November 15, we have no class. November 22, we do a SAGE (computer) session in WN-S329. 1. Material covered
(sketch)
These notes are not meant as course notes and are not carefully written. They serve mainly as a summary and/or reminder for what we have done in class.
On November 8, we covered what corresponds to Chapters 11 and 12 of Cassels and Sections II.3-5 and IV.3 of Silverman-Tate.
Throughout, we will assume that we have an elliptic curve E over Q, given by y2= x3+ ax2+ bx + c,
with a, b, c ∈ Z. We also have a prime p.
• We discussed the p-adic valuation vp on Q, and the discrete valuation ring Z(p), which
is called R in Silverman-Tate, section II.4. See exercises 2.6-8 of Silverman-Tate and the beginning of chapter 2 in Cassels. Cassels also defines the p-adic numbers Qp in section
chapter 2, but we have not used those. • We defined the reduction ˜E over Fp by
y2= x3+ ax2+ bx + c,
with a, b, c the images of a, b, c in Fp, and let ˜Ens(Fp) be the set of nonsingular points over
Fp.
• We let E0(Q) ⊂ E(Q) be the set of points of nonsingular reduction, and E1(Q) the kernel
of reduction. Then we have
E1(Q) = {O} ∪ { (x, y) ∈ E(Q) : vp(x) < 0 or vp(y) < 0}.
• We proved the following statement: for any affine point (x, y) in E(Q) with vp(x) < 0 or
v(y) < 0, there is an integer n > 0 such that vp(x) = −2n and vp(y) = −3n.
• We concluded that for every point (x, y) in E(Q) there are integers m, n, e with gcd(m, e) = gcd(n, e) = 1 and x = m/e2 and y = n/e3.
• We defined
En(Q) = {O} ∪ { (x, y) : vp(x) ≤ −2n, vp(y) ≤ −3n },
which is called G(n)in Cassels and C(pn) in Silverman-Tate.
• Our elliptic curve has a projective model given by
Y2Z = X3+ aX2Z + bXZ2+ cZ3, of which another affine part, with Y 6= 0 and coordinates
s = Z Y = 1 y and t = X Y = x y, is given by s = t3+ at2s + bts2+ cs3. • Since x3+ ax2
+ bx + c is monic, every root in Q is an integer, so no element of E(Q)[2], besides O, is contained in E1(Q); these are the points at infinity for the affine part given
by Y 6= 0 with coordinates (s, t), so in terms of these coordinates we have En(Q) = {(s, t) ∈ E(Q) : vp(t) ≥ n and vp(s) ≥ 3n}
= {(s, t) ∈ E(Q) : vp(t) ≥ n and vp(s) > 0}.
• Stated the following proposition (proof at the end of class, see below).
Proposition. Suppose n ≥ 1 and P1, P2, P3∈ E(Q) collinear with P1, P2∈ En(Q). Then
we have vp t(P1) + t(P2) + t(P3) ≥ 3n. 1
2
• Concluded from the proposition that for each n ≥ 1, the subset En(Q) is a group and
there is an injective homomorphism
En(Q)/E3n(Q) ,→ pnZ(p)/p3nZ(p) ∼= pnZ/p3nZ∼= Z/p2nZ of abelian groups.
• We have a filtration
. . . ⊂ En+1(Q) ⊂ En(Q) ⊂ . . . ⊂ E1(Q) ⊂ E0(Q) ⊂ E(Q),
and we say that the level of a point P ∈ E1(Q) is that integer n for which we have
P ∈ En(Q) \ En+1(Q), i.e., the level is n if and only if vp(x) = −2n and vp(y) = −3n,
which is equivalent to n = vp(t) (assuming we already know P ∈ E1(Q)).
• We showed how it follows that E1(Q) is torsion-free.
• We concluded that torsion points have integral coordinates.
• We also concluded that if ˜E is nonsingular, then there is an injective homomorphism (Silverman-Tate, IV.3)
E(Q)tors,→ ˜E(Fp).
• We also showed how the theorem of Nagell-Lutz follows: if (x, y) ∈ E(Q) is a torsion point, then x, y ∈ Z and y|D, where the discriminant D equals −4a3c + a2b2+ 18abc − 4b3− 27c2.
In fact, one can also prove the stronger fact y2|D (exercise).
2. Proof of Proposition
The proposition above can be found in both Cassels (Lemma 2 of chapter 11) and Silverman-Tate (page 50-54, ending with t1+ t2+ t3 ∈ p3νR). Cassels uses a clever trick (as always) that
makes the computation very efficient. However, he does not really explain what’s behind it. It avoids the lengthy computation that Silverman-Tate needs on page 52. Here I will split it up into two steps so that you see what actually happens.
Lemma. Suppose p divides a, b, and c and P1, P2, P3∈ E(Q) are collinear with P1, P2∈ E0(Q).
Then we have vp t(P1) + t(P2) + t(P3) ≥ min vp(a), vp(b), vp(c).
Proof. First note that the reduction ˜E has a singular point at (0, 0), as a = b = c = 0. Let the line that contains P1, P2, P3 be given by the linear equation αx + βy = γ, with vp(α), vp(β), vp(γ) ≥ 0
and one of the three valuations equal to 0. The first part of Cassels’ trick uses that E0(Q) is a
group, so P3= −P1− P2∈ E0(Q). Therefore, all three points have nonsingular reduction. As the
line does not intersect the curve ˜E in any more than the three points ˜Pi, the reduction of the line
does not go through (0, 0). This shows vp(γ) = 0 and after dividing the equation for the line by
γ, we may assume γ = 1. In terms of the coordinates s and t, the line is then given by s = αt + β. The values t(Pi) (for i = 1, 2, 3) are the roots of the equation we get by substituting s = αt + β
in the equation for E, so of
αt + β = t3+ at2(αt + β) + bt(αt + β)2+ c(αt + β)3. This can be written as h(t) = 0 for h(t) = c3t3+ c2t2+ c1t + c0, with
c3= 1 + aα + bα2+ cα3 and c2= β(a + 2bα + 3cα2)
and two coefficients c1, c2 that do not matter. Since the values t(Pi) are roots of h, we have
h(t) = c3(t − t(P1))(t − t(P2))(t − t(P3)).
Comparing coefficients of t2 gives
t(P1) + t(P2) + t(P3) = −
c2
c3
.
From the fact that p divides a, b, c we get vp(c3) = 0, so from vp(α), vp(β) ≥ 0, we find
vp t(P1) + t(P2) + t(P3) = vp(c2) ≥ vp a + 2bα + 3cα2 ≥ min vp(a), vp(b), vp(c).
3
Proof of the proposition above. Suppose n ≥ 1 and let E0 be the elliptic curve given by
y02= x03+ a0x02+ b0x0+ c0 with
a0 = p2na, b0= p4nb, c0= p6nc. Then there is a morphism
τ : E → E0, (x, y) 7→ (x0, y0) = (p2nx, p3ny).
For any P = (x, y) ∈ En(Q) we have vp(x) ≤ −2n and vp(y) ≤ −3n, so for τ (P ) = (x0, y0) we
have vp(x0) ≤ 0 and vp(y0) ≤ 0, so τ (P ) does not reduce to (0, 0) and we have τ (P ) ∈ E00(Q). This
implies that for P1, P2, P3∈ E(Q) with P1, P2∈ En(Q), we have τ (P1), τ (P2), τ (P3) ∈ E0(Q) and
τ (P1), τ (P2) ∈ E00(Q). If P1, P2, P3are collinear, then so are τ (P1), τ (P2), τ (P3), so with t0= x0/y0
we conclude vp t0(τ (P1)) + t0(τ (P2)) + t0(τ (P3)) ≥ min vp(a0), vp(b0), vp(c0) ≥ 2n
from the lemma. From t(P ) = pnt0(τ (P )), we conclude
vp t(P1) + t(P2) + t(P3) = vp pn t0(τ (P1)) + t0(τ (P2)) + t0(τ (P3) ≥ n + 2n = 3n. 3. Homework
In exercise 2.12 of Silverman-Tate, find the full torsion subgroup for all examples for which there is no nontrivial 2-torsion.
Furthermore, do three of the exercises below, except for problems you already did last time of course. Do not choose only the easiest ones, and do not choose problems from different sources that are almost the same problems!
(1) Cassels, chapter 12: everything except for exercise 4. (2) Silverman-Tate, chapter 2: 1,2,4,5,11.
