Proof. Assume that M is a pseudo-null submodule of W. Then, by the very definition of what it means to be pseudo-null, we have that $M_{\mathfrak{p}}=0$ for all prime ideals $\mathfrak{p}$ of R of height less than or equal to one. We now show that for any prime ideal $\mathfrak{p}$ of R of height one and any $n >0$ the natural map $\psi\,{:}\, R/\mathfrak{p}^n \to S^{-1}(R/\mathfrak{p}^n)$ where $S=R \setminus \mathfrak{p}$ is an injection. Applying this to the prime ideals $\mathfrak{p}_i$ appearing in the decomposition of W shows that we must have $M=0$ .

To show the claim, assume that for some $x \in R$ we have $\psi(x + \mathfrak{p}^n)=0$ . Then for some $s \in S$ we have $sx \in \mathfrak{p}^n$ . Since R is a UFD, therefore $\mathfrak{p}= \langle y \rangle$ for some $y \in R$ (see [ Reference Matsumura43 , Theorem 20·1]). The element y is necessarily an irreducible(=prime) element of R. So $y^n \mid sx$ . Since $y \nmid s$ and R is a UFD, therefore $y^n \mid x$ which shows that $x \in \mathfrak{p}^n$ .

Proof. To simplify notation, we will denote $X(E/K_{\infty})$ by simply $X_{\infty}$ . Assume that $f_{\infty} \neq 0$ . Taking into account Lemmas 2·9 and 2·1, we have by [ Reference Bourbaki5 , chapter VII, section 4·4 Theorem 5], that there exist irreducible power series $f_j \in {{\mathbb{Z}_p}}[[T, U]]$ , integers $m_i, n_j$ and an exact sequence

$$0 \to W \to X_{\infty} \to B \to 0, $$

where $W=\bigoplus_{i=1}^s \Lambda_2/p^{m_i} \oplus \bigoplus_{j=1}^t \Lambda_2/f_j^{n_j}$ and B is a pseudo-null $\Lambda_2$ -module. From this exact sequence, we get another exact sequence

(1) \begin{equation}B^{\Upsilon=0} \to W/\Upsilon \to X_{\infty}/\Upsilon \to B/\Upsilon \to 0.\end{equation}

Since B is a pseudo-null $\Lambda_2$ -module, therefore it has Krull dimension at most one. So the $\Lambda_2/\Upsilon$ -modules $B/\Upsilon$ and $B^{\Upsilon=0}$ also have Krull dimension at most one and hence they are $\Lambda_2/\Upsilon$ -torsion.

Therefore, it follows from the sequence (1) that $X_{\infty}/\Upsilon$ and $W/\Upsilon$ have the same $\Lambda_2/\Upsilon$ -rank. Hence we must show that $W/\Upsilon$ is $\Lambda_2/\Upsilon$ -torsion if and only if $f_{\infty}$ and $\Upsilon$ are relatively prime. Clearly if $D=\Lambda_2/p^{m_i}$ , then $D/\Upsilon$ is a torsion $\Lambda_2/\Upsilon$ -module. So we see that $W/\Upsilon$ has positive $\Lambda_2/\Upsilon$ -rank if and only if $\Lambda_2/\langle f_j^{n_j}, \Upsilon \rangle$ has positive $\Lambda_2/\Upsilon$ -rank for some j and this is the case, if and only if, for some j, $f_j$ divides $\Upsilon$ (see [ Reference Matsumura43 , Theorem 17·4]).

Proof. If $L \in \mathcal{E}^{\subseteq K_\infty}(K)$ , consider the map (induced by restriction)

$$s\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/L) \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{{{\rm Gal}}(K_{\infty}/L)}.$$

We have that $\ker s$ injects into

\[ H^1({{\rm Gal}}(K_{\infty}/L), E(K_{\infty})[p^{\infty}])=\ker(H^1(G_S(L), E[p^{\infty}]) \to H^1(G_S(K_{\infty}), E[p^{\infty}])).\]

Since $H^1({{\rm Gal}}(K_{\infty}/L), E(K_{\infty})[p^{\infty}])$ is clearly cofinitely generated over ${{\mathbb{Z}_p}}$ , so is $\ker s$ . Therefore by considering the dual of the map $s$ , we see that $X(E/L)$ will be a torsion $\Lambda$ -module if this is the case for $X(E/K_{\infty})_{{{\rm Gal}}(K_{\infty}/L)}$ .

If $f_{\infty}=0$ , then $X(E/K_{\infty})$ is a pseudo-null $\Lambda_2$ -module. Hence it has Krull dimension at most one. Therefore for any $L \in \mathcal{E}^{\subseteq K_\infty}(K)$ , we have that $X(E/K_{\infty})_{{{\rm Gal}}(K_{\infty}/L)}$ has Krull dimension at most one and hence is a torsion ${{\mathbb{Z}_p}}[[{{\rm Gal}}(K_{\infty}/L)]]$ -module. Therefore in this case for all $L \in \mathcal{E}^{\subseteq K_\infty}(K)$ , $X(E/L)$ is a torsion $\Lambda$ -module.

Now assume that $f_{\infty} \neq 0$ . From Lemma 2·2 (and since $\Lambda_2$ is a UFD), we see that the desired result follows from the following lemma.

Proof. Let $\tilde{G}$ be the subgroup topologically generated by $\sigma^a\tau^b$ and $\sigma^c\tau^d$ . Then $\tilde{G}$ has finite index in G. Let $I_{\tilde{G}}$ (resp. $I_G$ ) be the ideal of $\Lambda(\tilde{G})$ (resp. $\Lambda(G)$ ) generated by $\sigma^a\tau^b-1, \sigma^c\tau^d-1$ and p. Since $\tilde{G}$ is topologically generated by $\sigma^a\tau^b$ and $\sigma^c\tau^d$ , therefore $I_{\tilde{G}}$ has finite index in $\Lambda(\tilde{G})$ . As $\tilde{G}$ has finite index in G therefore $\Lambda(G)$ is a finitely generated $\Lambda(\tilde{G})$ -module. These two facts imply that $I_G$ has finite index in $\Lambda(G)$ . By [ Reference Matsumura43 , Theorem 17·4], this in turn implies that the sequence $\sigma^a\tau^b-1, \sigma^c\tau^d-1, p$ is regular. In particular $\sigma^a\tau^b-1$ and $\sigma^c\tau^d-1$ are relatively prime.

Proof. Let ${F = K_\infty^H}$ with ${H \in \mathcal{H}}$ , and let L be given. Suppose $v \in S$ splits completely in $L/K$ . Since v does not split completely in $F/K$ , therefore we see that the decomposition group $D_v$ for a prime of $K_{\infty}$ over v has ${{\mathbb{Z}_p}}$ -rank one. It follows from this that v does not split completely for any $L' \in \mathcal{E}^{\subseteq K_\infty}(K)$ with $L' \neq L$ . Since the set S is finite, (a) excludes finitely many ${L \in \mathcal{E}^{\subseteq K_\infty}(K)}$ . Now let $\mathfrak{p}$ be a prime of K above p. Since $\mathfrak{p}$ ramifies in $F/K$ , therefore the inertia group $I_{\mathfrak{p}}$ of a prime of $K_{\infty}$ above $\mathfrak{p}$ must have ${{\mathbb{Z}_p}}$ -rank greater than or equal to one. It follows from this that (b) excludes at most finitely many ${{\mathbb{Z}_p}}$ -extensions. Proposition 2·3 shows that (c) excludes finitely many extensions. Putting all these results together gives the proposition.

Proof. We must show that:

1. (a) no prime in S splits completely in $K_{cyc}/K$ ;

2. (b) every prime of K above p ramifies in $K_{cyc}/K$ ;

3. (c) $X(E/K_{cyc})$ is a torsion $\Lambda$ -module.

Proof. This follows from [ Reference Greenberg20 , proposition 3·2(ii)].

Proof. For simplicity, let $F_n=K_{\infty}^{H_n}$ and $F = F_0$ . For any n we have a commutative diagram with vertical maps induced by restriction:

By the snake lemma we have an exact sequence

$$0 \to \ker s_n \to \ker g_n \to \ker h_n \cap {\text{img}}\ \theta_n \to {\text{coker}}\ s_n \to {\text{coker}}\ g_n. $$

From this exact sequence we see that $\ker s_n$ (resp. ${\text{coker}}\ s_n$ ) will be finite and bounded independently of n if we show that $\ker g_n$ (resp. ${\text{coker}}\ g_n$ and $\ker h_n$ ) is finite and bounded independently of n.

Let $S_n$ be all the primes of $F_n$ above those in S and $S_{p,n}$ be the primes of $S_n$ above p. We define the following groups:

1. (1) $A_n\,:\!=H^1(H_n, E(K_{\infty})[p^{\infty}])$ ;

2. (2) $B_n\,:\!=H^2(H_n, E(K_{\infty})[p^{\infty}])$ ;

3. (3) $C_n\,:\!=\prod_{w \in S_n \setminus S_{p,n}} H^1(H_{n,w}, E(K_{\infty,w})[p^{\infty}])$ ;

4. (4) $D_n\,:\!=\prod_{w \in S_{p,n}} H^1(H_{n,w}, \tilde{E}(k_{\infty,w})[p^{\infty}])$ .

In (3) and (4) we have also written w for a fixed prime of $K_{\infty}$ above w and $H_{n,w}$ is the decomposition group. $k_{\infty,w}$ is the residue field and $\tilde{E}$ is the reduction of E over the residue field.

Since all primes of K above p ramify in F (because $H \in \mathcal{H}$ ), the results of [ Reference Coates and Greenberg7 ] allow us to write

$$\bigoplus_{v \in S} J_v(E/F_n)= \prod_{w \in S_n\setminus S_{p,n}} H^1(F_{n,w}, E[p^{\infty}]) \times \prod_{w \in S_{p,n}} H^1(F_{n,w}, \tilde{E}[p^{\infty}]). $$

Here $\tilde{E}$ denotes the reduction of the elliptic curve over the residue field. We can similarly do the same for $\oplus_{v \in S} J_v(E/K_{\infty})$ .

From this and Shapiro’s lemma we can write $\ker h_n=C_n \times D_n$ . Also, we have $\ker g_n=A_n$ and ${\text{coker}}\ g_n$ injects into $B_n$ . Therefore, we only need to show that the groups $A_n$ , $B_n$ , $C_n$ and $D_n$ are all finite and bounded independently of n.

First we deal with $A_n$ : Let $W=E(K_{\infty})[p^{\infty}]$ . If $\gamma$ is a topological generator of H, then $A_n=W/(\gamma^{p^n}-1)W$ . The kernel of $\gamma^{p^n}-1$ acting on W is $E(F_n)[p^{\infty}]$ . This is a finite group by Lemma 2·7.

Let $W_{\text{div}}$ be the maximal divisible subgroup of W. Since $W_{\text{div}}$ has finite ${{\mathbb{Z}_p}}$ -corank and the kernel of $\gamma^{p^n}-1$ acting on W is finite, it follows that ${(\gamma^{p^n}-1)W_{\text{div}}=W_{\text{div}}}$ . Therefore we see that $H^1(H_n, W)$ has order bounded by $[W\,{:}\, W_{\text{div}}]$ . This takes care of the group $A_n$ . As for $B_n$ , this group is zero for all n because $cd_p(H_n)=1$ .

Now we deal with $C_n$ : let $w \in S_n \setminus S_{p,n}$ . As primes in S do not split completely in $F/K$ (because $H \in \mathcal{H}$ ), $F_{n,w}/(K_{H,n})_w$ is a ${{\mathbb{Z}_p}}$ -extension. This together with the fact that $(K_{H,n})_w$ has no $\mathbb{Z}_p^2$ -extension implies that $H_{n,w}=0$ . Therefore $C_n=0$ for all n.

Finally we deal with $D_n$ : Let $w \in S_{p,n}$ . Suppose first that the decomposition group of a prime of $K_{\infty}$ above w is an open subgroup of ${{\rm Gal}}(K_{\infty}/K)$ so $H_{n,w}$ is isomorphic to ${{\mathbb{Z}_p}}$ . Since primes of K above p ramify in $F/K$ , we see that the $H_{n,w}$ -invariants of $\tilde{E}(k_{\infty,w})[p^{\infty}]$ are finite. Therefore, as in the proof for $A_n$ , we see that $H^1(H_{n,w}, \tilde{E}(k_{\infty,w})[p^{\infty}])$ is finite and bounded independently of n. Now let ${w \in S_{p,n}}$ be a prime such that the decomposition subgroup of any prime of $K_\infty$ above w is not open in ${{\rm Gal}}(K_\infty/K)$ . Since $\mathcal{H}$ is non-empty, the decomposition group will have ${{\mathbb{Z}}}_p$ -rank one. As there are only finitely many primes in $F_n$ above v (because ${H \in \mathcal{H}}$ ), w splits completely in the ${{\mathbb{Z}}}_p$ -extension $K_\infty/F_n$ , and thus ${H_{n,w} = 0}$ .

Proof. Choose some ${H \in \mathcal{H}}$ , and recall that $X(E/K_\infty^H)$ is a torsion $\Lambda$ -module. In view of Proposition 2·8, it follows that $X(E/K_\infty)_H$ is $\Lambda$ -torsion. Therefore the intended result follows from [ Reference Lim41 , lemma 4·7].

Proof. By Lemma 2·7 we have that $E(K_{\infty}^{H_n})[p^{\infty}]$ is finite for all n. Taking this into account, we have that if $X(E/K_{\infty}^{H_n})$ is a torsion $\Lambda_{K_n}$ -module, then from [ Reference Hachimori and Venjakob25 , Theorem 7·2] we get that the sequence in the statement of the lemma is exact and also $H^2(G_S(K_{\infty}^{H_n}), E[p^{\infty}])=0$ (note that loc. cit. Theorem 7·2 requires p to be odd. However as K was assumed to be totally imaginary if $p=2$ , the proof also works for $p=2$ ). Since for $H \in \mathcal{H}$ we have that $X(E/K_{\infty}^H)$ is a torsion $\Lambda$ -module, we get the desired result for $n=0$ .

Suppose that $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ . To complete the proof, from what we just observed, it will suffice to show that $X(E/K_{\infty}^{H_n})$ is a torsion $\Lambda_{K_n}$ -module. To prove this, we can proceed as in [ Reference Coates and Sujatha10 , proposition 2·5]. Consider the following commutative diagram with exact rows

The map $\hat{s}_n$ is the dual of the map $s_n$ in Proposition 2·8 and $\theta_n$ is induced by $\hat{s}_n$ . By Proposition 2·8, ${\text{coker}}\ \hat{s}_n$ is finite and hence so is ${\text{coker}}\ \theta_n$ . Since $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ and $H_n$ has finite index in H, therefore $(X(E/K_{\infty})_f)_{H_n}$ is finitely generated over ${{\mathbb{Z}_p}}$ . Thus we get that $X(E/K_{\infty}^{H_n})_f$ is finitely generated over ${{\mathbb{Z}_p}}$ . This implies that $X(E/K_{\infty}^{H_n})$ is a torsion $\Lambda_{K_n}$ -module which completes our proof.

Proof. Now proceed as in [ Reference Coates, Schneider and Sujatha8 , lemma 2·3]. Let $H \in \mathcal{H}$ and $n \geq 0$ . For simplicity, let $F_n=K_{\infty}^{H_n}$ . Let $S_n$ be the set of all the primes of $F_n$ above those in S and $S_{p,n}$ be the primes of $S_n$ above p.

Consider the map induced by restriction

$$h_n\,{:}\, \bigoplus_{v \in S} J_v(E/F_n) \to \bigoplus_{v \in S} J_v(E/K_{\infty})^{H_n}. $$

We claim that $h_n$ is surjective. By the same arguments as in the proof of Proposition 2·8, we have that ${\text{coker}}\ h_n$ is a submodule of

$$M_n\,:\!=\prod_{w \in S_n\setminus S_{p,n}} H^2(H_{n,w}, E(K_{\infty,w})[p^{\infty}]) \times \prod_{w \in S_{p,n}} H^2(H_{n,w}, \tilde{E}(k_{\infty,w})[p^{\infty}]). $$

In each factor above, we have also written w for a fixed prime of $K_{\infty}$ above w and $H_{n,w}$ is the decomposition group. $k_{\infty,w}$ is the residue field and $\tilde{E}$ is the reduction of the elliptic curve. Since ${cd_p(H_{n,w}) \le 1}$ , we therefore see that ${M_n = \{0\}}$ , whence $h_n$ is surjective.

Consider the commutative diagram

The exactness of the sequence in the statement of the lemma is equivalent to the surjectivity of $\rho_n$ . Since $h_n$ is surjective, the commutative diagram shows that the surjectivity of $\rho_n$ will follow if $\lambda_n$ is surjective. Therefore, the desired result follows from Lemma 2·10.

Proof. Let $F = K_\infty^H$ for some ${H \in \mathcal{H}}$ . From Lemma 2·10, it follows that $H^2(G_S(F), E[p^{\infty}])=0$ . As $cd_p(H)=1$ , the Hochschild-Serre spectral sequence implies that we have a surjection ${H^2(G_S(F), E[p^{\infty}]) \twoheadrightarrow H^2(G_S(K_\infty), E[p^{\infty}])^{H}}$ . This implies $H^2(G_S(K_{\infty}), E[p^{\infty}])=0$ .

Proof. We proceed as in [ Reference Coates, Schneider and Sujatha8 , lemma 2·4]. Combining [Reference Neukirch, Schmidt and Wingberg48, proposition 8·3·18] and the previous lemma, we get that

$$H^m(G_S(K_{\infty}), E[p^{\infty}])=0, \,\,\, m \geq 2. $$

In view of this result, the Hochschild–Serre spectral sequence (see [ Reference Neukirch, Schmidt and Wingberg48 , lemma 2·1·3]) implies that for any $i \geq 1$ we have an exact sequence

\begin{align}H^{i+1}(G_S(K_{\infty}^{H_n}), E[p^{\infty}]) \to H^i(H_n, H^1(G_S(K_{\infty}), E[p^{\infty}])) \to H^{i+2}(H_n, E(K_{\infty}^H)[p^{\infty}]). \end{align}

The group on the right vanishes because $cd_p(H_n)=1$ . For $i \geq 2$ the group on the left vanishes by [ Reference Neukirch, Schmidt and Wingberg48 , proposition 8·3·18]. By Lemma 2·10, for $i=1$ the group on the left vanishes for $n=0$ and assuming that $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ it vanishes for all n. This implies the result.

Proof. We proceed as in [ Reference Coates, Schneider and Sujatha8 , lemma 2·6]. Let $n \geq 0$ and assume that $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ if $n >0$ . Now consider the sequence

(2) \begin{equation}\displaystyle 0 \longrightarrow {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n} \longrightarrow H^1(G_S(K_{\infty}), E[p^{\infty}])^{H_n} \longrightarrow \bigoplus_{v \in S} J_v(E/K_{\infty})^{H_n} \longrightarrow 0.\end{equation}

We will show that $H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))=0$ if the sequence (2) is exact. Therefore, the lemma will follow from Lemma 2·11.

Assume that (2) is exact. Let

$$A_{\infty}={\text{img}}\,(H^1(G_S(K_{\infty}), E[p^{\infty}]) \to \bigoplus_{v \in S} J_v(E/K_{\infty})), $$

i.e. we have an exact sequence

$$\displaystyle 0 \longrightarrow {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \longrightarrow H^1(G_S(K_{\infty}), E[p^{\infty}]) \longrightarrow A_{\infty} \longrightarrow 0. $$

This exact sequence and Lemma 2·13 imply that we have an exact sequence

\begin{align*} \displaystyle 0 \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n} \to H^1(G_S(K_{\infty}), E[p^{\infty}])^{H_n} \to A_{\infty}^{H_n} \to H^1(H_n, {{{{\rm Sel}}_p}}(E/K_{\infty})) \to 0. \end{align*}

The exactness of the sequence (2) implies that

$$A_{\infty}^{H_n}=\bigoplus_{v \in S} J_v(E/K_{\infty})^{H_n}, $$

whence it is clear that $H^1(H_n, {{{{\rm Sel}}_p}}(E/K_{\infty}))=0$ , as required.

Proof. We proceed as in [ Reference Coates, Schneider and Sujatha8 , proposition 2·9]. Let

$$A_{\infty}={\text{img}}\,(\lambda_{\infty}\,{:}\, H^1(G_S(K_{\infty}), E[p^{\infty}]) \to \oplus_{v \in S} J_v(E/K_{\infty}))$$

i.e. we have an exact sequence

$$\displaystyle 0 \longrightarrow {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \longrightarrow H^1(G_S(K_{\infty}), E[p^{\infty}]) \longrightarrow A_{\infty} \longrightarrow 0. $$

Choose an element ${H \in \mathcal{H}}$ , and let ${F = K_\infty^H}$ . It then follows from the above sequence and from Lemma 2·13 that

$$H^1(H, A_{\infty}) \cong H^2(H, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})), $$

whence $H^1(H, A_{\infty})=0$ because $cd_p(H)=1$ .

Now let $B_{\infty}={\text{coker}}\ \lambda_{\infty}$ . Taking the $H_{cyc}$ -cohomology of the sequence

$$0 \to A_{\infty} \to \bigoplus_{v \in S} J_v(E/K_{\infty}) \to B_{\infty} \to 0$$

and using Lemma 2·11, we get that $B_{\infty}^H$ injects into ${H^1(H, A_{\infty})=0}$ . Therefore, $B_{\infty}=0$ . This implies that $\lambda_{\infty}$ is surjective.

Proof. This follows from a similar argument to the one used in the proof of Proposition 2·8.

Proof. Suppose that $E(K_{\infty})[p^{\infty}]$ is finite. Assume $H \in \mathcal{H}$ and $n \geq 0$ . Denote $K_{\infty}^{H_n}$ by $L_{\infty}$ and $K_{H,n}$ by L. Then $L_{\infty}/L$ is a ${{\mathbb{Z}_p}}$ -extension with tower fields $L_n$ . Let $s_n\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/L_n) \to {{{{\rm Sel}}_{p^{\infty}}}}(E/L_{\infty})^{\Gamma_n}$ be the map induced by restriction. Since $E(L_{\infty})[p^{\infty}]$ is finite with $\#E(L_{\infty})[p^{\infty}] \leq \#E(K_{\infty})[p^{\infty}]$ , we get from Lemma 3·1 that $\ker s_n$ is finite of order at most $\#E(K_{\infty})[p^{\infty}]$ . From this we see that the desired result follows from Theorem 3·2 and Lemma 3·3.

Proof. (1) follows from Zarhin’s Theorem [ Reference Zarhin58 ]. For (2) assume that E has complex multiplication by the ring of integers $\mathcal{O}_L$ of a quadratic imaginary field L, ${[K\,{:}\, {{\mathbb{Q}}}]\le 8}$ and ${p \geq 5}$ . To prove the desired result it will suffice to show that $E(LK_{\infty})[p^{\infty}]=0$ .

Proof. For simplicity, let $S_n={{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}^{H_n})$ and $S_{\infty}={{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})$ . Moreover, we let ${m \ge 0}$ be arbitrary but fixed, and we abbreviate $s_{n,m}'$ to $s_n'$ for convenience. Consider the following commutative diagram

By Proposition 2·8, ${\text{coker}}\ s_n$ is finite and bounded independently of n. Since we have a surjection ${\text{coker}}\ s_n \twoheadrightarrow {\text{coker}}\ s'_n$ , therefore the same is true for ${\text{coker}}\ s'_n$ .

By the snake lemma, we have an exact sequence $\ker s_n \to \ker s'_n \to {\text{coker}}\ \tilde{s}_n$ . Since $\ker s_n$ is finite and bounded independently of n by Proposition 2·8, therefore to show that $\ker s'_n$ is finite and bounded independently of n, we only need to show that ${\text{coker}}\ \tilde{s}_n$ is finite and bounded independently of n. To this end, note that ${\text{img}}\ \tilde{s}_n=p^m\, {\text{img}}\ s_n$ . Therefore the multiplication by $p^m$ map induces a surjection from ${\text{coker}}\ s_n$ onto ${\text{coker}}\ \tilde{s}_n$ . This implies that ${\text{coker}}\ \tilde{s}_n$ is finite and the order is bounded independently of n, because this property is true for ${\text{coker}}\ s_n$ by Proposition 2·8.

Proof. Assume that $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ . For simplicity, let $F_n=K_{\infty}^{H_n}$ . Fix ${n \in {{\mathbb{N}}}}$ and let ${m > 0}$ be such that ${X(E/K_{\infty})[p^{\infty}]=X(E/K_{\infty})[p^m]}$ and ${X(E/F_n)[p^{\infty}]=X(E/F_n)[p^m]}$ . Consider the commutative diagram with exact rows

From the snake lemma applied to this diagram, we get an exact sequence

(3) \begin{equation}\ker \phi_n \to \ker \hat{s}_n \to \ker \theta_n \to {\text{coker}}\ \phi_n \to {\text{coker}}\ \hat{s}_n \to {\text{coker}}\ \theta_n \to 0.\end{equation}

By Proposition 2·8, both $\ker \hat{s}_n$ and ${\text{coker}}\ \hat{s}_n$ are finite. This, together with the exact sequence (3), implies that ${\text{coker}}\ \theta_n$ is finite. Also since $\ker \hat{s}_n$ is finite, the exact sequence (3) shows that in order to show that $\ker \theta_n$ is finite it will suffice to show that ${\text{coker}}\ \phi_n$ is finite. We now proceed to show this.

Since $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ and $H_n$ has finite index in H, therefore $(X(E/K_{\infty})_f)_{H_n}$ is finitely generated over ${{\mathbb{Z}_p}}$ . Therefore $\ker \theta_n$ is finitely generated over ${{\mathbb{Z}_p}}$ . From this fact and the fact that ${\text{coker}}\ \hat{s}_n$ is finite, we see from the exact sequence (3) that ${\text{coker}}\ \phi_n$ is actually finite (it is finitely generated over ${{\mathbb{Z}_p}}$ and annihilated by $p^m$ ).

From now on suppose that condition $\mathcal{C}_{H,m}$ is met, and choose m accordingly. We want to show that the orders of $\ker \theta_n$ and ${\text{coker}}\ \theta_n$ are bounded independently of n. By Proposition 2·8, the cardinalities of $\ker \hat{s}_n$ and ${\text{coker}}\ \hat{s}_n$ are bounded independently of n. Therefore the exact sequence (3), implies that the order of ${\text{coker}}\ \theta_n$ is bounded independently of n. Also since $\ker \hat{s}_n$ is finite and bounded independently of n, the exact sequence (3) shows that in order to show that the order of $\ker \theta_n$ is bounded independently of n, it will suffice to show that the order of ${\text{coker}}\ \phi_n$ is bounded independently of n.

Dualising $\phi_n$ , we get

$$\hat{\phi}_n\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/F_n)/p^m \to ({{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})/p^m)^{H_n}. $$

We will show that the order of $\ker \hat{\phi}_n$ is bounded independently of n. Consider the exact sequence:

$$0 \to p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})/p^m \to 0. $$

For any n, this sequence gives us an exact sequence

$$0 \to (p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))^{H_n} \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n} \to ({{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})/p^m)^{H_n}. $$

Therefore we have an injection

$$\varphi_n\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}/(p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))^{H_n} \hookrightarrow ({{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})/p^m)^{H_n}. $$

We define the map

$$\psi_n\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/F_n)/p^m \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}/(p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))^{H_n}. $$

Since $\varphi_n$ is an injection, therefore $\ker \psi_n=\ker (\varphi_n \circ \psi_n) = \ker \hat{\phi}_n$ . So we see that we need to show that the order of $\ker \psi_n$ is bounded independently of n. Consider the exact sequence

$$0 \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m] \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \xrightarrow{\times p^m} p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}) \to 0. $$

This sequence induces an exact sequence

\begin{align*} 0 \to p^m({{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}) \to (p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))^{H_n} \to H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m]) \to 0. \end{align*}

We have a zero on the right because $H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))=0$ from Lemma 2·14. This exact sequence gives us an exact sequence

(4) \begin{equation}0 \to H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m]) \to \frac{{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}}{p^m} \xrightarrow{\pi_n} \frac{{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}}{(p^m{{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty}))^{H_n}} \to 0. \end{equation}

Let $s'_n\,{:}\, {{{{\rm Sel}}_{p^{\infty}}}}(E/F_n)/p^m \to {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})^{H_n}/p^m$ be the map in Proposition 4·1. Then $\psi_n = \pi_n \circ s'_n$ . It is easy to see that we have an exact sequence

$$0 \to \ker s'_n \to \ker \psi_n \to \ker \pi_n \to {\text{coker}}\ s'_n. $$

Taking the exact sequence (4) into account, the above exact sequence becomes

$$0 \to \ker s'_n \to \ker \psi_n \to H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m]) \to {\text{coker}}\ s'_n. $$

Recall that $\ker \psi_n$ is finite for all n. We need to show its order is bounded. According to Proposition 4·1, both $\ker s'_n$ and ${\text{coker}}\ s'_n$ are finite and bounded independently of n. Therefore by the above exact sequence, $H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m])$ is finite for all n and we need to show that its order is bounded independently of n. The Pontryagin dual of $H^1(H_n, {{{{\rm Sel}}_{p^{\infty}}}}(E/K_{\infty})[p^m])$ is $(X(E/K_{\infty})/p^m)^{H_n}$ . Since $X(E/K_{\infty})$ is finitely generated over the Noetherian ring $\Lambda_2$ , therefore it follows that $X(E/K_{\infty})/p^m$ is a Noetherian $\Lambda_2$ -module. The $\Lambda_2$ -submodules $(X(E/K_{\infty})/p^m)^{H_n}$ form an increasing nested chain, the chain must stabilise and so the orders are bounded. This completes the proof.

Proof. Let $H \in \mathcal{H}$ and assume that the hypotheses in the statement of the proposition are true. As usual, let $F_n=K_{\infty}^{H_n}$ . From Lemma 2·10 we get that $X(E/F_n)$ is a torsion $\Lambda_{K_n}$ -module for all n. From Theorem 3·4 it follows that there exists $t >0$ such that for all n the maximal finite $\Lambda_{K_n}$ -submodule of $X(E/F_n)$ is annihilated by $p^t$ . Also since $X(E/K_{\infty})$ is finitely generated over the Noetherian ring $\Lambda_2$ , therefore $X(E/K_{\infty})[p^{\infty}]=X(E/K_{\infty})[p^m]$ for some $m \geq t$ . We will show that $X(E/F_n)[p^{\infty}]=X(E/F_n)[p^{2m}]$ for all n.

For any $n \geq 0$ , let $Y(E/F_n)=X(E/F_n)/X(E/F_n)[p^m]$ . Then we have a commutative diagram

To see that the first map on the top row is an injection, we first note that since $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ we have by Lemma 2·14 that $H_1(H_n, X(E/K_{\infty}))=0$ . But $X(E/K_{\infty})_f \cong p^mX(E/K_{\infty})$ which is a submodule of $X(E/K_{\infty})$ . It follows that $H_1(H_n, X(E/K_{\infty})_f)=0$ because $cd_p(H_n)=1$ . This shows that the first map on the top row is in fact an injection. From the snake lemma applied to above diagram, we get an exact sequence

(5) \begin{equation}0 \to \ker \phi_n \to \ker \hat{s}_n \to \ker \theta_n \to {\text{coker}}\ \phi_n \to {\text{coker}}\ \hat{s}_n \to {\text{coker}}\ \theta_n \to 0.\end{equation}

By Proposition 2·8, $\ker \hat{s}_n$ is finite. This, together with the exact sequence (5), implies that $\ker \phi_n$ is finite. Since $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ and $H_n$ has finite index in H, therefore $(X(E/K_{\infty})_f)_{H_n}$ is finitely generated over ${{\mathbb{Z}_p}}$ . Therefore, for all n, $\ker \theta_n$ is finitely generated over ${{\mathbb{Z}_p}}$ . Combining this with the fact that ${\text{coker}}\ \hat{s}_n$ is finite by Proposition 2·8, the exact sequence (5) shows that ${\text{coker}}\ \phi_n$ is finitely generated over ${{\mathbb{Z}_p}}$ and hence is finite since the target group is annihilated by $p^m$ .

So we have shown that the map

$$\phi_n\,{:}\, X(E/K_{\infty})[p^m]_{H_n} \to X(E/F_n)[p^m]$$

has finite kernel and cokernel. In an identical way we can show that the map

$$\phi'_n\,{:}\, X(E/K_{\infty})[p^{\infty}]_{H_n} \to X(E/F_n)[p^{\infty}]$$

has finite kernel and cokernel. Combining this with the fact that $X(E/K_{\infty})[p^{\infty}]=X(E/K_{\infty})[p^m]$ , we get that $\mu_{G_n/H_n}(X(E/F_n)[p^{\infty}])=\mu_{G_n/H_n}(X(E/F_n)[p^m])$ .

Taking Lemma 2·1 into account, we have an exact sequence

(6) \begin{equation}0 \to \bigoplus_{i=1}^s \Lambda(G_n/H_n)/p^{m'_i} \to X(E/F_n)[p^{\infty}] \to B \to 0,\end{equation}

where B is finite. This exact sequence induces the sequence

(7) \begin{equation}0 \to \bigoplus_{i=1}^s (\Lambda(G_n/H_n)/p^{m'_i})[p^m] \to X(E/F_n)[p^m] \to B[p^m].\end{equation}

Since we have $\mu_{G_n/H_n}(X(E/F_n)[p^{\infty}])=\mu_{G_n/H_n}(X(E/F_n)[p^m])$ we see from the exact sequences (6) and (7) that we must have $m'_i \le m$ for $i=1,...,s$ .

The exact sequence (6) induces an exact sequence

(8) \begin{equation}0 \to A' \to X(E/F_n)[p^{\infty}] \to \bigoplus_{i=1}^s \Lambda(G_n/H_n)/p^{m'_i} \to B' \to 0,\end{equation}

where A’ and B’ are finite. By assumption, $p^t$ annihilates A’ and $m \geq t$ . Also $m'_i \le m$ for $i=1,...,s$ . Therefore it follows that $p^{2m}$ annihilates $X(E/F_n)[p^{\infty}]$ , i.e. $X(E/F_n)[p^{\infty}]=X(E/F_n)[p^{2m}]$ . This completes the proof.

Proof. We follow the proof of [ Reference Coates, Schneider and Sujatha8 , proposition 2·12] closely. For simplicity, let $F_n=K_{\infty}^{H_n}$ with $F=F_0$ . Also let $X_{\infty}=X(E/K_{\infty})$ . First, recall from [ Reference Howson27 , corollary 1·7] that if M is a finitely generated $\Lambda(G)$ -module, then

$$p^{\mu_G(M)}=\prod_{i \geq 0} \# H_i(G, M[p^{\infty}])^{(-1)^i} = \chi(G, M[p^\infty]). $$

We have a similar formula when M is a finitely generated $\Lambda(G/H)$ -module. From the Hochschild–Serre spectral sequence it is easy to prove that

$$\chi(G, X_{\infty}[p^{\infty}])=\prod_{i=0,1}\chi(G/H, H_i(H, X_{\infty}[p^{\infty}]))^{(-1)^i}. $$

From this we get

(9) \begin{equation} \mu_G(X_{\infty})=\mu_{G/H}(H_0(H, X_{\infty}[p^{\infty}]))-\mu_{G/H}(H_1(H, X_{\infty}[p^{\infty}])).\end{equation}

Consider the exact sequence

(10) \begin{equation}0 \to X_{\infty}[p^{\infty}] \to X_{\infty} \to X_{\infty, f} \to 0.\end{equation}

Since $cd_p(H)=1$ , therefore the functor $H_1(H,-)$ is left exact. On the other hand, one has that $X[p^{\infty}] \subseteq X_{\infty}$ and $X_{\infty,f} \cong p^t X_{\infty} \subseteq X_{\infty}$ for some big enough t. Therefore, combining the above observations with Lemma 2·14, we have that $H_1(H, X_{\infty}[p^{\infty}])=H_1(H, X_{\infty,f})=0$ . This fact implies that we have a short exact sequence

(11) \begin{equation}0 \to H_0(H, X_{\infty}[p^{\infty}]) \to H_0(H, X_{\infty}) \to H_0(H, X_{\infty,f}) \to 0.\end{equation}

Taking (9) into account, we obtain

(12) \begin{equation}\mu_G(X_{\infty})=\mu_{G/H}(H_0(H, X_{\infty}))-\mu_{G/H}(H_0(H, X_{\infty,f})).\end{equation}

But the quantity $\mu_{G/H}(H_0(H, X_{\infty}))$ is precisely $\mu_{G/H}(X(E/F))$ by virtue of the dual descent map $(X_{\infty})_H \to X(E/F)$ having finite kernel and cokernel (cf. Proposition 2·8). Therefore we get the desired formula for $n=0$ .

Now assume that $X_{\infty, f}$ is finitely generated over $\Lambda(H)$ . Then from Lemma 2·14 we get that $H_1(H_n, X_{\infty})=0$ . Also, Lemma 2·10 gives that $X(E/F_n)$ is a torsion $\Lambda(G_n/H_n)$ -module. Therefore, the same proof above may be adapted to show that for any $n \geq 0$ that

(13) \begin{equation}\mu_{G_n}(X_{\infty})=\mu_{G_n/H_n}(X(E/F_n))-\mu_{G_n/H_n}(H_0(H_n, X_{\infty,f})).\end{equation}

But since $X_{\infty,f}$ is finitely generated over $\Lambda(H)$ and $H_n$ has finite index in H, therefore $(X_{\infty,f})_{H_n}$ is finitely generated over ${{\mathbb{Z}_p}}$ . In particular, this quotient is a torsion $\Lambda(G_n/H_n)$ -module and has $\mu_{G_n/H_n}$ -invariant zero. We see from this that the formula (13) becomes

$$\mu_{G_n}(X_{\infty})=\mu_{G_n/H_n}(X(E/F_n)). $$

We also have kind of a converse to Theorem 5·1.

Proof. According to the previous theorem, this equality implies that $\mu_{G/H}((X(E/K_{\infty})_f)_H)=0$ . According to Proposition 2·8, the map $\alpha\,{:}\, X(E/K_{\infty})_H \to X(E/K_{\infty}^H)$ has finite kernel and cokernel. Since by definition of $\mathcal{H}$ , $X(E/K_{\infty}^H)$ is a torsion $\Lambda(G/H)$ -module, we get that $X(E/K_{\infty})_H$ is also a torsion $\Lambda(G/H)$ -module. It follows that $(X(E/K_{\infty})_f)_H$ is also a torsion $\Lambda(G/H)$ -module. Therefore, since $\mu_{G/H}((X(E/K_{\infty})_f)_H)=0$ , we get that $(X(E/K_{\infty})_f)_H$ is finitely generated over ${{\mathbb{Z}_p}}$ . This implies that $X(E/K_{\infty})_f$ is finitely generated over $\Lambda(H)$ .

Remark.

1. (1) We stress that the sum in (b) is taken over all ${H \subseteq G}$ of dimension 1, not only over ${H \in \mathcal{H}}$ . This is one of two main reasons for us to prove the asymptotic formulas (14) and (15) also for ${H \not\in \mathcal{H}}$ (under the assumption 2), although this will require to prove a stronger control theorem (see Proposition 6·8 below). The second motivation to study also the case ${H \not\in \mathcal{H}}$ comes from Section 10 below: we will apply Theorem 6·3 to the anticyclotomic ${{\mathbb{Z}}}_p$ -extension $K_{ac}$ of an imaginary quadratic number field. However, we cannot ensure that the corresponding subgroup H of G, fixing $K_{ac}$ , is contained in $\mathcal{H}$ , because some of the primes in S might split completely in $K_{ac}$ . Therefore it will be very useful to have Theorem 6·3(a) available also for ${H \not\in \mathcal{H}}$ .

2. (2) If $X_f$ is finitely generated as a $\Lambda(H)$ -module, as in part (c) of the theorem, then the general hypothesis that each $X_n$ is finitely generated and torsion over $\Lambda_{n}$ is automatically satisfied, by Lemma 2·10.

3. (3) Suppose that ${H \in \mathcal{H}}$ and that $X_f$ is finitely generated over $\Lambda(H)$ . Then Theorem 5·1 implies that actually $\mu_n = \mu_G(X) p^n$ , and in particular ${m_1 = 0}$ , in equation (14). Moreover, we will show in Theorem 8·1 below that in this case $\lambda_n = l p^n$ in equation (15).

Proof. This is [ Reference Cuoco12 , lemma 2·8].

Proof. This follows from [ Reference Cuoco12 , proposition 2·1 and remark on p. 430].

Proof. Fix a pseudo-isomorphism ${\varphi\,{:}\, X \longrightarrow M}$ , where M is an elementary torsion $\Lambda_2$ -module. We let N and R denote the kernel and the image of $\varphi$ . Then N and $M/R$ are pseudo-null $\Lambda_2$ -modules.

Recall the definitions (16) and (17) of $\eta_n$ and ${\alpha_{n,0} = {\eta_n}/{U}}$ . Now we define ${Y_n = \eta_n \cdot X}$ for every ${n \in {{\mathbb{N}}}}$ , and we let ${W_n = \varphi(Y_n)}$ , ${n \in {{\mathbb{N}}}}$ . Then

\[ W_n = \alpha_{n,0} \cdot W_0\]

for each ${n \in {{\mathbb{N}}}}$ , since $\varphi$ is a $\Lambda_2$ -module homomorphism. The map $\varphi$ induces surjections

\[ \overline{\varphi_n} \,{:}\, X/Y_n \twoheadrightarrow R/W_n \]

with kernel $(N + Y_n)/Y_n$ for each ${n \in {{\mathbb{N}}}}$ .

We will prove a control theorem below (see Proposition 6·8) which is independent from the results of Cuoco and which implies that the kernels and the cokernels of the canonical maps

\[ X/Y_n = X/\eta_n X \longrightarrow X(E/F_n) = X_n \]

are torsion ${{\mathbb{Z}}}_p[[T]]$ -modules, where we recall that $X(E/F_n)$ denotes the Pontryagin dual of the Selmer group of E over ${F_n = K_\infty^{H_n}}$ (when $H \in \mathcal{H}$ it follows from Proposition 2·8 that the kernels and cokernels are even finite and of bounded order). In particular, since $X(E/F_n)$ is a torsion ${{\mathbb{Z}}}_p[[T]]$ -module by assumption, the same holds true for the quotient $X/Y_n$ , ${n \in {{\mathbb{N}}}}$ . Therefore also $R/W_n$ and $(N + Y_n)/Y_n$ (i.e. the kernel and the image of $\overline{\varphi_n}$ ) are torsion ${{\mathbb{Z}}}_p[[T]]$ -modules. Moreover, the additivity of $\mu$ - and $\lambda$ -invariants on short exact sequences of finitely generated torsion ${{\mathbb{Z}}}_p[[T]]$ -modules imply that

\[ \mu(X/Y_n) = \mu(R/W_n) + \mu((N + Y_n)/Y_n)\]

and

\[ \lambda(X/Y_n) = \lambda(R/W_n) + \lambda((N + Y_n)/Y_n). \]

Now we can apply Lemma 6·5 in order to conclude that the $\mu$ -and $\lambda$ -invariants of the quotients $(N+ Y_n)/Y_n$ stabilise. Moreover,

\[ \mu(R/W_n) = \mu(R/W_0) + \mu(W_0/\alpha_n W_0)\]

for every ${n > 0}$ , and a similar equation holds for the $\lambda$ -invariants.

Note that ${W_0 \subseteq M}$ is a finitely generated and torsion $\Lambda_2$ -module, and ${W_0/\alpha_n W_0 \subseteq R/W_n}$ is a finitely generated and torsion ${{\mathbb{Z}}}_p[[T]]$ -module for each ${n > 0}$ . Therefore by Lemma 6·6 and the above we conclude that for sufficiently large n we have

\[\mu(X/Y_n) = \mu_G(X) p^n + m_1 n + c_1\]

and

\[\lambda(X/Y_n) = l p^n + c_2\]

for suitable constants $m_1$ , l, $c_1$ and $c_2$ .

By Proposition 6·8 we have that

\[ |\lambda(X/Y_n) - \lambda(X_n)|\]

and

\[ |\mu(X/Y_n) - \mu (X_n)|\]

are both bounded independently of n. Combining this fact with the equations for $\mu(X/Y_n)$ and $\lambda(X/Y_n)$ above, we get equations (14) and (15).

Proof. If $H \in \mathcal{H}$ , we get the desired result from Proposition 2·8. Otherwise if $H \notin \mathcal{H}$ , we assume that $\mathcal{H} \neq \emptyset$ and that the decomposition group of any prime of $K_{\infty}$ above p is an open subgroup of ${{\rm Gal}}(K_{\infty}/K)$ . In this more general setting the only concern is that one has to take more care to prove that $C_n$ and $D_n$ (using the notation in the proof of Proposition 2·8) are finite when primes in $S_n \setminus S_{p,n}$ split completely in $F_n/k_n$ and when not all primes in $S_{p,n}$ ramify in $F_n/k_n$ .

Now consider the map

$$h_n\,{:}\, \bigoplus_{v \in S} J_v(E/F_n) \to \bigoplus_{v \in S} J_v(E/K_{\infty}).$$

By the proof of Proposition 2·8 we see that:

1. (1) the Pontryagin dual of $H^1({{\rm Gal}}(K_{\infty}/F_n), E(K_{\infty})[p^{\infty}])$ surjects onto ${\text{coker}}\ \phi_n$ ;

2. (2) $\ker \phi_n$ injects into a quotient of the Pontryagin dual of $\ker h_n$ .

Let $\gamma$ be a topological generator of ${{\rm Gal}}(K_{\infty}/F_n)$ . We have $H^1({{\rm Gal}}(K_{\infty}/F_n), E(K_{\infty})[p^{\infty}])=E(K_{\infty})[p^{\infty}]/(\gamma-1)E(K_{\infty})[p^{\infty}]$ . Therefore ${\text{corank}}_{{{\mathbb{Z}_p}}}(H^1({{\rm Gal}}(K_{\infty}/F_n), E(K_{\infty})[p^{\infty}])) \leq 2$ . It then follows from (1) above that ${\text{coker}}\ \phi_n$ is a torsion ${{\mathbb{Z}_p}}[[T]]$ -module with $\mu=0$ and $\lambda \leq 2$ .

Now we may write $\ker h_n=\bigoplus_{v \in S} \ker h_{n, v}$ . We now prove that $\ker \phi_n$ is a torsion ${{\mathbb{Z}_p}}[[T]]$ -module whose $\mu$ - and $\lambda$ -invariants are bounded as n varies. From (2) above it will suffice to show that for each $v \in S$ the Pontryagin dual of $\ker h_{n,v}$ is a torsion ${{\mathbb{Z}_p}}[[T]]$ -module whose $\mu$ - and $\lambda$ -invariants are bounded as n varies.

First, consider a prime $v \in S$ that does not divide p. Assume that v does not split completely in $F/k$ . Then for any sufficiently large n, no prime of $k_n$ above v splits completely in $F_n/k_n$ . It follows then by the same argument as in Proposition 2·8 that we have that $\ker h_{n,v}=0$ .

Now assume that v splits completely in $F/k$ . Let $n \geq 0$ . We have that every prime of $k_n$ above v splits completely in $F_n/k_n$ . Recall from the introduction that we defined $J_v(E/F_n)$ as $J_v(E/F_n)={\mathop{\varinjlim}\limits} \bigoplus_{w|v}H^1(L_w,E)[p^{\infty}]$ where the direct limit runs over finite extensions L of $k_n$ contained in $F_n$ . Had there been a finite number of primes of $F_n$ above v, we could then write $J_v(E/F_n)=\oplus_{w | v} H^1(F_{n,w}, E)[p^{\infty}]$ , where the sum runs over all primes w of $F_n$ above v. This was the case in the proof of Proposition 2·8 and so as in the proof we were able to express $\ker h_{n,v}$ as a finite direct sum of cohomology groups. However as v splits completely in $F_n/k_n$ we cannot write $J_v(E/F_n)$ as a direct sum. This will make the expression of $\ker h_{n,v}$ more complicated. Let l be the rational prime below v. If L is a finite extension of ${{\mathbb{Q}_l}}$ , then by Mattuck’s theorem $E(L) \cong {{\mathbb{Z}_l}}^{[L\,{:}\, {{\mathbb{Q}_l}}]} \times T$ where T is a finite group. Since $l \neq p$ , we get from this that $E(L) \otimes {{\mathbb{Q}_p}}/{{\mathbb{Z}_p}}=0$ . Therefore we have an isomorphism $H^1(L, E)[p^{\infty}] \cong H^1(L, E[p^{\infty}])$ . By taking direct limits, we have such an isomorphism for any algebraic extension L of ${{\mathbb{Q}_l}}$ . Now let $\{F_{n,m}\}_{m \in {{\mathbb{N}}}}$ be the layers of the ${{\mathbb{Z}_p}}$ -extension $F_n/k_n$ . Using the isomorphism just mentioned as well as Shapiro’s lemma we can write

(18) \begin{equation}\ker h_{n,v}= {\mathop{\varinjlim}\limits}_m \bigoplus_{w_m | v} H^1({{\rm Gal}}(K_{\infty,w_m}/(F_{n,m})_{w_m}), E(K_{\infty,w_m})[p^{\infty}]),\end{equation}

where the direct sum runs over the primes $w_m$ of $F_{n,m}$ above v. In the above we have also written $w_m$ for a fixed place of $K_{\infty}$ above $w_m$ . Now let $\{w'_1, w'_2,..., w'_c\}$ be the set of primes of $F_{n,0}=k_n$ above v. Let $1 \leq i \leq c$ . Since $k_{n,w'_i}$ is a finite extension of ${{\mathbb{Q}_l}}$ we have that $E(k_{n,w'_i})[p^{\infty}]$ is finite. So it follows by the same argument as in the proof of Proposition 2·8 that $H^1({{\rm Gal}}(K_{\infty,w'_i}/k_{n,w'_i}), E(K_{\infty,w'_i})[p^{\infty}])$ is finite. We can write

(19) \begin{equation}H^1({{\rm Gal}}(K_{\infty,w'_i}/k_{n,w'_i}), E(K_{\infty,w'_i})[p^{\infty}]) \cong \bigoplus_{j=1}^{t_i} {{\mathbb{Z}}}/p^{m_{i,j}}{{\mathbb{Z}}}.\end{equation}

For any $m \in {{\mathbb{N}}}$ and $w_m | w_0 |v$ , $H^1({{\rm Gal}}(K_{\infty,w_m}/(F_{n,m})_{w_m}), E(K_{\infty,w_m})[p^{\infty}])$ is isomorphic to $H^1({{\rm Gal}}(K_{\infty,w_0}/(F_{n,0})_{w_0}), E(K_{\infty,w_0})[p^{\infty}])$ . Letting $U_{n,v}$ be the Pontryagin dual of $\ker h_{n,v}$ we therefore see that from (18), (19) and [ Reference Kidwell37 , corollary A·8] that we have an isomorphism

$$ U_{n,v} \cong \bigoplus_{i=1}^c\big(\bigoplus_{j=1}^{t_i} {{\mathbb{Z}_p}}[[T]]/p^{m_{i,j}}\big).$$

We see from this that $U_{n,v}$ is a torsion ${{\mathbb{Z}_p}}[[T]]$ -module with $\lambda$ -invariant $\lambda_{n,v}=0$ and $\mu$ -invariant $\mu_{n,v}=\sum_w \text{ord}_p(\#H^1({{\rm Gal}}(K_{\infty,w}/F_{n,w}), E(K_{\infty,w})[p^{\infty}]))$ , where the sum runs over all the primes w of $k_n$ dividing v (for any such w we have also written w for a fixed place of $K_{\infty}$ (and $F_n$ ) above w). Since $\mathcal{H} \neq \emptyset$ , therefore it follows that the decomposition group of a prime v of $K_{\infty}$ in ${{\rm Gal}}(K_{\infty}/K)$ has ${{\mathbb{Z}_p}}$ -rank at least one. So as v splits completely in $F/k$ , it follows that the number of primes of $k_n$ above v is bounded independently of n. Therefore we see from this that $\mu_{n,v}$ is bounded independently of n.

Now let $v \in S$ be a prime above p. By hypothesis, the decomposition group of any prime of $K_{\infty}$ above p is an open subgroup of ${{\rm Gal}}(K_{\infty}/K)$ . Therefore we can write $\ker h_{n,v}=\oplus_{w | v} \ker h_{n,w}$ where $h_{n,w}\,{:}\, H^1(F_{n,w}, E)[p^{\infty}] \to H^1(K_{\infty,w}, E)[p^{\infty}]$ is the restriction map (as above, we have also written w for a fixed place of $K_{\infty}$ above w). We now determine the structure of $\ker h_{n,w}$ as a ${{\mathbb{Z}_p}}$ -module for any such w above v. Let