introduction to elliptic curvesachter/ucr/lect/cr1.pdf · diophantine equations 1 diophantine...

Introduction to Elliptic Curves

Jeff Achter

[email protected]

Department of MathematicsColorado State University

January 2011Universidad de Costa Rica

Jeff Achter ([email protected]) Introduction to Elliptic Curves Jan 2011 1 / 31

Diophantine equations

1 Diophantine equationsBachet’s equationQuadratic equationsReturn to Bachet

2 Points on cubicsLines and cubicsCombining pointsInterludeElliptic curvesStructure


Diophantine equations

Diophantine equationsFix an interesting ring R, like

R = Z, the integers;R = Q, the rational numbers { a

b : a, b ∈ Z}and a polynomial equation f (X, Y) = 0, like

X3 + Y3 = 1.

Basic QuestionDescribe the R-solutions to f (X, Y) = 0:

{(x, y) : f (x, y) = 0, x, y ∈ R}.

Is this set empty? Finite? Infinite?


Diophantine equations Bachet’s equation

Bachet’s equation

Fix c ∈ Q, and studyY2 − X3 = c.

Bachet (1621) turns one solution into many solutions:



Bachet: exampleConsider

Y2 − X3 = −2.

It has the “easy” solution (3, 5), since

52 − 33 = −2.

It has “harder” solutions, too: −2 can be written as(383

1000

)2

−(

129100

)3

or(113259286337279

449455096000

)2

−(

234092288158675600

)3

or

( 1644557217519796256439143766866676956618981558720105932815223934923525719974563641453744978655831227509874752000 )2−( 30037088724630450803382035538503505921

3010683982898763071786842993779918400 )3...



(Partial) Explanation

TheoremBachet If (x, y) is a solution to Y2 − X3 = c, then so is(

x4 − 8cx4y2 ,

−x6 − 20cx3 + 8c2

8y3

).


Diophantine equations Quadratic equations

Quadratic equations

1400 years earlier, Diophantus of Alexandria considered quadraticequations in two variables, like

x2 + 2xy + 3y2 + x + 2y = 0

He shows:

Theorem (Diophantus)

If the quadratic equation F(X, Y) = 0 has one rational solution, then it hasinfinitely many solutions.



Geometry of Quadratics

For the result of Diophantus:Let C be the graph of F(X, Y) = 0.Let P = (x, y) be the known rational solutio.For m ∈ Q:

I Lm is line through P with slope m.I Lm ∩ C = {P, Qm}.I Qm = (xm, ym) has rational coordinates.



ExampleConsider

x2 + 2xy + 3y2 + x + 2y = 0

and P = (−1, 0). Then Lm is:

y− 0 = m · (x− (−1))y = mx + m

For Lm ∩ C, solve

y = mx + m

x2 + 2xy + 3y2 + x + 2y = 0

x2 + 2x(mx + m) + 3(mx + m)2 + x + 2(mx + m) = 0

x2 + 2mx2 + 2mx + 3m2x2 + 6m2x + 3m2 + x + 2m = 0

(x + 1)(x + 2mx + 3m2x + 3m2 + 2m) = 0



Since(x + 1)(x + 2mx + 3m2x + 3m2 + 2m) = 0,

one solution is x = −1; other is

xm = − m(3m + 2)1 + 2m + 3m2

ym = m(x + 1)

=m

1 + 2m + 3m2


Diophantine equations Return to Bachet

Initial attempt

Intersect the curveC : y2 = x3 − 2

with the line Lm through (3, 5) of slope m.Can we produce more points this way?



Rationality problemsTry m = 3:

y2 = x3 − 2y− 5 = 3(x− 3)

so that

y = 3x− 4

(3x− 4)2 = x3 − 2

x3 − 9x2 + 24x− 18 = 0

(x− 3)(x2 − 6x + 6) = 0

x ∈ {3, 3 +√

3, 3−√

3}.

We get solutions, but not in Q.Jeff Achter ([email protected]) Introduction to Elliptic Curves Jan 2011 12 / 31


Tangents

Instead, try the tangent line:Since y2 = x3 − 2,

dydx

=3x2

2y

and the tangent line is given by

y− 3 =2110

(x− 5)

y =2710

x− 3110

.



Tangents

Compute intersection:

y2 = x3 − 2

(2710

x− 3110

)2 = x3 − 2

x3 − 441100

x2 +65150

x− 1161100

= 0

(x− 3)(x− 3)(x− 129100

) = 0

New, rational solution x− 129100 , y = 27

10 x− 3110 = 383

100 .



Bachet’s Theorem, revisited

To derive

TheoremBachet If (x, y) is a solution to Y2 − X3 = c, then so is(

x4 − 8cx4y2 ,

−x6 − 20cx3 + 8c2

8y3

).

Compute the tangent line to the curve y2 − x3 = c at the givenpoint.Compute the intersection with the original curve.There will be a unique new solution.


Points on cubics

1 Diophantine equationsBachet’s equationQuadratic equationsReturn to Bachet

2 Points on cubicsLines and cubicsCombining pointsInterludeElliptic curvesStructure


Points on cubics Lines and cubics

Lines and cubics

Now consider a general cubic polynomial

f (x, y) = 0

and the curve C it defines.If you like, keep working with y2 − x3 − 2.Suppose P = (x0, y0) and Q = (x1, y1) are solutions:

Construct the line L between P and Q.Take the intersection of L with C.

Hopefully, this will give another solution!



Slightly wishful thinking

Hopeful ClaimIf L is a line and C a cubic curve, then # L ∩ C = 3.

There may be fewer points than we want:L : y = 3x− 4 intersects C in only one Q solution.Solution: be willing to use complex numbers!M : y = 27

10 x− 3110 in only two points:

Solution: Count (3, 5) twice, since the line is tangent to the curve!N : x = 3

√2 (a vertical line).

If x = 3√

2 and y2 = 0, get solution ( 3√

2, 0) with multiplicity 2.Solution: Projective plane!





There may be fewer points than we want:L : y = 3x− 4 intersects C in only one Q solution.

Solution: be willing to use complex numbers!M : y = 27




If x = 3√







There may be fewer points than we want:L : y = 3x− 4 intersects C in only one Q solution.Solution: be willing to use complex numbers!

M : y = 2710 x− 31

10 in only two points:

{(3, 5), (129/100, 383/1000)}.



If x = 3√









{(3, 5), (129/100, 383/1000)}.



If x = 3√





Slightly wishful thinkingHopeful ClaimIf L is a line and C a cubic curve, then # L ∩ C = 3.



{(3, 5), (129/100, 383/1000)}.

Solution: Count (3, 5) twice, since the line is tangent to the curve!We can see this algebraically; x-values satisfy

(x− 3)2(x− 129100

) = 0.

N : x = 3√

2 (a vertical line).If x = 3

√2 and y2 = 0, get solution ( 3

√2, 0) with multiplicity 2.

Solution: Projective plane!









If x = 3√


2, 0) with multiplicity 2.

Solution: Projective plane!









If x = 3√





Intersections of lines

We would like the statement

Hopeful ClaimAny two distinct lines meet in a single point.

Problem: Parallel lines don’t meet.Solution: Add a “point at infinity” in each direction.

We have declared that two parallel lines then meet “at infinity” inexactly one way.Can make sense of this geometry; the resulting object is called the projectiveplane P2.



Cubics in P2

Recall our equationy2 = x3 − 2;

the slope of the tangent line at (x, y) is dydx = 3x2

2y .

As x → ∞, the tangent line becomes vertical.Declare that our curve contains the point at infinity correspondingto all vertical lines.Then x = 3

√2 meets our curve at ( 3

√2, 0) (twice) and the point at

infinity.


Points on cubics Combining points

Combining points

f (x, y) a nonsingular cubic polynomialC the corresponding curveP0 = (x0, y0) and P1 = (x1, y1) solutions to

f (x, y) = 0.

Then we can define a point

P2 = P0 ∗ P1 :

Compute the intersection of C and the line P0P1.Look for the “third” point.



Rationality

TheoremSuppose f has coefficients in a field K, and P0, P1 defined over K.Then P0 ∗ P1 exists, and is defined over K.

May have to allow a “point at infinity”.Rationality can be seen from explicit formulae.



Calculation of ∗Return to y2 = x3 − 2.Suppose P0, P1 defined over some field K, and x0 6= x1. Then

LP0P1 : y = mx + b

m =y1 − y0

x1 − x0

b = y0 −µx0

Solve y = mx + b and y2 = x3 − 2:

x3 − 2− y2 = 0

x3 − 2− (mx + b)2 = 0

x3 −m2x2 − 2mbx− (b2 − 2) = 0

But we know two solutions, x0 and x1; thus

x3 −m2x2 − 2mbx− (b2 − 2) = (x− x0)(x− x1)(x− x2)

x2 = m2 − x0 − x1 and y2 = mx2 + b.



Special cases

What if...

P0 = P1?

For L, use tangent line at P0.x0 = x1?Tangent line is vertical; intersection is O, the point at infinity.One of the points is O?For L, use vertical line through the other point.



Special cases

What if...

P0 = P1?For L, use tangent line at P0.

x0 = x1?Tangent line is vertical; intersection is O, the point at infinity.One of the points is O?For L, use vertical line through the other point.



Special cases

What if...

P0 = P1?For L, use tangent line at P0.x0 = x1?

Tangent line is vertical; intersection is O, the point at infinity.One of the points is O?For L, use vertical line through the other point.



Special cases

What if...

P0 = P1?For L, use tangent line at P0.x0 = x1?Tangent line is vertical; intersection is O, the point at infinity.

One of the points is O?For L, use vertical line through the other point.



Special cases

What if...

P0 = P1?For L, use tangent line at P0.x0 = x1?Tangent line is vertical; intersection is O, the point at infinity.One of the points is O?

For L, use vertical line through the other point.



Special cases

What if...

P0 = P1?For L, use tangent line at P0.x0 = x1?Tangent line is vertical; intersection is O, the point at infinity.One of the points is O?For L, use vertical line through the other point.


Points on cubics Interlude

CoordinatesGiven a cubic polynomial

ax3 + bx2 y + cx2 y + dy3 + ex2 + f xy + gy2 + hx + iy + j,

can (eventually) change coordinates so that it looks like

y2 + a1xy + a3 y = x3 + a2x2 + a4x + a6;

O at infinity corresponds to vertical lines.Except in characteristic two or three, can complete the square:

y 7→ 12(y− a1x− a3)

and then rescale to get

y2 = x3 + ax + b,

the short Weierstrass form of a cubic.


Points on cubics Interlude

Smoothness

Giveny2 = f (x) = x3 + ax + b.

Thenthe curve is smooth⇐⇒ it doesn’t intersect itself⇐⇒ both derivatives never simultaneously vanish⇐⇒ f (x) has distinct roots⇐⇒ the discriminant ∆ = −16(4a3 + 27b2) is nonzero.

Henceforth, we assume this.


Points on cubics Elliptic curves

Elliptic curves defined

An elliptic curve E over K...abstract ...is a smooth projective irreducible curve of genus one equipped

with a rational point O.concrete ...in short Weierstrass form is an equation

y2 = f (x),

and the point O at infinity.

E(K) is the set of points on the curve.E(L) is the set of points on E with coordinates in L, if L is anyextension of K.



Combinations, revisited

Suppose P0, P1 ∈ E(K).We have defined P2 = (x2, y2) = P0 ∗ P1, the third point ofintersection of E and LP0P1 .Define

P0 + P1 = (x2,−y2).

TheoremThe operation + turns E(K) into an abelian group:

identity element is O;−P = −(x, y) = (x,−y).



Geometric version

TheoremThe set E(K), with:

identity element O, andgroup law specified by

is an abelian group.


Points on cubics Structure

C

E/C admits complex uniformization.There is a lattice Λ ⊂ C, andThere is a lattice Λ ⊂ C and an analytic isomorphism

C/Λ - E(C)

(The isomorphism is given by elliptic functions, which are doublyperiodic meromorphic functions.)


Points on cubics Structure

QTheorem (Mordell)The group E(Q) is finitely generated.

There are generators P1, · · · , Pr:If P ∈ E(Q), then there are a1, · · · , ar:

P = a1P1 + a2P2 + · · ·+ arPr.

Open problemIs

minE/Q

min(r : E(Q) = 〈P1, · · · , Pr〉)

unbounded?


introduction to elliptic curvesachter/ucr/lect/cr1.pdf · diophantine equations 1 diophantine...

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