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.
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
- Understand the proof of Mordell-Weil
- See where it is non-effective.
- From the proof, extract a strategy to sometimes compute the rank (define Selmer groups, Shafarevich-Tate group).
Outline proof of Mordell-Weil.
Part 1: Prove that
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
is finite and that \(E\) has a “height function” then \(E(K)\) is finitely generated.
Theorem 1.1. Descent theorem (see Thm. VIII 3.1).
Let \(A\) be an abelian group, suppose that there exists a function
with the following properties:
- Let \(Q \in A\) then there is a constant \(c_1\) depending on \(Q\) and \(A\) such that\begin{equation*} h(P+Q) = 2h(P) + c_1,\,\forall P \in A\text{.} \end{equation*}
- There is an integer \(m \ge 2\) and a constant \(c_2\) depending on \(A\) s.t.\begin{equation*} h(mP) \ge m^2 h(P) - c_2,\,\forall P\in A\text{.} \end{equation*}
- For every constant \(c_3\text{,}\) the set\begin{equation*} \{P\in A : h(P) \le c_3\} \end{equation*}is finite.
suppose further that for the \(m\) in \(2.\) we have \(A/mA\) is finite. Then \(A\) is finitely generated.
Proof.
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
for some \(1 \le i \le r\text{.}\) Repeat this for
by property 2. of \(h\) we have
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 \(P\text{.}\)
since \(m \ge 2\text{.}\) Hence for \(n\) sufficiently large (to make \(\frac 1{2^n} h(P) \le 1\)) we have
Since \(P\) is a linear combination of \(P_n\) and \(Q_i\)
it follows that every \(P \in A\) is a linear combination of points in
Remark 1.2.
On \(E/\QQ\) the height function
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:
Theorem 1.4. Weak Mordell-Weil.
Let \(K\) be a number field \(E/K\) an elliptic curve, \(m \ge 2\) then
We will prove this under the assumption that \(E\lb m \rb \subseteq E(K)\text{.}\) This is WLOG since:
Lemma 1.5.
Let \(L/K\) be a finite Galois extension, if
is finite then so is
Proof.
induced by
and prove that \(\phi\) is finite. Kernel \(\phi\) is given by
take \(P \in \phi\text{.}\) We can choose \(Q_P \in E(L)\) such that \(Q_P = P\text{.}\) Define a map of sets
Note that
Now we show that the association
is 1 to 1.
Suppose that \(P,P' \in E(K) \cap m E(L)\) satisfying \(\lambda_P = \lambda_{P'}\) then
for all \(\sigma \in G_{L/K}\) so \(Q_P - Q_{P'} \in E(K)\) and hence
hence
\(G_{L/K}\) 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
where \(Q\) is a choice of point in \(E(\overline K)\) such that \(mQ = P\text{.}\)
Proposition 1.7.
\(\kappa\) is well defined, bilinear, the kernel in the first argument is \(m E(K)\) and in the second argument is \(G_{\overline K/L}\) where \(L = K(\lb m \rb \inv E(K))\) is the compositum of all fields \(\kappa(x(Q), y(Q))\) as \(Q\) ranges over all the points of \(E(\overline K)\) s.t. \(m Q \in E(K)\text{.}\)
Hence the Kummer pairing induces a perfect bilinear pairing
i.e. the map
is an isomorphism.
Proof.
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.
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 \(F\text{.}\)
Proposition 1.9.
Let \(E/K\) be an elliptic curve
for \(P \in E(K)\) have \(\frac 12 P \in E(\overline K)\) s.t. \(\frac 12 P \oplus \frac 12 P = P\text{.}\)
- \(K(\frac 12 P) /K\) is a Galois extension and \(\Gal{K(\frac 12 P)}{K} = C_2 \times C_2\) from Lemma 1.
-
\begin{equation*} \phi_P \colon \absgal{K} \to E(K)[2] \end{equation*}\begin{equation*} g \mapsto Q^\sigma - Q = g(\frac 12 P) - \frac 12 P \end{equation*}is well defined and has kernel \(\Gal{K}{K(\frac 12 P)}\text{.}\)
-
\begin{equation*} \phi \colon E(K) / 2E(K) \to \Hom_{cts}(\absgal{K}, E(K)[2]) \end{equation*}\begin{equation*} P \mapsto \phi_P \end{equation*}is well defined and injective. Now \(\phi_P\) is continuous by \(2.\) and so\begin{equation*} \phi_{P\oplus Q} (g) = g(\frac 12 (P\oplus Q)) - (\frac 12 P\oplus \frac 12 Q) \end{equation*}\begin{equation*} = g(\frac 12 P) \oplus g(\frac 12Q) - \frac 12 P \ominus \frac 12 Q \end{equation*}\begin{equation*} = \phi_P(g) \oplus \phi_Q(g) \end{equation*}a homomorphism.\begin{equation*} \phi_{2Q}(g) = g(\frac 12 2 Q)) - \frac12 2(Q) = g(Q) - Q = 0 \end{equation*}for all \(g\in \absgal K\) if \(Q \in E(K)\) so this is well defined. For injectivity:\begin{equation*} \phi_P(g) = 0 \implies g(\frac 12 P) = \frac 12 P \forall g \in \absgal K \end{equation*}\begin{equation*} \implies \frac 12 P \in E(K) \implies P \in 2 E(K) \end{equation*}which gives injectivity.
-
\begin{equation*} \eta \colon \Hom_{cts}(\absgal K , E(K) [2] ) \to K^\times/{K^\times}^2 \times K^\times/{K^\times}^2 \times K^\times/{K^\times}^2 \end{equation*}\begin{equation*} \psi \mapsto\psi_\alpha ,\psi_\beta, \psi_\gamma \end{equation*}\begin{equation*} \psi(g )\in \{0, (\alpha,0)\} \subseteq E(K) \iff g \in \Gal{\overline K }{K(\sqrt {\psi_\alpha})} \end{equation*}then \(\eta\) is an injective homomorphism. It is an isomorphism to the subgroup of triples \(a,b,c\) s.t. \(abc \in {K^\times}^2\text{.}\) Proof:\begin{equation*} \Hom_{cts}(\absgal K, C_2) \simeq K^\times/{K^\times}^2 \end{equation*}with \(\psi\) s.t. \(\ker \psi = \Gal{\overline K}{ K\sqrt d} \leftrightarrow d\text{.}\) It is an isomorphism:\begin{equation*} \ker \psi_i = \Gal{\overline K}{ K(\sqrt {d_i})},\,i =1,2 \end{equation*}\begin{equation*} \ker \psi_1 \psi_2 = \Gal{\overline K}{ K(\sqrt {d_1d_2})} \end{equation*}Now apply this to \(E(K) \lb 2\rb = C_2 \times C_2\) to get an isomorphism to \(K^\times / {K^\times}^2 \times K^\times / {K^\times}^2 \text{.}\) Record this third homomorphism to get \(\eta\text{.}\)
- If \(P = (x_0, y_0) \in E(K)\) then\begin{equation*} \eta(\phi_P) = (x_0 - \alpha, x_0 - \beta, x_0 -\gamma)\text{.} \end{equation*}Proof sketch: If\begin{equation*} E\colon y^2 = x^3 + Ax^2 + Bx \end{equation*}then for \(Q = (x_0, y_0) \in E(K)\text{.}\)\begin{equation*} 2Q = \left(\left( \frac {x_0 - B}{2y_0}\right)^2 , \ldots\right) \end{equation*}Hence if \(2Q = P = (x_1, y_1)\) then \(\sqrt{x_1} \in K(\frac 12 P)\text{.}\) So if\begin{equation*} E \colon y^2 = (x-\alpha)(x-\beta)(x-\gamma) \end{equation*}then\begin{equation*} P = (x_2, y_2) \end{equation*}then\begin{equation*} \sqrt{x_2 - \alpha} , \sqrt{x_2 - \beta} , \sqrt{x_2 - \gamma} \in K(\frac 12 P) \end{equation*}\begin{equation*} K(\sqrt{x_2 - \alpha}) , K(\sqrt{x_2 - \beta}) , K(\sqrt{x_2 - \gamma}) \subseteq K(\frac 12 P) \end{equation*}\begin{equation*} \implies K(\frac 12 P) = K(\sqrt{x_2 - \alpha},\sqrt{x_2 - \beta} , \sqrt{x_2 - \gamma}) \end{equation*}
Example 1.10.
Let
for \(P \in E(\QQ), \QQ(\frac 12 P ) / \QQ\) can only ramify at 2.
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 | ? |
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 \(2m^2/ n^2 - 1 = B^2\)
and
if \(m \equiv 0(2) \implies -1 = \square \pmod 8\) a contradiction.
So \(2 - n^2 = \square \pmod 8 \implies n^2 \equiv1 \pmod 8\)
4) Use the group structure!
Theorem 1.12. Complete 2-decent.
Let \(K\) be a field of characteristic 0 and
The map
replacing \(x_0 -\alpha\) with \((x_0 - \beta) (x_0 - \gamma)\) if 0.
Triples \((a,b,c)\) that lie in the image satisfy \(abc \in {K^\times}^2\text{.}\) A triple \(a,b,c\) with \(abc \in {K^\times}^2\) lies in the image iff it is in the image of \(E(K)\lb 2\rb\) or
is soluble with \(z_i \in K^\times\text{.}\) In which case
iii) If \(K\) is a number field and \((a,b,c)\) is in the image then
only ramifies at primes dividing \(2(\alpha - \beta)(\alpha - \gamma)(\beta - \gamma)\text{.}\)
Exercise 1.13.
Recall:
where \(\phi_P \colon \sigma \mapsto Q^\sigma - Q\) where \(Q = 2P\text{.}\) Which is well-defined and injective.
Elements of
Lemma 1.14.
Let \(n \ge 1\)
-
\begin{equation*} \psi \colon E(K)/ nE(K) \to \{ K \subseteq F \subseteq \overline K\} \end{equation*}\begin{equation*} P\mapsto K(\frac 1n P, E[n]) \end{equation*}is well defined.
- \(K(\frac 1n P, E\lb n \rb ) / K\) only ramifies at \(\ideal p | n \Delta_E\text{.}\)
- \begin{equation*} \Gal{K(\frac 1n P, E [n] )}{K} \le \ZZ/n \times \ZZ/n \end{equation*}
- There are only finitely many fields satisfying 2. and 3. so \(\image \psi\) is finite.
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
Remark 1.16.
If \(G\) acts trivially on \(M\) then
When \(G\) is profinite then we want that the skew homomorphisms factor through finite Galois groups. We will prove that
Theorem 1.17.
If
is an exact sequence of \(G\)-modules then
Lemma 1.18.
- \(\psi\) is finite-to-one (gives Mordell-Weil)
- Let\begin{equation*} \phi_P \colon G_K \to E[n] \end{equation*}\begin{equation*} \phi_P (gh) = \phi_P(g) + g \phi_P(h) \end{equation*}is a skew (or crossed) homomorphism. If \((\frac 1n P ) '\) is another choice of \(\frac 1n P\) and \(\phi_P'\) is the corresponding skew homomorphism, then\begin{equation*} \phi_P - \phi_P ' \end{equation*}is of the form\begin{equation*} g\mapsto T \ominus gT \end{equation*}where \(T \in E\lb n\rb\text{.}\)
- \(\phi_P\) factors through\begin{equation*} \Gal{K(\frac 1n P, E\lb n \rb)}{K}\text{.} \end{equation*}
-
\begin{equation*} \phi\colon E(K) / nE(K) \to Z/B \end{equation*}\begin{equation*} P \mapsto \phi_P \end{equation*}is an injective homomorphism. Where\begin{equation*} Z = \{\text{skew homs } G_K \to E[n]\} \end{equation*}\begin{equation*} B = \{\text{skew homs } G_K \to E[n] \text{ of the form }g\mapsto T\ominus gT,\,T\in E\lb n\rb\}\text{.} \end{equation*}
Proof.
- 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.
-
\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 \(G= G_K\)
to get
which gives the long exact sequence
Problem:
is infinite. What subgroup of
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
We have from the theorem
we want to understand \(\image \delta\) i.e. the subgroup
this is as hard as finding \(E(K)\text{,}\) here is why:
Claim 1.19.
corresponding to principal homogeneous spaces for \(E\) (genus 1 curves whose jacobian is \(E\))
Finding
is equivalent to finding which PHS coming from \(H^1\) have a rational point. ??? Hensel's lemma.
Let \(C\) be a curve
\(C\) is a PHS for \(E\) iff \(E\) is the jacobian of \(C\text{.}\)
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.
Theorem 1.21.
The twists of \(C/K\) up to \(K\)-isomorphism are in 1-1 correspondence with elements of
where
Proof.
Let \(C'/K\) be a twist of \(C/K\) then there exists an isomorphism \(/\overline K\)
associate the following map
Check that \(\xi\) is a cocycle
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
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
and translations
Example 1.23.
\(E/K\) elliptic, consider
a quadratic extension and \(\chi\) the associated character
The group \(\pm 1\) can be viewed as automorphisms of \(C\text{.}\) So use \(\chi\) to define the cocycle
Let \(C/K\) be the corresponding twist of \(E/K\text{,}\) we find an equation for \(C/K\text{.}\) Choose
and write
since \(\lb - 1 \rb(x,y) = (x,-y)\) the action of \(\sigma \in G_K\) on
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
is defined over \(K\) and defines an elliptic curve. Moreover
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 \(K\text{.}\)
morphism defined over \(K\) satisfying
- \begin{equation*} \mu(P, 0) = P \,\forall P \in C \end{equation*}
- \begin{equation*} \mu(\mu(p,P), Q) = \mu(p,P+Q) \,\forall P \in C \end{equation*}
-
\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.
Proposition 1.25.
Let \(E/K\) and \(C/K\) be a principal homogeneous space for \(E/K\text{.}\) Fix a point \(p_0 \in C\) and define a map
- \(\theta\) is an isomorphism over \(K(p_0)\text{.}\) In particular \(C/K\) is a twist of \(E/K\text{.}\)
- \(\forall p,q \in C\)\begin{equation*} q- p = \theta\inv(q) - \theta \inv (p)\text{.} \end{equation*}
- \(\theta\) is a morphism over \(K\text{.}\)
Definition 1.26.
Two homogeneous space \(C/K\) and \(C'/K\) for \(E/K\) are equivalent if there is an isomorphism
defined over \(K\) and is compatible with the action of \(E\) on \(C\) and \(C'\text{.}\)
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
Proposition 1.27.
Let \(C/K\) be a PHS for \(E/K\) then \(C/K\) is in the trivial class \(\iff C(K) \ne \emptyset\text{.}\)
Theorem 1.28.
Let \(E/K\) then there is a natural bijection after fixing \(p_0 \in C\)
Proof.
Well-definedness:
is a cocycle. Suppose that \(C'/K\) and \(C/K\) are two equivalent PHS then
and
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
and view
From the theorem on
there exists a curve \(C/K\) and a \(\overline K\)-isomorphism
s.t.
Define a map \(\mu \colon C\times E \to C\)
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.
for \(\alpha \in \Aut(E)\text{.}\)
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 \(\xi\colon G_K \to E\)
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 \(E/K\)
then the translation by \(T\) map is given by
for
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. \(x^\id \mapsto (x^\id)^\sigma\text{.}\)
need to find the subfield of \(K(\sqrt d)(x,y)_\xi\) fixed by \(\sigma\text{.}\) Note:
are invariant, take
and find relations between \(z\) and \(w\) to get
Claim: \(C/K\) is the PHS of \(E/K\) corresponding to \(\left\{\xi\right\}\text{.}\) There is a natural map
so that
- 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.
Definition 1.31. \(m\)-Selmer groups.
The \(m\)-Selmer group of \(E/K\) is the subgroup of
defined by
Definition 1.32. The Shafarevich-Tate group.
The Shafarevich-Tate group of \(E/K\) is the subgroup of
defined by
Theorem 1.33.
There is an exact sequence
- \begin{equation*} 0 \to E(K)/mE(K) \to \Sel^m(E/K) \to \Sha(E/K)[m] \to 0 \end{equation*}
- \(\Sel^m(E/K)\) is finite.
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
is torsion.
So we may write
where for each prime \(p\)
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
So
where \(T_p\) is a finite abelian \(p\)-group.
The group
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
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
Definition 1.34. \(p^\infty\)-Selmer group.
Consider \(\Sel_{p^n}(E/K)\) and take the direct limit
to define the \(p^\infty\)-Selmer group.
One shows that
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
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
then BSD 1 says that
functional equation for \(L(E,s)\text{.}\)
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:
a conjecture based on conjecture is bad so we go one step further.
Theorem 1.36.
The sign in the functional equation of \(L(E,s)\) is equal to the global root number of \(E\text{.}\)
This is defined by
the local root numbers defined in terms of the local galois representations. Non-trivial to understand, but manageable.
Conjecture 1.37. Parity conjecture.
Example 1.38.
\(w_v = 1\) if \(v\nmid \infty 7 \cdot 13\)
(in general \(-1^{g}\) where \(g\) is dimension of the abelian variety).
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.
Theorem 1.39.
Assume \(\Sha\) is finite, let \(\phi \colon E \to E'\) be an isogeny whose degree is not divisible by \(\characteristic(K)\text{,}\) then
Remark 1.40.
In fact this is true for all abelian varieties over \(K\text{.}\)
Example 1.41.
Let
http://www.lmfdb.org/EllipticCurve/Q/65/a/1
. \(\Delta_E = 5\cdot 13\text{,}\) it has a 2-isogenous curve \(E'\text{.}\)
Compute
then
So \(\Reg_E \ne 1\text{,}\) \(\Reg_{E'} \ne 1\) so \(E\) has at least one rational point of infinite order, so \(\rk \ge 1\text{.}\)
Lemma 1.42.
Assume \(\Sha\) is finite, let
be a \(K\)-rational isogeny of degree \(d\text{.}\)
Write \(n = \rk_E = \rk_{E'}\text{.}\) Pick a basis \(\Lambda = \langle P_1, \ldots, P_n\rangle\) for
write \(\Lambda'\) for a basis of \(E'(K)/\tors\text{.}\) Write \(\phi^\vee \colon E' \to E\) for the dual isogeny s.t. \(\phi\phi^\vee = \lb d\rb\text{.}\)
using the following fact
Then
Back to the example
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 \(p\text{.}\)
Conjecture 1.43. \(p\)-parity.
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
of abelian varieties over \(K\text{.}\) Let
Recall \(\rk_p = \rk + \delta_p\) where
and
Strategy, we show that for \(\Psi\) an isogeny s.t. \(\Psi \Psi^\vee = \lb p \rb\text{.}\) Then
Remark 1.45.
Let \(A^\vee\) be the dual of \(A\text{.}\) \(A^\vee = \Pic^0(A)\text{.}\)
So
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
Aside: Generalisation of the definition of \(\Sel^n(E/\QQ)\).
Consider
an isogeny of abelian varieties. We have
from which we extract
we then define
so
We want to show:
Theorem 1.46.
Let \(E/K\) be an elliptic curve, \(K\) a number field, if \(\Psi \) is s.t. \(\Psi \Psi^\vee = \lb p \rb\) then
We will show this in 3 parts, first the left, then the right, then the equality with the global root number.
Step 1.
Proposition 1.47.
Proof.
Note that
hence
now
Proof of this: For each generator \(R\) of \(E(K)/E(K)_\tors\) then
are not in the image of \(\lb p\rb\) which implies the size is \(p^{\rk(E/K)}\text{.}\) Also
since
and since \(\lb p \rb\) is trivial on all
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
Step 2.
We show that
Theorem 1.48.
Let \(A,B / K \) be abelian varieties given with a non-zero global exterior form \(\omega_A, \omega_B\text{.}\) Suppose
is an isogeny and
its dual.
Let \(\Sha_0(A/K)\) denote \(\Sha(A/K)\) mod its divisible part. And
Then
Remark 1.49.
If \(A = E\text{,}\) \(B =E'\) with \(\Psi\) s.t. \(\Psi \Psi^\vee = \lb p \rb \) then
and \(|\Sha_0| = \square\text{.}\)
Sketch proof of theorem.
We show how to obtain the quotient of Tamagawa numbers, for a sufficiently large set of places \(S\) of \(K\)
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
Snake lemma gives
and
Also
Moreover since \(E, E'\) are isogenous we have
hence since
similarly for \(E'\text{.}\) We have
Hence
Step 3.
We need to show that
i.e. we need to show that
Strategy:
and relate
to \(w_v\) for \(v\nmid \infty\) and
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
\(v\) missing \(p\text{.}\) More precisely
where \(\psi_v\) is the map induced by \(\psi\) on \(E(K_v)\text{.}\)
What about \(v | p\) to extract
from
at finite places we can use a diagram involving
If \(v\nmid p \) then \(| \coker | = 1\) since then on the level of the formal group \(\psi\) induces a map
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 \(\in \ints_K\text{.}\)
If \(v | p\) then \(\coker\) contributes to the snake lemma and at that place
for a particular choice of \(\omega\text{.}\)
Proving \(p\)-parity.
To prove the \(p\)-parity conjecture
We will show that
by relating
and \(w_v\) at some place \(v\nmid p\infty\)
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
if \(v\) is complex \(|\ker \psi_v| = p\) \(|\coker \psi_v | = 1\text{.}\) so
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.
- \(|\ker \psi_v | = p\) (the \(p\)-tors in \(\ker \psi\) are real)
- \(|\ker \psi_v | = 1\) (the \(p\)-tors in \(\ker \psi\) are not real)
Moreover \(|\coker \psi | = 1\) always, \(\sgn(\Delta_E) = \sgn(\Delta_{E'})\)
More generally if \(\deg \Psi\) is odd then
since \(\lb \CC:\RR\rb = 2 \) is coprime to \(E\lb \psi\rb\) (see Atiyah's book).
In the first case
In the second case
For \(K\) a local field let \(F = K(\ker \psi_v)\) noting that
from its action on points in \(\ker \psi = F/K\) is cyclic.
Consider the composition
and denote
the image of \(-1\) under the above map.
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
with a 2-isogeny \(\psi/K\)
by translating 2-torsion to (0,0)
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 \(E'\)
by snakey
let \(n(E), n(E')\) be the number of real connected components \(n = E(\RR) / E^0(\RR)\)
By the third column
now \(| \coker \psi_v^0| = 1\) as the map on identity component is surjective. hence
Recall: to prove the \(2\)-parity conjecture for \(E/K\)
missed
Notation
\(\disc(q_1(x)) = \delta\) \(\disc(q_2(x)) = 16b\)
a) If \(\delta \gt 0\text{,}\) \(b \gt 0\) then \(E,E'\) both have two real components, \(n(E) = n(E') =2\text{.}\)
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 \(a \gt 0\text{.}\)
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 \(n(E) = 2,n(E') = 1\text{.}\)
since \(b \lt 0\) and \(b = \alpha\beta\text{.}\)
so no correction if \(\delta \gt 0, b \lt 0\text{.}\)
c) If \(\delta \lt 0,b \gt 0\text{,}\) \(n(E) = 1\text{,}\) \(n(E') =2\text{.}\)
and
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.
First guess
Recall: let \(K\) be a local field
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
if \(n\) odd and unit is not a square.
So guess
works over \(\RR\text{.}\)
\(v\nmid 2 \infty\) need to show that
if \(E\) has good reduction at \(v\text{.}\)
Need to show that
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
by Tates algorithm
we show that
Recall
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
here \(w_v = -1\) and
need to show that
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 \((0,0)\)
so \(s_1 = -s_2\) and \(s_1 s_2 = -a\) implies \(s_1^2 = a\text{.}\)
so \(s_1 \in k^\times\) then \(a\in {k^{\times}}^2\)
as both are units.
Now \(b = \alpha\beta \equiv 0 \pmod {\pi_K}\) so
same Taylor expansion gives
so \(s_3 = -s_4 \) and \(s_3 s_4 = 2a\text{,}\) \(s_3 ^2 = -2a\) hence
split multiplicative
So we should use this Hilbert symbol instead, it doesn't change the real case.
If \(E\) has non-split multiplicative reduction
done since \(a,-2a\) precisely not squares.
What are these invariants purely in theory?