Algorithmic Number Theory: 5th International Symposium, by Manjul Bhargava (auth.), Claus Fieker, David R. Kohel (eds.)

By Manjul Bhargava (auth.), Claus Fieker, David R. Kohel (eds.)

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"The e-book includes 39 articles approximately computational algebraic quantity concept, mathematics geometry and cryptography. … The articles during this publication replicate the vast curiosity of the organizing committee and the contributors. The emphasis lies at the mathematical idea in addition to on computational effects. we suggest the ebook to scholars and researchers who are looking to examine present examine in quantity idea and mathematics geometry and its applications." (R. Carls, Nieuw Archief voor Wiskunde, Vol. 6 (3), 2005)

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Additional info for Algorithmic Number Theory: 5th International Symposium, ANTS-V Sydney, Australia, July 7–12, 2002 Proceedings

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Then there is an isomorphism E1 (Kp ), given by z → (x(z), y(z)) where x(z) = z −2 + . . and F(pOp ) −3 y(z) = −z + . . are Laurent series with coefficients in OK . It follows that Gpn is the set of points in rE(K) lying in the image of F(p n/2 Op ). In particular Gpn is a subgroup of rE(K). The “if” part of the lemma follows from the preceding paragraph. Now we prove the “only if” part. Let G = Gden(x(P )) . Then G is a subgroup of rE(K) Z, so G is free of rank 1. Let Q be a generator of G. By definition of G, we have P ∈ G, so P is a multiple of Q.

C. R. ): ANTS 2002, LNCS 2369, pp. 33–42, 2002. c Springer-Verlag Berlin Heidelberg 2002 34 Bjorn Poonen For example, the subset Z≥0 := {0, 1, 2, . . } of Z is diophantine, since for a ∈ Z, we have a ∈ Z≥0 ⇐⇒ (∃x1 , x2 , x3 , x4 ∈ Z) x21 + x22 + x23 + x24 = a. One can show using “diagonal arguments” that there exists a listable subset L of Z whose complement is not listable. It follows that for this L, there is no algorithm that takes as input an integer a and decides in a finite amount of time whether a belongs to L; in other words, membership in L is undecidable.

In particular, S4 = OF . Also, S is diophantine over OK , so each Si is diophantine over OK . In particular, OF = S4 is diophantine over OK . 6 Questions 1. Is it true that for every number field K, there exists an elliptic curve E over Q such that rk E(Q) = rk E(K) = 1? The author would conjecture so. If so, then Hilbert’s Tenth Problem over OK is undecidable for every number field K. 2. Can one weaken the hypotheses of Theorem 1 and give a diophantine definition of OF over OK using any elliptic curve E over K with rk E(K) = 1, not necessarily defined over F ?

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