Kolyvagin's Conjectures

What about curves $E$ with $r_{E,{\rm an}}\geq 2$? Suppose that $E$ is an elliptic curve over $\mathbb{Q}$ with $r_{E,{\rm an}}\geq 2$. In the short paper [Kol91], Kolyvagin states an amazing structure theorems for Selmer groups assuming the following unproved conjecture, which is the appropriate generalization of the condition that $P_1$ has infinite order.

Conjecture 3.32 (Kolyvagin [Kol91])   Let $E$ be any elliptic curve over $\mathbb{Q}$ and fix a prime $\ell\in B(E)$ and a prime power $M=\ell^n$ of $\ell$. Then there is at least one cohomology class $c_{\lambda} \in \H^1(K, E[M])$ that is nonzero.

So far nobody has been able to show that Conjecture 3.32 is satisfied by every elliptic curve $E$ over $\mathbb{Q}$, though several people are currently working hard on this problem (including Vatsal and Cornut). Proposition 3.28 above implies that Conjecture 3.32 is true for elliptic curves with $r_{E,{\rm an}}\leq 1$.

Kolyvagin also goes on in [Kol91] to give a conjectural construction of a subgroup

$$V \subset E(K)/E(K)_{\tor }$$

for which $\rank (E(\mathbb{Q})) = \rank (V)$. Let $\ell$ be an arbitrary prime, i.e., so we do not necessarily assume $\ell\in B(E)$. One can construct cohomology class $c_{\lambda} \in \H^1(K, E[M])$, so long as $\lambda \in \Lambda^{n + k_0}$, where $\ell^{k_0/2} E(\mathbf{K})[\ell^\infty] = 0$, and $\mathbf{K}$ is the compositum of all class field $K_{\lambda}$ for $\lambda \in \Lambda$. For any $n\geq 1$, $k\geq k_0$, and $r\geq 0$, let

$$V_{n,k}^r \subset \varinjlim_{m}\H^1(K,E[\ell^m])/E(K)_{\tor }$$

be the subgroup generated by the images of the classes $\tau_{\lambda} = \tau_{\lambda,n}\in\H^1(K,E[\ell^n])$ where $\lambda$ runs through $\Lambda_{n+k}^r$.

Conjecture 3.33 (Kolyvagin)   Let $E$ be any elliptic curve over $\mathbb{Q}$. Then for all prime numbers $\ell$, there exists an integer $r$ such that for all $k\geq k_0$ there is an $n$ such that $V_{n,k}^r\neq 0$.

Recall that

$$n(p) = \ord _\ell(\gcd(p+1, a_p)) \geq 1$$

and

$$n(\lambda) = \min_{p\mid \lambda} n(p).$$

Let $m'(\lambda)$ be the maximal nonnegative integer such that $P_{\lambda} \in \ell^{m'(\lambda)} E(K_{\lambda})$. Let $m(\lambda) = m'(\lambda)$ if $m'(\lambda) < n(\lambda)$, and $m(\lambda) = \infty$ otherwise. For any $r\geq 0$, let

$$m_r = \min \{ m(\lambda) : \lambda \in \Lambda^r \},$$

and let $f$ be the minimal $r$ such that $m_r$ is finite.

Proposition 3.34   We have $f=0$ if and only if $y_K$ has infinite order.

Let $SD = \ell^n S$, where

$$S = \varinjlim_{n} \Sel ^{(\ell^n)}(K,E[\ell^n]) .$$

If $A$ is a $\mathbb{Z}[1,\sigma]$-module and $\varepsilon = (-1)^{r_{E,{\rm an}}-1}$. then

$$A^v = \{b \in A : \sigma(b) = (-1)^{v+1} \varepsilon b\}$$

Assuming his conjectures, Kolyvagin deduces that for every prime number $\ell$ there exists integers $k_1$ and $k_2$ such that for any integer $k\geq k_1$ we have

$$\ell^{k_2} SD^{(f+1)}[M] \subset V_{n,k}^f \subset SD^{(f+1)}[M].$$

Here the exponent of $f+1$ means the $+1$ or $-1$ eigenspace for the conjugation action.

Conjecture 3.35 (Kolyvagin)   Let $E$ be any elliptic curve over $\mathbb{Q}$ and $\ell$ any prime. There exists $v\in \{0,1\}$ and a subgroup

$$V \subset (E(K)/E(K)_{\tors })^{(v)}$$

such that

$$1 \leq \rank (V) \equiv v\pmod{2}.$$

Let $a = \rank (V) - 1$. Then for all sufficiently large $k$ and all $n$, one has that

$$V_{n,k}^a \equiv V \mod{\ell^n (E(K)/E(K)_{\tor })}.$$

Assuming the above conjecture for all primes $\ell$, the group $V$ is uniquely determined by the congruence condition in the second part of the conjecture. Also, Kolyvagin proves that if the above conjecture is true, then the rank of $E^v(\mathbb{Q})$ equals the rank of $V$, and that ${\mbox{{\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}\selectfont Sh}}}(E^v/\mathbb{Q})[\ell^\infty]$ is finite. (Here $E^v$ is $E$ or its quadratic twist.)

When $P_1$ has infinite order, the conjecture is true with $v = 1$ and $V=\mathbb{Z}P_1$. (I think here $E$ has $r_{E,{\rm an}}=0$.)