How to calculate Gal$(F(\mu_{p^\infty})/F(\mu_p))$ for a number field $F$?

Question

Let $F$ be a number field. Recall that we define $$F(\mu_{p^\infty})=\bigcup_{n=1}^{\infty}F(\mu_{p^n}).$$ I want to calculate the group Gal$(F(\mu_{p^\infty})/F(\mu_p))$. I know that this is supposed to be equal to $\mathbb{Z}_p$, but I am unable to prove it. Every reference I have seen just states this directly, which means that this is probably really simple to do, but I am unable to see it.

Answer 1

I wouldn't say this isomorphism is "really simple" when learning these things for the first time. It involves an isomorphism in infinite Galois theory and an isomorphism between a certain subgroup of $\mathbf Z_p^\times$ with $\mathbf Z_p$. In particular, the group ${\rm Gal}(F(\mu_{p^\infty})/F(\mu_{p}))$ is not really "directly" isomorphic to $\mathbf Z_p$ is any natural way, but rather this comes out as the result of composing some isomorphisms.

To start off, the statement you made is not quite true since it has counterexamples when $p = 2$: $$ {\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q(\mu_2)) = {\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q) \cong \mathbf Z_2^\times, $$ which is not isomorphic to $\mathbf Z_2$ since it has nontrivial torsion, namely the number $-1$. That $-1$ corresponds to complex conjugation acting on $\mathbf Q(\mu_{2^\infty})$. More generally, when $F$ is a number field with a real embedding, then viewing $F$ in $\mathbf R$ shows complex conjugation is a nontrivial element in ${\rm Gal}(F(\mu_{2^\infty})/F(\mu_{2})) = {\rm Gal}(F(\mu_{2^\infty})/F)$, so this Galois group has an element of order $2$ and thus can't be isomorphic to $\mathbf Z_2$. Maybe you meant to tell us $p$ is odd?

Anyway, the result you ask about is related to how Galois groups behave when forming composite fields: think about $F(\mu_{p^{\infty}})$ as a composite field such as $F \,\mathbf Q(\mu_{p^{\infty}})$, or perhaps better as $F(\mu_p) \,\mathbf Q(\mu_{p^{\infty}})$

When $K$ is a field, $L/K$ is Galois extension, and $E$ is an arbitrary extension of $K$ that lies in a common field with $L$ (e.g., $E/K$ is algebraic with $L$ and $E$ both in a common algebraic closure of $K$), the extension $LE/E$ is Galois and restricting elements of ${\rm Gal}(LE/E)$ to $L$ is an injective homomorphism ${\rm Gal}(LE/E) \hookrightarrow {\rm Gal}(L/K)$. The image is ${\rm Gal}(L/K \cap E)$, so $$ {\rm Gal}(LE/E) \cong {\rm Gal}(L/L \cap E). $$ For infinite Galois groups, this is an isomorphism not just of groups, but of topological groups.

Let's apply this to $K = \mathbf Q(\mu_p)$, $L = \mathbf Q(\mu_{p^\infty})$, and $E = F(\mu_p)$ where $F$ is a number field. Then $LE = F(\mu_{p^\infty})$, so the above isomorphism says $$ {\rm Gal}(F(\mu_{p^\infty})/F(\mu_{p})) \cong {\rm Gal}(\mathbf Q(\mu_{p^\infty})/M) $$ where $M = L \cap E = \mathbf Q(\mu_{p^\infty}) \cap F(\mu_p)$, which is a subfield of $F(\mu_p)$ and thus is a number field. What does the group ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/M)$ look like?

The field $M$ is not just a number field: since $\mu_p \subset M$, we have $\mathbf Q(\mu_p) \subset M$, so ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/M)$ is a subgroup of ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/\mathbf Q(\mu_p))$. Moreover, it's an open subgroup since $[M:\mathbf Q(\mu_p)]$ is finite (that's infinite Galois theory at work: closed subgroups of finite index are also open subgroups). What is ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/\mathbf Q(\mu_p))$ and what are its open subgroups?

Given the context of the question, surely you understand how ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/\mathbf Q) \cong \mathbf Z_p^\times$ for all primes $p$. Inside that group, ${\rm Gal}(\mathbf Q(\mu_{p^\infty})/\mathbf Q(\mu_p))$ corresponds to $1+p\mathbf Z_p$: a $p$-adic unit being used as an exponent on $p$-power roots of unity fixes $\mu_p$ exactly when the exponent is $1 \bmod p$, meaning the unit is in $1 + p\mathbf Z_p$. Thus $$ {\rm Gal}(F(\mu_{p^\infty})/F(\mu_{p})) \cong {\rm Gal}(\mathbf Q(\mu_{p^\infty})/M) = {\rm open \ subgroup \ of \ } 1 + p\mathbf Z_p. $$ This is true for all primes $p$, including $p = 2$. What are the open subgroups of $1+p\mathbf Z_p$? This is where a distinction arises between $p = 2$ and $p > 2$. (I gave a counterexample at the start when $p = 2$, so things have to break at $p = 2$ somewhere, and we've now reached that step.)

When $p$ is odd, $1 + p\mathbf Z_p$ is isomorphic to $\mathbf Z_p$. Here are two ways to set that up: (i) if $u \equiv 1 \bmod p$ and $u \not\equiv 1 \bmod p^2$, e.g., $u = 1 + p$, the map $\mathbf Z_p \to 1 + p\mathbf Z_p$ where $x \mapsto u^x$ is an isomorphism, or (ii) the $p$-adic logarithm is an isomorphism $1+p\mathbf Z_p \to p\mathbf Z_p$, and the latter group is obviously isomorphic to $\mathbf Z_p$. Now the key point about $\mathbf Z_p$ is that its open subgroups are the groups $p^k \mathbf Z_p$ where $k \geq 0$ and those are all isomorphic to $\mathbf Z_p$. So anything isomorphic to an open subgroup of $\mathbf Z_p$ is itself isomorphic to $\mathbf Z_p$. Thus, when $p$ is odd, $$ {\rm Gal}(F(\mu_{p^\infty})/F(\mu_{p})) \cong \mathbf Z_p. $$

When $p = 2$ the previous paragraph breaks down since $1+2\mathbf Z_2 = \mathbf Z_2^\times \not\cong \mathbf Z_2$ (there is a nontrivial element of finite order in $1 + 2\mathbf Z_2$ but not in $\mathbf Z_2$). We can fix things in this case by digging a little deeper: $1+4\mathbf Z_2 \cong \mathbf Z_2$ (e.g., the $2$-adic log is an isomorphism $1+4\mathbf Z_2 \to 4\mathbf Z_2 \cong \mathbf Z_2$) and ${\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q(\mu_4)) = 1 + 4\mathbf Z_2$, which is isomorphic to $\mathbf Z_2$. Thus all open subgroups of $1+4\mathbf Z_2$ are isomorphic to $\mathbf Z_2$, so the correct thing to say when $p = 2$ is that $$ {\rm Gal}(F(\mu_{2^\infty})/F(\mu_{4})) \cong {\rm open \ subgroup \ of \ } {\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q(\mu_4)) \cong \mathbf Z_2, $$ where the last isomorphism can be read as saying ${\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q(\mu_4)) \cong \mathbf Z_2$ or as saying open subgroups of ${\rm Gal}(\mathbf Q(\mu_{2^\infty})/\mathbf Q(\mu_4))$ are isomorphic to $\mathbf Z_2$: both of those statements are true.

The $F$ that are counterexamples to your statement when $p = 2$ (such as those $F$ with a real embedding) and the way it gets corrected confirms the useful advice that often in the $p$-adics, "if $p = 2$ then $p = 4$".