Completion of a number field is a finite extension of the same completion of rational numbers Let $K$ be a number field and $v$ an absolute value in $K$. Then, the restriction $v|_{\mathbb Q}$ is an absolute value in $\mathbb Q$ and we can consider the completions $K_v$ and $\mathbb Q_v$. My question is: is it true that the extension $K_v/\mathbb Q_v$ is also finite?
I'm reading Silverman's book "Arithmetic on Elliptic Curves" and he uses this fact to define the local degree of $K$ as $n_v = [K_v: Q_v]$ and for his purpose, it seems that this number needs to be fnite. I've never studied Algebraic Number Theory and would like to know if this is true and have some explanation about it. Thank you very much!
 A: Yes. This is pretty standard in a not-first course in algebraic number theory. In particular, for a finite algebraic extension $K$ of a number field (or even separable extension $K$ of "global field") $k$, for a "place/completion" $v$ of $k$, and for $w$ ranging over the extensions of $v$ to $K$,
$$
[K:k] \;=\; \sum_{w|v} [K_w:k_v]
$$
As they say, "the sum of the local degrees is the global degree". :)
The proof of this stronger assertion is not too complicated:
$$
\dim_k K \;=\; \dim_{k_v}(K\otimes_k k_v)
$$
by general facts about extending scalars on vector spaces, and
with $K=k(\alpha)$ (a "primitive element", which exists because the extension is finite and separable), with $f$ the irreducible monic poly in $k[x]$ with $\alpha$ a $0$,
$$
K\otimes_k k_v \;\approx\; \big(k[x]/\langle f(x)\rangle\big)\otimes_k k_v
\;\approx\; k_v[x]/\langle f(x)\rangle
\;\approx\; \bigoplus k_v[x]/\langle f_i\rangle
$$
where the $f_i$ are the irreducible monic factors of $f$ in $k_v[x]$.
In particular, no gritty stuff. :)
