Southwestern Center for Arithmetical Algebraic Geometry 1998 Arizona Winter School Workshop 13–18 March 1998 in Tucson, Arizona

Diophantine Geometry Related to the ABC Conjecture

STUDENT PRESENTATION

STUDENTS:  FACULTY:
Sang Yook An  Minhyong Kim
Seog Young Kim  Barry Mazur
David Marshall  William McCallum
Susan Marshall  Dinesh Thakur

Alexander Perlis

Theorem: Assuming the ABC, Szpiro, and B&SD conjectures, all semistable elliptic curves of fixed rank have

|ILLI| ≪ N^{3/4+ε} .

From: Dorian Goldfeld and Lucien Szpiro, Bounds for the order of the Tate–Shafarevich group, Compositio Mathematica 97, 1995, pp. 71–87.

ASSUMPTIONS

ABC Conjecture For all A, B, C ∈ Z with GCD(A, B, C) = 1 and A + B = C,

SUP (|A| , |B| , |C|) ≪ N_{0}(ABC)^{1+ε} ,

where N_{0} is the conductor of the product ABC:

N_{0}(ABC) = _{p|ABC}^{∏} p.

Szpiro’s Conjecture For all elliptic curves E/Q,

|D| ≪ N^{6+ε} ,

where D is the minimal discriminant of E, and N is the conductor of E.

Remark The above conjectures are equivalent. We will make use of both formulations.

ASSUMPTIONS

Birch & Swinnerton-Dyer Conjecture Let L(s) ≡ L-series of E, and r ≡ RANK(E). Then

ORD_{s=1}L(s) = r,

and

\frac{L(s)}{(s − 1)^{r}} |_{s=1} = \frac{2^{r} · |ILLI| · Ω · R · ∏_{p|N} c_{p}}{|E(Q)_{TOR}|^{2}} ,

where Ω is the real period, R is the regulator, and the c_{p} are the local periods (Tamagawa numbers).

THEOREM

Assuming the previous three conjectures, we have:

For all semistable elliptic curves of fixed rank r,

|ILLI| ≪ N^{3/4+ε} .

Precise statement: For each rank r and each ε > 0, there exists a constant C_{r,ε} > 0 so that, for all semistable elliptic curves of rank r, we have

|ILLI| < C_{r,ε} · N^{3/4+ε} .

OVERVIEW OF PROOF

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , to be shown
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , to be shown
• \frac{1}{R} ≪ C_{r} (constant depending on r), to be shown
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

4-1

OVERVIEW OF PROOF

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , to be shown
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , to be shown
• \frac{1}{R} ≪ C_{r} (constant depending on r), to be shown
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

OVERVIEW OF PROOF

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , to be shown
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , to be shown
• \frac{1}{R} ≪ C_{r} (constant depending on r), to be shown
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

5-a

OVERVIEW OF PROOF

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , to be shown
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , to be shown
• \frac{1}{R} ≪ C_{r} (constant depending on r), to be shown
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

5-b

OVERVIEW OF PROOF

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , to be shown
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , to be shown
• \frac{1}{R} ≪ C_{r} (constant depending on r), to be shown
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

5-c

Establishing \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε}

Recall, for ℜ(s) > 3/2, and semistable E:

L(s) = _{p|N}^{∏} \frac{1}{1 ± p^{−s}} _{p∤N}^{∏} \frac{1}{1 − a_{p} p^{−s} + p^{1−2s}} ,

where a_{p} = (1 + p) − #E_{p}(F_{p}). To bound \frac{L(s)}{(s−1)^{r}} |_{s=1} , we proceed indirectly. As usual, set

Λ(s) = N^{s/2}(2π)^{−s}Γ(s)L(s),

so that Λ(2 − s) = ±Λ(s). We have the bounds:

• |\frac{Λ(s)}{(s−1)^{r}} | ≪ N^{3/4+ε} for ℜ(s) = 3/2 + ε.

• |\frac{Λ(s)}{(s−1)^{r}} | ≪ N^{3/4+ε} for ℜ(s) = 1/2 − ε.

Establishing \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε}

Now we will transfer these bounds to the strip 1/2 − ε ≤ ℜ(s) ≤ 3/2 + ε.

By a generalized maximum modulus principle, it is sufficient to show, throughout the strip,

|\frac{Λ(s)}{(s−1)^{r}} | is bounded (by something).

Establishing \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε}

Obtaining an arbitrary bound on |\frac{Λ(s)}{(s−1)^{r}} |:

• Some upper bound on Λ(s):

Λ(s) = N^{s/2} \underbrace{(2π)^{−s}Γ(s)L(s)}_{Mellin transform of the weight 2 cusp form ∑ a_{n}q^{n}}  (semistable)

• Some lower bound on (s − 1)^{r}: B&SD implies that \frac{Λ(s)}{(s−1)^{r}} does not blow up at s = 1.

Establishing \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε}

Thus our bounds transfer to give

|\frac{Λ(s)}{(s − 1)^{r}} | ≪ N^{3/4} + ε

for 1/2 − ε ≤ ℜ(s) ≤ 3/2 + ε. In particular, substituting s = 1 gives

N^{1/2} (2π)^{−1}Γ(1) \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{3/4+ε} ,

so that

\frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} .

Establishing \frac{1}{Ω} ≪ N^{1/2+ε}

Recall that the real period is

Ω = ∫_{E(R)} |ω| ,

where ω is the associated invariant differential.

Upon choosing a Weierstrass equation, E(R) has one of two possible forms:

Let γ denote the infinite component, and let δ denote the number of components. The integral along either component is the same; thus

Ω = δ ∫_{γ} |ω| .

Establishing \frac{1}{Ω} ≪ N^{1/2+ε}

Starting with a global minimal equation Y^{2} + a_{1}XY + a_{3}Y = X^{3} + a_{2}X^{2} + a_{4}X + a_{6},

we have ω = \frac{dX}{2Y + a_{1}X + a_{3}} .

The usual transformations (see Tate’s article or Silverman’s book) give the convenient form:

y^{2} = x^{3} − 27c_{4}x − 54c_{6}
ω = \frac{3dx}{y} .

Furthermore (we’ll need this later):

1728D = c_{4}^{3} − c_{6}^{2},
GCD(c_{4}, D) = 1.  ← semistable

Establishing \frac{1}{Ω} ≪ N^{1/2+ε}

Then Ω = δ ∫_{γ} |ω| becomes

Ω = δ ∫_{γ} |\frac{3 dx}{y} | = 2δ ∫_{r_{0}}^{∞} \frac{3 dx}{√x^{3} − 27c_{4}x^{2} − 54c_{6}} ,

where r_{0} is the largest real root of the cubic.

Knowing how to compute Ω, we now show how to bound 1/Ω:

\frac{1}{Ω} ≪_{↑_{Calculus}}SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} ) ≪_{↑_{ABC}} N^{1/2+ε} .

Calculus: \frac{1}{Ω} ≪ SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} )

We must study

Ω = 2δ ∫_{r_{0}}^{∞} \frac{3 dx}{√x^{3} − 27c_{4}x − 54c_{6}} .

We will consider two cases based on the value

k = |c_{4}| / |c_{6}|^{2/3} .

If |c_{6}|^{2} ≥ |c_{4}|^{3}, i.e., if 0 < k ≤ 1, then write the integral in the form

\frac{1}{|c_{6}|^{1/2}} ∫_{r_{0}}^{∞}_{\frac{3 dx}{√(\frac{x}{|c_{6}|^{1/3}} )^{3}±27(\frac{|c_{4}|}{|c_{6}|^{2/3}} )(\frac{x}{|c_{6}|^{1/3}} )±54}} . The substitution u = x/ |c_{6}|^{1/3} gives

\frac{1}{|c_{6}|^{1/6}} ∫ ∞ r(k) ↑_{largest real root of u^{3} ± 27ku ± 54} \frac{3 du}{√u^{3} ± 27ku ± 54} .

Calculus: \frac{1}{Ω} ≪ SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} )

Three observations:

• The integral

\frac{1}{|c_{6}|^{1/6}} ∫_{r(k)}^{∞} \frac{3 du}{√u^{3} ± 27ku ± 54}

converges when u^{3} ± 27ku ± 54 has distinct roots, i.e., when k ≠ 1.

• The integral goes to ∞ as k → 1, which is okay since we seek a lower bound.

• The integral is continuous in k for k ∈ [0, 1 − ε].

Calculus: \frac{1}{Ω} ≪ SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} )

Hence a lower bound must exist for the integral, so that

\frac{1}{|c_{6}|^{1/6}} ≪ Ω.

The other case, |c_{6}|^{2} ≤ |c_{4}|^{3}, similarly gives

\frac{1}{|c_{4}|^{1/4}} ≪ Ω.

Combining the two cases gives

\frac{1}{Ω} ≪ SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} ) .

ABC: SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} ) ≪ N^{1/2+ε} .

Recall

GCD(c_{4}, D) = 1,
c_{4}^{3} − c_{6}^{2} = 1728D.

Let d = GCD(c_{4}^{3}, 1728). Applying ABC to

\frac{c_{4}^{3}}{d} − \frac{c_{6}^{2}}{d} = \frac{1728D}{d} .

gives

SUP (\frac{|c_{4}|^{3}}{d} , \frac{|c_{6}|^{2}}{d} ) ≪ N_{0} (\frac{c_{4}^{3}c_{6}^{2}1728D}{d^{3}} )^{1+ε}
≪ N_{0} (c_{4}c_{6}2 · 3 · D) ^{1+ε}
≪ |c_{4}|^{1+ε} |c_{6}|^{1+ε} N_{0}(D)^{1+ε}
= |c_{4}|^{1+ε} |c_{6}|^{1+ε} N^{1+ε} .

Absorb the d’s into the “≪” constant: SUP (|c_{4}|^{3} , |c_{6}|^{2}) ≪ |c_{4}|^{1+ε} |c_{6}|^{1+ε} N^{1+ε} .

ABC: SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} ) ≪ N^{1/2+ε} .

There are two cases to consider. If |c_{6}|^{2} ≤ |c_{4}|^{3}, then our inequality

SUP (|c_{4}|^{3} , |c_{6}|^{2}) ≪ |c_{4}|^{1+ε} |c_{6}|^{1+ε} N^{1+ε} becomes |c_{4}|^{3} ≪ |c_{4}|^{5/2+5/2ε} N^{1+ε} . Thus |c_{4}|^{1/2−5/2ε} ≪ N^{1+ε} , or simply |c_{4}|^{1/2} ≪ N^{1+ε} , so that |c_{4}|^{1/4} ≪ N^{1/2+ε} .

The other case, |c_{4}|^{3} ≤ |c_{6}|^{2}, similarly gives |c_{6}|^{1/6} ≪ N^{1/2+ε} .

Combining the cases gives SUP (|c_{4}|^{1/4} , |c_{6}|^{1/6} ) ≪ N^{1/2+ε} .

PROGRESS REPORT

Rearranging B&SD gives:

|ILLI| = \frac{L(s)}{(s − 1)^{r}} |_{s=1} · \frac{|E(Q)_{TOR}|^{2}}{2^{r} Ω R _{p|N}^{∏}c_{p}} ≪ N^{3/4+ε} ; (rank fixed)

Thus, to bound |ILLI|, we seek bounds for:

• \frac{L(s)}{(s − 1)^{r}} |_{s=1} ≪ N^{1/4+ε} , DONE
• |E(Q)_{TOR}|^{2} ≤ 144, by Mazur’s theorem
• \frac{1}{Ω} ≪ N^{1/2+ε} , DONE
• \frac{1}{R} ≪ C_{r} (constant depending on r), TO BE SHOWN
• \frac{1}{_{p|N}^{∏}c_{p}} ≤ 1, since each c_{p} = |\frac{E(Q_{p})}{E_{0}(Q_{p})} | ≥ 1

Establishing \frac{1}{R} ≪ C_{r}

Recall that the regulator is

R = DET [⟨P_{i}, P_{j}⟩] ,

where ⟨·, ·⟩ is the canonical height pairing, and {P_{i}} is a basis for E(Q)/E(Q)_{TOR}.

Viewing the free Z-module E(Q)/E(Q)_{TOR} as a lattice (of rank r) inside the R-vectorspace

\frac{E(Q)}{E(Q)_{TOR}} ⊗_{Z} R (of dimension r) ,

the regulator is simply

R = COVOL(lattice)^{2},

where volume is measured with respect to the length

ℓ(P) = √⟨P, P ⟩.

Among all nonzero lattice points, let h_{0} be the minimal canonical height. Three results will help us bound 1/R:

• Minkowski’s Theorem from the geometry of numbers will give us:

\frac{1}{R} ≪ c_{r} (\frac{1}{h_{0}} )^{r} .

• Hindry–Silverman will give us:

\frac{1}{h_{0}} ≪ something involving S.

• Szpiro’s conjecture will give us:

S ≤ 6 + ε + c_{ε}.

Establishing \frac{1}{R} ≪ c_{r} (\frac{1}{h_{0}} )^{r}

Theorem (Minkowski) Let B_{n} denote the ball of radius n. If

VOL(B_{n}) ≥ 2^{r} COVOL(lattice),

then B_{n} contains a nonzero lattice point. In particular, the minimal nonzero length ℓ_{0} satisfies ℓ_{0} ≤ n.

Corollary Since VOL(B_{n}) = n^{r} VOL(B_{1}), the theorem requires

n^{r} ≥ 2^{r} COVOL(lattice)/ VOL(B_{1}),

and gives ℓ_{0}^{r} ≤ n^{r}. In particular, the possibility

n^{r} = 2^{r} COVOL(lattice)/ VOL(B_{1})

gives us the unconditional conclusion

ℓ_{0} ≤ 2^{r} COVOL(lattice)/ VOL(B_{1}).

Establishing \frac{1}{R} ≪ c_{r} (\frac{1}{h_{0}} )^{r}

The corollary gave,

ℓ_{0} ≤ 2^{r} COVOL(lattice)/ VOL(B_{1}),

which for us becomes

ℓ_{0}^{r} ≤ 2^{r}√R/ VOL(B_{1}).

Length and canonical height satisfy the identity ℓ(P)^{2} = 2ĥ(P); in particular, ℓ_{0}^{2} = 2h_{0}. Thus we have

(2h_{0})^{r} ≤ 2^{2r}R/ VOL(B_{1})^{2},

so that

\frac{1}{R} ≤ \frac{2^{r}/ VOL(B_{1})^{2}}{h_{0}^{r}} .

In short, we have established

\frac{1}{R} ≪ c_{r} (\frac{1}{h_{0}} )^{r} .

Theorem (Hindry–Silverman) Letting S = LOG |D| / LOG N denote the “Szpiro number”, we have

\frac{1}{h_{0}} ≤ \frac{(20S)^{8}10^{1.1+4S}}{LOG |D|} .

Conjecture (Szpiro) We have |D| ≪ N^{6+ε} . Hence we have

LOG |D| ≤ c_{ε} + (6 + ε) LOG N,

or

S = \frac{LOG |D|}{LOG N} ≤ 6 + ε + \frac{c_{ε}}{LOG N} ≤ 6 + ε + c_{ε}.

Now combine the pieces:

\frac{1}{R} ≪ c_{r} \frac{1}{h_{0}^{r}}
≤ c_{r} \frac{(20S)^{8r}10^{1.1r+4rS}}{(LOG |D|)^{r}}
≤ c_{r} \frac{(20(6 + ε + c_{ε}))^{8r}10^{1.1r+4r(6+ε+c_{ε})}}{(LOG |D|)^{r}}

Since |D| is always at least 3, we know

\frac{1}{LOG |D|} < 1,

and thus we may conclude

\frac{1}{R} ≪ C_{r}.

THANKS

• To Barry Mazur and Minhyong Kim, for suggesting this material.

• To Minhyong Kim, Dinesh Thakur, Kirti Joshi, and Bill McCallum, for answering all our questions.

These slides were last modified in March 1999. The most recent version is available for anonymous retrieval from the website:

         www.math.arizona.edu/∼aprl

Send comments and corrections to Alex Perlis:

            aprl@math.arizona.edu