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

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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 ?