Iwasawa Theory

That the quotient of ${\mathbb{Z}}_p^{\times }$ by its group of torsion elements is isomorphic to ${\mathbb{Z}}_p$ follows from (3.1.1). We then need only remark that any open subgroup of ${\mathbb{Z}}_p$ has the form $p^n{\mathbb{Z}}_p$ for some $n\ge 0$, so is itself isomorphic to ${\mathbb{Z}}_p$. □

The inertia subgroup of $v$ in $\mathrm{\Gamma }=\mathrm{Gal}(F_{\infty }∕F)$ is a closed subgroup of $\mathrm{\Gamma }$ and therefore equal to ${\mathrm{\Gamma }}^{p^n}$ for some $n\ge 0$, unless it is trivial. In the case that $v$ is archimedean, only the latter case is possible as an inertia group at $v$ has order at most $2$ in general. In general, in the former case, $F_n$ is its fixed field, and the completion of $F_n$ at a prime over $v$ has a tamely, totally ramified ${\mathbb{Z}}_p$-extension that is the completion of $F_{\infty }$. On the other hand, the completion of $F_n$ being a characteristic zero local field, such an extension does not exist. □

By Proposition 3.1.7, no prime not over $p$ ramifies in $F_{\infty }∕F$, so if no primes over $p$ ramify, then $F_{\infty }∕F$ would be unramified everywhere. However, the Hilbert class field of $F$ is of finite degree, so this is not possible. That is, there exists a $v$ over $p$ such that that the inertia group at $v$ in $\mathrm{\Gamma }$ is nontrivial, hence equal to some ${\mathrm{\Gamma }}^{p^n}$. □

If $v$ split completely in $F_{\mathrm{cyc}}∕F$, then it would also have to split completely in the extension $F({\mu }_{p^{\infty }})∕F({\mu }_q)$, since $F({\mu }_{p^{\infty }})=F_{\infty }({\mu }_q)$, where $q=p$ for $p$ odd and $q=4$ for $p=2$. But this means that $F_v({\mu }_{p^{\infty }})=F_v({\mu }_q)$, which is to say that $F_v({\mu }_q)$ contains ${\mu }_{p^{\infty }}$, which is impossible.

On the other hand, we know that ${\mathbb{Q}}_{\mathrm{cyc}}∕\mathbb{Q}$ is totally ramified at $p$, so the resulting local extension ${\mathbb{Q}}_{p,\mathrm{cyc}}∕{\mathbb{Q}}_p$ is totally ramified as well. But then the completion of $F_{\mathrm{cyc}}$ at a prime above $v$ is simply the compositum $F_v\cdot {\mathbb{Q}}_{p,\mathrm{cyc}}$, and therefore its intersection with the maximal unramified extension of ${\mathbb{Q}}_p$ must be of finite degree over ${\mathbb{Q}}_p$. In particular, $F_v\cdot {\mathbb{Q}}_{p,\mathrm{cyc}}∕F_v$ has an infinite inertia group, which therefore must have the form ${\mathrm{\Gamma }}_{\mathrm{cyc}}^{p^n}$ for some $n\ge 0$, where ${\mathrm{\Gamma }}_{\mathrm{cyc}}=\mathrm{Gal}(F_{\mathrm{cyc}}∕F)$. □

Since $F_{v,\mathrm{cyc}}∕F_v$ is an unramified ${\mathbb{Z}}_p$-extension by Propositions 3.1.7 and 3.1.9, it is the maximal unramified pro-$p$ extension of $F_v$. It follows that for any prime $w$ of $E$ lying over $v$, we must have $E_w=F_{v,\mathrm{cyc}}$, since the Galois closure of $E_w∕F_v$ is a pro-$p$ extension of $F_v$ containing $F_{v,\mathrm{cyc}}$. □

This is a consequence of Theorem 1.5.7, since Proposition 3.1.7 tells us that the ${\mathbb{Z}}_p$-rank of the maximal abelian $V_{\mathit{𝑝\infty }}$-ramified extension of $F$ is the ${\mathbb{Z}}_p$-rank of $\mathrm{Gal}(\tilde{F}∕F)$. □

Since $L_n∕F_n$ is an $S$-ramified abelian pro-$p$ extension, so is $L_n{F}_{\infty }∕F_{\infty }$. Therefore, $L_n\subseteq L_{\infty }$. We claim that ${\bigcup }_n{L}_n=L_{\infty }$. Let $x\in L_{\infty }$. Then $F_{\infty }(x)∕F_{\infty }$ is an $S$-ramified abelian $p$-extension. Let $y$ be a field generator of the Galois closure of $F(x)$ as an extension of $F$. To show that $x\in L_n$ for some $n$, it therefore suffices to show that $y\in L_n$. Let $m$ be such that $F_n(y)\cap F_{\infty }=F_m$. Then $F_m(y)\cap F_{\infty }=F_m$ as well, and the restriction map

is surjective, so $F_m(y)∕F_m$ is abelian.

Since $L_{\infty }∕F_{\infty }$ is $S$-ramified, and $F_{\infty }∕F$ is $V_p$-ramified, we have that $F_m(y)∕F_m$ is $S\cap V_p$-ramified. If $v$ is a place over $p$ in $F_m$ that is not in $S_{F_m}$, then since $F_{\infty }(y)∕F_{\infty }$ is unramified over $v$, the same must be true of $F_n(y)∕F_n$ for some $n$, and therefore $y\in L_{\infty }$.

It now follows that the inverse limit of restriction maps

is an isomorphism, and since ${\bigcup }_n(F_{\infty }\cap L_n)=F_{\infty }={\bigcup }_n{F}_n$, we have that

is an isomorphism as well, as desired. □

For $n\ge m$, set ${\mathrm{\Sigma }}_{n,m}={\mathrm{\Sigma }}_{F_n∕F_m}$ in the notation of Theorem 1.3.14, which also provides exact sequences fitting into commutative diagrams

of ${\mathbb{Z}}_p[{\mathrm{\Gamma }}_m]$-modules for $n^{\prime }\ge n$. Let $I_v^{(m)}$ denote the inertia group at $v$ in ${\mathrm{\Gamma }}^{p^m}$, which can only be nontrivial for $v\in V_p$ which do not split completely in $F_{\infty }∕F$, and let

be the natural map given by inclusion and product. In the inverse limit over $n$, we obtain exact sequences

Note that $\ker {\mathrm{\Sigma }}^{(m)}$ is finitely generated over ${\mathbb{Z}}_p$ and $A_m$ is finite. By Nakayama’s Lemma, we see that $X_{\infty }$is a finitely generated $\mathrm{\Lambda }$-module. Moreover, we see that ${(X_{\infty })}_{{\mathrm{\Gamma }}^{p^m}}$ is of bounded ${\mathbb{Z}}_p$-rank for all $m$. Were $X_{\infty }$ to have nontrivial $\mathrm{\Lambda }$-rank, then since there would exist a pseudo-isomorphism from $X_{\infty }$ to the direct sum $M$ of ${\mathrm{\Lambda }}^r$ and a torsion module, the ranks of ${(X_{\infty })}_{{\mathrm{\Gamma }}^{p^m}}$ would necessarily have been unbounded, since ${\mathrm{\Lambda }}_{{\mathrm{\Gamma }}^{p^m}}\cong {\mathbb{Z}}_p[{\mathrm{\Gamma }}_m]$, and

has finite cokernel. □

Let $N_n:X_{\infty }\to A_n$ be the inverse limit of norm maps $N_{n^{\prime },n}$ for $n^{\prime }\ge n$. Let us use $Y_n$ to denote the kernel of $N_n$, which is a $\mathrm{\Lambda }$-submodule of $X_{\infty }$ that is pseudo-isomorphic to $X_{\infty }$.

Fix $m$ sufficiently large such that every prime over $p$ that ramifies in $F_{\infty }∕F_m$ is totally ramified. In particular, we have that $N_m$ is surjective. We consider $n\ge m$. Let $S_n$ be the set of primes (over $p$) in $F_n$ that ramify in $F_{\infty }$, and hence are totally ramified, and note that $|S_n|=|S_m|\ne 0$ by Lemma 3.1.8. Then the inertia group at $v\in S_n$ in ${\mathrm{\Gamma }}^{p^n}$ is ${\mathrm{\Gamma }}^{p^n}$ itself.

Let $L_n$ be the maximal unramified abelian pro-$p$ extension of $F_n$, and let $L_{\infty }$ be the maximal unramified abelian pro-$p$ extension of $F_{\infty }$. We have $X_{\infty }\cong \mathrm{Gal}(L_{\infty }∕F_{\infty })$ and $A_n\cong \mathrm{Gal}(L_n∕F_n)$, so $Y_n\cong \mathrm{Gal}(L_{\infty }∕L_n{F}_{\infty })$. Let $E$ be the maximal unramified $p$-extension of $F_n$ in $L_{\infty }$. Since any $v\in S_n$ is totally ramified in $F_{\infty }∕F_n$, we have $E\cap F_{\infty }=F_n$. Since $E{F}_{\infty }∕F_{\infty }$ is abelian, this tells us that $E∕F_n$ is abelian as well. Thus, $E$ is equal to the maximal unramified abelian $p$-extension of $F_n$ in $L_{\infty }$. Consequently, $\mathrm{Gal}(L_{\infty }∕L_n)$ is topologically generated by the inertia groups $J_v^{(n)}$ in $\mathrm{Gal}(L_{\infty }∕F_n)$ for primes $v\in S_n$, and $Y_n$ is the intersection of the latter group with $\mathrm{Gal}(L_{\infty }∕F_{\infty })$, i.e., it consists of those elements which restrict trivially to $\mathrm{\Gamma }$.

In other words (for $n=m$), we have that $Y_m$ is topologically generated as a pro-$p$ group by elements $g=\sigma {\tau }^{-1}\in \mathrm{Gal}(L_{\infty }∕F_{\infty })$, where $\sigma \in J_v^{(m)}$ and $\tau \in J_w^{(m)}$ for primes $v,w\in S_m$ are such that $\sigma$ and $\tau$ both restrict to ${\gamma }^{p^m}$ for a fixed topological generator $\gamma$ of $\mathrm{\Gamma }$. We can compute the action of the element ${\omega }_{n,m}={\sum }_{i=0}^{p^{n-m}-1}{\gamma }^{p^{m}i}$ on $g$ as follows:

As the elements ${\sigma }^{p^{n-m}}{\tau }^{-p^{n-m}}$ topologically generate $Y_n$, this implies that ${\omega }_{n,m}{Y}_m=Y_n$.

Since $A_n=X_{\infty }∕Y_n$, we conclude that

for all $n\ge m$. Since $Y_m$ is pseudo-isomorphic to $X_{\infty }$, we have $\lambda =\lambda (Y_m)$ and $\mu =\mu (Y_m)$. Since $|X_{\infty }∕Y_m|$ is a constant power of $p$, Theorem 2.4.7 yields the result. □

As in the proof of Theorem 3.2.13, we let $Y_n$ denote the kernel of the inverse limit of norm maps $N_n:X_{\infty }\to A_n$ for each $n\ge 0$. We showed that there exists $m\ge 0$ sufficiently large so that $N_n$ is surjective and ${\omega }_{n,m}{Y}_m=Y_n$ for all $n\ge m$. We consider a directed system of short exact sequences with morphisms as in the following diagram

for $n^{\prime }\ge n\ge m$. Since $X_{\infty }∕Y_m$ is a finite $\mathrm{\Lambda }$-module, in the direct limit we obtain isomorphisms

Since $Y_m$ injects into $X_{\infty }$ with finite cokernel, Proposition 2.6.10 yields that the natural map $\alpha (X_{\infty })\to \alpha (Y_m)$ is an injective pseudo-isomorphism. Since $A_{\infty }^{\vee }\mathit{\cong 𝛼}(Y_m)$, the final statement follows from Theorem 2.6.13. □

Let $M_{\infty }$ be such that ${𝔛}_{\infty }=\mathrm{Gal}(M_{\infty }∕F_{\infty })$, and define $M_n$ for $n\ge 0$ by

We then have that ${({𝔛}_{\infty })}_{{\mathrm{\Gamma }}_n}\cong \mathrm{Gal}(M_n∕F_{\infty })$, and therefore we have an exact sequence

Since any archimedean place splits completely in a ${\mathbb{Z}}_p$-extension, we have $r_2(F_n)=p^n{r}_2(F)$ and hence ${\mathrm{rank}}_{{\mathbb{Z}}_p}{𝔛}_n=p^n{r}_2(F)+1+{\delta }_n$. It follows that ${\mathrm{rank}}_{{\mathbb{Z}}_p}{({𝔛}_{\infty })}_{{\mathrm{\Gamma }}_n}=p^n{r}_2(F)+{\delta }_n$ for all $n$. Since ${\delta }_n$ is bounded in $n$, the result then follows from Proposition 2.4.12. □

Assume first that $F$ contains ${\mu }_{2p}$. Let $r={\mathrm{rank}}_{{\mathbb{Z}}_p}{E}_n$, and choose units such that ${\alpha }_1,\dots ,{\alpha }_r\in {𝒪}_{F_n}^{\times }$ generate $E_n$ modulo its torsion subgroup as a ${\mathbb{Z}}_p$-module and such that the images of ${\alpha }_{{\delta }_n+1},\dots ,{\alpha }_r$ under

generate ${\iota }_n(E_n)$ modulo its torsion subgroup.

Let $p^k$ be the exponent of the $p$-power torsion in ${\iota }_n(E_n)$. Then, for each $1\le i\le {\delta }_n$, there exist $a_{\mathit{𝑖𝑗}}\in {\mathbb{Z}}_p$ for each ${\delta }_n+1\le j\le r$ such that

has trivial $p^k$th power. Fix $l\ge 1$. For every $i$ and $j$ as above, choose $b_{\mathit{𝑖𝑗}}\in \mathbb{Z}$ such that

and then set

It follows that ${\iota }_n{({\beta }_i)}^{p^k}\in {\iota }_n{(E_n)}^{p^{k+l}}$ for each $i$.

Since ${\alpha }_1,\dots ,{\alpha }_r$ form a ${\mathbb{Z}}_p$-linear basis of the maximal $p$-torsion-free quotient of $E_n$, the images of the elements ${\beta }_1,\dots ,{\beta }_{{\delta }_n}$ in $F_n^{\times }∕F_n^{\times p^l}$ generate a subgroup isomorphic to ${(\mathbb{Z}∕p^l\mathbb{Z})}^{{\delta }_n}$. By Kummer theory, the group $F_n^{\times }\cap F_{\infty }^{\times p}$ is exactly ${\mu }_{p^{\infty }}{F}_n^{\times p}$, and since the closed subgroup of $E_n$ generated by ${\beta }_1,\dots ,{\beta }_{{\delta }_n}$ is $p$-torsion-free, the images of these elements generate a subgroup of $F_{\infty }^{\times }∕F_{\infty }^{\times p^l}$ that is also isomorphic to ${(\mathbb{Z}∕p^l\mathbb{Z})}^{{\delta }_n}$.

Now consider

and note that $\mathrm{Gal}(K∕F_{\infty })$ is isomorphic to ${(\mathbb{Z}∕p^l\mathbb{Z})}^{{\delta }_n}$. Since ${\iota }_n{({\beta }_i)}^{p^k}\in {\iota }_n{(E_n)}^{p^{l+k}}$ and ${\beta }_i^{1∕p^l}$ is a $p^{l+k}$th root of ${\beta }_i^{p^k}$, we have that every prime of $v$ over $p$ splits completely in this extension. Since $\mathrm{Gal}(K∕F_{\infty })$ is already a quotient of ${𝔛}_{\infty }$, it is then a quotient of $X_{\infty }^{\prime }$. In other words, we have surjections

for every $l$. Since $X_{\infty }^{\prime }$ is $\mathrm{\Lambda }$-torsion, Proposition 2.2.13 tells us that ${\delta }_n\le \lambda (X_{\infty }^{\prime })$.

Now, if $F$ does not contain ${\mu }_{2p}$, we still have $\delta (F_n)\le \delta (F_n({\mu }_{2p}))$, and since the latter numbers are bounded, we have the result. □

This is simply the inverse limit of the sequences of Theorem 1.5.4 for the fields $F_n$, which remains exact as the modules in question are profinite. The assumptions are simply to insure that the number of terms in the direct sum of local unit or multiplicative groups is finite: otherwise, one need merely replace the direct sums by inverse limits of direct sums at the finite level. □

Recall from Theorem 1.4.5 that we have exact sequences

where $A_n^{\prime }=A_{F_n,S}$. The direct limit as $n$ heads towards infinity yields

For any abelian group $B$, with respect to the maps $B∕p^{m}B\to B∕p^{m+1}B$ induced by multiplication by $p$, we have

where the maps are the natural inclusion maps on the right-hand side of the middle term. Applying this to $B={𝒪}_{F_{\infty },S}^{\times }$ and noting that

since $A_{\infty }^{\prime }$ is $p$-power torsion, we have the lower exact sequence. (Note that this did not require $F_{\infty }$ to be the cyclotomic ${\mathbb{Z}}_p$-extension).

Now, note that $H^1(G_{F_{\infty },S},{\mu }_{p^{\infty }})$ is isomorphic via Kummer theory to the direct limit of the groups $B_{n,m}∕F_n^{\times p^m}$, where $B_{n,m}$ is the subgroup of $x\in F_n^{\times }$ such that $x{𝒪}_{F_n,S}={𝔞}^{p^m}$ for some fractional ideal $𝔞$ of ${𝒪}_{F_n,S}$. We then have maps