 ) ad = 0)(1 O(d 2 q min( d/p +1,j) )) = 1 O(d 2 q min( d/p +1,j) ) Now we deal with the probability that H f is singular at Q for some Q A 1 A 1, which happens if and only if f(q) = f s (Q) = f t (Q) = 0. Again we choose f 0 A 3,d g 1, g 2, h A 0, d/p uniformly at random, but this time we put f = f 0 + g p 1t + g p 2s + h p The distribution of f is uniform over A 3,d. The rest of the proof is rather analogous to what we did before, so we only give a sketch. Define W 2 = {f t = 0}, W 1 = W 2 {f s = 0} and W 0 = W 1 {f = 0}. W 2 depends only on the choices of f 0, g 1 and W 1 only on f 0, g 1, g 2. By applying the same arguments as in Step 1 and 2, we may show that Prob(dim (W 1) ad = 0) = 1 O(dq min( d/p +1,j) ) By applying the same arguments as in Step 3, we may bound the probability that f does not vanish at any of the irreducible components of (W 1) ad given that dim (W 1) ad = 0, so that Prob((W 0) ad = ) 1 O(d 2 q min( d/p +1,j) ) Now the proof of the lemma is complete. 14

15 Lemma 2.9. Prob(f Q high ) = 0 Proof. Apply Lemma 2.7 with U = X and j = d/p. Then we take the limit as d. Now we have gathered all the ingredients needed to prove Theorem 2.1. Proof of Theorem 2.1. For each, we have that Therefore P low D P low Q med Q high Prob(f Pe low 0 ) Prob(f D) Prob(f Pe low 0 Q med Q high ) Now take, Lemma 2.2, 2.4, 2.5, and 2.9 combine to give that Prob(f D) = lim Prob(f Plow e 0 ) = Prob(f is good at all points Q (H f ) P ) P P 1 = (1 q 2deg P ) 2 P P 1 = ζ P 1 Fq (2) 2 The statement on the conditional probability in Theorem 2.1 now follows directly from Erman and Wood s result [[3], Theorem 9.5] that the asymptotic probability for a randomly chosen curve of bidegree (3, d) to be non-singular is ζ P 1 Fq (2) 1 ζ P 1 Fq (3) 1. 3 Elliptic Surfaces in P 2 P Introduction In this section we study non-singular hypersurfaces in P 2 P 1 given by sections in H 0 (P 2 P 1, O(3, d)). Such a surface can be viewed as a P 1 -family of plane cubics that may contain degenerate (singular) fibers. We first discuss some geometric properties. In the space PH 0 (P 2, O(3)) which parametrizes plane cubics, curves with worse than nodal singularities (reducible or cuspidal) comprise a closed subset of codimension 2. A P 1 -family of plane cubics corresponds to a morphism P 1 PH 0 (P 2, O(3)). A Bertini type argument shows that those non-singular hypersurfaces whose fibers have at most nodal singularities comprise an open dense subset of the parameter 15

16 space PH 0 (P 2 P 1, O(3, d)) P 9. We say these hypersurfaces are good. When the base field κ is infinite, we see that the density of good hypersurfaces is 1. However, when κ = F q is a finite field, H 0 (P 2 P 1, O(3, d)) is a finite set, so sections that do not give good hypersurfaces may have a nonzero density. In the rest of this section, we assume that char F q > 3. In a similar vein to section 2, we may show that as d, this density converges to an infinite product over P 1. We may explicitly compute the terms in this infinite product and then show that this infinite product converges. Before introducing the result, we first introduce some notations. Notation Let us label the coordinates of P 2 P 1 as ((X 0 : X 1 : X 2 ), (T 0 : T 1 )) and let S 3,d F q [X 0, X 1, X 2, T 0, T 1 ] be the set of bi-homogeneous polynomials of degree 3 in X i s and d in T j s. Then S 3,d can be identified with H 0 (P 2 P 1, O(3, d)) in the natural way. If B d S 3,d is a subset, then we define the asymptotic probability for a randomly chosen f S 3,d to belong to B, or simply probability, as Prob(f B) = lim d #B S 3,d #S 3,d if it exists. Given a section f S 3,d, we denote the corresponding hypersurface defined by the vanishing of f as H f. Let P P 1 F q be a point. We say that a hypersurface H f, or sometimes f, is good at the point P if H f is non-singular at all points Q π 1 (P ), and the fiber (H f ) P P 2 κ(p ) is either a non-singular or a nodal curve. Otherwise, we say H f, or f, is bad at P. Of course, a hypersurface is good if and only if it is good at every P P 1 F q. Define S = d {f S 3,d : H f is good} and for P P 1 F q, S P = d {f S 3,d : H f is good at P } We will show that Lemma 3.1. Let e = deg P. Then Prob(f S P ) = (1 q 2e )(1 2q 2e q 3e + 2q 5e q 6e 1 3 q 7e ) In particular, we have inequalities (1 q 2e )(1 q 3e/2 ) < Prob(f S P ) < (1 q 2e ) 2 when char F q > 3. 16

17 Now we may state our main theorem for elliptic surfaces over finite fields: Theorem 3.2. Assume char F q > 3. The density of good hypersurfaces of bidegree (3, d) in P 2 P 1 over F q converges to a number in (0, 1) as d. More precisely, Prob(f S) = Prob(f S P ) P P 1 Fq Combined with the previous lemma, we obtain a bound for Prob(S) in terms of Zeta functions: 3.2 Plane Cubics Classification ζ P 1 Fq (2) 1 ζ P 1 Fq (3/2) 1 < Prob(f S) < ζ P 1 Fq (2) 2 We first make some easy observations on plane cubics. The following is a table made by B. Poonen that classifies all plane cubics. For convenience we are going to use the same terminology. smooth cubic e = 0 nodal cubic e = 1 cuspidal cubic e = 2 conic + line e = 2 conic + tangent e = 3 line + double line e = 3 three lines e = 3 concurrent lines e = 4 triple line e = 2 Figure 1: Cubic plane curves and their Euler characteristics [2] 17

18 We may use tangent cones to differentiate the nodal cubics from singular cubics of other types. Suppose we find a singular point Q on a cubic C. We ask two questions: 1. Does the tangent cone of C at Q contain a double line? If yes, then we have found a cuspidal cubic or a reducible cubic of the type conic + tangent, line + double line, or a triple line. 2. Does the tangent cone of C at Q share an irreducible component, which has to be a line, with C? If yes, then we have found a reducible cubic. If both answers are no, then we have found a nodal cubic. Now we explain how to perform the above tests using partial derivatives. Let us label the coordinates of P 2 as X 0, X 1, X 2 and let Q P 2 be a closed point. Without loss of generality, we assume X 0 (Q) 0 and dehomogenize a homogeneous cubic polynomial in κ[x 0, X 1, X 2 ] at X 0 to a polynomial g in x 1 = X 1 /X 0, x 2 = X 2 /X 0. The cubic C = {g = 0} is singular at Q if and only if g(q) = g (Q) = g (Q) = 0 x 1 x 2 Suppose Q is a singularity. Note that deg Q has to be 1 since a plane cubic cannot be singular at a point of higher degree. For m = 2, 3, we define polynomials T m g(u 0, U 1 ) = ( ) m m g (Q)U r x r 1x s 0 r U1 s κ[u 0, U 1 ] 2 r+s=m which are homogeneous in U 0, U 1. If T 2 g(u 0, U 1 ) 0, then the tangent cone of C at Q must have degree 3, in which case C must be a triple line, a line + double line, or concurrent lines (see Figure 1). Otherwise, T 2 g(x 1 x 1 (Q), x 2 x 2 (Q)) = 0 describes the (affine) tangent cone of C at Q, which we denote by T C Q (C). T C Q (C) contains a double line if and only if T 2 g(u 0, U 1 ) vanishes at a closed point in P 1 = Proj κ[u 0, U 1 ] with multiplicity 2. For convenience, we say a triple line contains a double line, and if T 2 g(u 0, U 1 ) 0, it vanishes at all points in Proj κ[u 0, U 1 ] with multiplicity. Taylor expansion of g at Q says g = 1 2 T 2 g(x 1 x 1 (Q), x 2 x 2 (Q)) T 3 g(x 1 x 1 (Q), x 2 x 2 (Q)) we see that the T C Q (C) shares an irreducible component with C if and only if there T 2 g(u 0, U 1 ) and T 3 g(u 0, U 1 ) share a zero on Proj κ[u 0, U 1 ]. Therefore if we fix a particular closed point in Proj κ[u 0, U 1 ], then both conditions on T m (U 0, U 1 ) become linear in partial derivatives. 18

19 We may summarize the above observations into a geometric statement: Let A 2 P 2 be the affine chart as above. Consider two subschemes of A 2 Proj κ[u 0, U 1 ]: W 1 = {g = g x 1 = g x 2 = T 2 g(u 0, U 1 ) = 0} W 2 = {g = g x 1 = g x 2 = T 2 g(u 0, U 1 ) = T 3 g(u 0, U 1 ) = 0} 1. If there does not exist a closed point Q A 2 such that the fiber of W 1 over Q is non-reduced, then C cannot be a cuspidal curve with the singularity in this affine chart A If W 2 =, then C cannot be a reducible curve with a singularity in this affine chart A 2. These statements are weaker than our observations, but they suffice for our purposes. We are going to use them in the proof of Lemma Counting Plane Cubics of Different Types In this section we count the number of reducible, cuspidal and nodal cubics in the projective plane P 2 over a finite field F q. We will apply results in this section with q = q e for e = 1, 2,. Let S i = H 0 (P 2, O(i)) for i = 1, 2, 3. Plane cubics are given by sections in S 3 = H 0 (P 2, O(3)). We may identify S i with the subset of F q [X 0, X 1, X 2 ] consisting of homogeneous polynomials of degree i. Given f i S i, we denote its vanishing locus in P 2 by Z(f). Define R = {f S 3 : Z(f) is reducible} C = {f S 3 : Z(f) is cuspidal} N = {f S 3 : Z(f) is nodal} Let H j F q [U 0, U 1 ] denote the set of homogeneous polynomials of degree j. Given g H j, denote its vanishing locus in P 1 = Proj F q [U 0, U 1 ] by Z(g). Note that dim Fq H j = j + 1. If g H j and W P 1 is a finite subscheme, let g W be the element of H 0 (W, O W ) that on each connected component W i equals the restriction of U j i g, where i is the smallest index such that U i 0 at W i. Lemma 3.3. If g 2 H 2, g 3 H 3 are randomly chosen, then Prob(g 2 is coprime to g 3 g 2 is a nonzero square) = 1 q 1 19

20 Proof. Let L be the closed point contained in Z(g 2 ). g 2 and g 3 are coprime if and only if L Z(g 3 ). Since deg L = 1, the F q -linear map H 3 κ(l) given by g 3 g 3 L is surjective. Therefore the probability that a randomly chosen g 3 H 3 vanishes at L is 1/#κ(L) = 1/q. Lemma 3.4. If g 2 H 2, g 3 H 3 are randomly chosen, then Prob(g 3 is coprime to g 2 g 2 is not a square) = 1 (q + 1)(q 1)2 q3 Proof. Now we have to divide it into two cases. Case 1: g 2 is an reducible polynomial, so that Z(g 2 ) contains two distinct closed points L 1, L 2, both of which have to be of degree 1. g 3 is coprime to g 2 if and only if L 1, L 2 Z(g 3 ). Since H 3 κ(l 1 ) κ(l 2 ) is surjective, events {g 3 (L 1 ) = 0} and {g 3 (L 2 ) = 0} are independent. Hence Prob(g 3 is coprime to g 2 g 2 is reducible) = (1 q 1 ) 2 By counting the number of pairs (L 1, L 2 ), L 1 L 2, we easily see that #{g 2 H 2 : g 2 is reducible with two distinct roots} = 1 q(q + 1)(q 1) 2 where the factor q 1 comes from the fact that g 2 is determined by L 1, L 2 up to scalar multiplication by F q. Case 2: g 2 is an irreducible polynomial. Then g 3 fails to be coprime to g 2 if and only if g 2 divides g 3 if and only if g 3 vanishes at L 2 := Z(g 2 ), which is a closed point of degree 2. Since H 3 κ(l 2 ) given by g 3 g 3 L2 is surjective, Prob(g 3 is coprime to g 2 g 2 is irreducible) = 1 q 2 Since there are (q 2 q)/2 possible choices for L 2 s and g 2 is determined by L 2 up to a nonzero constant in F q, we see that #{g 2 H 2 : g 2 is irreducible} = 1 2 (q2 q)(q 1) Therefore the conditional probability in the lemma equals to q + 1 2q (1 q 1 ) 2 + q 1 2q (1 q 2 ) = 1 (q + 1)(q 1)2 q3 Lemma 3.5. Let Q P 2 be a closed point of degree 1, then for a randomly chosen f S 3, Prob(f is a cuspidal curve with a cusp at Q) = 1 q 7 (q 1)2 (q + 1) Prob(f is a nodal curve with a node at Q) = 1 q 7 (q 1)3 (q + 1) 20

21 Proof. Without loss of generality, assume that X 0 0 at Q, so that Q A 2, on which we may use coordinates x 1 = X 1 /X 0, x 2 = X 2 /X 0. We further assume that x 1 (Q) = x 2 (Q) = 0, i.e. Q is the origin of the affine chart X 0 0. We may identify f S 3 with its dehomogenization f/x 3 0 = g(x 1, x 2 ) on A 2. We may write g = g 0 + g 1 + g 2 + g 3 where g j is the homogeneous part of degree j of g. If the g j s are chosen uniformly and independently at random, then the distribution of f is uniform over S. Z(g) is a cuspidal curve with a cusp at Q if and only if i. g 0 = g 1 = 0 (Z(f) is singular at Q.) ii. g 2 0 and g 2 is a square. (The tangent cone of Z(f) at Q is a doubled line.) iii. g 2 and g 3 are coprime. (Z(f) is irreducible.) We easily see that Prob(i) = 1/q 3, Prob(ii) = (q 2 1)/q 3 and Lemma 3.3 gives Prob(iii ii). Therefore the probability that Z(f) has a cusp at Q is 1 q 3 q 2 1 q 3 (1 1 q ) = 1 q 7 (q 1)2 (q + 1) Similarly, Z(g) is a nodal curve with a node at Q if and only if i. g 0 = g 1 = 0 (Z(f) is singular at Q.) ii. g 2 0 and g 2 is not a square. (The tangent cone of Z(f) at Q is two distinct lines.) iii. g 2 and g 3 are coprime. (Z(f) is irreducible.) This time Prob(i) = 1/q 3, Prob(ii) = (q 3 q 2 )/q 3 and Lemma 3.4 gives Prob(iii ii). Therefore the probability that Z(f) has a node at Q is 1 q 3 q 3 q 2 q 3 1 q 3 (q + 1)(q 1)2 = 1 q 7 (q 1)3 (q + 1) We use the following lemmas to treat reducible cubics. We will use the multiset formula twice. It is a well-known formula, but for readers convenience we give the statement below. Lemma 3.6. (Multiset Formula) The number of k-multisets of a set of n elements is ( ) n + k 1 k 21

22 Lemma 3.7. In S 2 {0}, the number of reducible polynomials is 1 2 (q 1)(q2 + q + 1)(q 2 + q + 2) and the number of irreducible polynomials is 1 2 q(q 1)2 (2q + 1)(q 2 + q + 1) Proof. If f S 2 {0} is reducible, then Z(f) is a union of two lines, which may or may not coincide. By considering the dual projective plane ˇP 2, we see that there are q 2 + q + 1 lines in total. By applying the multiset formula, we have that there are 1 2 (q2 + q + 1)(q 2 + q + 2) choices for a union of two lines which are allowed to coincide. Since the correspondence f Z(f) is (q 1)-to-1, there are 1 2 (q 1)(q2 + q + 1)(q 2 + q + 2) non-zero reducible polynomials in S 2. The number of irreducible polynomials is deduced from the fact that dim Fq S 2 = 6. Lemma 3.8. The number of f S 3 {0} that can be factored into three linear terms is 1 6 (q 1)(q2 + q + 1)(q 2 + q + 2)(q 2 + q + 3) The number of f S 3 {0} that can be factored into a irreducible quadratic polynomial and a linear polynomial is 1 2 q(q 1)2 (2q + 1)(q 2 + q + 1) 2 Proof. To count f S 3 {0} that can be factored into three linear terms, we first count the number of possible Z(f) s. Z(f) is a union of three lines, which do not have to be distinct. The multiset formula again gives us that there are 1 6 (q2 + q + 1)(q 2 + q + 2)(q 2 + q + 3) choices for Z(f). The number of f is obtained by adding a factor of (q 1) as before. If f S 3 {0} can be factored into an irreducible quadratic polynomial and a linear polynomial, then Z(f) is the union of a conic and a line, which may or may not intersect tangentially. We can infer from Lemma 3.7 that there are 1 2 q(q 1)(2q + 1)(q2 + q + 1) 22

23 conics. There are q 2 + q + 1 choices for the line, so there are in total choices for f. Lemma q(q 1)2 (2q + 1)(q 2 + q + 1) 2 #R = 1 3 (q2 + q + 1)(q 1)(q + 1)(3q 4 q 3 + 2q 2 2q + 3) #C = q 3 (q 2 + q + 1)(q 1) 2 (q + 1) #N = q 3 (q 2 + q + 1)(q 1) 3 (q + 1) Proof. We obtain the number #R by taking the sum of the two numbers in the statement of the previous lemma. #C and #N follow directly from Lemma 3.5 and the facts that there are (q 2 + q + 1) closed points of degree 1 in P 2 over F q and dim Fq S = A Density Result over Finite Fields In a similar vein as in section 2, we are going to work with the following sets: For a fixed N, we define P low = d 0{f S 3,d : H f is good at all P P 1 F q,deg P < } Q med = d 0{f S 3,d : H f is bad at some P P 1 F q,deg P [, d/p ]} Q high = d 0{f S 3,d : H f is bad at some P P 1 F q,deg P > d/p} Points of Low and Medium Degree We first do some local analysis on a fiber. Let P P 1 = Proj F q [T 0, T 1 ] be a fixed point and let e = deg P. Suppose T i 0 at P, so P Spec F q [T 1 i /T i ] Proj F q [T 0, T 1 ]. Let t = T 1 i /T i and r(t) F q [t] be an irreducible polynomial such that the second infinitesimal neighborhood P (2) = Spec F q [t]/r(t) 2. Let B := F q [t] and m := (r(t)) B be the maximal ideal. Denote the restriction map B/m 2 B/m by g g and extend it to a map H 0 (P 2 B/m, O(3)) 2 H 0 (P 2 B/m, O(3)) by applying the reduction map to each coefficient. Similarly we denote a F q -linear section of the reduction map B/m 2 B/m by h h and extend it to a map H 0 (P 2 B/m, O(3)) H0 (P 2 B/m, O(3)). Let ϕ 2 P : S 3,d H 0 (P 2 B/m, O(3)) be the restriction map. H 2 f is non-singular at a point Q π 1 (P ) if and only if ϕ P (f) does not vanish at Q (2) P 2 B/m. 2 Consider the vector space H 0 (P 2 F q e, O(3)) 2, we say that a pair (F 1, F 2 ) is good if, it satisfies one the following conditions: 23

24 1. F 1 describes a non-singular curve in P 2 F q e. 2. F 1 describes a nodal curve but F 2 does not vanish at the node. This terminology is explained by the following lemma, which is an analogous statement to Lemma 2.3. Lemma The map defined by ψ : H 0 (P 2 B/m, O(3)) 2 H 0 (P 2 B/m2, O(3)) (F 1, F 2 ) F 1 + r(t)f 2 is an isomorphism of B/m-vector spaces. Moreover, f is good at P if and only if ψ 1 (ϕ P (f)) in H 0 (P 2 B/m, O(3))2 is a good pair. Proof. The above map has an inverse F (F, (F F ))/r(t)). For f to be good at P, we need ϕ P (f) to describe a non-singular or nodal curve. Let Q P 2 B/m be a closed point and suppose ϕ 2 P (f) corresponds to (F 1, F 2 ). If F 1 is non-singular at Q, then H f must also be non-singular at Q. Otherwise, H f is non-singular at Q if and only if F 2 does not vanish at Q. The following lemma assumes the same role as lemma 2.2 in Section 2, and the proof is similar. Lemma If {P 1, P 2,, P s } is a set of finitely many closed points in P 1, then for f randomly chosen from d S 3,d, Prob(f is good at all P i ) = s Prob(f is good at P i ) i=1 In particular, we have Prob(f P low ) = deg (P )< Prob(f is good at P ) since there are only finitely many points with degree <. Proof. Denote P 2 P 1 by X. Let W := P (2) i Then we have an exact sequence and δ := deg W = s i=1 2deg P i. 0 O X (3, d δ) O X (3, d) O X (3, d) XW 0 which induces an exact sequence H 0 (X, O X (3, d)) H 0 (X W, O X (3, d) XW ) H 1 (X, O X (3, d δ)) 24

25 By Künneth formula, H 1 (X, O X (3, d δ)) H i (P 2, O(3)) Fq H j (P 1, O(d δ)) i+j=1 When d δ 0, Serre Duality implies H 1 (P 1, O(d δ)) H 0 (P 1, O( 2 (d δ))) = 0. We also have H 1 (P 2, O(3)) = 0. Therefore H 1 (X, O X (3, d δ)) = 0 and the restriction map S 3,d = H 0 (P 2 P 1, O(3, d)) H 0 (X W, O X (3, d) XW ) = s i=1 H 0 (X (2) P, O(3)) i is surjective for d i 2deg P i. This surjectivity, together with our observation that whether f is good at all Q (H f ) Pi is completely determined by ϕ Pi (f) H 0 (P 1, O(3)), implies that when d δ, we have P (2) i #S 3,d Pe low s 0 #ϕ Pi ({f S 3,d : f is good at P i }) = #S 3,d #H 0 (X (2) i=1 P, O(3)) i The lemma follows by taking d on both sides. Lemma The fraction of good pairs in H 0 (P 2 F q e, O(3)) 2 is (1 q 2e )(1 2q 2e q 3e + 2q 5e q 6e 1 3 q 7e ) By the previous lemma, if e = deg P, then this fraction is precisely Prob(f S P ), defined in Section 3.1. Proof. A randomly chosen (F 1, F 2 ) H 0 (P 2 F q e, O(3)) 2 fails to be good if and only if one of the following four events happen: 1. F 1 describes a nodal curve, and F 2 vanishes at the node of F F 1 describes a cuspidal curve. 3. F 1 describes a reducible curve. 4. F 1 0. We denote the above events by E 1, E 2, E 3, E 4 respectively. To simplify notation, we denote q e by a bold q. Our results on the number of cubics of different types readily imply Prob(E 1 ) = Prob(F 1 is a nodal curve)prob(f 2 vanishes at the node of F 1 ) = 1 q 7 (q2 + q + 1)(q 1) 3 (q + 1) 1 q = 1 q 8 (q2 + q + 1)(q 1) 3 (q + 1) 25

26 and Prob(E 2 ) = 1 q 7 (q2 + q + 1)(q 1) 2 (q + 1) Prob(E 3 ) = 1 3q 10 (q2 + q + 1)(q 1)(q + 1)(3q 4 q 3 + 2q 2 2q + 3) It is clear that Prob(E 4 ) = q 10. The events E j s are clearly disjoint, so the fraction of good pairs in the lemma is computed by 1 4 Prob(E j ) j=1 Proof of Lemma 3.1. The first statement is simply Lemma We check the inequalities in the second statement. First we show that when x R and x 5, (1 x 2 ) (1 2x x 3 + 2x 5 x x 7 ) > 0 Denote the function of x on the left hand side as a(x). x 2 a(x) = x 2 ( 1 3 x 7 + x 6 2x x 3 + x 2 ) = 1 3 x 5 + x 4 2x x > 2x x > 0 Therefore a(x) > 0 when x 5. Since we have assumed char F q 5, we have q 5 and (1 q 2e )(1 2q 2e q 3e + 2q 5e q 6e 1 3 q 7e ) < (1 q 2e ) 2 Now we show that when x 5, (1 x 3/2 ) (1 2x x 3 + 2x 5 x x 7 ) < 0 26

27 Let us call the function of x on the left hand side b(x). x 3/2 b(x) = x 3/2 ( 1 3 x 7 + x 6 2x x 3 + 2x 2 x 3/2 ) = 1 3 x 5.5 x x x 1.5 2x > 1 3 x 5.5 x 4.5 2x > 0 Therefore when x 5, we have b(x) < 0 and (1 q 2e )(1 q 3e/2 ) < (1 q 2e )(1 2q 2e q 3e + 2q 5e q 6e 1 3 q 7e ) Points of High Degree We are going to bound the probability that a randomly chosen f S 3,d is bad at some point P with high degree (> d/p ). Therefore the following lemma is to be applied with j = d/p. The main idea used here has been explained in section Let A 3,d = F q [x, y] Fq F q [t]. For each f A 3,d we define polynomials T 2 f(u 0, U 1, x, y, t) = f xx U f xy U 1 U 0 + f yy U 2 0 T 3 f(u 0, U 1, x, y, t) = f xxx U f x 2 yu 2 1 U 0 + 3f xy 2U 1 U f y 3U 3 0 Lemma Let U be an open subscheme of P 2 P 1 and j > 2 be an integer. The probability that for a randomly chosen f S 3,d there exists a closed point Q U that satisfies the following two conditions 1. P := π(q) has degree > j. 2. The fiber (H f ) P is a cuspidal or reducible cubic with a singularity at Q, or the hypersurface H f has a singularity at Q. is bounded by O(d 4 q min( d/p +1,j) ) Proof. It suffices to show the lemma for an affine chart A 2 A 1 since P 2 P 1 can be covered by six such charts. Without loss of generality, assume that X 0, T 0 0 on U = A 2 A 1 and let x = X 1 /X 0, y = X 2 /X 0 and t = T 1 /T 0, such that U = Spec F q [x, y] Fq F q [t]. Hence we may identify S 3,d with A 3,d. Following the ideas presented in section 3.2.1, we only have to bound the density of f A 3,d such that there exists a closed point Q A 2 A 1 that satisfies the following three conditions 27

28 1. P := π(q) has degree > j. 2. f(q) = f x (Q) = f y (Q) = 0 3. T 2 f(u 0, U 1, Q) and T 3 f(u 0, U 1, Q) have a common zero on Proj F q [U 0, U 1 ] (Type I), T 2 f(u 0, U 1, Q) vanishes at a point on Proj F q [U 0, U 1 ] with multiplicity 2 (Type II), or f t (Q) = 0 (Type III). To bound the density of polynomials of Type I or Type II, we are naturally led to consider the scheme Proj F q [U 0, U 1 ] A 2 A 1. f is of Type I only if there exists a closed point Q in the sub- Type I: scheme W 0 := {f = f x = f y = T 2 f = T 3 f = 0} Proj F q [U 0, U 1 ] A 2 A 1 such that deg π(q ) j, where we still denote the projection to the last A 1 component by π. On the affine chart U = Spec F q [u] A 2 A 1 where u = U 1 /U 0, we dehomogenize T 2 f, T 3 f to: T 2 f(u, x, y, t) = f xx u 2 + 2f xy u + f yy T 3 f(u, x, y, t) = f xxx u 3 + 3f x 2 yu 2 + 3f xy 2u + f yyy We call a closed point Q Proj F q [U 0, U 1 ] A 2 A 1 admissible if deg π(q) j and a subscheme V Proj F q [U 0, U 1 ] A 2 A 1 admissible if it contains an admissible point. We denote the union of admissible irreducible components of (V ) red by V ad, and the collection of its irreducible components by Irr V ad. Following Poonen s idea, we write f in a form so that the derivatives involved are largely independent. If h A 3,d, g i A 0, d/p for 0 i 4 uniformly and independently at random, then the distribution of f = h + g p 4x 3 + g p 3y 2 + g p 2x + g p 1y + g p 0 is also uniform over A 3,d. Direct computation shows T 3 f = u 3 (h xxx + 6g p 4) + 3u 2 h x 2 y + 3uh xy 2 + h yyy T 2 f = u 2 (h xx + 6xg p 4) + 2uh xy + h yy + 2g p 3 f x = h x + 3g p 4x 2 + 2g p 3x + g p 2 f y = h y + g p 1 We define a sequence of subschemes W 0 W 4 U by W 4 = {T 3 f = 0}, W 3 = W 4 {T 2 f = 0} W 2 = W 3 {f x = 0}, W 1 = W 2 {f y = 0} 28

29 As the reader may check, we deliberately set up the supscripts so that for each k = 0,, 4, W k only depends on the choice of h, g k,, g 4. Let E denote the event that a. dim (W 1 ) ad = 0 b. f does not vanish at any of Irr(W 1 ) ad. Clearly if E holds for f, then W 0 does not contain any admissible point. Therefore it suffices to show that for randomly chosen h, g 1,, g 4, Prob(E) = 1 O(d 4 q min( d/p +1,j) ) Step 1: Conditioned on a choice of h, the probability that dim W 4 = 4 is at most q ( d/p +1). Proof of Step 1: Indeed, dim W 4 = 4 if only if T 3 f vanish identically on U, which happens only when T 3 f = 0 as a polynomial in F q [u, x, y, t], and in particular h xxx + 6g p 4 = 0. For each h, there is at most one choice for g 4 such that h xxx + 6g p 4 = 0 as polynomials, and hence the claim follows. Step 2: For k = 3, 2, 1, conditioned on a choice of h, g k,, g 4 for which dim W k+1 k, the probability that dim (W k ) ad k is at most O(d 4 k q min( d/p +1,j) ). Proof of Step 2: We only show this for k = 3, since the proofs for k = 2, 1 are completely analogous. Let V 1,, V l be all the irreducible components of (W 4 ) ad, where l = #{Irr(W 4 ) ad }. By Bèzout s theorem, l = O(d). We need to bound the set G bad i = {g 3 A 0, d/p : T 2 f vanishes identically on V i } If g, g G bad i, then g g vanishes identically on V i. Hence if G bad i a coset of the kernel of the linear map, it is ϕ i : A 0, d/p H 0 (V i, O Vi ) Now we apply Lemma 2.6 by identifying U = A 1 A 2 A 1 with A 3 A 1 to obtain that #{Im ϕ i } q min( d/p +1,j). Therefore the probability that a randomly chosen g 3 lies in G bad i is at most q min( d/p +1,j). Since there are O(d) such V i s, the claim follows. Step 3: Conditioned on choice h, g 4,, g 1 such that dim (W 1 ) ad = 0, the probability that (W 0 ) ad is at most O(d 4 q min( d/p +1,j) ). Proof of Step 3: By Bézout s theorem again, we see that #{W 0 } = O(d 4 ). For an admissible point Q W 0, the set H bad of g 0 A 0, d/p for which f vanishes at Q forms a coset of the kernel of the map ϕ Q : A 0, d/p H 0 (Q, O Q ). We 29

30 apply Lemma 2.6 again with W = Q to prove claim 3. Finally Step 1 and 2 together give that Prob(dim (W 1 ) ad ) = 0) (1 q ( d/p +1) ) and Step 3 gives 3 (1 d k q min( d/p +1,j) ) k=1 = 1 O(d 3 q min( d/p +1,j) ) Prob(E) Prob(dim (W 1 ) ad ) = 0)(1 O(d 4 q min( d/p +1,j) ) = 1 O(d 4 q min( d/p +1,j) ) Type II: We can check whether f is of Type II locally. It suffices to bound the density of f A 3,d such that there exists a closed point Q U in the subscheme W 0 = {f = f x = f y = T 2 f = (T 2 f) u = 0} U with deg π(q ) j. The rest of the arguments is rather analogous to what we did before, so we only give a sketch. We define W 4 = {(T 2 f) u = 0}, W 3 = W 4 {T 2 f = 0} W 2 = W 3 {f x = 0}, W 1 = W 2 {f y = 0} This time we choose f A 3,d by choosing h A 3,d, g i A 0, d/p for 0 i 4 uniformly and independently at random and putting Direct computation shows f = h + g p 4xy + g p 3y 2 + g p 2x + g p 1y + g p 0 (T 2 f) u = 2uh xx + 2(h xy + g p 4) T 2 f = u 2 h xx + 2u(h xy + g p 4) + h yy + g p 3 f x = h x + g p 4y + g p 2 f y = h y + g p 1 By applying the same argument as in Step 1 and 2 of the Type I case, we may show that Prob(dim (W 1) ad = 0) = 1 O(d 3 q min( d/p +1,j) ) By applying the same arguments as in Step 3, we may bound the probability that f does not vanish at any of the irreducible components of (W 2) ad conditioned on dim (W 2) ad = 0, so that Prob((W 0) ad = ) 1 O(d 4 q min( d/p +1,j) ) 30

31 Type III: This is the easiest case to deal with. There is no need to involve U i s. The proof is completely analogous to the second part of the proof of Lemma 2.7, where we treated the case in which the curve H f has a singularity on A 1 A 1, except that we have A 2 A 1 instead of A 1 A 1 and one more variable in this case. Now the proof of the lemma is complete Proof of the Main Result Lemma Proof. By definition, lim Prob(f Qmed e 0 ) = 0 Prob(f Q med ) = lim Prob(f d Q med e d 0 ) for f d randomly chosen from S 3,d. It suffices to show that Prob(f d Q med ) is universally bounded by O(q ), where the implied constant is independent of d or. Let P be a point of degree e d/p on P 1. By the proof of Lemma 3.11, the restriction map S 3,d H 0 (X P (2), O(3)) is surjective since p > 2 and e < d/p < d/2. By Lemma 3.12, there exists a constant c, such that the probability that f d is bad at P is cq 2e for each e. d/p Prob(f d Q med ) (number of points of degree e in P 1 )cq 2e e= d/p c (q e + 1)q 2e q = O( 1 q ) = 1 O(q ) e= Lemma Prob(f Q high ) = 0 Proof. Apply Lemma 3.13 with j = d/p and U = P 2 P 1. Then let d. For each, we have P low S P low 31 Q med Q high