On the maximal abelian pro-$p$ extension unramified outside $p$ and the Leopoldt's conjecture

Solution 1:

Now I have nearly clearing everything up. Let me write them as an answer.

Problem: Let $K$ be a number field and $p$ be prime number. Let $M$ be the maximal abelian pro-$p$ extension of $K$ unramified outside $p$. Describe $\mathrm{Gal}(M|K)$ and calculate its $\mathbb{Z}_p$-rank.

For any modulus $\mathfrak{m}$, we have the exact sequence (given in this post) $$ 0 \to (\mathcal O_K/\mathfrak m)^{\times}/\mathcal O_K^{\times} \xrightarrow{\alpha} \mathrm{Cl}_{\mathfrak m} \to \mathrm{Cl}_K \to 0 \quad\quad (\dagger). $$

Here the map $\alpha$ can be regarded as the composition $$ \mathcal O_K^{\times} \xrightarrow{\alpha^{\flat}} (\mathcal O_K/\mathfrak m)^{\times} \rightarrow (\mathcal O_K/\mathfrak m)^{\times} / \mathrm{image}(\alpha^{\flat}). $$

Now consider the modulus $\mathfrak{m}_n := p^n$ in $K$ for each integer $n$. Let $H_{p,n}$ be the class field of $\mathfrak{m}_n$ over $K$. Then $\mathrm{Cl}_{\mathfrak{m}_n} \cong \mathrm{Gal}(H_{p,n} | K)$ and every places outside the ones lying above $p$ in $K$ is unramified in $H_{p,n}$. Hence $(\dagger)$ becomes $$ 0 \to (\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times} \to \mathrm{Gal}(H_{p,n} | K) \to \mathrm{Cl}_K \to 0 $$

Then we take the $p$-primary part of each object in the exact sequence and get $$ 0 \to ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}] \to \mathrm{Gal}(H_{p,n}| K)[p^{\infty}] \to \mathrm{Cl}_K[p^{\infty}] \to 0 \quad\quad (\star) $$ Here firstly, the functor $[p^{\infty}]$ is left exact since it is the right adjoint of the inclusion functor from the category of $p$-primary abelian groups to the category of abelian groups. Then shrinking the category of abelian groups to torsion abelian groups, we see that it is right exact as well. (This follows from the first comment by @Mindlack, or by the observation that in the category of torsion abelian groups, $-[p^{\infty}] = -\otimes_{\mathbb{Z}} \mathbb{Z}_p$, the latter one is right exact. See this post.)

Then we take the inverse limit in the exact sequence $(\star)$ and get $$ 0 \to \lim ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}] \to \lim (\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]) \to \lim \mathrm{Cl}_K[p^{\infty}] \to 0 \quad\quad (\star\star). $$

Indeed, the functor $\lim$ is left exact. Moreover, since the leftmost term in $(\star)$ is finite (being included in $\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]$) , it satisfies the Mittag-Leffler condition. Hence $\lim$ is right exact here as well.

And note that $\lim \mathrm{Cl}_K[p^{\infty}] = \mathrm{Cl}_K[p^{\infty}]$, once we write explicity the inverse system out. Hence $$ 0 \to \lim ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}] \to \lim (\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]) \to \mathrm{Cl}_K[p^{\infty}] \to 0 \quad\quad (\star\star). $$

So it remains to deal with the remaining two terms.

• For the $\lim (\mathrm{Gal}(H_{p,n}| K)[p^{\infty}])$, we break it step by step:

1. Dealing with $\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]$. Note that $G_n := \mathrm{Gal}(H_{p,n}| K)$ is a finite abelian groups, we can decompose it as $$ G_n \cong G_n[p^{\infty}] \cup_{\ell \neq p} G_n[\ell^{\infty}], $$ where in this case, $G_n[p^{\infty}]$ is the $p$-Sylow subgroup of $G_n$. Consider the subextension $M_{p,n}$ of $H_{p,n}$ fixed by $\cup_{\ell \neq p} G_n[\ell^{\infty}]$. Then $M_{p,n} | K$ has Galois group isomorphic to $G_n[p^{\infty}]$. This is then the maximal $p$-subextension of $H_{p,n}$. Then all places in $K$ outside the ones lying over $p$ in $K$ are unramified in $M_{p,n}$ as they are in $H_{p,n}$. In this way, we have obtained a maximal $p$-subectension $M_{p,n}$ of $H_{p,n}$ such that every place outside $p$ is unramified, and its Galois group $\mathrm{Gal}(M_{p,n}|K) = \mathrm{Gal}(H_{p,n}| K)[p^{\infty}]$.

2. Now we take the limit of $\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]$ with respect to $n$, by Galois theory, we obtain that $$ \lim (\mathrm{Gal}(H_{p,n}| K)[p^{\infty}]) = \lim \mathrm{Gal}(M_{p,n}|K) = \mathrm{Gal}(M|K). $$ Therefore, the middle term in $(\star\star)$ is merely $\mathrm{Gal}(M|K)$.

• For the limit $\lim ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}]$, I turn to the following exact sequence first $$ \mathcal O_K^{\times} \to (\mathcal O_K/(p^n \mathcal{O}_K))^{\times} \to (\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times} \to 1. $$

Here we have invoked the convention in the shaded part in this answer. So the sequence may not be left exact.

Then tensoring this with $\mathbb{Z}_p$ over $\mathbb{Z}$ we get $$ \mathcal O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p \to (\mathcal O_K/(p^n \mathcal{O}_K))^{\times}\otimes_{\mathbb{Z}} \mathbb{Z}_p \to ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})\otimes_{\mathbb{Z}} \mathbb{Z}_p \to 1. $$ Since it is a right exact functor. Again by this post, we see that for torsion groups $M$, $M[p^{\infty}] = M \otimes_{\mathbb{Z}} \mathbb{Z}_p$. So the right two terms can be modified (while $O_K^{\times}$ may not be a torsion group by Dirichlet's unit theorem, it will remain unchanged) as $$ \mathcal O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p \to (\mathcal O_K/(p^n \mathcal{O}_K))^{\times}[p^{\infty}]\to ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}] \to 1. $$ And taking inverse limit, we get $$ \mathcal O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p \to \lim(\mathcal O_K/(p^n \mathcal{O}_K))^{\times}[p^{\infty}] \to \lim ((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}/\mathcal O_K^{\times})[p^{\infty}] \to 1. \quad\quad (\star\star\star) $$ Again we note that the leftmost term $O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p$ satisfies the Mittag-Leffler condition, since it is finite (as the right two terms are). Hence here taking inverse limit preserves right exactness.

So to describe the third term, we shall compute the second term in $(\star\star\star)$. We claim that $$\lim((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}[p^{\infty}]) = \prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1}, $$ where $U_{\mathfrak p}^{1}$ is the subgroup of $\mathcal{O}_{K, \mathfrak p}^{\times}$ consisting of elements congruent to $1$ modulo $p$.

Proof of the claim: we compute \begin{align*} \lim((\mathcal O_K/(p^n \mathcal{O}_K))^{\times}[p^{\infty}]) &= \lim\left( \prod_{\mathfrak{p} | p} \left((\mathcal O_K/(\mathfrak{p}^{e_{\mathfrak{p}}n}))^{\times}[p^{\infty}] \right) \right) \\ &= \prod_{\mathfrak{p} | p} \lim\left((O_K/(\mathfrak{p}^{e_{\mathfrak{p}}n}))^{\times}[p^{\infty}]\right) \\ &= \prod_{\mathfrak{p} | p} \lim((O_K/(\mathfrak{p}^{e_{\mathfrak{p}}n}))^{\times}) [p^{\infty}] \\ &= \prod_{\mathfrak{p} | p} (\lim (O_K/(\mathfrak{p}^{e_{\mathfrak{p}}n})))^{\times} [p^{\infty}] \\ &= \prod_{\mathfrak{p} | p} (\mathcal{O}_{K,\mathfrak{p}})^{\times} [p^{\infty}] \\ &= \prod_{\mathfrak{p} | p} (1+ \varpi_{\mathfrak{p}} \mathcal{O}_{K,\mathfrak{p}})) \\ &= \prod_{\mathfrak{p} | p} U_{\mathfrak{p}}^{1}. \end{align*} Here:

• 1st equality: Use the multiplicative version Chinese remainder theorem and interchange $\prod$ and $[p^{\infty}]$,
• 2nd equality: Interchange $\lim$ and $\prod$,
• 3rd equality: Interchange $\lim$ and $[p^{\infty}]$. Since $[p^{\infty}]$ is right adjoint, this can be done (note that the resulting $\lim$ does not escape from the category of $p$-primary abelian groups),
• 4th equality: Interchange $\lim$ and $(-)^{\times}$. Note that the functor $(-)^{\times}$ from the category of commutative rings to abelian groups is right adjoint to the functor taking any abelian group $A$ to its group ring $\mathbb{Z}[A]$. So it indeed preserves $\lim$,
• 5th equality: From the definition of completion. Note that the ramification index $e_{\mathfrak{p}} := e(\mathfrak{p} | p)$ will not affect the result. The rest two equalities follow from definitions. So the claim is proved.

So to sum up, we obtain $$ 1 \rightarrow \prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1} / (O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p) \rightarrow \mathrm{Gal}(M|K) \rightarrow \mathrm{Cl}_K[p^{\infty}] \rightarrow 1. $$ and $$ O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p \rightarrow \prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1} \rightarrow \mathrm{Gal}(M|K) \rightarrow \mathrm{Cl}_K[p^{\infty}] \rightarrow 1. $$ There are no major distinction among the two. The first one is neater and the second one is helpful when calculating $\mathbb{Z}_p$-rank and concrete examples.

Calculating $\mathbb{Z}_p$-ranks: It suffices to wandering around the second exact sequence above. Let $r_p$ be the $\mathbb{Z}_p$-rank of the image of $O_K^{\times} \otimes_{\mathbb{Z}} \mathbb{Z}_p$ in $\prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1}$. Then $$ \mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M|K)) = \mathrm{rank}_{\mathbb{Z}_p}(\prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1}) - r_p + \mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Cl}_K[p^{\infty}]). $$ Now, $\mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Cl}_K[p^{\infty}])=0$ since $$ \mathrm{Cl}_K[p^{\infty}] \otimes_{\mathbb{Z}_p} \mathbb{Q}_p = 0. $$ Moreover, \begin{align*} \left(\prod_{\mathfrak p \mid p} U_{\mathfrak p}^{1}\right) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p &= \prod_{\mathfrak p \mid p} \left( (1+ \varpi_{\mathfrak{p}} \mathcal{O}_{K,\mathfrak{p}}) \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \right) \\ &\cong \prod_{\mathfrak p \mid p} \left( \mathcal{O}_{K,\mathfrak{p}} \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \right) \\ &\cong \prod_{\mathfrak p \mid p} \left( \mathbb{Z}_p^{e_{\mathfrak{p}}f_{\mathfrak{p}}} \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \right) \\ &=\mathbb{Q}_p^{\sum_{\mathfrak p \mid p}e_{\mathfrak{p}}f_{\mathfrak{p}}} \\ &=\mathbb{Q}_p^{[K:\mathbb{Q}]}. \end{align*} Here in the first line, we interchanged $\prod$ and ${-}_{\mathbb{Z}_p} \mathbb{Q}_p$. In the last line, we used the fundamental equality $\sum_{\mathfrak p \mid p}e_{\mathfrak{p}}f_{\mathfrak{p}} = [K:\mathbb{Q}]$. Hence, to sum up, $$ \mathrm{rank}_{\mathbb{Z}_p}(\mathrm{Gal}(M|K)) = [K:\mathbb{Q}] - r_p, $$ as desired!