Ranks of Elliptic Curvesmolsson/KConrad-ranks.pdfElliptic curves x y P P0 P + P0 x y P 2P An elliptic curve, for our needs, is a smooth curve E of the form y2 = x3 + ax + b. Since

Ranks of Elliptic Curves

Keith Conrad

April 15, 2018

Elliptic Curve Addition and theRank

Generalized integers

In number theory, many basic ideas (primes factorization, modulararithmetic, etc.) can be applied in settings beyond Z. The firstexample of this goes back to Gauss: the “complex integers” orGaussian integers

Z[i ] = {x + yi : x , y ∈ Z}.

This is closed under addition and multiplication. Other examples,using real numbers, are

Z[√

2] = {x + y√

2 : x , y ∈ Z}Z[

3√

2] = {x + y3√

2 + z3√

4 : x , y , z ∈ Z}Z[

4√

2] = {x + y4√

2 + z4√

4 + w4√

8 : x , y , z ,w ∈ Z}...

Z[d√

2] = {x0 + x1d√

2 + x2d√

22

+ · · ·+ xd−1d√

2d−1

: xi ∈ Z}.

Each of these is closed under addition and multiplication.

Units among generalized integers

Invertible elements under multiplication are called units. Productsof units are units: uu′ = 1 and vv ′ = 1⇒ (uv)(u′v ′) = 1.

In Z, the units are ±1: boring!

In Z[√

2], x + y√

2 is a unit when x2 − 2y2 = ±1 and all units are

±(1 +√

2)n, n ∈ Z.

In Z[ 3√

2], the units are

±(1− 3√

2)n, n ∈ Z.

In Z[ 4√

2], the units are

±(1 +√

2)n(1 +4√

2)n′, n, n′ ∈ Z,

where 1 +√

2 and 1 + 4√

2 are multiplicatively independent: if(1 +

√2)n(1 + 4

√2)n

′= 1 then n = n′ = 0.

We say Z has unit rank 0, Z[√

2] and Z[ 3√

2] have unit rank 1, andZ[ 4√

2] has unit rank 2.

Dirichlet’s unit theorem (special case)

Theorem (Dirichlet, 1846): For d ≥ 1, the units in Z[ d√

2] arefinitely generated under multiplication: there is a finite list of unitsε1, . . . , εr in Z[ d

√2] such that the set of all units in Z[ d

√2] is

±εn11 · · · εnrr , ni ∈ Z.

When d is even the least r is d/2, and when d is odd the least r is(d − 1)/2.

This says units in Z[ d√

2] for d = 2 and 3 are ±εn, for d = 4 theunits are ±εn11 ε

n22 .

The least r is called the unit rank of Z[ d√

2], and the theorem givesa simple formula for it. Every positive integer is a unit rank, e.g., ris unit rank of Z[ 2r

√2].

Contrast: units of Q are nonzero rationals: ±pn11 · · · pnkk where

k ∈ {0, 1, 2, . . .} and pi ∈ {2, 3, 5, . . .} are primes. This is notfinitely generated: 2, 3, 5, 7, . . . are multiplicatively independent.

Elliptic curves

x

y

P

P ′

P + P ′

x

y

P

2P

An elliptic curve, for our needs, is a smooth curve E of the formy2 = x3 + ax + b.

Since degree is 3, line through points P and P ′

on E (if P = P ′, use tangent at P) has a third point on E : wheny = mx + b, (mx + b)2 = x3 + ax + b has sum of roots equal tom2, so for two known roots r and r ′, the third root is m2 − r − r ′.Set reflection of 3rd point to be P + P ′: comm. and associative.

Elliptic curves

x

y

P

P ′

P + P ′

x

y

P

2P

An elliptic curve, for our needs, is a smooth curve E of the formy2 = x3 + ax + b. Since degree is 3, line through points P and P ′

on E (if P = P ′, use tangent at P) has a third point on E : wheny = mx + b, (mx + b)2 = x3 + ax + b has sum of roots equal tom2, so for two known roots r and r ′, the third root is m2 − r − r ′.

Set reflection of 3rd point to be P + P ′: comm. and associative.

Elliptic curves

x

y

P

P ′

P + P ′

x

y

P

2P

An elliptic curve, for our needs, is a smooth curve E of the formy2 = x3 + ax + b. Since degree is 3, line through points P and P ′

on E (if P = P ′, use tangent at P) has a third point on E : wheny = mx + b, (mx + b)2 = x3 + ax + b has sum of roots equal tom2, so for two known roots r and r ′, the third root is m2 − r − r ′.Set reflection of 3rd point to be P + P ′: comm. and associative.

Elliptic curves over Q

When E : y2 = x3 + ax + b has a, b ∈ Q and P and P ′ are inE (Q) then P + P ′ ∈ E (Q) because calculations only need Q.

Example. Consider y2 = x3 − 2x + 2 and P = (1, 1).

x

y

P

2P

y2 = x3 − 2x + 2

⇒ 2yy ′ = 3x2 − 2

⇒ 2dy

dx

∣∣∣∣(1,1)

= 1

⇒ dy

dx

∣∣∣∣(1,1)

=1

2.

Tangent line at P is y = (1/2)(x − 1) + 1 = (x + 1)/2. Solutionsof ((x + 1)/2)2 = x3 − 2x + 2 are x = 1, 1, and −7/4. Then(−7/4, y) ∈ E (Q) for y = (−7/4 + 1)/2 = −3/8, which makes2P = (−7/4, 3/8).

Elliptic curves over Q

When E : y2 = x3 + ax + b has a, b ∈ Q and P and P ′ are inE (Q) then P + P ′ ∈ E (Q) because calculations only need Q.

Example. Consider y2 = x3 − 2x + 2 and P = (1, 1).

x

y

P

2P

y2 = x3 − 2x + 2

⇒ 2yy ′ = 3x2 − 2

⇒ 2dy

dx

∣∣∣∣(1,1)

= 1

⇒ dy

dx

∣∣∣∣(1,1)

=1

2.

Tangent line at P is y = (1/2)(x − 1) + 1 = (x + 1)/2. Solutionsof ((x + 1)/2)2 = x3 − 2x + 2 are x = 1, 1, and −7/4. Then(−7/4, y) ∈ E (Q) for y = (−7/4 + 1)/2 = −3/8, which makes2P = (−7/4, 3/8).

Elliptic curves over Q: finite and infinite order points

Say P ∈ E (Q) has infinite order if {P, 2P, 3P, 4P, . . .} does notrepeat, and finite order if {P, 2P, 3P, 4P, . . .} cycles.

Example. On E : y2 = x3 − 2x + 2, let P = (1, 1). Then P hasinfinite order: already saw 2P = (−7/4, 3/8), and can show

3P = (97/121,−1271/1331), 4P = (13729/144,−1608463/1728).

x

y

P

2P

3P

Nagell–Lutz theorem (1930s) says if (x , y) ∈ E (Q) has finite orderthen x , y ∈ Z: 2P = (−7/4, 3/8) has infinite order, so P does too!

Elliptic curves over Q: Mordell(-Weil) theorem

If E has rational coefficients then E (Q) has finitely many points offinite order (Nagell–Lutz). Poincare (1901) suggested, and Mordell(1922) proved, that E (Q) is finitely generated.

Geometric meaning: there is a finite set of points in E (Q) suchthat the rest of E (Q) is obtained from it by repeated intersectionsof secant and tangent lines with E .

Algebraic meaning: there is a finite list P1, . . . ,Pr in E (Q), eachwith infinite order, such that

E (Q) = {Tj + n1P1 + · · ·+ nrPr : ni ∈ Z}

where Tj has finite order (finitely many of these) and P1, . . . ,Pr

are independent:

m1P1 + · · ·+ mrPr = n1P1 + · · ·+ nrPr =⇒ mi = ni for all i .

Compare to units in Z[ 4√

2]: ±(1 +√

2)n(1 + 4√

2)n′

for n, n′ ∈ Z.

Examples of minimal generating sets for E (Q)

E : y2 = x3 + 1 has rank 0: E (Q) = {n(2, 3) : n = 0, 1, . . . , 5}.

E : y2 = x3 − 2x + 2 has rank 1: E (Q) = {n(1, 1) : n ∈ Z}.

E : y2 = x3 − 4x + 1 has rank 2:

E (Q) = {n (0, 1)︸︷︷︸P

+n′ (−1, 2)︸︷︷︸Q

: n, n′ ∈ Z}.

P + Q = (2, 1), 2P = (4, 7), 2Q = (33/16,−79/64).

x

y

P

Q

P + Q

2Q

Finding the Rank

How to find the rank of an elliptic curve over Q?

Unlike Dirichlet’s theorem about the rank of units, there is noknown simple formula for the rank of E (Q) from E ’s coefficients.

In the 1960s, a rank “formula” was conjectured using behavior of

Np(E ) = |{(x , y) mod p : y2 ≡ x3 + ax + b mod p}|+ 1.

Example. E : y2 = x3 − 4x + 1.

p 2 3 5 7 11 13 17 19 23 29 31 37

Np(E ) 3 8 10 13 16 19 26 26 31 23 31 45

Conjecture. For each E ,∏p≤X

Np(E )

p?∼ C (logX )rk(E(Q)).

Original data used X ≤ 1000.

Graphs are for y2 = x3 − d2x ,taken from Rubin & SilverbergBAMS, 2002.


Due to Tate’s influence, conjecture about∏

p≤X Np(E )/p turnedinto a conjecture about how much L(E , s) vanishes at s = 1, where

L(E , s) =∏p

1

1− ap(E )/ps + p/p2s=∑n≥1

an(E )

ns

for Re(s) >3

2, with ap(E ) = p + 1− Np(E ). Formally,

L(E , 1) “=”∏p

1

1− (p + 1− Np(E ))/p + 1/p

=∏p

1

Np(E )/p=∏p

p

Np(E ),

so believing∏

p≤X Np(E )/p ∼ C (logX )rank(E(Q)) suggests that

rank(E (Q))?= ords=1(L(E , s)) (BSD conjecture)


Why does E : y2 = x3 − 4x + 1 have rank 2? Upper/lower bounds.

Lower: show P = (0, 1) and Q = (−1, 2) are of infinite order andindependent, so E (Q) has rank at least 2.

Upper: use more advanced techniques (e.g., Galois cohomology) toshow the rank is at most 2.

Prove points independent by ideas from linear algebra.

Toy example. Show 1 +√

2 and 1 + 4√

2 are mult. independent:

(1 +√

2)n(1 +4√

2)n′

= 1⇒ n log(1 +√

2) + n′ log(1 +4√

2) = 0.

Replacing 4√

2 with − 4√

2 sends√

2 = 4√

22

to (− 4√

2)2 =√

2, so

(1 +√

2)n(1 +4√

2)n′

= 1 ⇒ (1 +√

2)n(1− 4√

2)n′

= 1

⇒ (1 +√

2)n|1− 4√

2|n′ = 1

⇒ n log(1 +√

2) + n′ log |1− 4√

2| = 0.


Why does E : y2 = x3 − 4x + 1 have rank 2? Upper/lower bounds.

Lower: show P = (0, 1) and Q = (−1, 2) are of infinite order andindependent, so E (Q) has rank at least 2.

Upper: use more advanced techniques (e.g., Galois cohomology) toshow the rank is at most 2.

Prove points independent by ideas from linear algebra.

Toy example. Show 1 +√

2 and 1 + 4√

2 are mult. independent:

(1 +√

2)n(1 +4√

2)n′

= 1⇒ n log(1 +√

2) + n′ log(1 +4√

2) = 0.

Replacing 4√

2 with − 4√

2 sends√

2 = 4√

22

to (− 4√

2)2 =√

2, so

(1 +√

2)n(1 +4√

2)n′

= 1 ⇒ (1 +√

2)n(1− 4√

2)n′

= 1

⇒ (1 +√

2)n|1− 4√

2|n′ = 1

⇒ n log(1 +√

2) + n′ log |1− 4√

2| = 0.


The relations

n log(1 +√

2) + n′ log(1 +4√

2) = 0

n log(1 +√

2) + n′ log |1− 4√

2| = 0

can be written as(log(1 +

√2) log(1 + 4

√2)

log(1 +√

2) log |1− 4√

2|

)(n

n′

)=

(0

0

)and the matrix has determinant −2.158 . . . 6= 0, so n = n′ = 0.

Similar ideas prove independence of points on E (Q), replacinglogarithms of units with a height pairing on points to linearize theproblem.

This is how Elkies showed in 2006 that an E with coefficients ofover 80 digits has rank(E (Q)) ≥ 28: find 28 points and check the28× 28 height pairing matrix has nonzero determinant. (In 2016,Klagsbrun, Sherman, and Weigandt proved this example wouldhave rank exactly 28 if GRH is true.)

How to find high rank elliptic curves over Q?

Let ET : y2 = x3 + A(T )x + B(T ) be elliptic curve over Q(T ).Then E (Q(T )) is finitely generated (Lang–Neron) and for all butfinitely many t ∈ Q, Et : y2 = x3 + A(t)x + B(t) is elliptic curveover Q and

rank(Et(Q)) ≥ rank(ET (Q(T ))).

Ex: On ET : y2 = x3 − T 2x + 1 let PT = (0, 1), QT = (−1,T ).

PT + QT = (T 2 − 2T + 2,T 3 − 3T 2 + 4T − 3),

2PT =

(T 4

4,T 6 − 8

8

),

2QT =

(T 4 + 2T 2 + 9

4T 2,T 6 − 5T 4 − 9T 2 − 27

8T 3

).

At T = 2 get E2 : y2 = x3 − 4x + 1 with points P2 = (0, 1) andQ2 = (−1, 2) in E2(Q), and recover earlier sums of these points:

P2 + Q2 = (2, 1), 2P2 = (4, 7), 2Q2 =

(33

16,−79

64

).

How to find high rank elliptic curves?

Nagao’s heuristic: if E is an elliptic curve over Q then a heuristicformula for the rank of E (Q) is

1

2− 1

X

∑p≤X

ap(E ) log p =1

2− 1

X

∑p≤X

(p + 1− Np(E )) log p

for large X . This is motivated by calculations with L′(E , s)/L(E , s)based on BSD (residue at s = 1 should equal the rank of E (Q):f (s) = c(s − 1)r + · · · ⇒ f ′(s)/f (s) = r/(s − 1) + · · · ).

If ET is an elliptic curve over Q(T ) and, for a particular t ∈ Q,Nagao’s heuristic suggests

rank(Et(Q)) > rank(ET (Q(T ))),

then do an explicit search for more points on Et(Q) for thatparticular t and verify independence with height pairing matrix.

Nagao (1990s) used this idea to discover record (at the time)elliptic curves over Q with rank at least 17, 20, and 21.

The Parity Conjecture

Functional equation and its sign

For each elliptic curve E over Q, there is NE ∈ Z+ such that

Λ(E , s) := Ns/2E (2π)−sΓ(s)L(E , s)

for Re(s) > 3/2 extends to all s ∈ C (analytic) and there is afunctional equation

Λ(E , s) = wEΛ(E , 2− s), wE = ±1.

The piece Ns/2E (2π)−sΓ(s) is nonvanishing at s = 1.

The sign wE has a purely algebraic description usingrepresentation theory.Letting L(E , s) have order of vanishing r at s = 1, so

L(E , s) = cr (s − 1)r + cr+1(s − 1)r+1 + · · · , cr 6= 0

we have cr = L(r)(E , 1)/r ! 6= 0.Differentiating the functional equation k times (any k),

Λ(k)(E , s) = (−1)kwEΛ(k)(E , 2− s),

Thus Λ(r)(E , 1) = (−1)rwEΛ(r)(E , 1)⇒ wE = (−1)r .

Parity conjecture

Since wE can be computed purely algebraically, the formula

wE = (−1)r = (−1)ords=1(L(E ,s))

tells us ords=1(L(E , s)) mod 2, so knowing wE should also tell usrank(E (Q)) mod 2. That this works is called the parity conjecture.

The parity conjecture is weaker than BSD, but already gives usinteresting information: if wE = −1 then ords=1(L(E , s)) is odd, sowe expect rank(E (Q)) is odd and thus positive, so E (Q) is infinite.

The parity conjecture was proved by Nekovar for elliptic curvesover Q, conditional on finiteness of XE/Q (still an open problem).

That the condition wE = −1 should imply E (Q) is infinite will beseen later to have applications to several elementary questions thatare not initially about elliptic curves.

Applications of ranks

Application of ranks: congruent number problem

For which n in Z+ is there a rational right triangle with area n?Below are n = 5, 6, 7.

If a2 + b2 = c2 with 12ab = n then y2 = x3 − n2x where

x =nb

c − a, y =

2n2

c − a6= 0

and this is a bijection preserving rationality and positivity.

Ex. For area 5, triangle (a, b, c) = (3/2, 20/3, 41/6) correspondsto (x , y) = (25/4, 75/8) with y2 = x3 − 25x .


Let En : y2 = x3 − n2x = x(x + n)(x − n). Points (x , y) on En(Q)with y 6= 0 have infinite order, so one rational right triangle witharea n implies infinitely many such right triangles.

Ex (n = 5). From a triangle to a point, doubling it, and then back,(3

2,

20

3,

41

6

) P =

(25

4,

75

8

),

2P =

(1681

144,−62279

1728

)

(−1519

492,−4920

1519,−3344161

747348

).

In 1975, Stephens showed for squarefree n > 0 that

wEn =

{1, if n ≡ 1, 2, 3 mod 8,

−1, if n ≡ 5, 6, 7 mod 8,

so if n ≡ 5, 6, 7 mod 8 then we expect En(Q) to have odd (sopositive) rank. Monsky (1990) proved this unconditionally forprime p ≡ 5, 7 mod 8.


Example. The number 157 is prime and 157 ≡ 5 mod 8. Thusthere is definitely a rational right triangle with area 157.

The triangle corresponds to a rational point on y2 = x3 − 1572xwith y 6= 0.

Here are the simplest rational point and triangle, which were foundby Zagier (1990):

x =69648970982596494254458225

166136231668185267540804,

y =538962435089604615078004307258785218335

67716816556077455999228495435742408,

a =411340519227716149383203

21666555693714761309610,

b =6803298487826435051217540

411340519227716149383203,

c =224403517704336969924557513090674863160948472041

8912332268928859588025535178967163570016480830.


Example. The number 157 is prime and 157 ≡ 5 mod 8. Thusthere is definitely a rational right triangle with area 157.

The triangle corresponds to a rational point on y2 = x3 − 1572xwith y 6= 0.

Here are the simplest rational point and triangle, which were foundby Zagier (1990):

x =69648970982596494254458225

166136231668185267540804,

y =538962435089604615078004307258785218335

67716816556077455999228495435742408,

a =411340519227716149383203

21666555693714761309610,

b =6803298487826435051217540

411340519227716149383203,

c =224403517704336969924557513090674863160948472041

8912332268928859588025535178967163570016480830.

Application of ranks: Sylvester’s conjecture

Which n in Z+ are x3 + y3 for x , y ∈ Q? Yes for n = 1 and 2, nofor n = 3. Sylvester (really Selmer) conjectured for primes p ≥ 5,

if p ≡ 2, 5 mod 9 then it is not x3 + y3 (false at p = 2),

if p ≡ 4, 7, 8 mod 9 then it is x3 + y3.

If p ≡ 1 mod 9 it can go either way. Data for p < 200: 19, 37,127, and 163 are x3 + y3 while 73, 109, 181, and 199 are not.

On Ep : x3 + y3 = p each (x , y) ∈ Ep(Q) has infinite order, so if pis x3 + y3 with x , y ∈ Q at least once then it is infinitely often.

It can be shown that

wEp =

{−1, if p ≡ 4, 7, 8 mod 9,

1, otherwise,

p ≡ 2, 5 mod 9 =⇒ rank(Ep(Q)) = 0,

p ≡ 4, 7, 8 mod 9 =⇒ rank(Ep(Q)) = 1,

p ≡ 1 mod 9 =⇒ rank(Ep(Q)) = 0 or 2.

Application of ranks: Hilbert’s 10th problem

H10: is there an algorithm that, for every polynomial equationwith coefficients in Z (in finitely many variables), determines if ithas a solution in Z?

Answer: NO. This is the MRDP theorem (1970).

What if we replace Z with ring of integers of a number field?

Denef (1980) showed that for each totally real number field K(e.g., K = Q(

√2)), if there is elliptic curve E over Q such

that E (K ) and E (Q) have rank 1, then H10 has answer NOfor the integers of K . The argument uses denominators ofx-coordinates of points on E .

By work of Poonen, Shlapentokh, et al. if there is an ellipticcurve E over Q such that E (K ) and E (Q) have same positiverank, then H10 has answer NO for integers of K .

Existence of suitable elliptic curves follows from conjecturesabout Tate–Shafarevich groups (Mazur–Rubin) or from rankaspects of the BSD conjecture (Murty–Pasten).



















Application of ranks: class numbers of imaginary quadratic fields

For squarefree d > 0, let hd be class number of Q(√−d). By

Hecke and Heilbronn, hd →∞ as |d | → ∞, but not effectively.

1935: Siegel showed that for all ε > 0 there’s cε > 0 such thathd > cεd

1/2−ε for all d , but cε can’t be explicitly determined.

1952/1966-67: Heegner and then Baker and Stark determinewhen hd = 1: 9 examples.

1971: Baker & Stark determine when hd = 2: 18 examples.

1976: Goldfeld showed hd ≥ cε(log d)1−ε with computable cε,provided there is some elliptic curve E over Q such that

ords=1(L(E , s)) ≥ 3.

This condition makes the exponent on log d at least 1− ε.

Mid-1980s: Gross and Zagier constructed an explicit exampleof the elliptic curve Goldfeld needed: an E where wE = −1and provably L′(E , 1) = 0 (not just L′(E , 1) ≈ .000000001).









ords=1(L(E , s)) ≥ 3.











ords=1(L(E , s)) ≥ 3.



Behavior of ranks

Elevated rank

Let’s return to elliptic curves ET over Q(T ). Recall from before

rank(Et(Q)) ≥ rank(ET (Q(T )))

for all but finitely many t ∈ Q. Question: Could this inequality bestrict all the time? If so, say ET has elevated rank. The parityconjecture suggests a strategy to find examples of elevated rank.

Example (Cassels, Schinzel) The elliptic curve

ET : y2 = x3 − (7(1 + T 4))2x

over Q(T ) has the following properties:

ET (Q(T )) has rank 0 (it is finite), for all t ∈ Q, wEt = −1.

The parity conjecture would imply Et(Q) has odd rank fromwEt = −1, so we should have

rank(Et(Q)) > rank(ET (Q(T ))) for all t ∈ Q.

An elliptic curve y2 = x3 − A(T )2x is not “truly” defined overQ(T ): by the change of variables x 7→ A(T )x and y 7→ A(T )3/2ythe curve becomes y2 = x3 − x , which is defined over Q.

Elevated rank

Let’s return to elliptic curves ET over Q(T ). Recall from before

rank(Et(Q)) ≥ rank(ET (Q(T )))

for all but finitely many t ∈ Q. Question: Could this inequality bestrict all the time? If so, say ET has elevated rank. The parityconjecture suggests a strategy to find examples of elevated rank.

Example (Cassels, Schinzel) The elliptic curve

ET : y2 = x3 − (7(1 + T 4))2x

over Q(T ) has the following properties:

ET (Q(T )) has rank 0 (it is finite), for all t ∈ Q, wEt = −1.

The parity conjecture would imply Et(Q) has odd rank fromwEt = −1, so we should have

rank(Et(Q)) > rank(ET (Q(T ))) for all t ∈ Q.

An elliptic curve y2 = x3 − A(T )2x is not “truly” defined overQ(T ): by the change of variables x 7→ A(T )x and y 7→ A(T )3/2ythe curve becomes y2 = x3 − x , which is defined over Q.

Elevated rank

Assuming standard conjectures about elliptic curves over Q andpolynomials in Z[T ], every elliptic curve over Q(T ) with elevatedrank can be turned into an elliptic curve over Q by a change ofvariables.

Now consider the “function field” case: Z Fp[u], Q Fp(u).

Z Fp[u]

units: ±1 units: F×p = Fp − {0}prime irreducible|m| deg g

Div. Algorithm Div. AlgorithmZ/(m) Fp[u]/(g)

Theorem. For m 6=0, |Z/(m)|= |m|; for g 6=0, |Fp[u]/(g)|=pdeg g .

Theorem. For m ≥ 1, Z/(m) is a field if and only if m is prime.For any g 6= 0, Fp[u]/(g) is a field if and only if g is irreducible.

Theorem. For every prime p, F×p is cyclic. For every irreducible π,(Fp[u]/(π))× is cyclic.

Elevated rank

Assuming standard conjectures about elliptic curves over Q andpolynomials in Z[T ], every elliptic curve over Q(T ) with elevatedrank can be turned into an elliptic curve over Q by a change ofvariables.

Now consider the “function field” case: Z Fp[u], Q Fp(u).

Z Fp[u]

units: ±1 units: F×p = Fp − {0}prime irreducible|m| deg g

Div. Algorithm Div. AlgorithmZ/(m) Fp[u]/(g)

Theorem. For m 6=0, |Z/(m)|= |m|; for g 6=0, |Fp[u]/(g)|=pdeg g .

Theorem. For m ≥ 1, Z/(m) is a field if and only if m is prime.For any g 6= 0, Fp[u]/(g) is a field if and only if g is irreducible.

Theorem. For every prime p, F×p is cyclic. For every irreducible π,(Fp[u]/(π))× is cyclic.

Elevated rank

Recall: standard conjectures about elliptic curves over Q andpolynomials in Z[T ] imply every elliptic curve over Q(T ) withelevated rank can be turned into an elliptic curve over Q by achange of variables.

However. . . the analogue for Fp[u][T ] of one of the conjecturesabout Z[T ] is false.

Theorem (BC, KC, HH) For each odd prime p there is an ellipticcurve ET over Fp(u)(T ) that is not convertible into an ellipticcurve over Fp(u) by a change of variables and has the followingproperties:

ET (Fp(u)(T )) has rank 1,

For all t ∈ Fp(u), wEt = 1 and rank(Et(Fp(u))) > 0.

The parity conjecture over Fp(u), if true, would imply here that

rank(Et(Fp(u))) > rank(ET (Fp(u)(T )))

for all t ∈ Fp(u) since the left side should be even, right side is 1.

Elevated rank









Elevated rank









Values of ranks

Unlike Dirichlet’s theorem about the rank of units, we don’t knowwhich integers are ranks of some E (Q).

1950s: Neron conjectured ranks of E (Q) are bounded.

1960s-recently: believe ranks of E (Q) are unbounded.

Why did the conventional wisdom change?

1 Known examples of lower bound on rank(E (Q)) kept gettingbigger: at least 4 (1945), 7 (1975), 12 (1982), 20 (1993), and28 (2006).

2 For elliptic curves over Fp(u), where Mordell–Weil theorem istrue, examples of Tate and Shafarevich, and later Ulmer, haveunbounded ranks.

In 2017, Park, Poonen, Voight, and Wood introduced a heuristicprobabilistic model for elliptic curve ranks that suggests ranks arebounded

: finitely many elliptic curves over Q have rank ≥ 22. Thecurrent record for infinitely many ranks is 19.

Values of ranks







In 2017, Park, Poonen, Voight, and Wood introduced a heuristicprobabilistic model for elliptic curve ranks that suggests ranks arebounded: finitely many elliptic curves over Q have rank ≥ 22. Thecurrent record for infinitely many ranks is 19.

Values of ranks







In 2017, Park, Poonen, Voight, and Wood introduced a heuristicprobabilistic model for elliptic curve ranks that suggests ranks arebounded: finitely many elliptic curves over Q have rank ≥ 22. Thecurrent record for infinitely many ranks is 19.

Thanks!

Documents

Ranks of Elliptic Curvesmolsson/KConrad-ranks.pdfElliptic curves x y P P0 P + P0 x y P 2P An elliptic curve, for our needs, is a smooth curve E of the form y2 = x3 + ax + b. Since