Elliptic Curves and Public Key Cryptography
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1 Elliptic Curves and Public Key Cryptography Jeff Achter January 7, Introduction to Elliptic Curves 1.1 Diophantine equations Many classical problems in number theory have the following form: Let R be an interesting ring, such as: Z, the integers; or Q = { a b : a, b Z} the rationals; or Z/p, the integers modulo p. Let f (X, Y) be a polynomial with coefficients in R. We would like to describe the set of R-solutions to f (X, Y) = 0, i.e., the set {(x, y) : x, y R, f (x, y) = 0}. Is this set empty? Finite? Infinite? For example, like Bachet in 1621, we might consider the equation y 2 x = 0, and look for solutions in Q. It has the easy solution (3, 5), since = 2. It also has harder solutions, since 2 can also be written as ( ) ( ) or 100 ( ) ( ) or ( ) 2 ( ) 3... How did he come up with this? It s actually easier to back up before we go forwards. About 1400 years earlier, Diophantus of Alexandria developed a number of techniques for dealing with the equations which now bear his name. For example, he looked at quadratic equations in two variables, like x 2 + 2xy + 3y 2 + x + 2y = 0 and was able to show: 1
2 1.1 Diophantine equations 1 INTRODUCTION TO ELLIPTIC CURVES Theorem 1.1. If the quadratic equation F(X, Y) = 0 has one rational solution, then it has infinitely many solutions. His proof is quite geometric. A picture has been left out here. Sketch. Let C be the graph of F(X, Y) = 0. Start with the known rational solution P. For each rational number m, let L m be the line through P with slope m. Then L m and C intersect in two points, P and a second point, Q m ; and the point Q m has rational coordinates. For example, if we start with the point ( 1, 0) satisfying x 2 + 2xy + 3y 2 + x + 2y = 0, the line through ( 1, 0) with slope m is given by So to find the points of intersection, solve y 0 = m (x ( 1)) y = mx + m x 2 + 2xy + 3y 2 + x + 2y = 0 x 2 + 2x(mx + m) + 3(mx + m) 2 + x + 2(mx + m) = 0 x 2 + 2mx 2 + 2mx + 3m 2 x 2 + 6m 2 x + 3m 2 + x + 2m = 0 (x + 1)(x + 2mx + 3m 2 x + 3m 2 + 2m) = 0 One solution is x = 1, the solution we already knew about; the other is and thus m(3m + 2) x m = 1 + 2m + 3m 2 y m = m(x + 1) m = 1 + 2m + 3m 2 y = mx + m Note that if m is rational, then both x m and y m are rational numbers. A little more work then shows that distinct values of m give rise to distinct solutions to f (x, y) = 0; and that every solution arises in this way. We may now return to Bachet s problem y 2 = x 3 2, and the known point (3, 5). We may construct the line of slope m through (3, 5) and intersect it with Bachet s curve: For instance, if we take the line of slope m = 3, we try to solve y 2 = x 3 2 y 5 = 3(x 3) 2
3 1.1 Diophantine equations 1 INTRODUCTION TO ELLIPTIC CURVES and find that y = 3x 4 (3x 4) 2 = x 3 2 x 3 9x x 18 = 0 (x 3)(x 2 6x + 6) = 0 x {3, 3 + 3, 3 3}. But we ve lost rationality! However, if we compute the tangent line to Bachet s curve at our known point, we find it has slope y 2 = x 3 2 2y dy = 3x 2 dy dx = 3x2 2y and so at (3, 5), it has slope /2 5 = ; the tangent line is given by and we obtain y 3 = 21 (x 5) 10 x x x = 0 (x 3)(x 3)(x ) = 0 and we have found a new, rational x-coordinate In general, we can explain Bachet s result: Theorem 1.2. Bachet If (x, y) is a solution to Y 2 X 3 = c, then so is as follows: ( x 4 8cx 4y 2, x6 20cx 3 + 8c 2 ) 8y 3. Compute the tangent line to the curve y 2 x 3 = c at the given point. Compute the intersection with the original curve. There will be a unique new solution. 3
4 1.2 Points on cubics 1 INTRODUCTION TO ELLIPTIC CURVES 1.2 Points on cubics Intersections Now consider a general cubic polynomial f (x, y) = 0 and the curve C it defines. Suppose P = (x 0, y 0 ) and Q = (x 1, y 1 ) are points on the curve. We would like to implement the following strategy for generating new points: Construct the line L = L PQ. Intersect L with C. And hopefully, this yields another solution. This will work, provided that #L C = 3. But there may be fewer points than we want, for a couple of different reasons. For example, consider the curve y 2 x 3 + 2: The line L : y = 3x 4 intersects C in only one Q solution. This is just because we haven t used enough numbers a statement like the one we re after can only be true if we re willing to work with K alg, instead of K. The line M : y = x intersects C in only two points, namely, (3, 5) and ( 100, ). In point of fact, we should have counted (3, 5) twice, since the line M is tangent to C (to order 2) at this point. We can see this algebraically; recall that the x-values of intersections satisfied the polynomial (x 3) 2 (x ) = 0. The vertical line x = 3 2 only intersects C at ( 3 2, 0). Multiplicity won t get us out of trouble; the algebraic equation just becomes y 2 = 0, and thus the intersection N C contains (3, 5) with multiplicity two. Fixing this last problem requires a little geometric ingenuity. If we replaced C with some other line, the sought-for claim would be: Two distinct lines meet in a single point. Which is clearly false, since parallel lines don t meet. The solution is to add a point at infinity in each direction; and two parallel lines are declared to intersect in the corresponding point at infinity. We can make sense of this geometry, and the resulting object is called the projective plane, P 2. Return to our continuing example y 2 = x 3 2; recall that the slope of the tangent line at (x, y) is 3x2 2y. As x, this tangent line becomes vertical. So we declare that C contains the point at infinity corresponding to the vertical direction. Now, in our alleged counterexample, N C consists of ( 3 2, 0) with multiplicity two and the point at infinity in the vertical direction; henceforth, this last point will be denoted O Combining points Let f (x, y) be a nonsingular cubic polynomial, C the correspondign curve, and P i = (x i, y i ) i {0, 1} be solutions to f (x, y) = 0. Then we can define a point as follows: P 2 = P 0 P 1 4
5 1.2 Points on cubics 1 INTRODUCTION TO ELLIPTIC CURVES Compute the intersection of C and the line P 0 P 1. Look for the third point. Moreover, we have: Theorem 1.3. Suppose f has coefficients in a field K, and P 0, P 1 defined over K. Then P 0 P 1 exists, and is defined over K. May have to allow a point at infinity. The rationality can be seen from explicit formulae, if desired. We compute in the special case of y 2 = x 3 2. Suppose P 0, P 1 defined over some field K, and x 0 = x 1. Then Solve y = µx + β and y 2 = x 3 2: But we know two solutions, x 0 and x 1 ; thus and L P0 P 1 : y = µx + β µ = y 1 y 0 x 1 x 0 β = y 0 µx 0 x 3 ax b y 2 = 0 x 3 ax b (µx + β) 2 = 0 x 3 µ 2 x 2 (a + 2µβ)x (b + β 2 ) = 0 x 3 µ 2 x 2 (a + 2µβ)x (b + β 2 ) = (x x 0 )(x x 1 )(x x 2 ) x 2 = µ 2 x 0 x 1 y 2 = µx 2 + β Loose ends Given a cubic polynomial ax 3 + bx 2 y + cx 2 y + dy 3 + ex 2 + f xy + gy 2 + hx + iy + j, can (eventually) change coordinates so that it looks like y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 ; and O at infinity corresponds to vertical lines. Moreover, except in characteristic two or three, can complete the square: y 1 2 (y a 1x a 3 ) 5
6 1.3 Elliptic curves 1 INTRODUCTION TO ELLIPTIC CURVES and then rescale to get y 2 = x 3 + ax + b, the short Weierstrass form of a cubic. Even though characteristic 2 is especially interesting for computer applications, in these notes I am going to ignore characteristic two, so that the formulas stay relatively manageable. So, given y 2 = f (x) = x 3 + ax + b, we have: the curve is smooth it doesn t intersect itself both derivatives never simultaneously vanish f (x) has distinct roots the discriminant = 16(4a b 2 ) is nonzero. 1.3 Elliptic curves An elliptic curve (in short Weierstrass form) is an equation y 2 = f (x),, f a smooth cubic as above. (More intrinsically, an elliptic curve E is a smooth projective irreducible curve of genus one equipped with a rational point O.) We ll use the notation E(K) for the set of points on the curve. More generally, if L/K is any extension, let E(L) be the set of points on E with coordinates in L. If P 0 P 1 = (x 2, y 2 ), define P 0 + P 1 = (x 2, y 2 ), the point obtained by reflecting P 0 P 1 across the x-axis. (Somewhat more intrinsically, this is (P 0 P 1 ) O.) Theorem 1.4. The operation + turns E(K) into an abelian group: identity element is O; P = (x, y) = (x, y). There s a clean, geometric characterization of the group law: Theorem 1.5. The set E(K), with: identity element O, and group law specified by P + Q + R = O P, Q and R collinear is an abelian group. 6
7 2 PUBLIC KEY CRYPTOGRAPHY Problems (1.1). Let E/Q be the elliptic curve y 2 = x What is ( 2, 3) + ( 1, 4)? In the next problems, we work with an elliptic curve over a field K in which 2 is invertible. (1.2). Suppose E is given in the form y 2 = f (x). (a) Describe all points P such that 2P = P + P = O. (HINT: Note that P + P = O P = P.) (b) Show that 2P = O P = O or x(p) is a root of f (x). (c) Suppose K is algebraically closed. Show that is isomorphic to Z/2 Z/2. E[2](K) = {P E(K) : 2P = O} (1.3). Suppose E is given by a equation y 2 = x 3 + ax + b, and P = (x, y) E(K) is not O, y = y(p) = 0. Find a formula for 2P: (a) What is the slope of the tangent line L to E at P? (b) Find a formula for L. (c) Find all points of intersection of L and E. You should be able to get a formula for 2P of the form (g(x, y), h(x, y)), where g and h are rational functions. If the denominator of g vanishes for some particular point P 0 = (x 0, y 0 ), we interpret this as 2P 0 = O. (1.4). (a) Find a formula for 3P. (HINT: Use problem (1.3), and the fact that 3P = P + 2P.) (b) Suppose K is algebraically closed, char(k) = 3. Given your formula, how many points P do you think satisfy 3P = O? 2 Public key cryptography 7
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