Let $K$ and $L$ be number fields, $v$ a place of $K$ (either archimedean or non-archimedean) and $\theta:K\simeq L$ a ring isomorphism.
I am trying to show that $\theta$ induces an isomorphism $K_v\simeq L_{\theta(v)}$ where the subscripts denote completions.
Am I right in saying that $\mid x\mid_v=\mid\theta(x)\mid_{\theta(v)}$ for all $x\in K$, so that if $\alpha\in K_v$ is represented by a Cauchy sequence $\{\alpha_n\}\subset K$ then $\{\theta(\alpha_n)\}$ is a Cauchy sequence in $L$ which converges in $L_{\theta(v)}$?
Then would $\alpha \mapsto \operatorname{lim}_n \theta(\alpha_n)$ be the isomorphism I want?
How does one define $\theta(v)$ if $v$ is archimedean?
Many thanks.
Yes that's right. The isomorphism is indeed given by the formula you suggest, and is extended by continuity via limits. The archimedean case works exactly the same. The infinite primes are just embeddings of $K$ (or $L$) into $\Bbb C$ and any isomorphism between them is canonically induced by an element of $\operatorname{Gal}(K/\Bbb Q)$ which is also itself canonically the set of embeddings $K\to\Bbb C$, so $\theta(v)$ for a place $v$ infinite, is really just $\theta\circ v$, i.e. left multiplication in the Galois group (which is a group under composition of functions).