Asking myself how to construct extensions of fields where primes split completely, I came across this answer. I have problems to understand one part of the proof of the Lemma in the answer. Let's state it already in a more general way:
Lemma. Let $V$ be a DVR with quotient field $K$. Let $p \in V$ be a prime element of $V$ and let $K_p$ be the completion of $K$ with respect to the $p$-adic valuation on $K$. Let $L$ be a finite Galois extension of $K$ such that there exists a $K$-embedding $\sigma: L \to K_p$. Then $p$ splits completely in $L$.
My proof attempt goes along the same lines as the proof in the answer above: Let $G$ be the Galois group of $L$ over $K$ and $n = [L:K] = |G|$. Then $\{ \sigma \circ g \mid g \in G \}$ is a set of $n$ distinct $K$-embeddings of $L$ into $K_p$, each of which induces a prime of $L$ lying over $p$.
What I cannot see: Why are all these primes distinct so that $p$ splits completely? Of course, conversely, each of these primes $P$ induces a $K$-embedding of $L$ into $K_p$. But why is this $K$-embedding equal to the initial $K$-embedding $\sigma \circ g$ that induced $P$?
Thank you for your help!
First of all, there is no need to single out a choice of prime element. I'd prefer to speak of the valuation $v$ on $K$ associated to $V$ and write the completion as $K_v$.
In any case, for a Galois extension $L/K$, all extensions of $v$ to valations on $L$ have the same $e$ (ramification index) and same $f$ (residue field degree). Therefore it suffices to show there is some valuation $w$ on $L$ extending $v$ with $e(v|w) = 1$ and $f(v|w) = 1$ to know that's true of all valuations on $L$ extending $v$. When you have a $K$-embedding $L \to K_v$, the $v$-adic closure of $L$ in $K_v$ (under this embedding) is $K_v$: you have $K \subset L \subset K_v$ and $K$ is $v$-adically dense in $K_v$, so the $v$-adic closure of $L$ in $K_v$ is forced to be $K_v$.
The valuation $w$ imposed on $L$ from its embedding into $K_v$ is an example of a valuation on $L$ that extends $v$. And since $L_w = K_v$, $[L_w:K_v] = 1$. At the same time, $[L_w:K_v] = e(v|w)f(v|w)$, so $e(v|w) = 1$ and $f(v|w) = 1$. Note $e$ and $f$ are unchanged by completing, so we can check they are 1 by computing them for the extension of completions at the corresponding valuations.