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

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

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

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Read or Download Algorithmic Number Theory: 5th International Symposium, ANTS-V Sydney, Australia, July 7–12, 2002 Proceedings PDF

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

Example text

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 ? Such elliptic curves may be easier to find. But our proof of Theorem 1 seems to require the fact that E is defined over F and has rk E(F ) = 1, since Lemma 5 fails if the ideal I of OF is instead assumed to be an ideal of OK .

We first show that for any ideal I ⊆ OK , the set GI := { Q ∈ rE(K) : I | den(x(Q)) } is a subgroup of rE(K). ) Since an intersection of subgroups is a subgroup, it suffices to prove this when I = pn for some prime p and some n ∈ Z≥1 . Let Op be the completion of OK at p. Let F ∈ OK [[z1 , z2 ]] denote the formal group of E with respect to the parameter z := −x/y, as in Chapter 4 of [Sil92]. 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 + .

Yn ) in K m+n such that the fractional ideal (x1 , . . , xm ) divides the fractional ideal (y1 , . . , yn ) is diophantine over OK . 2. The set of (t, u) ∈ K × × K × such that den(t) | den(u) is diophantine over OK . 3. The set of (t, u) ∈ K × × K such that den(t) | num(u) is diophantine over OK . 4. The set of (t, u) ∈ OK × K × such that t | den(u) is diophantine over OK . Proof. Statement 1 is clear, since the condition is that there exist cij ∈ OK such that yj = i cij xi for each j. Statement 2 follows from statement 1, since den(t) | den(u) if and only if the fractional ideal (u, 1) divides (t, 1).

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