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Heegner-point constructions for curves x^3 + y^3 = k
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a talk presented 8 November 2003, by Noam D. Elkies at the
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Princeton Workshop on the conjecture of Birch and
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Swinnerton-Dyer,
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========================================================================
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0) General Introduction: How little we still know about the BSD conjecture
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1) Specific Introduction: the curves x^3+y^3=k
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2) Why the case of prime k (and sign -1) requires a new idea
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3) The Heegner-point variant that works 2/3 of the time for k=p and k=p^2
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------------------------------------------------------------------------
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[Thank organizers. I was asked to speak about my work on x^3+y^3=k --
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which serves me right for procrastinating on writing it up during the
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past ~10 years. It's a rather small part of the BSD story; but, in
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some sense/direction we don't know that much more about BSD now than
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we did in 1993. So I'll start by putting the x^3+y^3=k case in this
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context.]
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0) Let E/Q be an elliptic curve.
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["Random" elliptic curves over "random" number fields
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are still hopeless.]
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We'll heavily use a modular parametrization of E by X_0(N); thanks to
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Wiles, Taylor, Breuil, B.Conrad, and Diamond, we now know that such a
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parametrization exists. This is essentially the only new result
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towards BSD in the past decade -- it shows that the analytic rank
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exists! (But for CM families like x^3+y^3=k and Dy^2=x^3-x we knew it
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already.)
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Recall that the Birch and Swinnerton-Dyer conjecture asserts that
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analytic rank = arithmetic rank, and the refined conjecture relates
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the leading term of L(E,s) at s=1 with numerical invariants of the
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curve.
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How to check it for E? First must compute arithmetic and analytic
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ranks.
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@ Arithmetic: descents and point searches will "eventually"
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determine the rank, if (as conjectured) Sha(E/Q) is finite. For
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family of curves, generally even this might fail unless we're really
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lucky and the upper and lower bounds agree... No method known for
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proving that E(Q) contains enough independent points without actually
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finding/constructing them.
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@ Analytic: generally hard, even for individual E. One can compute
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any derivative of L(E,s) at s=1 to arbitrary precision; this can prove
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that the derivative isn't zero, but how to prove it *is* zero?
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Analytic rank 0: true by Kolyvagin up to a bounded factor; can be
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checked effectively (and often practically) for any given case.
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Analytic rank 1: L(E,1)=0 by parity. If L'(E,1)>0 then again OK by
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Kolyvagin up to an effectively removable finite fudge factor.
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[Both of these cases heavily use Heegner points. In particular, in
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the rank-1 case, Gross-Zagier formula says L'(E,1) is proportional to
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the canonical height of a Heegner point, so if L'(E,1)>0 this point
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generates a full-rank subgroup of E(Q). This is effective not just in
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theory but also in practice; e.g., M.Watkins recently did a case of
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conductor 66157667 and no special structure, finding a rational point
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of canonical height 12557+ (thousands of digits) that is almost
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certainly a generator of the M-W group, in under 3 days on Ahtlon MP
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1600.]
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Analytic rank 2: L(E,1)=0 because it's an integer multiple of period/M
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for some effective M. If we can find 2 independent points, we've
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checked BSD. But this might not be computationally feasible --
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without Heegner points, searching for rational points is
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exponential-time, and the only recent progress is lowering the base of
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the exponent. If we do have enough generators, we can then check
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refined BSD numerically to some precision and guess the size of
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Sha(E/Q), but we have no idea how to either prove that this analytic
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|Sha| is actually the square integer it's approximating, nor prove
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that |Sha| actually is that integer, or indeed that Sha(E) is finite.
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Analytic rank 3: L'(E,1)=0 because it's proportional to the height of
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a Heegner point and there are no non-torsion points of very small
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canonical height. (In practice it's sometimes easier to test whether
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L'(E,1)=0 by Heegner computation, but that's another story! [ANTS-V])
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Difficulty of testing BSD and refined BSD as in the case of analytic
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rank 2.
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Analytic rank 4 and higher: we can check numerically that L''(E,1) or
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L'''(E,1) seems to vanish, but we can't prove it, so we can't prove
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BSD, let alone refined BSD (though we can still test numerically).
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[Note that even if rank 2 and higher are "rare", they do occur;
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e.g. for "congruent number" curves Dy^2=x^3-x:
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r D
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0 1
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1 5
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2 34
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3 1254
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4 29274
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5 48272239
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6 6611719866
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7 797507543735
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The r=7 curve was obtained just last week by Nicholas Rogers, and the
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r=5 and r=6 examples are also new improvements on his previous
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computation. Is r unbounded? I don't think we have enough evidence,
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but I've seen no convincing hints that r=O(1). Current record is 24
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(Martin-McMillen 2000).]
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Families of curves: even worse. For example: if Dy^2=x^3-x has odd
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sign (this is known to be equivalent to: squarefree part of |D| is
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congruent mod 8 to one of 5,6,7), must it have positive arithmetic
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rank? Even if rank is at most 1 by 2-descent, how to prove it? The
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difficulty is that we can neither show directly that L'(E,1)>0, nor
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prove in general that the Heegner point does not vanish. This has
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been done only in a few simple cases, such as D a prime (one of
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Heegner's original applications of Heegner points!). Even there one
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uses not Heegner points for the modular parametrization of Dy^2=x^3-x,
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a curve of conductor 16D^2 or 32D^2 (how to tell that the resulting
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point is nontrivial?), but starting from a curve y^2=x^3-4x or
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y^2=x^3+4x of conductor 32 or 64, constructing a point P rational over
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Q(sqrt(-|D|)), and showing that its image under Galois is *not* -P but
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T-P for a suitable 2-torsion point T -- which proves that P is not
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itself torsion.
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------------------------------------------------------------------------
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The next case past Dy^2=x^3-x and other families of quadratic twists
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is cubic, quartic, or sextic twists. These share some of the nice
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behavior of Dy^2=x^3-x (constant CM j-invariant) but are more
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complicated. We'll consider the curves we'll call E(k): x^3 + y^3 =
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k. This is another classical family, studied intensively by Sylvester
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(1879), Selmer (1951), Birch-Stephens (1966) ... Here too the ranks
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can get quite high; the same N.Rogers, working with me on his doctoral
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thesis, recently found the first examples known of ranks 8,9,10,11 --
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but that too is material for another talk. Here we'll again
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concentrate on families of E(k) of odd analytic rank when one can
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prove using a descent argument that the rank is at most 1, and ask to
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prove it by actually constructing a point. For Dy^2=x^3-x we used the
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2-isogenies with Dy^2=x^3+4x; here we'll descend via the 3-isogenies
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between E(k) and the curve E'(k): xy(x+y)=k. Again the complexity of
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the descent depends on the number of (distinct) prime factors of k.
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The simplest case -- past the classical k=1, r=0 -- is k=p or k=p^2,
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where p is a prime other than 3 (it's also classical that E(3) and
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E(9) have ranks 0 and 1 respectively). In that case, the 3-descent
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yields an upper bound on the arithmetic rank of E(k) that depends on
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the residue of p mod 9 as follows:
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p mod 9 | 1 2 4 5 7 8
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--------+------------------
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bound | 2 0 1 0 1 1
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in particular, when p mod 9 is 4, 7, or 8, the curve has odd sign and
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Birch-Stephens (following Selmer, but apparently not Sylvester!)
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conjectured that in each of these cases E(p) actually has rank 1. The
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same should hold for E(p^2) with p in the same residue classes. To
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prove this we must find a recipe for constructing a non-torsion
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rational point on E(k) or E'(k). We have such a recipe, of course --
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the Heegner construction on E(k), considered as a modular curve in its
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own right -- but again we have no idea how to prove in general that
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thie resulting point is non-torsion. So we again look to find a point
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P on some initial curve like E(1) that's rational over Q(cbrt(p)), or
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better yet K(cbrt(p)) where K is the CM field
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K = Q(sqrt(-3)) = Q(rho) [rho a generator of mu_3]
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and prove that P transforms correctly under Galois. Note that if we
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put the origin of E(k) at the obvious point (1:-1:0) then the CM of
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E(k) is given by (x:y:z) --> (x:y:rho*z), so P = (x:y:cbrt(k)*z) can
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again be expressed in Galois terms.
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------------------------------------------------------------------------
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But this didn't work. One can write down candidates: start from E(1),
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which is the modular curve X_0(27), and use the 27-isogenies between
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CM curves of discriminant -9k^2 or -27k^2. But it always seems to
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produce torsion points.
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Closest approach: Satge' (1987) proved that E(2p) has a rational
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point, and thus rank 1, for odd p congruent to 2 mod 9, starting from
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E(2) (which happens to be the modular curve X_0(36)=X(6)) and using
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36-isogenies between CM curves of discriminant -12p^2.
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In retrospect Gross explained: the Gross-Zagier formula actually gives
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the height of the Heegner point not just in terms of L'(E,1), but in
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terms of the derivative at s=1 of the product of L-functions of two
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curves. In the usual setting of quadratic twists, the other
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L-function is the L-function of the initial curve, so we're OK if (as
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with Dy^2=x^3-x) that initial curve has rank zero. But for cubic
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twists -- where there's as yet no G-Z formula but one can formulate a
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conjectural analogue -- the companion elliptic curve is a different
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twist. If we go from E(k_0) to E(ck_0), the companion is E(c^2 k_0).
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So k_0=1, c=p yields the companion curve E(p^2), which also has sign
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-1, so the product vanishes to order at least 2 at s=1, its derivative
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vanishes, and we expect the Heegner point to be torsion. For Satge's,
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k_0=2, c=p, and E(2p^2) has rank zero by 3-descent, so one expects to
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succeed -- Satge' proved it.
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------------------------------------------------------------------------
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For p=9k+4 or 9k+7, I proved in 1993 that E(p) and E(p^2) do have
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positive rank, using yet another variant of the Heegner-point method.
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Let pi be a factor of p in Z[rho] determined up to conjugation by the
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condition that 3|pi-1. So, p = pi pi' where pi' is the conjugate. We
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construct a point P on E(pi) or E(pi^2), show that P is rational over
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K(cbrt(pi')), and prove that the image of P under the appropriate
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generator of Gal(K(cbrt(pi'))/K) is T+rho*P where T is a nontrivial
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3-torsion point. This yields a K-rational point on the plane cubic pi
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X^3 + pi' Y^3 = Z^3 or pi^2 X^3 + pi'^2 Y^3 = Z^3, which is a
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principal homogeneous space for E(p) or E(p^2), and thus a nontrivial
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K-rational point on E(p) or E(p^2). Since K is the CM field, it
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follows that these curves have a nontorsion point also over Q. (In
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fact the point we construct turns out to be Q-rational already,
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because X,Y are also conjugate over Q.) In Gross's interpretation,
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when c and k_0 are in K but not necessarily in Q, the companion twist
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of E(c k_0) is not E(k_0/c) but E(k_0/c') -- so when k_0=pi and c=pi'
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the companion is just E(1), which has rank zero as desired.
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To construct our point P, we use a modular parametrization of E(pi) or
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E(pi^2). Since this is not an elliptic curve over Q, we can't use any
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X_0(N). We'll use a curve we'll call X'_0(9p) or X'_0(27p), the X'
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meaning that for the p-level structure we use matrices that reduce mod
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p to upper triangular _with cubic residues on the diagonal_. [For k=p
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or p^2, the 3-part of the level is 9 if k is 4 mod 9, and 27 if k is 7
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mod 9.] We then use a point P_0 on this curve parametrizing a cyclic
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isogeny of degree 9p or 27p between two CM curves of discriminant -27,
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chosen so that the kernel includes ker(pi'). We then use stanadard CM
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theory/techniques to prove the point is defined over K(cbrt(pi')), and
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that its image P on E(pi) is mapped by Galois to T+rho*P for some
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(sqrt(-3))-torsion point T. [NB we don't need to take a Galois trace
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of P from some class field to get the desired rational point, which
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would be needed for most Heegner-point constructions.] The hard part
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is showing that T is nontrivial, which we do by directly computing the
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modular symbols that go into T!
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