$\sigma \in$ Gal$(\bar{\mathbb{Q}}/\mathbb{Q})$ induces an isomorphism of local fields $K_v \longrightarrow \sigma(K)_{v^{\sigma}}$ clarification

Question

Let $K$ be a number field and $\sigma \in$ Gal$(\bar{\mathbb{Q}}/\mathbb{Q})$, then it induces an isomorphism $\sigma: K \longrightarrow \sigma(K)$.

This $\sigma$ also identifies sets of absolute values of $K$, $M_K$ and of $\sigma(K)$, $M_{\sigma(K)}$ as $ v \mapsto v^{\sigma}$.

Then the book(Silverman, AEC) says that it is clear $\sigma$ also induces an isomorphism $$ K_v \longrightarrow \sigma(K)_{v^{\sigma}}. $$

Here $K_v$ is the completion of $K$ with respect to $v$ and $\sigma(K)_{v^{\sigma}}$ is the completion of $\sigma(K)$ with respect to $v^{\sigma}$.

I don't understand why it is clear that local fields are also isomorphic. And how is this isomorphism even defined?

I would appreciate it if anyone could help explain this last isomorphism.

Thank you.

Answer 1

The key is the universal property of completions of valuation fields:

Let $(K, v)$ be a valuation field. Then there exists a complete valuation field $(K_v, V)$ and a map $i: (K, v) \longrightarrow (K_v, V)$ (i.e. a field homomorphism respecting the valuation) with dense image. This map is universal in the sense that if $\sigma: (K, v) \longrightarrow (L, w)$ is a map of valued fields with $(L, w)$ complete then there is a unique lift $\Sigma: (K_v, V) \longrightarrow (L, w)$ of $\sigma$ to $K_v$. In other words, $\Sigma \circ i = \sigma$ and is unique with respect to this property.

It's worth pausing to believe this universal property. The map $i: K \longrightarrow K_v$ should be known to you. Take now this map $\sigma: K \longrightarrow L$. Then as $K$ has dense image in $K_v$, every element of $K_v$ is a limit $a = \lim i(a_n)$ of a sequence in $K$. We can then define $\Sigma(a) = \lim \sigma(a_n)$. $\sigma(a_n)$ is Cauchy as all the maps in question preserve valuation, so by completeness $\Sigma(a)$ is defined. Checking well definition and that this is a map of valued fields is something I won't do here.

Now let's return to the problem at hand. We see that $\sigma:K \longrightarrow \sigma[K]$ is an isomorphism of fields. In fact, we put a valuation $v^\sigma$ on $\sigma[K]$ via $v^\sigma(\alpha) = v(\sigma^{-1}(\alpha))$. $\sigma$ will then be a map of valued fields $(K, v) \longrightarrow (\sigma[K], v^\sigma)$. It's annoying to keep writing the latter field, so I'll set $(L, w) = (\sigma[K], v^\sigma)$. In fact, I'll prove in general that any isomorphism of valued fields $\sigma: (K, v) \longrightarrow (L, w)$ induces an isomorphism of their completions.

Consider the composition $K \longrightarrow L \longrightarrow L_w$. This is a map of valued fields, and $L_w$ is complete. Hence, by the universal property stated above, this induces a map $K_v \longrightarrow L_w$. Of course, as $\sigma$ is an isomorphism, we can apply this process to $L \xrightarrow{\sigma^{-1}} K \longrightarrow K_v$ and get a map $L_w \longrightarrow K_v$. These will be inverses by the uniqueness part of the universal proeprty. If you've dealt with universal properties before, this is completely standard. If not, I'll write a bit below to explain this last step.

Let's name these maps. I'll denote $i: K \longrightarrow K_v$, $j: L \longrightarrow L_w$. Let's also call $\overline{\sigma}: K_v \longrightarrow L_w$ and $\overline{\sigma^{-1}}: L_w \longrightarrow K_v$. Now, by the universal property, we have $\overline{\sigma} \circ i = j \circ \sigma$ and $\overline{\sigma^{-1}} \circ j = i \circ \sigma^{-1}$. We first want to show that $\overline{\sigma} \circ \overline{\sigma^{-1}}: L_w \longrightarrow L_w$. By the uniqueness in the universal property, it suffices to show that $(\overline{\sigma} \circ \overline{\sigma^{-1}}) \circ j = id \circ j$. Less abstractly, this latter condition tells us that $\overline{\sigma} \circ \overline{\sigma^{-1}}$ is equal to the identity on $j[L]$, which is dense. Indeed, $(\overline{\sigma} \circ \overline{\sigma^{-1}}) \circ j = \overline{\sigma} \circ i \circ \sigma^{-1} = j$ as desired.

Concretely, the map $\overline{\sigma}$ can be described as follows: let $a = \lim i(a_n) \in K_v$. Then $\overline{\sigma}(a) = \lim j(\sigma(a_n))$. We can act as if $i$ and $j$ were simply the inclusions, which simplifies this definition to be $\overline{\sigma}\left(\lim a_n\right) = \lim \sigma(a_n)$.

EDIT: Here's a diagram to keep all of the compositions straight: