Elliptic Curves Dr. Carmen Bruni University of Waterloo November - - PowerPoint PPT Presentation

Elliptic Curves

• Dr. Carmen Bruni

University of Waterloo

November 4th, 2015

• Dr. Carmen Bruni

Elliptic Curves

Revisit the Congruent Number Problem

Congruent Number Problem Determine which positive integers N can be expressed as the area

• f a right angled triangle with side lengths all rational.

For example 6 is a congruent number since it is the area of the 3 − 4 − 5 right triangle.

• Dr. Carmen Bruni

Elliptic Curves

From Triangles to Curves

Now, we’re going to take the information about our triangle and get a new equation which will turn out to represent a curve in the real plane. Let x2 + y2 = z2 and xy = 2N for rationals x, y, z and some congruent number N. Adding and subtracting 2xy = 4N to the first equation gives x2 + 2xy + y2 = z2 + 4N x2 − 2xy + y2 = z2 − 4N Factoring and dividing by 4 gives the two equations x + y 2 2 = (z/2)2 + N x − y 2 2 = (z/2)2 − N

• Dr. Carmen Bruni

Elliptic Curves

From Triangles to Curves

With the equations x + y 2 2 = (z/2)2 + N x − y 2 2 = (z/2)2 − N we multiply these two equations together gives x + y 2 2 · x − y 2 2 = ((z/2)2 + N)((z/2)2 − N) x + y 2

• ·

x − y 2 2 = (z/2)4 + N(z/2)2 − N(z/2)2 − N2 (x − y)(x + y) 4 2 = (z/2)4 − N2 x2 − y2 4 2 = (z/2)4 − N2

• Dr. Carmen Bruni

Elliptic Curves

From Triangles to Curves

Letting u = z/2 and v = (x2 − y2)/4, the previous equation becomes v2 = u4 − N2 Multiplying by u2 gives (uv)2 = (u2)3 − N2u2 Finally, we let y = uv and x = u2 which gives us the equation y2 = x3 − N2x We call such curves where y2 equals a cubic in x an Elliptic Curve (provided the discriminant is nonzero; this is the case for cubics associated to the Congruent Number Problem).

• Dr. Carmen Bruni

Elliptic Curves

Examples of an Elliptic Curve

Let’s look at examples of elliptic curves. What do they look like on the real plane? Let’s try to draw y2 = x3 − x first by drawing y = x3 − x and then trying to draw the elliptic curve.

• Dr. Carmen Bruni

Elliptic Curves

Drawing y = x3 − x

First, note that y = x3 − x = x(x − 1)(x + 1) and so the equation has three zeroes at x = 0, ±1. Now let’s break this curve into four intervals and see what happens in each interval y = x3 − x = x(x − 1)(x + 1). Between −∞ and −1, the function is negative. Between −1 and 0, the function is positive. Between 0 and 1, the function is negative. Between 1 and ∞, the function is positive. Lastly, the curve should look smooth with no breaks.

• Dr. Carmen Bruni

Elliptic Curves

The Cubic Curve y = x3 − x

Here is the picture (Using Desmos.com)

• Dr. Carmen Bruni

Elliptic Curves

The Elliptic Curve y 2 = x3 − x

What changes when we make the left hand side y2 instead of y? For almost all values of x, we will get not 1 but 2 output values (the exceptions are the roots). This means that we no longer have a function, rather a curve. The cubic must be positive to have a real root! So all the areas where the picture is negative are gone. The curve still has no breaks and is symmetric about the x-axis, that is, if I reflect the top half of the picture, it should match the bottom half. The function should still be smooth (even at 1).

• Dr. Carmen Bruni

Elliptic Curves

The Cubic Curve y = x3 − x

Here is the picture (All graphs courtesy of Desmos.com) Notice that the curve has two connected components!

• Dr. Carmen Bruni

Elliptic Curves

Connected Components

Note: In general, not all elliptic curves have two components. Some have one like y2 = x3 − 1: However, the elliptic curves associated to the Congruent Number Problem always have two connect components.

• Dr. Carmen Bruni

Elliptic Curves

Points on an elliptic curve

Elliptic curves have infinitely many real points. As an example, y2 = x3 − x has infinitely many real points by noticing that the cubic on the right is always positive when x > 1 and hence we can find a y value by taking the square root. So if we take x = 2, then we see that y2 = 23 − 2 = 6 and so the point P = (2, √ 6) and Q = (2, − √ 6) are on the curve.

• Dr. Carmen Bruni

Elliptic Curves

Points on an elliptic curve

From the perspective of Diophantine equations, it is interesting to ask: How many integer points are on elliptic curves? For the example y2 = x3 − x, it turns out that (±1, 0) and (0, 0) are the only integer points, though this is hardly

• bvious.

How many rational points are on elliptic curves? Above, the only rational points are also the integral ones. More on this later.

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

With an elliptic curve, we can actually describe a way to, given two rational points P and Q, create a third rational point R. Let’s begin with the elliptic curve y2 = x3 − x + 1 for illustrative purposes. y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

Let’s take the points P = (−1.324, 0) and Q = (0, 1) (correct to three decimal places). y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

Draw the line between P and Q. It intersects the curve in a third point as shown in the picture at coordinates (1.895, 2.43). y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

Draw the vertical line through the point which must intersect the curve in a third point, in our case, R = (1.895, −2.43) (this is the same as reflecting about the x-axis). Define P + Q = R for points

• n an elliptic curve (note

that this isn’t just adding the coordinates!) y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

If P = Q, then we can still add points. Here, we use the tangent line to find a third point of intersection. To the right, we start with the point P = (−1, 1) on the same elliptic curve. y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

Using calculus, we can calculate the tangent line at P to be y = x + 2. This intersects the elliptic curve at the point (3, 5). y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

Reflecting as before gives us that 2P = P + P = (3, −5). y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Group Law of an Elliptic Curve

What about if the line between P and Q is vertical? We define a “point at infinity” and call it R = O. This point intersects all vertical lines. In this case, we also call Q = −P (this is the reflection of P about the x-axis). Thus P − P = P + Q = R = O y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Your Turn!

Try an example. Add the points P = (0, 1) and Q = (3, 5). y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Your Turn!

The slope of the line between P and Q is m = 5 − 1 3 − 0 = 4 3 and the y intercept is b = 1 since P = (0, 1) is on the line y = 4

3x + 1.

Thus the equation of the line between P and Q is y = 4

3x + 1.

y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Your Turn!

Where does the line y = 4

3x + 1 intersect y2 = x3 − x + 1?

Plug the equation of the line into the elliptic curve to get: ( 4

3x + 1)2 = x3 − x + 1 16 9 x2 + 8 3x + 1 = x3 − x + 1

x3 − 16

9 x2 − 11 3 x = 0

x(x2 − 16

9 x − 11 3 ) = 0

The last quadratic must have x = 3 as a root since we know the line intersects at the points P = (0, 1) and Q = (3, 5). So factoring the above gives x(x − 3)(x + 11

9 ) = 0

Thus the other point of intersection occurs when x = −11

9 .

The corresponding y value is y = 4

3( −11 9 ) + 1 = −44 27 + 1 = −17 27

• Dr. Carmen Bruni

Elliptic Curves

Your Turn!

This line intersects the elliptic curve at the point ( −11

9 , −17 27 ).

Then finally, reflecting (negating the y-coordinate) gives the point R = ( −11

9 , 17 27)

y2 = x3 − x + 1

• Dr. Carmen Bruni

Elliptic Curves

Formulas For Adding Points

Let’s summarize the above for adding two points P = (x1, y1) and Q = (x2, y2) on the elliptic curve y2 = x3 + Cx + D. Let ℓ be the line connecting P and Q and suppose ℓ is defined by y = mx + b We can describe the slope m and the y-intercept b via m = y2−y1

x2−x1

If P = Q

3x2

1 +C

2y1

If P = Q and b = y1 − mx1 where again we used calculus to compute the tangent line in the case when P = Q.

• Dr. Carmen Bruni

Elliptic Curves

Formulas For Adding Points

As in our example, we can find the intersection of y2 = x3 + Cx + D and y = mx + b by solving (mx + b)2 = x3 + Cx + D m2x2 + 2mxb + b2 = x3 + Cx + D 0 = x3 − m2x2 + (C − 2mb)x + D − b2

• Dr. Carmen Bruni

Elliptic Curves

Formulas For Adding Points

This new polynomial has x1 and x2 as solutions since P and Q are

• n both the line and the curve. Hence,

0 = x3 − m2x2 + (C − 2mb)x + D − b2 = (x − x1)(x − x2)(x − x3) = x3 − (x1 + x2 + x3)x2 + (x1x2 + x1x3 + x2x3)x − x1x2x3 which must hold for all values of x. Hence the coefficients on either side match up. Thus, comparing the x2 coefficients on either side gives −m2 = −(x1 + x2 + x3) ⇒ x3 = m2 − x1 − x2 and y3 = mx3 + b. Hence reflecting gives P + Q = (x3, −y3)

• Dr. Carmen Bruni

Elliptic Curves

Formulas For Adding Points on y 2 = x3 − N2x

When we add P = (x, y) = Q on the elliptic curve y2 = x3 − N2x with N squarefree, the formula for the x-coordinate of P + P becomes: (x2 − N2)2 (2y)2 (see the problem set). Notice here that the x-coordinate is a square, has an even denominator and the numerator shares no common factor with N provided P = (0, 0) or (±N, 0) (see the problem set).

• Dr. Carmen Bruni

Elliptic Curves

Revisit the Congruent Number Problem

Congruent Number Problem Determine which positive integers N can be expressed as the area

• f a right angled triangle with side lengths all rational.
• Dr. Carmen Bruni

Elliptic Curves

Key Theorem 1

Theorem 1. Let (x, y) be a point with rational coordinates on the elliptic curve y2 = x3 − N2x where N is a positive squarefree integer. Suppose that x satisfies three conditions:

1 x is the square of a rational number 2 x has an even denominator 3 x has a numerator that shares no common factor with N

Then there exists a right angle triangle with rational sides and area N, that is, N is congruent.

• Dr. Carmen Bruni

Elliptic Curves

Key Theorem 2

Theorem 2. A number N is congruent if and only if the elliptic curve y2 = x3 − N2x has a rational point P = (x, y) distinct from (0, 0) and (±N, 0). Thus, determining congruent numbers can be reduced to finding rational points on elliptic curves!

• Dr. Carmen Bruni

Elliptic Curves

Next Time

We prove these theorems. We figure out how to go from a rational point on an elliptic curve to a rational right triangle with area N. We revisit Don Zagier’s example. We discuss some tricks for finding rational points on elliptic curves.

• Dr. Carmen Bruni

Elliptic Curves