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Section 1 Elliptic curves and their ranks

Sources: Silverman I, V. Dokchitser's lectures.

Subsection 1.1 Mordell-Weil

Let \(K\) be a number field and let \(E/K\) be an elliptic curve. The group \(E(K)\) is finitely generated.

\begin{equation*} E(K) \simeq E(K)_{\text{tors}} \oplus \ZZ^r\text{.} \end{equation*}

Where \(E(K)_{\text{tors}}\) is a finite subgroup and \(r\) is the rank, a non-negative integer.

Assuming that we can compute the torsion subgroup, computing the rank would completely determine \(E(K)\) and hence solve the associated diophantine problem.

Plan

  1. Understand the proof of Mordell-Weil
  2. See where it is non-effective.
  3. From the proof, extract a strategy to sometimes compute the rank (define Selmer groups, Shafarevich-Tate group).

Part 1: Prove that

\begin{equation*} E(K)/mE(K) \end{equation*}

is finite for some \(m \ge 2\text{.}\)

Part 2: use a descent argument with heights of points.

Of these two parts of the proof, part 1 is the challenging/interesting one.

For part 2: Assuming that

\begin{equation*} E(K)/mE(K) \end{equation*}

is finite and that \(E\) has a “height function” then \(E(K)\) is finitely generated.

Choose elements \(Q_1, \ldots, Q_r\in A\) to represent the finitely many cosets in \(A/mA\text{.}\) Let \(P\) be a point in \(A\text{.}\) We show that \(P\) can be generated by \(Q_1, \ldots, Q_r\) plus a set of finitely many points of bounded height.

First write

\begin{equation*} P = mP_1 + Q_{i_1} \end{equation*}

for some \(1 \le i \le r\text{.}\) Repeat this for

\begin{equation*} P_1 = m P_2 + Q_{i_2} \end{equation*}
\begin{equation*} P_2 = m P_3 + Q_{i_3} \end{equation*}
\begin{equation*} \vdots \end{equation*}
\begin{equation*} P_{n-1} = m P_n + Q_{i_n} \end{equation*}

by property 2. of \(h\) we have

\begin{equation*} h(P_j) \le \frac{1}{m^2} (h(m P_j) + c_2) \end{equation*}
\begin{equation*} \frac{1}{m^2} (h(P_{j-1}) - Q_{i_j}) + c_2) \end{equation*}
\begin{equation*} \le \frac{1}{m^2}(2 h(P_{j-1}) + c'_1 + c_2) \end{equation*}

by 1. Where \(c'_1\) is the maximum of the constants from \(i\) for \(Q\) in \(\{-Q_1, \ldots, -Q_r\}\text{.}\) Note that \(c_1'\) and \(c_2\) do not depend on \(P\) and that \(h(P) \ge0\text{.}\) We repeat this inequality starting from \(P_n\) and working back to \(P\text{.}\)

\begin{equation*} h(P_n) \le \left( \frac 2{m^2} \right)^n h(P) + \frac 1{m^2} \left(1+ \frac 2{m^2} + \left(\frac 2{m^2}\right)^2 + \cdots+ \left(\frac 2{m^2}\right)^{n-1}\right)(c_1' + c_2) \end{equation*}
\begin{equation*} = \left( \frac 2{m^2} \right)^n h(P) + \frac 1{m^2} \left(1+ \frac 2{m^2} + \left(\frac 2{m^2}\right)^2 + \cdots+ \left(\frac 2{m^2}\right)^{n-1}\right)(c_1' + c_2) \end{equation*}
\begin{equation*} \lt \left( \frac 2{m^2} \right)^n h(P) + \frac{c_1' + c_2}{m^2 - 2} \end{equation*}
\begin{equation*} \le \frac 1{2^n} h(P) + \frac{c_1' + c_2}{2}\text{,} \end{equation*}

since \(m \ge 2\text{.}\) Hence for \(n\) sufficiently large (to make \(\frac 1{2^n} h(P) \le 1\)) we have

\begin{equation*} h(P_n) \le 1 + \frac 12 (c_1' + c_2)\text{.} \end{equation*}

Since \(P\) is a linear combination of \(P_n\) and \(Q_i\)

\begin{equation*} P = m^n P_ n + \sum_{j=1}^n m^{j-1} Q_{i_j}\text{,} \end{equation*}

it follows that every \(P \in A\) is a linear combination of points in

\begin{equation*} \{Q_1, \ldots, Q_r\}\cup \{Q \in A : h(Q) \le 1 + \frac 12 (c_1' + c_2) \}\text{.} \end{equation*}
Remark 1.2.

On \(E/\QQ\) the height function

\begin{equation*} h \colon E(\QQ) \to \QQ \end{equation*}
\begin{equation*} P \mapsto \begin{cases} \log(\max\{|p|,|q|\}),\,x(P) = \frac pq, \amp P\ne 0,\\ 0, \amp P= 0. \end{cases} \end{equation*}

satisfies the conditions of Theorem 1.1.

Remark 1.3.

The above proof is effective. To find generators of \(E(\QQ)\) first compute \(c_1 = c_1(Q_i)\) for each \(i\text{,}\) then compute \(c_2\text{.}\) Find points of bounded height. Note that we need \(Q_1,\ldots ,Q_r\) to start with.

It remains to show part 1:

We will prove this under the assumption that \(E\lb m \rb \subseteq E(K)\text{.}\) This is WLOG since:

\begin{equation*} 0 \to \phi \to E(K)/ m E(K) \xrightarrow{\varphi} E(L)/mE(L) \to 0 \end{equation*}

induced by

\begin{equation*} E(K) \subseteq E(L)\text{,} \end{equation*}

and prove that \(\phi\) is finite. Kernel \(\phi\) is given by

\begin{equation*} \frac{E(K) \cap mE(L)}{mE(K)}\text{,} \end{equation*}

take \(P \in \phi\text{.}\) We can choose \(Q_P \in E(L)\) such that \(Q_P = P\text{.}\) Define a map of sets

\begin{equation*} \lambda_P \colon G_{L/K} \to E\lb m \rb \end{equation*}
\begin{equation*} \sigma \mapsto Q_P^\sigma - Q_P\text{.} \end{equation*}

Note that

\begin{equation*} \lb m \rb (Q_P^\sigma - Q_P) = (\lb m \rb Q_P)^\sigma - \lb m \rb Q_P = 0\text{.} \end{equation*}

Now we show that the association

\begin{equation*} \phi \to \operatorname{Map}(G_{L/K}, E\lb m \rb) \end{equation*}
\begin{equation*} P \mapsto \lambda_P \end{equation*}

is 1 to 1.

Suppose that \(P,P' \in E(K) \cap m E(L)\) satisfying \(\lambda_P = \lambda_{P'}\) then

\begin{equation*} (Q_P - Q_{P'})^\sigma = Q_P - Q_{P'} \end{equation*}

for all \(\sigma \in G_{L/K}\) so \(Q_P - Q_{P'} \in E(K)\) and hence

\begin{equation*} P - P' = \lb m \rb Q_P - \lb m \rb Q_{P'} \in m E(K) \end{equation*}

hence

\begin{equation*} P = P' \pmod {m E(K)}\text{.} \end{equation*}

\(G_{L/K}\) and \(E\lb m \rb\) are both finite, hence so is \(\phi\text{.}\)

Now we will prove the weak Mordell-Weil theorem. Using the above lemma we can reduce to the case where \(E\lb m \rb \subseteq E(K)\text{,}\) so we assume this going forwards.

Definition 1.6. The Kummer pairing.

The Kummer pairing is

\begin{equation*} \kappa \colon E(K) \times G_{\overline K/K} \to E\lb m \rb \end{equation*}
\begin{equation*} P,\sigma \mapsto Q^\sigma - Q \end{equation*}

where \(Q\) is a choice of point in \(E(\overline K)\) such that \(mQ = P\text{.}\)

Of part 4.

Take

Remark 1.8.

A homomorphism \(\phi \colon \absgal{K} \to G\) for a finite group \(G\) is continuous if it comes from a finite Galois extension, i.e.

\begin{equation*} \exists F/K \text{ finite Galois },\,\tilde \phi \colon \Gal{F}{K} \to G \end{equation*}

s.t. \(\phi\) is the composition \(\absgal{K} \to \Gal FK \xrightarrow{\tilde \phi} G\text{.}\) So \(\phi(g) \) only cares about what \(g\) does to \(F\text{.}\)

Example 1.10.

Let

\begin{equation*} E \colon y^2 = x(x-1)(x+1) \end{equation*}

for \(P \in E(\QQ), \QQ(\frac 12 P ) / \QQ\) can only ramify at 2.

\begin{equation*} \QQ(\frac 12 P ) \subseteq \QQ(i, \sqrt2) \end{equation*}
\begin{equation*} P = (x_0, y_0) \mapsto x_0, x_0 - 1, x_0 + 1 \in \QQ^\times/ {\QQ^\times}^2 \end{equation*}

is a homomorphism so \(x_0 , x_0 -1 , x_0 + 1 \) are \(\pm 1, \pm 2\) up to square.

\(x_0\) \(x_0 - 1\) \(x_0 + 1\) rat?
1 1 1 1) rat
1 -1 -1 2) non-rat
1 2 2 1) rat
1 -2 -2 2) non-rat
-1 1 -1 2) non-rat
-1 -1 1 1) rat
-1 2 -1 2) non-rat
-1 -2 2 1) rat
2 1 2 3) non-rat
2 -1 -2 2) non-rat
2 2 1 4) rat
2 -2 -1 2) non-rat
-2 1 -2 ?
-2 -1 2 ?
-2 2 -1 ?
-2 -2 1 ?
Table 1.11. Images

1) The 2-torsion points \(P = 0 ,(0,0), (1,0), (-1,0) \in E(\QQ)\) give us some rows. 2) As we have \(x_0 \gt -1\) we get \(x_0 + 1 \gt 0\) so \(x_0(x_0 - 1) \gt 0\) for the product to be a square (and hence \(\gt 0\)). 3) \(x_0 = 2A^2\text{,}\) \(x_0-1 = B^2\text{,}\) \(x_0 + 1 = 2C^2\) with \(A,B,C \in \QQ \smallsetminus \{ 0\}\text{.}\) Let \(A = m/n\) so \(2m^2/ n^2 - 1 = B^2\)

\begin{equation*} 2m^2 - n^2 = (Bn)^2 \end{equation*}

and

\begin{equation*} 2m^2+ n^2 = 2(Cn)^2 \end{equation*}

if \(m \equiv 0(2) \implies -1 = \square \pmod 8\) a contradiction.

\begin{equation*} m \equiv 1 \pmod2 \implies m^2 \equiv1 \pmod 8\text{.} \end{equation*}

So \(2 - n^2 = \square \pmod 8 \implies n^2 \equiv1 \pmod 8\)

\begin{equation*} 2 + n^2 = 2\square \pmod 8 \implies n^2 \equiv 0 \pmod 8 \end{equation*}
\begin{equation*} |E(\QQ) / 2 E(\QQ)| = 4 \end{equation*}
\begin{equation*} | E(\QQ)[2]| = 4 \implies \rk = 0 \end{equation*}
\begin{equation*} E(\QQ) \cong E(\QQ) [2 ]\text{.} \end{equation*}

4) Use the group structure!

\begin{equation*} E \colon y^2 = x(x-5)(x+5)\text{.} \end{equation*}

Recall:

\begin{equation*} \phi \colon E(K)/2E(K) \to \Hom_{cts}(G_K, E(K) [2]) \end{equation*}
\begin{equation*} P \mapsto \phi_P \end{equation*}

where \(\phi_P \colon \sigma \mapsto Q^\sigma - Q\) where \(Q = 2P\text{.}\) Which is well-defined and injective.

Elements of

\begin{equation*} \Hom_{cts}(G_K, E[2]) \leftrightarrow a,b,c \in (K^\times / {K^\times}^2) \text{ s.t. } abc \in {K^\times}^2 \end{equation*}
\begin{equation*} (x_0,y_0) \mapsto (x_0 - \alpha, x_0 - \beta, x_0 -\gamma)\text{.} \end{equation*}

To do descent, need more than \(\psi\) (i.e. injection).

Definition 1.15.

Let \(G\) be a group and \(M\) a \(G\)-module then let

\begin{equation*} H^0(G, M) = M^G = \{m \in M : gm = m \forall g\in G\} \end{equation*}
\begin{equation*} H^1(G, M) = \{\text{skew homs } G \to M\}/\{\text{skew homs } G \to M \text{ of the form }g\mapsto g(t) - t,\, t\in M\}\text{.} \end{equation*}
Remark 1.16.

If \(G\) acts trivially on \(M\) then

\begin{equation*} H^0(G, M) = M \end{equation*}
\begin{equation*} H^1(G, M) = \Hom(G,M)\text{.} \end{equation*}

When \(G\) is profinite then we want that the skew homomorphisms factor through finite Galois groups. We will prove that

\begin{equation*} E(K)/nE(K) \hookrightarrow H^1(G_K, E[n])\text{.} \end{equation*}
  1. There are finitely many skew homomorphisms
    \begin{equation*} \Gal{K(\frac 1nP, E[n])}{K} \to E[n] \end{equation*}
    and by 4.
    \begin{equation*} P \mapsto \{\phi_P, K(\frac 1n P, E\lb n \rb)\} \end{equation*}
    is injective. So \(\psi\colon P\mapsto K(\frac 1n P, E\lb n \rb)\) is finite to one by 3.
  2. \begin{equation*} \phi_P(gh) = \frac 1n P \ominus gh \frac 1n P \end{equation*}
    \begin{equation*} = \left((\frac 1n P) \ominus g(\frac 1n P)\right) \oplus \left( g(\frac 1n P) \ominus g(h(\frac 1n P))\right) \end{equation*}
    \begin{equation*} = \phi_P \oplus g(\phi_P(h))\text{.} \end{equation*}
    Remark: If \(E\lb n \rb \subseteq E(K)\) then \(\phi_P\) is a homomorphism. Recall for \(n=2\)
    \begin{equation*} \phi_P(gh) = \frac 12 P \ominus gh(\frac 12 P) \end{equation*}
    \begin{equation*} = \frac12 P \ominus h(\frac 12P) \oplus h(\frac 12 P) \ominus g(h( \frac 12 P)) \end{equation*}
    \begin{equation*} = \phi_P(h) \oplus \phi_P(g) \end{equation*}
    since \(2h(\frac 12 P) = h(P)= P\text{.}\) Consider now
    \begin{equation*} \frac 1n P = \frac 1n P' \oplus T \end{equation*}
    for some \(T\in E\lb n \rb\)
    \begin{equation*} (\phi_P \ominus \phi_{P}')(g) = \phi_P(g) - \phi_{P}'(g) = \frac 1n P \ominus g(\frac 1n P) - [(\frac 1n P ) \oplus T \ominus g(\frac 1n P) \oplus gT] \end{equation*}
    \begin{equation*} =T \ominus gT\text{.} \end{equation*}

Take \(G= G_K\)

\begin{equation*} B = E(\overline K), A= E[n], C = E(\overline K) \end{equation*}

to get

\begin{equation*} 0 \to E[n] \to E(\overline K) \xrightarrow{\cdot n} E(\overline K) \to 0 \end{equation*}

which gives the long exact sequence

\begin{equation*} 0 \to E(K)[n] \to E(K) \xrightarrow{\cdot n} E(K) \xrightarrow{\delta} H^1(G_K, E[n]) \to H^1(G_K, E(\overline K)) \to \end{equation*}
\begin{equation*} \implies E(K)/nE(K) \hookrightarrow H^1(G_K, E[n])\text{.} \end{equation*}

Problem:

\begin{equation*} H^1(G_K, E[n]) \end{equation*}

is infinite. What subgroup of

\begin{equation*} H^1(G_K, E[n]) \end{equation*}

do we land in?

Notation: When \(v\) is a place of \(K\) we have \(G_{K_v} \subseteq G_K\text{,}\) for any module \(M\) have \(M^{G_K} \le M^{G_{K_v}}\) and

\begin{equation*} \Res \colon H^1(G_K, E[n]) \to H^1 (G_{K_v} , E[n])\text{.} \end{equation*}

We have from the theorem

\begin{equation*} \xymatrix{ 0 \ar[r] & E(K)/ nE(K)\ar[d] \ar[r]^\delta & H^1(G_K, E[n]) \ar[d]^{\Res} \ar[r] & H^1(G_K, E(\overline K))[n] \ar[d]^{\Res}\ar[r]&0\\ 0 \ar[r] & \prod_v E(K_v)/ nE(K_v) \ar[r]^\delta & \prod_v H^1(G_{K_v}, E[n]) \ar[r] & \prod_v H^1(G_{K_v}, E(\overline K))[n] \ar[r]&0 } \end{equation*}

we want to understand \(\image \delta\) i.e. the subgroup

\begin{equation*} \ker \{ H^1(G_K, E[n]) \to H^1 (G_K, E(\overline K))\} \end{equation*}

this is as hard as finding \(E(K)\text{,}\) here is why:

Finding

\begin{equation*} \ker \{ H^1(G_K, E[n]) \to H^1 (G_K, E(\overline K))\} \end{equation*}

is equivalent to finding which PHS coming from \(H^1\) have a rational point. ??? Hensel's lemma.

Let \(C\) be a curve

\begin{equation*} \operatorname{Isom}(C) \leftrightarrow C(\overline K) \times \Aut(C) \end{equation*}
\begin{equation*} \tau_p \circ \alpha \leftrightarrow (P,\sigma) \end{equation*}
\begin{equation*} \operatorname{Twist}(E/K) \leftrightarrow H^1(G_K, \operatorname{Isom}(C)) \end{equation*}
\begin{equation*} C\simeq_{\overline K} E \end{equation*}
\begin{equation*} PHS \leftrightarrow H^1(G_K, E(\overline K)) \end{equation*}

\(C\) is a PHS for \(E\) iff \(E\) is the jacobian of \(C\text{.}\)

\begin{equation*} \xymatrix{ 0 \ar[r] & E(K)/ nE(K)\ar[d] \ar[r]^\delta & H^1(G_K, E[n]) \ar[d]^{\Res} \ar[r] & H^1(G_K, E(\overline K))[n] \ar[d]^{\Res}\ar[r]&0\\ 0 \ar[r] & \prod_v E(K_v)/ nE(K_v) \ar[r]^\delta & \prod_v H^1(G_{K_v}, E[n]) \ar[r] & \prod_v H^1(G_{K_v}, E(\overline K))[n] \ar[r]&0 } \end{equation*}
Definition 1.20. Twists of curves.

A twist of \(C/K\) is a smooth curve \(C'/K\) that is isomorphic to \(C\) over \(\overline K\text{.}\)

If \(C_1, C_2\) are twists of \(C/K\) and \(C_1\simeq_K C_2\) then we say that \(C_1\) and \(C_2\) are equivalent modulo \(K\)-isomorphism.

We denote \(\operatorname{Twist}(C/K)\) - the set of twists of \(C/K\) modulo \(K\)-isomorphism.

Let \(C'/K\) be a twist of \(C/K\) then there exists an isomorphism \(/\overline K\)

\begin{equation*} \phi\colon C' \to C \end{equation*}

associate the following map

\begin{equation*} \xi \colon G_K \to \operatorname{Isom}(C) \end{equation*}
\begin{equation*} \sigma\mapsto \phi^\sigma \phi\inv\text{.} \end{equation*}

Check that \(\xi\) is a cocycle

\begin{equation*} \xi_{\sigma\tau } = (\xi_{\sigma})^\tau \xi_\tau \end{equation*}

for all \(\sigma,\tau\in G_K\text{.}\) Denote \(\{\xi\}\) the associated class in \(H^1\text{.}\) \(\{\xi\}\) is determined by the \(K\)-isomorphism class of \(C'\) independent of the choice \(\phi\text{.}\)

The map

\begin{equation*} \operatorname{Twist}(C/K) \leftrightarrow H^1(G_K, \operatorname{Isom}(C)) \end{equation*}
\begin{equation*} C' \mapsto \{\xi\} \end{equation*}

is a bijection.

Injective, trace through.

Surjectivity, define the function field using the curve.

Remark 1.22.

If \(C\) is an elliptic curve then \(\operatorname{Isom}(C)\) is generated by

\begin{equation*} \Aut(C) (\text{fixing }0) \end{equation*}

and translations

\begin{equation*} \tau_P \colon C \to C \end{equation*}
\begin{equation*} Q\mapsto Q+P\text{.} \end{equation*}
Example 1.23.

\(E/K\) elliptic, consider

\begin{equation*} K(\sqrt d) \end{equation*}

a quadratic extension and \(\chi\) the associated character

\begin{equation*} \chi\colon G_K \to \{\pm 1\} \end{equation*}
\begin{equation*} \sigma \mapsto \sigma(\sqrt d) / \sqrt d\text{.} \end{equation*}

The group \(\pm 1\) can be viewed as automorphisms of \(C\text{.}\) So use \(\chi\) to define the cocycle

\begin{equation*} \xi \colon G_K \to \operatorname{Isom}(C) \end{equation*}
\begin{equation*} \sigma \mapsto [\chi (\sigma)]\text{.} \end{equation*}

Let \(C/K\) be the corresponding twist of \(E/K\text{,}\) we find an equation for \(C/K\text{.}\) Choose

\begin{equation*} y^2 = f(x) \text{ for } E/K \end{equation*}

and write

\begin{equation*} \overline K(E) = \overline K(x,y) \end{equation*}
\begin{equation*} \overline K(C) = \overline K(x,y)_\xi \end{equation*}

since \(\lb - 1 \rb(x,y) = (x,-y)\) the action of \(\sigma \in G_K\) on

\begin{equation*} \overline K(x,y)_\xi \text{ is given by }\sqrt d ^\sigma = \chi(\sigma) \sqrt d \end{equation*}
\begin{equation*} x ^\sigma = x,\,y = \chi(\sigma) y \end{equation*}

note that the function \(x' = x\) and \(y' = y /\sqrt d\) are in \(\overline K(x,y)_\xi\) and are fixed by \(G_K\text{.}\) Now \(x',y'\) satisfy

\begin{equation*} d{y'}^2 =f(x')/K \end{equation*}

is defined over \(K\) and defines an elliptic curve. Moreover

\begin{equation*} (x,y) \mapsto (x', y' \sqrt{d}) \end{equation*}

is an isomorphism over \(K(\sqrt d)\text{.}\)

Note \(C/K\) is not a principal homogeneous space for \(E/K\text{.}\)

Definition 1.24. Homogenous spaces.

Let \(E/K\) be an elliptic curve, a principal homogeneous space for \(E/K\) is a smooth curve \(C/K\) together with a simply transitive algebraic group action of \(E\) on \(C\) defined over \(K\text{.}\)

\begin{equation*} \mu \colon C\times E \to C \end{equation*}

morphism defined over \(K\) satisfying

  1. \begin{equation*} \mu(P, 0) = P \,\forall P \in C \end{equation*}
  2. \begin{equation*} \mu(\mu(p,P), Q) = \mu(p,P+Q) \,\forall P \in C \end{equation*}
  3. \begin{equation*} \forall p,q \in C,\,\exists ! P\in E \text{ s.t. } \end{equation*}
    \begin{equation*} \mu(p,P) = q \end{equation*}
    so we may define a subtraction map
    \begin{equation*} \nu \colon C\times C \to E \end{equation*}
    \begin{equation*} p,q \mapsto P \end{equation*}
    as above.
Definition 1.26.

Two homogeneous space \(C/K\) and \(C'/K\) for \(E/K\) are equivalent if there is an isomorphism

\begin{equation*} \phi \colon C \to C' \end{equation*}

defined over \(K\) and is compatible with the action of \(E\) on \(C\) and \(C'\text{.}\)

\begin{equation*} \xymatrix{ C \ar[r]^\theta \ar[d]_\phi & E \ar[d] \\ C' \ar[r] & E' } \end{equation*}

The equivalence class of PHS for \(E/K\) containing \(E/K\) acting on itself via translation is called the trivial class.

The collection of equivalence classes of PHS for \(E/K\) is called the Weil-Châtelet group, denoted

\begin{equation*} WC(E/K)\text{.} \end{equation*}

Well-definedness:

\begin{equation*} \sigma \mapsto p_0^\sigma - p_0 \end{equation*}

is a cocycle. Suppose that \(C'/K\) and \(C/K\) are two equivalent PHS then

\begin{equation*} p_0^\sigma - p_0 \end{equation*}

and

\begin{equation*} {p_0'}^\sigma - p_0' \end{equation*}

are cohomologous.

Injective, suppose that \(p_0^\sigma - p_0\) and \({p_0'} ^\sigma - p_0'\) corresponding to \(C/K\) and \(C'/K\) that are cohomologous and prove that \(C\simeq_K C'\text{.}\)

Surjective: let \(\xi \colon G_K \to E(\overline K)\) be a cocycle representing an element in \(H^1(G_K, E)\text{.}\) Embed

\begin{equation*} E(\overline K) \hookrightarrow \operatorname{Isom}(E) \end{equation*}
\begin{equation*} P \mapsto \tau_P \end{equation*}

and view

\begin{equation*} \xi \in H^1(G_K, \operatorname{Isom} E)\text{.} \end{equation*}

From the theorem on

\begin{equation*} \operatorname{Twist}(E/K) \leftrightarrow H^1(G_K, \operatorname{Isom}(E)) \end{equation*}

there exists a curve \(C/K\) and a \(\overline K\)-isomorphism

\begin{equation*} \phi\colon C \to E \end{equation*}

s.t.

\begin{equation*} \forall \sigma \in G_K : \phi^\sigma \phi\inv = \text{translation by } -\xi_\sigma\text{.} \end{equation*}

Define a map \(\mu \colon C\times E \to C\)

\begin{equation*} (p,Q) \mapsto \phi\inv (\phi(p) + Q)\text{.} \end{equation*}

Show that \(\mu\) is simply transitive.

Show \(\mu\) defined over \(K\text{.}\) Compute the cohomology class associated to \(C/K\) and show it is \(\xi\text{.}\)

Remark 1.29.

For a given \(C/K\) of genus 1 one can define several structures of PHS.

\begin{equation*} \{C/K,\mu\}^\alpha = \{ C/K , \mu\circ(1\times \alpha)\} \end{equation*}
\begin{equation*} \mu^\alpha(p,Q) = \mu(p, \alpha Q) \end{equation*}

for \(\alpha \in \Aut(E)\text{.}\)

\begin{equation*} \xymatrix{ C \ar[r]^\mu & E \ar[d]_P \\ C' \ar[r]^{\mu^\alpha} & E' } \end{equation*}
Example 1.30.

\(E/K\) and \(K (\sqrt d)/K\) a quadratic extension. Let \(T \in E(K)\) be a non-trivial point of order 2. Then \(\xi\colon G_K \to E\)

\begin{equation*} \sigma \mapsto \begin{cases} 0 \amp\text{ if } (\sqrt d)^\sigma = \sqrt d,\\ T \amp\text{ if } (\sqrt d)^\sigma = -\sqrt d. \end{cases}\text{.} \end{equation*}

We construct the PHS corresponding to \(\{\xi \} \in H^1(G_K, E(\overline K))\text{.}\) Since \(T \in E(K)\) can choose a Weierstraß equation for \(E/K\)

\begin{equation*} E\colon y^2 = x^3 + ax^2 + bx \text{ with } T = (0,0) \end{equation*}

then the translation by \(T\) map is given by

\begin{equation*} \tau_T(P) = (x,y) + (0,0) = \left(\frac bx, -\frac{by}{x^2}\right) \end{equation*}

for

\begin{equation*} P = (x,y)\text{.} \end{equation*}

Thus if \(\sigma \in G_K\) is non-trivial, \(\sigma\) acts on \(\overline K(E)_\xi\text{,}\) which is isomorphic to \(\overline K(E)\) but \(\absgal K\) action is twisted by \(\xi\text{,}\) i.e. \(x^\id \mapsto (x^\id)^\sigma\text{.}\)

\begin{equation*} (\sqrt d )^\sigma = -\sqrt d \end{equation*}
\begin{equation*} x^\sigma = \frac bx ,\, y^\sigma = -\frac{by}{x^2} \end{equation*}

need to find the subfield of \(K(\sqrt d)(x,y)_\xi\) fixed by \(\sigma\text{.}\) Note:

\begin{equation*} \frac{\sqrt d x}{y},\,\sqrt d \left(x -\frac bx\right) \end{equation*}

are invariant, take

\begin{equation*} z = \frac{\sqrt d x }{y},\,w = \sqrt d\left( x - \frac bx\right)\left(\frac xy\right)^2 \end{equation*}

and find relations between \(z\) and \(w\) to get

\begin{equation*} C \colon dw^2 = d^2 - 2a dz^2 + (a^2 - 4b) z^4\text{.} \end{equation*}

Claim: \(C/K\) is the PHS of \(E/K\) corresponding to \(\left\{\xi\right\}\text{.}\) There is a natural map

\begin{equation*} \phi \colon E \to C \end{equation*}
\begin{equation*} (x,y) \mapsto (z,w) \end{equation*}
\begin{equation*} (x,y) \mapsto \left( \frac{ \sqrt d y}{x^2 + ax + b}, \frac{\sqrt d (x^2 - b)}{x^2 + ax + b}\right) \end{equation*}

so that

\begin{equation*} \phi (0,0) = (0,-\sqrt d) \end{equation*}
\begin{equation*} \phi (0) = (0,\sqrt d) \end{equation*}
  • Prove that \(\phi \) is an isomorphism so \(C\) is a twist.
  • \(C\) is the PHS corresponding to \(\{\xi\}\text{.}\) Take \(p \in C \) and compute
    \begin{equation*} \sigma\mapsto p^\sigma - p = \phi\inv(p^\sigma) - \phi\inv (p) \end{equation*}
    for example let \(p = (0,\sqrt d) \in C\text{,}\) if \(\sigma = \id\) then \(p^\sigma - p = 0 - 0 = 0\text{.}\) If \(\sigma = -\id\) then \(p^\sigma - p = T - 0 = T\text{.}\)

Back to Selmer, we want to have the image of our weak Mordell-Weil land in something finite.

\begin{equation*} \xymatrix{ 0 \ar[r] & E(K)/ mE(K)\ar[d] \ar[r]^\delta & H^1(G_K, E[m]) \ar[d]^{\Res} \ar[r] & WC(E/K)[m] \ar[d]^{\Res}\ar[r]&0 \\ 0 \ar[r] & \prod_v E(K_v)/ nE(K_v) \ar[r]^\delta & \prod_v H^1(G_{K_v}, E[n]) \ar[r] & \prod_v WC (E/ K_v)[m] \ar[r]&0 } \end{equation*}
Definition 1.31. \(m\)-Selmer groups.

The \(m\)-Selmer group of \(E/K\) is the subgroup of

\begin{equation*} H^1(G_K, E[m]) \end{equation*}

defined by

\begin{equation*} \Sel^m(E/K) = \ker \left\{H^1(G_K, E[m]) \to \prod_v WC(E/K_v) \right\}\text{.} \end{equation*}
Definition 1.32. The Shafarevich-Tate group.

The Shafarevich-Tate group of \(E/K\) is the subgroup of

\begin{equation*} WC(E/K) \end{equation*}

defined by

\begin{equation*} \Sha(E/K) = \ker\left\{WC(E/K) \to\prod_v WC(E/K_v)\right\}\text{.} \end{equation*}

Subsection 1.2 \(p^\infty\)-Selmer and the structure of \(\Sha\)

\(H^1(G_K, E(\overline K))\) is torsion for general galois cohomological reasons. So

\begin{equation*} \Sha (E/K) \subseteq H^1(G_K, E(\overline K)) \end{equation*}

is torsion.

So we may write

\begin{equation*} \Sha(E/K) = \bigoplus_p \Sha_{p^\infty}(E/K) \end{equation*}

where for each prime \(p\)

\begin{equation*} \Sha_{p^\infty} (E/K) \end{equation*}

denotes the \(p\)-primary part of \(\Sha(E/K)\text{.}\) (i.e. the subgroup of elements whose order is a power of \(p\text{.}\)) By descent

\begin{equation*} \Sha (E/K) [ m] \text{ is finite for all } m \ge 1\text{.} \end{equation*}

So

\begin{equation*} \Sha_{p^\infty} (E/K) \cong (\QQ_p/\ZZ_p)^{\delta_p} \oplus T_p,\, \delta_p \in \ZZ_{\ge 0} \end{equation*}

where \(T_p\) is a finite abelian \(p\)-group.

\begin{equation*} T_p \cong \ZZ/p^{s_1} \ZZ \oplus \cdots \oplus \ZZ/p^{s_l} \ZZ,\,s_i\in \ZZ_{\ge 0}\text{.} \end{equation*}

The group

\begin{equation*} \bigoplus_p (\QQ_p/\ZZ_p)^{\delta_p}\subseteq \Sha (E/K) \end{equation*}

is called the infinitely divisible subgroup of \(\Sha\) denoted \(\Sha_{div}\text{.}\)

The conjecture that \(\Sha\) is finite implies \(\delta_p = 0\) for all \(p\text{.}\) And \(T_p \ne 0 \) for only finitely many \(p\text{.}\)

There is a pairing called the Cassels-Tate pairing

\begin{equation*} \Sha(E/K) \times \Sha(E/K) \to \QQ/\ZZ \end{equation*}

which is bilinear and alternating, and the kernel on either side is the infinitely divisible group. If \(\Sha(E/K)\) is finite then the pairing is non-degenerate and hence

\begin{equation*} |\Sha(E/K)| = \square \in \ZZ\text{.} \end{equation*}
Definition 1.34. \(p^\infty\)-Selmer group.

Consider \(\Sel_{p^n}(E/K)\) and take the direct limit

\begin{equation*} \varinjlim_n \Sel_{p^n}(E/K) \end{equation*}

to define the \(p^\infty\)-Selmer group.

One shows that

\begin{equation*} X_p(E/K) = \Hom_{\ZZ_p} (\varinjlim_n \Sel_{p^n}(E/K), \QQ_p/\ZZ_p) \end{equation*}

called the Pontyragin dual of the \(p^\infty\) Selmer group is a finitely generated \(\ZZ_p\)-module. The associated \(\QQ_p\)-vector space, denoted \(\mathcal X_p(E/K) =X_p(E/K) \otimes_{\ZZ_p} \QQ_p\) has dimension \(\rk_p\text{.}\)

Definition 1.35.

\(\rk_p\) is called the \(p^\infty\)-Selmer rank of \(E/K\) and satisfies

\begin{equation*} \rk_p = \rk(E/K) + \delta_p\text{.} \end{equation*}

So if \(\Sha\) is finite then \(\delta_p =0\) for all \(p\text{.}\) Use BSD to compute parity of \(\rk_p\text{.}\)

Subsection 1.3 Consequences of BSD

Consider \(E/\QQ\text{:}\) Mordell-Weil implies that

\begin{equation*} E(\QQ) \simeq \ZZ^{\rk} \oplus \text{torsion} \end{equation*}

then BSD 1 says that

\begin{equation*} \underbrace{\ord_{s=1} L(E,s)}_{\rk_\an} = \rk\text{,} \end{equation*}

functional equation for \(L(E,s)\text{.}\)

\begin{equation*} L^*(E,s) = w L^*(E, 2-s) \end{equation*}

with \(w\in \{\pm 1\}\) the sign of the functional equation. If \(w = 1\) then \(L(E,s)\) is (essentially) symmetric at \(s=1\text{.}\) So \(\ord_{s=1} L(E,s)\) is even. If \(w = -1\) then \(\ord_{s=1} L(E,s)\) is odd.

We get BSD mod 2:

\begin{equation*} (-1)^{\rk} = w(\text{sign of f.e.}) \end{equation*}

a conjecture based on conjecture is bad so we go one step further.

Example 1.38.
\begin{equation*} E/\QQ \colon y^2 + y = x^3 + x^2 - 7x + 5 \end{equation*}
\begin{equation*} \Delta_E = -7 \cdot 13 \end{equation*}

\(w_v = 1\) if \(v\nmid \infty 7 \cdot 13\)

\begin{equation*} w_\infty = -1 \end{equation*}

(in general \(-1^{g}\) where \(g\) is dimension of the abelian variety).

\begin{equation*} w_7 = -1 \end{equation*}
\begin{equation*} w_{13} = -1 \end{equation*}

so \(w= -1\) and the rank is odd, hence there is a point of infinite order on this curve.

Problem.

On the one hand \(\prod_v w_v\) is computable. On the other hand \((-1)^{\rk}\) is precisely unknown.

\begin{equation*} (-1)^{\rk} = \prod _v w_v\text{.} \end{equation*}
Remark 1.40.

In fact this is true for all abelian varieties over \(K\text{.}\)

Example 1.41.

Let

\begin{equation*} E/\QQ \colon y^2 + xy = x^3 - x \end{equation*}

http://www.lmfdb.org/EllipticCurve/Q/65/a/1. \(\Delta_E = 5\cdot 13\text{,}\) it has a 2-isogenous curve \(E'\text{.}\)

Compute

\begin{equation*} c_5 = c_{13} = 1 \end{equation*}
\begin{equation*} c'_5 = c'_{13} = 2 \end{equation*}
\begin{equation*} \Omega_{E} = 2\Omega_{E'} \end{equation*}

then

\begin{equation*} \frac{\Reg_{E'}}{\Reg_E} = \frac{|\Sha_E| |E'_\tors|^2 \prod_p c_p \Omega_E}{|\Sha_{E'}| |E_\tors|^2 \prod_p c'_p \Omega_{E'}} \equiv \square \frac 24 \not \equiv 1 \square\text{.} \end{equation*}

So \(\Reg_E \ne 1\text{,}\) \(\Reg_{E'} \ne 1\) so \(E\) has at least one rational point of infinite order, so \(\rk \ge 1\text{.}\)

Back to the example

\begin{equation*} \frac{\Reg_E}{\Reg_{E'}} \equiv \frac 12 \square \end{equation*}

so by the lemma \(\rk\) is odd. Here we assumed that \(\Sha\) is finite for elliptic curves, one can drop the assumption of finiteness of \(\Sha\) to get unconditional results on the parity of \(\rk_p\) for all \(p\text{.}\)

This is known over \(\QQ\) and totally real fields.

How to compute the parity of \(\rk_p(E/K)\text{?}\) Need BSD-invariance for Selmer groups. (Details T. and V. Dokchitser “On the BSD quotients modulo squares”, and Milne “Arithmetic duality theorems”)

Definition 1.44.

For an isogeny

\begin{equation*} \Psi\colon A \to B \end{equation*}

of abelian varieties over \(K\text{.}\) Let

\begin{equation*} Q(\Psi) = | \coker(\Psi \colon A(K)/ A(K)_{\tors} \to B(K)/B(K)_{\tors})| \cdot | \ker(\psi \colon \Sha(A)_\divisible \to \Sha(B)_\divisible)|\text{.} \end{equation*}

Recall \(\rk_p = \rk + \delta_p\) where

\begin{equation*} \Sha = \bigoplus \Sha_{p^\infty} \end{equation*}

and

\begin{equation*} \Sha_{p^\infty} \simeq (\QQ_p/ \ZZ_p)^\delta_p \oplus T_p \end{equation*}
\begin{equation*} \Sha_{\divisible} = \bigoplus (\QQ_p / \ZZ_p)^{\delta_p}\text{.} \end{equation*}

Strategy, we show that for \(\Psi\) an isogeny s.t. \(\Psi \Psi^\vee = \lb p \rb\text{.}\) Then

\begin{equation*} p^{\rk_p(E/K)} \equiv \frac{Q(\Psi^\vee)}{Q(\Psi)} \equiv \frac{\prod_v c_p}{\prod_v c'_v} \frac{\Omega_E}{\Omega_{E'}} \pmod {{K^\times}^2}\text{.} \end{equation*}
Remark 1.45.

Let \(A^\vee\) be the dual of \(A\text{.}\) \(A^\vee = \Pic^0(A)\text{.}\)

So

\begin{equation*} (-1)^{\rk_p(E/K)} = (-1)^{\ord_p \left(\frac{\prod_v c_v \Omega_E}{\prod_v c'_v \Omega_{E'}}\right)} \end{equation*}

the parity of \(\rk_p(E/K)\) is computable from local invariants of \(E\) and \(E'\text{.}\)

To prove the \(p\)-parity conjecture it remains to prove

\begin{equation*} (-1)^{\ord_p \left(\frac{\prod_v c_v \Omega_E}{\prod_v c'_v \Omega_{E'}}\right)} = \prod_v w_v\text{.} \end{equation*}
Aside: Generalisation of the definition of \(\Sel^n(E/\QQ)\).

Consider

\begin{equation*} \Psi \colon A \to B \end{equation*}

an isogeny of abelian varieties. We have

\begin{equation*} 0 \to A(K)[\Psi] \to A(K) \xrightarrow \Psi B(K) \xrightarrow \delta H^1(G_K, A[\Psi]) \to H^1(G_K, A) \xrightarrow \Psi H^1(G_K, B) \end{equation*}

from which we extract

\begin{equation*} 0 \to B(K)/ \Psi(A(K)) \xrightarrow \delta H^1(G_K, A[\Psi]) \to H^1(G_K, A)[\Psi] \to 0 \end{equation*}
\begin{equation*} 0 \to \prod_v B(K_v)/ \Psi(A(K_v)) \xrightarrow \delta H^1(G_{K_v}, A[\Psi]) \to \prod_v H^1(G_K, A)[\Psi] \to 0 \end{equation*}

we then define

\begin{equation*} \Sel^{(\Psi)} (A/K) = \ker \left\{ H^1(G_K, A[\Psi]) \to \prod_v H^1(G_{K_v}, A) \right\} \end{equation*}
\begin{equation*} \Sha(A/K)= \ker\left\{ H^1(G_K, A) \to \prod_v H^1(G_{K_v}, A)\right\} \end{equation*}

so

\begin{equation*} 0 \to \underbrace{B(K)/ \Psi(A(K))}_{\coker(\Psi \colon A(K) \xrightarrow \Psi B(K))} \to \Sel^{(\Psi)} (A/K) \to \Sha(A/K) \to 0\text{.} \end{equation*}

We want to show:

We will show this in 3 parts, first the left, then the right, then the equality with the global root number.

Step 1.

Note that

\begin{equation*} Q(\Psi \circ \Psi^\vee) = Q(\Psi) Q(\Psi^\vee) \end{equation*}

hence

\begin{equation*} \frac{Q(\Psi)}{Q(\Psi^\vee)} \equiv \underbrace{Q(\Psi) Q(\Psi^\vee)}_{=Q([p])} \pmod{{K^\times}^2} \end{equation*}

now

\begin{equation*} | \coker( [p] \colon E(K)/ E(K)_\tors \to E'(K) / E'(K)_\tors ) | = p^{\rk(E/K)} \end{equation*}

Proof of this: For each generator \(R\) of \(E(K)/E(K)_\tors\) then

\begin{equation*} \frac1p R, \frac 2p R , \ldots, \frac{p-1}{p} R, R \end{equation*}

are not in the image of \(\lb p\rb\) which implies the size is \(p^{\rk(E/K)}\text{.}\) Also

\begin{equation*} | \ker([p] \colon \Sha(E/K)_\divisible \to \Sha(E'/K)_\divisible)| = p^{\delta_p} \end{equation*}

since

\begin{equation*} \Sha(E/K)_\divisible = \bigoplus_p (\QQ_p/\ZZ_p)^{\delta_p} \end{equation*}

and since \(\lb p \rb\) is trivial on all

\begin{equation*} (\QQ_l/\QQ_l)^{\delta_l},\, l\ne p \end{equation*}

then look at \(\lb p \rb \colon (\QQ_p/\ZZ_p)^{\delta_p} \to (\QQ_p/\ZZ_p)^{\delta_p}\) if \(x\in \QQ_p/\ZZ_p\) and \(\ker \lb p \rb\) then \(p x\in \ZZ_p \implies x = a/p\) for \(a\in \FF_p\text{.}\) so

\begin{equation*} p^{\delta_p}\text{.} \end{equation*}
Step 2.

We show that

\begin{equation*} \frac{Q(\Psi)}{Q(\Psi^\vee)} \equiv \frac{\prod_v c_p}{\prod_v c'_v} \frac{\Omega_E}{\Omega_{E'}} \pmod {{K^\times}^2}\text{.} \end{equation*}
Remark 1.49.

If \(A = E\text{,}\) \(B =E'\) with \(\Psi\) s.t. \(\Psi \Psi^\vee = \lb p \rb \) then

\begin{equation*} E\simeq E^\vee,\, E'\simeq {E'}^\vee \end{equation*}

and \(|\Sha_0| = \square\text{.}\)

\begin{equation*} \frac{Q(\Psi^\vee)}{Q(\Psi)} \equiv \frac{\prod_v c_p}{\prod_v c'_v} \frac{\Omega_E}{\Omega_{E'}}\pmod {{K^\times}^2}\text{.} \end{equation*}

We show how to obtain the quotient of Tamagawa numbers, for a sufficiently large set of places \(S\) of \(K\)

\begin{equation*} \frac{Q(\Psi^\vee)}{Q(\Psi)} \frac{| \Sha [\Psi^\vee]|}{| \Sha [\Psi]|} = \prod_{v\in S} \frac{|\ker \Psi_v|}{|\ker \Psi^\vee_v|} \end{equation*}

where \(\Psi_v\) is the induced map on \(E(K_v) \to E' (K_v)\text{.}\) If \(v\nmid \infty\) and \(v\in S\) what is

\begin{equation*} \frac{|\ker \Psi_v|}{|\coker \Psi_v|}? \end{equation*}
\begin{equation*} \xymatrix{ & & 0 \ar[d] & & \\ & 0\ar[d] & \ker \Psi_v \ar[d] & H_1\ar[d] & \\ 0 \ar[r] & E_1(K_v) \ar[r]\ar[d] &E(K_V) \ar[r]\ar[d] &E(K_v)/E_1(K_v) \ar[r]\ar[d] & 0 \\ 0 \ar[r] & E'_1(K_v) \ar[r]\ar[d] & E'(K_V)\ar[d] \ar[r]& E'(K_v)/E'_1(K_v) \ar[r]\ar[d] & 0\\ & 0 & \coker \Psi_v & H_2 & } \end{equation*}

Snake lemma gives

\begin{equation*} 0 \to \ker \Psi_v \to H_1 \to 0 \to \coker \Psi_v \to H_2 \to 0 \end{equation*}
\begin{equation*} \implies |\ker \Psi_v | = |H_1| \end{equation*}

and

\begin{equation*} |\coker \Psi_v| = |H_2|\text{.} \end{equation*}

Also

\begin{equation*} \left | \frac{E(K_v)/E_1(K_v)}{H_1}\right| = \left | \frac{E'(K_v)/E'_1(K_v)}{H_2}\right|\text{.} \end{equation*}

Moreover since \(E, E'\) are isogenous we have

\begin{equation*} | \widetilde E_\ns (\overline k) |=| \widetilde E'_\ns (\overline k) | \end{equation*}

hence since

\begin{equation*} 0 \to E_1(K_v) \to E_0 (K_v) \to \widetilde E_\ns (\overline k) \to 0 \end{equation*}

similarly for \(E'\text{.}\) We have

\begin{equation*} | E_0(K_v) / E_1(K_v) | = | E'_0(K_v) / E'_1(K_v) | \end{equation*}
\begin{equation*} \implies \left | \frac{E'(K_v) / E_1'(K_v)}{E(K_v) / E_1(K_v)} \right| = \left | \frac{E'(K_v) / E_0'(K_v)}{E(K_v) / E_0(K_v)} \right| = \frac{c_v}{c'_v}\text{.} \end{equation*}

Hence

\begin{equation*} (-1)^{\rk_p(E/K)} = (-1)^{\ord_p\left( \prod_v \left| \frac{\coker \Psi_v}{\ker \Psi_v}\right|\right)} \end{equation*}
\begin{equation*} = (-1)^{\ord_p\left( \frac{\prod_v c'_v}{\prod_v c_v} \underbrace{\prod_{v|\infty} \left| \frac{\coker \Psi_v}{\ker \Psi_v}\right|}_{\Omega_E/\Omega_{E'}} \right) }\text{.} \end{equation*}
Step 3.

We need to show that

\begin{equation*} (-1)^{\rk_p(E/K)}= w_E \text{ (p-parity)} \end{equation*}

i.e. we need to show that

\begin{equation*} (-1)^{\ord_p\left( \frac{\prod_v c'_v}{\prod_v c_v}\frac{\Omega_E}{\Omega_{E'}} \right)} = w_E \end{equation*}

Strategy:

\begin{equation*} (-1)^{\ord_p\left( \frac{\prod_v c'_v}{\prod_v c_v}\frac{\Omega_E}{\Omega_{E'}} \right)} = \prod_{v \nmid \infty} (-1)^{\ord_p \frac{c'_v}{c_v}} \prod_{v |\infty} (-1)^{\ord_p \left | \frac {\ker \Psi_v}{\coker \Psi_v} \right | } \end{equation*}

and relate

\begin{equation*} (-1)^{\ord_p \frac{c'_v}{c_v}} \end{equation*}

to \(w_v\) for \(v\nmid \infty\) and

\begin{equation*} (-1)^{\ord_p \left | \frac {\ker \Psi_v}{\coker \Psi_v} \right |} \end{equation*}

to \(w_v\) for \(v\mid \infty\text{.}\)

Then take product over all places.

Let \(E/ K \) be an elliptic curve admitting an isogeny \(\Psi\) of degree \(p\) (defined over \(K\)). Recall that we proved

\begin{equation*} p^{\rk_p(E/K)} = \prod_v \frac{c_v}{c_v'} \frac{\Omega_E}{\Omega_{E'}} \end{equation*}

\(v\) missing \(p\text{.}\) More precisely

\begin{equation*} p^{\rk_p(E/K)} \equiv \prod_{v|p\infty} \frac{c_v}{c'_{v}} \prod_{v|\infty} \left| \frac{\ker \psi_v}{\coker \psi_v}\right| \end{equation*}

where \(\psi_v\) is the map induced by \(\psi\) on \(E(K_v)\text{.}\)

What about \(v | p\) to extract

\begin{equation*} \frac{c_v}{c'_v} \end{equation*}

from

\begin{equation*} \left| \frac{\ker \psi_v}{\coker \psi_v}\right| \end{equation*}

at finite places we can use a diagram involving

\begin{equation*} 0 \to E_1(K_v) \to E'_1(K_v) \to \coker \to 0\text{.} \end{equation*}

If \(v\nmid p \) then \(| \coker | = 1\) since then on the level of the formal group \(\psi\) induces a map

\begin{equation*} \hat \psi \colon \hat E(\ideal m_K) \to \hat E'(\ideal m_K) \end{equation*}
\begin{equation*} T \mapsto aT + \cdots \end{equation*}

power series rep of \(\psi\) \(\psi(x,y) = (x', y')\) Silverman IV cor 4.3/ \(\omega' \circ \psi = \psi' \circ \omega\text{.}\) with leading \(a = \psi^* \omega'/ \omega \times \) unit \(\in \ints_K\text{.}\)

\begin{equation*} \implies aa' = p \in \ints_K^\times \implies \hat \psi\text{ isom}\text{.} \end{equation*}

If \(v | p\) then \(\coker\) contributes to the snake lemma and at that place

\begin{equation*} \frac{c_v}{c'_v} \left| \frac{\psi^* \omega'}{\omega}\right|_v = \frac{c_E}{c_E'} \left| \frac{\omega}{\omega_v^0} \right|_v \end{equation*}

for a particular choice of \(\omega\text{.}\)

Proving \(p\)-parity.

To prove the \(p\)-parity conjecture

\begin{equation*} (-1)^{\rk_p(E/K)} = w_E\text{.} \end{equation*}

We will show that

\begin{equation*} (-1)^{\ord_p \prod_v \frac{c_v}{c'_v} \frac{\Omega}{\Omega_{E'}}} = w_E \end{equation*}

by relating

\begin{equation*} (-1)^{\ord_p \frac{c_v}{c'_v}} \end{equation*}

and \(w_v\) at some place \(v\nmid p\infty\)

\begin{equation*} (-1)^{\ord_p \frac{\Omega_E}{\Omega_{E'} }}= (-1)^{\ord_p \left| \frac {\ker \psi_v}{\coker \psi_v}\right| } \end{equation*}

and \(w_v\) at \(v | \infty\text{.}\)

We only sketch these steps for \(v\nmid p\) and \(E \) is semistable at \(v\text{.}\)

The proofs of \(p\)-parity for \(p\) odd and \(p= 2\) are different.

\(p\) odd.

The \(p\)-parity conjecture is proven for principally polarized abelian varieties with a \(p\)-cyclic isogeny with \(p \ge 2g +2 \) or \(p \ge 2\) and semistable reduction and some local constraints at \(v|p\text{.}\) see Root numbers selmer groups and non-commutative Iwasawa theory, Coates, Fukaya, Kato, Sujatha

Sketch, for an elliptic curve with a \(p\)-isogeny \(\psi\) we look at \(v|\infty\) where \(w_v = -1\text{,}\) and

\begin{equation*} (-1)^{\ord_p \left | \frac{\ker \psi_v}{\coker \psi_v}\right |} \end{equation*}

if \(v\) is complex \(|\ker \psi_v| = p\) \(|\coker \psi_v | = 1\text{.}\) so

\begin{equation*} (-1)^{\ord_p \left | \frac{\ker \psi_v}{\coker \psi_v}\right |} = -1 = w_v\text{.} \end{equation*}

If \(v|\infty\) is real what does \(E(\RR)\) look like? Either there is a real period and so two real components, and all real \(p\)-torsion (if any) is on the identity component. Or there is no real period and only 1 real component that contains all real \(p\)-tors if any.

  1. \(|\ker \psi_v | = p\) (the \(p\)-tors in \(\ker \psi\) are real)
  2. \(|\ker \psi_v | = 1\) (the \(p\)-tors in \(\ker \psi\) are not real)
Figure 1.50.

Moreover \(|\coker \psi | = 1\) always, \(\sgn(\Delta_E) = \sgn(\Delta_{E'})\)

More generally if \(\deg \Psi\) is odd then

\begin{equation*} E'(\RR)/ \psi(E(\RR)) \hookrightarrow H^1(\Gal{\CC}{\RR}, E[\psi]) = 0 \end{equation*}

since \(\lb \CC:\RR\rb = 2 \) is coprime to \(E\lb \psi\rb\) (see Atiyah's book).

In the first case

\begin{equation*} (-1)^{\ord_p \left | \frac{\ker \psi_v}{\coker \psi_v}\right |} = -1 = w_v \end{equation*}

In the second case

\begin{equation*} (-1)^{\ord_p \left | \frac{\ker \psi_v}{\coker \psi_v}\right |} = 1 \ne w_v \end{equation*}

For \(K\) a local field let \(F = K(\ker \psi_v)\) noting that

\begin{equation*} \Gal{F}K \hookrightarrow (\ZZ/p\ZZ)^\times \end{equation*}

from its action on points in \(\ker \psi = F/K\) is cyclic.

Consider the composition

\begin{equation*} F^\times \xrightarrow{\text{local rec.}} \Gal F K \hookrightarrow(\ZZ/p)^\times\text{.} \end{equation*}

and denote

\begin{equation*} (-1, F/K) \end{equation*}

the image of \(-1\) under the above map.

\begin{equation*} (-1, F/K) = \begin{cases} 1 \amp \text{ if } -1 \text{ is a norm from }F\text{ to }K,\\ - 1\amp\text{otw}\end{cases} \end{equation*}

this is the Artin symbol.

This is perfect as they cancel out globally.

If \(v\) is complex then \(F = \CC\text{,}\) \(K = \CC\) and \((-1, F/K) = 1\)

If \(v\) is real and \(|\ker \psi_v| = p\) then \(F = \RR\text{,}\) \(K = \RR\) and \((-1, F/K) = 1\)

If \(v\) is real and \(|\ker \psi_v| = 1\) then \(F = \RR\text{,}\) \(K = \RR\) and \((-1, F/K) = -1\)

\(p =2\).

Note that \((-1, F/K) = 1\) for all places of \(K\) since if \(E\) admits a 2-isogeny \(\psi/K\) then is admits a 2-torsion point over \(K\text{.}\)

Hence \(F = K(\ker \psi_v) = K\)

set -up

\begin{equation*} E/K \end{equation*}

with a 2-isogeny \(\psi/K\)

\begin{equation*} E \colon y^2 = x(x+ax+ b) \end{equation*}

by translating 2-torsion to (0,0)

\begin{equation*} \psi\colon E \to E':y^2= x(x^2 - 2ax + \delta) \end{equation*}

where \(\delta = a^2 - 4b = \disc(x^2 + ax + b)\) if \(\delta \gt 0 \) then \(E(\RR) \) has two connected components. \(\delta \lt 0\) only 1. Have \(16b = \disc(x^2 - 2ax +\delta)\) likewise for \(E'\)

\begin{equation*} \xymatrix{ & & 0 \ar[d] & & \\ & \ker \psi_v^0 \ar[d] & \ker \psi_v \ar[d] & \ker \psi_/\ar[d] & \\ 0 \ar[r] & E^0(\RR) \ar[r]\ar[d] &E(\RR) \ar[r]\ar[d] &E(\RR)/E^0(\RR) \ar[r]\ar[d] & 0 \\ 0 \ar[r] & E'^0(\RR) \ar[r]\ar[d] & E'(\RR)\ar[d] \ar[r]& E'(\RR)/E'^0(\RR) \ar[r]\ar[d] & 0\\ & \coker \psi_v^0 & \coker \psi_v & \coker \psi_/ & } \end{equation*}

by snakey

\begin{equation*} \frac{ | \ker \psi_v^0 || \ker \psi_/ || \coker \psi_v|}{|\ker \psi_v || \coker \psi_v^0|| \coker \psi_/| } = 1 \end{equation*}
\begin{equation*} \implies \left| \frac{\coker \psi_v}{\ker \psi_v}\right| = \frac{| \coker \psi^0_v|| \coker \psi_/|}{|\ker \psi_v^0|| \ker\psi_/|} \end{equation*}

let \(n(E), n(E')\) be the number of real connected components \(n = E(\RR) / E^0(\RR)\)

By the third column

\begin{equation*} \frac{n(E')}{n(E)} \frac{|\ker \psi_/|}{|\coker \psi_/|} =1 \end{equation*}

now \(| \coker \psi_v^0| = 1\) as the map on identity component is surjective. hence

\begin{equation*} \left | \frac {\coker \psi_v}{\ker \psi_v}\right|= \frac{n(E')}{n(E) | \ker \psi_v^0|} \end{equation*}

Recall: to prove the \(2\)-parity conjecture for \(E/K\)

\begin{equation*} (-1)^{\rk_2(E)} = w\text{????????????????????} \end{equation*}

missed

Notation

\begin{equation*} E \colon y^2 = x(x^2+ ax+ b) = xq_1(x) \end{equation*}
\begin{equation*} E' \colon y^2 = x(x^2-2 ax+ \delta) = xq_2(x),\,\delta = a^2 - 4b \end{equation*}

\(\disc(q_1(x)) = \delta\) \(\disc(q_2(x)) = 16b\)

a) If \(\delta \gt 0\text{,}\) \(b \gt 0\) then \(E,E'\) both have two real components, \(n(E) = n(E') =2\text{.}\)

\begin{equation*} | \ker \psi_v^0| = \begin{cases} 1 \amp \text{ if } (0,0)\text{ is not on } E^0(\RR)\\ 2 \amp \text{ if } (0,0)\text{ is on } E^0(\RR)\\ \end{cases}= \begin{cases} 1 \amp \text{ if } a \lt 0\\ 2 \amp \text{ if } a \gt 0\\ \end{cases} \end{equation*}

write \(q_1(x) = x^2 + ax + b = (x-\alpha)(x-\beta)\) then if \((0,0) \in E^0(\RR)\text{,}\) \(\alpha,\beta\lt 0\) but \(a = -\alpha -\beta\) hence in this case \(a \gt 0\text{.}\)

\begin{equation*} (-1)^{\ord_2 \left| \frac{\ker \psi_v}{\coker \psi_v } \right|} = \begin{cases} 1 \amp \text{ if } a \lt 0\\ -1 \amp \text{ if } a \gt 0\\ \end{cases} \end{equation*}

so we need some correction if \(\delta \gt 0, b \gt 0, a \lt 0\text{.}\)

b) If \(\delta \gt 0, b \lt 0\) \(E\) has two real components and \(E'\) only 1 \(n(E) = 2,n(E') = 1\text{.}\)

\begin{equation*} |\ker \psi_v^0 | = 1 \end{equation*}

since \(b \lt 0\) and \(b = \alpha\beta\text{.}\)

\begin{equation*} (-1)^{\ord_2 \left| \frac {\ker \psi_v}{\coker \psi_v}\right|} = -1 \end{equation*}

so no correction if \(\delta \gt 0, b \lt 0\text{.}\)

c) If \(\delta \lt 0,b \gt 0\text{,}\) \(n(E) = 1\text{,}\) \(n(E') =2\text{.}\)

\begin{equation*} |\ker \psi_v^0| = 2 \end{equation*}

and

\begin{equation*} (-1)^{\ord_2 \left| \frac {\ker \psi_v}{\coker \psi_v}\right|} = 1 \end{equation*}

need correction if \(\delta \lt 0 , b \gt 0\text{.}\)

d) \(b \lt 0, \delta \lt 0\) contradiction, \(\delta = a^2 - 4b\text{.}\)

So in summary if \(\delta \gt 0, b \gt 0, a \lt 0\) or \(\delta \lt 0, b \gt 0\) need a correction, if \(\delta \gt 0, b \gt 0, a \gt 0\) or \(\delta \gt 0, b \lt 0\) no correction.

\begin{equation*} (-1)^{\ord_2 \left| \frac {\ker \psi_v}{\coker \psi_v}\right|} = ? w_v \end{equation*}

First guess

\begin{equation*} (a, -b)(-a, \delta) \end{equation*}

Recall: let \(K\) be a local field

\begin{equation*} K^\times \times K^\times \to \{ \pm 1\} \end{equation*}
\begin{equation*} (a,b) \mapsto \begin{cases} 1\amp\text{ if } a \text{ is a norm from } K(\sqrt b ) \to K,\\ -1\amp\text{ otw} \end{cases} \end{equation*}

If \(K\) is archimidean \((a,b) = -1 \iff a \lt 0,b \lt 0\text{.}\) If \(K\) is non-archimidean with odd residue characteristic then

\begin{equation*} (\text{unit},\text{unit}) = 1 \end{equation*}
\begin{equation*} (\text{unit},\pi^n) = -1 \end{equation*}

if \(n\) odd and unit is not a square.

\begin{equation*} (a,bc) = (a,b)(a,c)\text{.} \end{equation*}

So guess

\begin{equation*} (a, -b)(-a, \delta) \end{equation*}

works over \(\RR\text{.}\)

\(v\nmid 2 \infty\) need to show that

\begin{equation*} (-1)^{\ord_2 \frac{c_v}{c'_v}} = (a,-b)(-a, \delta) w_v \end{equation*}

if \(E\) has good reduction at \(v\text{.}\)

\begin{equation*} c_v = c'_v = 1\text{.} \end{equation*}

Need to show that

\begin{equation*} (a, -b)(-a,\delta) = 1\text{.} \end{equation*}

Since \(E,E'\) have good reduction at \(v\text{.}\) then \(b,\delta\) are units in \(K\text{.}\) If \(a \in \ints_K^\times\) then \((a,-b) (-a,b) = 1\) if \(a \equiv 0 \pmod {\pi_K}\) then since \(a^2 - 4b = \delta\) then \(\delta \equiv -4b \pmod{\pi_K}\text{.}\)

If \(E\) has split multiplicative reduction, (multiplicative reduction is when \(y^2 = f(x)\) and \(f(x)\) has a double root mod \(\pi_K\text{,}\) any two distinct tangents at the node, both defined over \(k\) (fixed by frob)). so \(E'\) also has split multiplicative reduction as \(\psi\) commutes with frobenius.

Need to compute

\begin{equation*} \frac{c_v}{c'_v} \end{equation*}

by Tates algorithm

\begin{equation*} c_E = v(\Delta_E) = n \end{equation*}

we show that

\begin{equation*} c_{E'} = v(\Delta_{E'}) = \begin{cases} 2n,\\ \frac12n \end{cases} \end{equation*}

Recall

\begin{equation*} E\colon y^2 = \overbrace{x(x^2 + ax + b)}^{f_E(x)} = x(x-\alpha)(x-\beta) = xq_1(x) \end{equation*}
\begin{equation*} \Delta_{f_E} = \alpha^2 \beta^2 ( \alpha- \beta)^2 = b^2 (\alpha - \beta)^2 = b^2 \delta \end{equation*}
\begin{equation*} {E'}\colon y^2 = \overbrace{x(x^2 -2ax + \delta)}^{f_{E'}(x)} = x(x-A)(x-B) = xq_2(x) \end{equation*}
\begin{equation*} \Delta_{f_{E'}} = A^2 B^2 ( A- B)^2 = \delta^2 (A - A)^2 = \delta^2 16b \end{equation*}

if \(v(\delta) = n\) then \(v(\Delta_{f_{E}}) = n\) so \(c_E = -n\) and \(v(\Delta_{f_{E'}}) = 2n\) so \(c_{E'} = 2n\) in general if \(E\) admits a \(p\)-isogeny and \(E\) has split multiplicative reduction then

\begin{equation*} \frac{c_E}{c_{E'}} = p^{\pm 1}\text{.} \end{equation*}

here \(w_v = -1\) and

\begin{equation*} (-1)^{\ord_2 \frac{c_E}{c_{E'}}} = -1 \end{equation*}

need to show that

\begin{equation*} (a,-b) (-a, \delta) = 1 \end{equation*}

if \(E\) has a double root at \((0,0)\) wlog \(\alpha \equiv 0 \pmod \pi_K\) then \(v(\delta) = 0\text{,}\) \(v(b) \gt 0 \) and both slopes of tangent at \((0,0)\) are defined over \(k\text{.}\)

Taylor expansion at \((0,0)\)

\begin{equation*} f(x,y) = y^2 - x^3 - ax^2 - bx \end{equation*}
\begin{equation*} = (y - s_1x)(y- s_2x) + h.o.t. \end{equation*}
\begin{equation*} = y^2 - xy(s_1 + s_2) + s_1s_2 x^2 + h.o.t. \end{equation*}

so \(s_1 = -s_2\) and \(s_1 s_2 = -a\) implies \(s_1^2 = a\text{.}\)

so \(s_1 \in k^\times\) then \(a\in {k^{\times}}^2\)

\begin{equation*} (a,-b) = 1\implies (-a,\delta) =1 \end{equation*}

as both are units.

Now \(b = \alpha\beta \equiv 0 \pmod {\pi_K}\) so

\begin{equation*} x^2 - 2ax + \delta \equiv (a-A)^2 \pmod {\pi_K} \end{equation*}

same Taylor expansion gives

\begin{equation*} f(x,y) = y^2 - x^3 + 2ax^2 - \delta x \end{equation*}
\begin{equation*} = f(x,y) - f(A,0) = (y - s_3(x-A))(y - s_4(x-A)) + h.o.t. \end{equation*}

so \(s_3 = -s_4 \) and \(s_3 s_4 = 2a\text{,}\) \(s_3 ^2 = -2a\) hence

\begin{equation*} (a, -b)(-2a,\delta) \end{equation*}

split multiplicative

\begin{equation*} -2a \in {K^\times}^2\text{,} \end{equation*}

So we should use this Hilbert symbol instead, it doesn't change the real case.

If \(E\) has non-split multiplicative reduction

\begin{equation*} \frac{c_E}{c_{E'}} = \begin{cases} 1,\amp\text{ if } v(\Delta_E),v(\Delta_{E'}) \text{ even}\\ 2,\amp\text{ if } v(\Delta_{E'}) \text{ odd}\\ \frac12\amp\text{ if } v(\Delta_E) \text{ odd} \end{cases} \end{equation*}
\begin{equation*} \implies (-1)^{\ord_2 \frac{c_E}{c_{E'}}} = \begin{cases} 1,\\ -1,\\ -1,\\ \end{cases} \end{equation*}

done since \(a,-2a\) precisely not squares.

What are these invariants purely in theory?