Algebraic Geometry and Its Applications: Dedicated to Gilles by Jean Chaumine, James Hirschfeld, Robert Rolland

By Jean Chaumine, James Hirschfeld, Robert Rolland

This quantity covers many issues together with quantity thought, Boolean services, combinatorial geometry, and algorithms over finite fields. This ebook includes many attention-grabbing theoretical and applicated new effects and surveys awarded by way of the easiest experts in those parts, resembling new effects on Serre's questions, answering a query in his letter to most sensible; new effects on cryptographic functions of the discrete logarithm challenge on the topic of elliptic curves and hyperellyptic curves, together with computation of the discrete logarithm; new effects on functionality box towers; the development of recent periods of Boolean cryptographic features; and algorithmic purposes of algebraic geometry.

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Extra resources for Algebraic Geometry and Its Applications: Dedicated to Gilles Lachaud on His 60th Birthday (Series on Number Theory and Its Applications)

Example text

However, the method given in [11] lacks an efficient procedure for testing whether a given α ∈ B is an endomorphism of J. 1) for determining whether an element of the set is an endomorphism of J. 8. We begin by observing that since K = Q(π), any α ∈ OK can be expressed as a polynomial f ∈ Q[π]. Since π satisfies a polynomial of degree 4 (the characteristic polynomial of Frobenius), f can be taken to have degree 3. We may thus write α= a0 + a1 π + a2 π 2 + a3 π 3 n (2) for some integers a0 , a1 , a2 , a3 , n.

Since α1σ and α2 both generate (α2 ) and √ have absolute value p, we deduce that α1σ = ±α2 . Step 4c ensures that this sign is positive, so α1 and α2 have the same characteristic polynomial hi (x), and thus the Frobenius element could be any of the elements output by Step 6. Since Aut(K/Q) is generated by σ and σ 2 is complex conjugation, we have output the Aut(K/Q)-orbit of the Frobenius element. If K is not Galois, then the Frobenius element must be either αi or αi . Since Aut(K/Q) in this case consists of only the identity and complex conjugation, Step 6 outputs the Aut(K/Q)-orbit of the Frobenius element.

Proof. The set B¯ is a Z/nZ-basis of OK /nOK , so if π k ≡ (1, 0, 0, 0) (mod n), then π k −1 ∈ nOK (since the first element of B¯ is 1). Since nOK is mapped to itself by Aut(K/Q), we have (π σ )k −1 ∈ nOK .

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