Computing the endomorphism ring of an ordinary elliptic curve

Transcription

1 Computing the endomorphism ring of an ordinary elliptic curve Massachusetts Institute of Technology April 3, 2009 joint work with Gaetan Bisson

2 Elliptic curves An elliptic curve E/F is a smooth projective curve of genus 1 with a distinguished rational point 0. The set E(F) of rational points on E form an abelian group. For char(f) 2, 3 we define E with an affine equation y 2 = x 3 + Ax + B, where 4A B 2 0. The j-invariant of E is j(e) = A 3 4A B 2. If F = F then j(e) uniquely identifies E (but not in F q ).

3 Elliptic curves over finite fields Consider F = F q. The size of the group E(F q ) is #E(F q ) = q + 1 t, for some integer t with t 2 q. The SEA algorithm computes t in polynomial time (very fast in practice). Typically t is nonzero in F q, in which case E is called ordinary. Some useful facts about t = t(e): 1. t(e 1 ) = t(e 2 ) E 1 and E 2 are isogenous. 2. j(e 1 ) = j(e 2 ) and t(e 1 ) = t(e 2 ) E 2 = E2. 3. j(e 1 ) = j(e 2 ) = t(e 1 ) = t(e 2 ) for j(e 1 ) / {0, 12 3 }.

4 Maps between elliptic curves An isogeny φ : E 1 E 2 is a rational map (defined over F) with φ(0) = 0. It induces a homomorphism from E 1 (F) to E 2 (F). The endomorphism ring End(E) contains all φ : E E. We have Z End E, but for F = F q, equality never holds. If E/F q is ordinary, then End(E) = O(D) where O(D) = Z + D + D Z 2 is the imaginary quadratic order of some discriminant D. We want to compute D.

5 The Frobenius endomorphism The endomorphism π : (x, y) (x q, y q ) on E(F q ) satisfies π 2 tπ + q = 0. If we set D π = t 2 4q and fix an isomorphism End E = O(D) we may regard π = t+ D π 2 as an element of O(D). Thus O(D π ) O(D), which implies D D π and that D and D π have the same fundamental discriminant D K. By factoring D π = v 2 D K we may determine D K and v. We then have D = u 2 D K for some u v. We want to compute u. This is easy if v is small (or smooth), but may be hard if not.

6 Computing isogenies We call a (separable) isogeny φ an l-isogeny if # ker φ = l. We restrict to prime l, in which case ker φ is cyclic. The classical modular polynomial Φ l Z[X, Y ] has the property Φ l ( j(e1 ), j(e 2 ) ) = 0 E 1 and E 2 are l-isogenous. The l-isogeny graph G l (F q ) has vertex set E(F q ) = {j(e/f q )} = F q, and edges (j 1, j 2 ) for Φ l (j 1, j 2 ) = 0 (note Φ l is symmetric). Φ l is big: O(l 3+ɛ ) bits.

7 The structure of the l-isogeny graph [Kohel] The connected components of G l (F q ) are l-volcanoes. An l-volcano of height h has vertices in level V 0,..., V h. Vertices in V 0 have endomorphism ring O(D 0 ) with l u 0. Vertices in V k have endomorphism ring O(l 2k D 0 ). 1. The subgraph on V 0 is a cycle (the surface). All other edges lie between V k and V k+1 for some k. 2. For k > 0 each vertex in V k has one neighbor in V k For k < h every vertex in V k has degree l + 1. See [Kohel 1996], [Fouquet-Morain 2002], or [S 2009] for more details.

8 A 3-volcano of height 2 with a 4-cycle

9 Algorithms to compute u Isogeny climbing: computes l-isogenies for prime l v to determine the power of l dividing u in. Probabilistic complexity O(q 3/2+ɛ ). Kohel s algorithm: computes the kernel of n-isogenies, where n = O(q 1/6 ) need not be a divisor of v. Deterministic complexity O(q 1/3+ɛ ) (GRH). New algorithm: computes the cardinality of smooth relations using isogenies of subexponential degree. Probabilistic complexity L[1/2, 3/2](q) (GRH+). ( L [α, c] (x) = exp (c + o(1)) (log x) α (log log x) 1 α). All algorithms have unconditionally correct output.

10 The action of the class group [CM theory] For an invertible ideal a O D = End(E), let E[a] be the subgroup of points annihilated by all a a. The map j(e) j(e/e[a]) corresponds to an isogeny of degree N(a). This defines a group action by the ideal group on the set {j(e/f q ) : End(E) = O(D)}. This action factors through the class group cl(o(d)) = cl(d). The action is faithful and transitive. See the books of [Cox], [Lang], or [Silverman] for more on CM theory.

11 Walking isogeny cycles ( ) If l v and D l = 1, the l-volcano containing j(e) is a cycle of length α, where α cl(d) contains an ideal of norm l. We can compute α (without knowing D) by walking a path j 0, j 1,... in G l (F q ) starting from j 0 = j(e): 1. Let j 1 be one of the two roots of Φ l ( X, j0 ) in Fq. 2. Let j k+1 be the unique root of Φ l ( X, jk ) /(X jk 1 ) in F q. The choice of j 1 is arbitrary (we cannot distinguish α and α 1 ). In either case, α (and α 1 ) is the least n for which j n = j 0. Step 2 finds the unique root of a degree l polynomial f (X) over F q. Complexity is T (l) = O(l 2 + M(l) log q) operations in F q.

12 Computing End(E) with class groups (naïvely) Given E/F q, let #E = q + 1 t and 4q = t 2 v 2 D K, so that End(E) = O(D) where D = u 2 D K for some u v. If u 1,..., u m are the divisors of v, then u = u i for some i. Pick any l v satisfying ( DK l ) = 1. For each D i = u 2 i D K there is an element α i cl(d i ) containing an ideal of norm l, but α i typically varies with i. We can compare α i to the length of the l-isogeny cycle containing j(e). These must be equal if u = u i. This is too slow, but we can exploit this idea.

13 Relations A relation R is a pair of vectors (l 1,..., l k ) and (e 1,..., e k ). We say R holds in cl(d) if for each i there is an α i cl(d) containing an ideal of norm l i such that α e 1 1 αe k k = 1. More generally, we define the cardinality of R in cl(d) by { #R/D = # τ {±1} k : } α τ i e i i = 1 in cl(d). #R/D does not depend on the choice of α i.

14 Counting relations Given a relation R with (l 1,..., l k ) and (e 1,..., e k ): 1. Set J 0 be a list containing the single element j(e). 2. For each element in J i walk e i steps in both directions of the l i cycle and append the two end points to the list J i #R/E is the number of times j(e) appears in the list J k. The complexity is k i=1 2i e i T (l i ) operations in F q.

15 The key lemma Lemma: If O(D 1 ) O(D 2 ) then #R/D 1 #R/D 2. Proof: There is a norm-preserving map from O(D 1 ) to O(D 2 ) that induces a group homomorphism from cl(d 1 ) to cl(d 2 ). Corollary: Let p v and set D 1 = (v/p) 2 D K and D 2 = p 2 D K. Let R be a relation with #R/D 1 > #R/D 2. If u is the conductor of O(D) = End(E) then Theorem: Such an R exists. p u #R/E < #R/D 1. Conjecture: Almost all R that hold in cl(d 1 ) don t hold in cl(d 2 ).

16 Algorithm to compute End(E) Given E/F q, the following algorithm computes D = u 2 D K, the discriminant of the order isomorphic to End(E). 1. Compute t = q + 1 #E, v, and D k, with 4q = t 2 v 2 D K. 2. For primes p v, find a relation R satisfying the corollary. Count #R/E in the isogeny graph to test whether p u. 3. Output u 2 D K. The algorithm above assumes v is square-free.

17 Finding smooth relations The following algorithm is adapted from Hafner/McCurley. We seek a smooth relation in cl(d 1 ). Pick a smoothness bound B and a small constant k 0 (say 3). 1. Let l 1,..., l n be the primes up to B with ( D1 l i ) = 1, and let α i cl(d 1 ) contain an ideal of norm l i. 2. Generate β = α x i i where all but k 0 of the x i are zero and the other x i are suitably bounded. 3. For each β, test whether N(b) is B-smooth, where b is a the reduced representative of β. 4. If so write α x i i = α y i i and compute R. Verify that #R/D 1 > #R/D 2 (almost always true). For suitable B, the complexity is L[1/2, 3/2]( D )

18 An example of cryptographic size (200 bits) We have 4q = t 2 v 2 D K where t = 212, D K = 7 and v = }{{} }{{} p 1 p 2. After finding 2 u and 127 u we test p 1 u by computing R 1 = (2 2533, , 29 2, 37 47, 79 1, 113 1, 149 1, 151 2, 347 1, ), which holds in cl(p 2 2 D K ) but not cl(p 2 1 D K ).We test p 2 u using R 2 = (2 23, 11 5, 43 1, 71 2 ), which holds in cl(p 2 1 D K ) but not in cl(p 2 2 D K ). Total time to compute End(E) is under 30 minutes

19 Certifying the endomorphism ring To verify a claimed value of u, it suffices to have a relation R p for each prime divisor of v such that: 1. For each prime p (v/u), we have #R p /E > #R p /p 2 D K. 2. For each prime p u, we have #R p /(u/p) 2 D K > #R p /E. Certificate size is O(log 2+ɛ q). Note that either D 1 u 2 D K or D 1 = (u/p) 2 D K. We always have D 1 D. Very useful when D D π. This yields an algorithm to compute u with complexity L [1/2 + o(1), 1] ( D ) + L [1/3, c] (q) which depends primarily on D, not q.