Relation between decomposition field and complete decomposition of a prime ideal In my lecture on Algebraic Number Theory in the chapter about Galois extensions we had the following statement:
$$ G_{\mathfrak{p}}= \{id\} \iff Z_{\mathfrak{p}}=L  \iff p~ \text{is completely decomposed} $$
where $A$ is a Dedekind domain, $K$ its quotient field, $L/K$ a finite Galois extension of degree $n$, $\mathcal{O}_L$ the integral closure of $A$ in $L$, $G=\mathrm{Gal}(L/K)$, $\mathfrak{p}$ a non-zero prime ideal in  $\mathcal{O}_L$ over $p$ which is a non-zero prime ideal in $A$. $G_{\mathfrak{p}}$ denotes the decomposition group of $\mathfrak{p}$ and $Z_{\mathfrak{p}}$ the corresponding decomposition field.
I do understand the first $\Leftrightarrow$, but have unfortunately some trouble to show (and understand) the second $\Leftrightarrow$.
My ideas so far were (for $\Rightarrow$):  If $G_{\mathfrak{p}}= \{id\}$, then the number of prime ideals in $\mathcal{O}_L$ above $p$ is $n$ as $(G:G_{\mathfrak{p}})=\vert G \vert= [L:K]=n$ and $G$ acts transitively on the set of all prime ideals over $p$. This gives $p \mathcal{O}_L= ({\mathfrak{p}_1} \cdot \dotsc \cdot \mathfrak{p}_n)^e$ as the decomposition of $p$ with $e$ the ramification degree. Now you should somehow get that $e$ equals one (as is required in the definiton of completely decomposed) but I just don't see why this follows? 
For the converse: no idea.
I would be very thankful for any hints and/or solutions!

Since $G_{\mathfrak p}$ is the stabilizer of $\mathfrak p$, we know that $G/G_{\mathfrak p}$ is in bijection with the orbit of $\mathfrak p$, so $[G:G_{\mathfrak p}]$ is the number $r$ of primes above $p$.
This number $r$ is $n=[L:K]$ iff $p$ is totally split. Moreover $r=[G:G_{\mathfrak p}]=n=|G|$ iff $G_{\mathfrak p} = \{1\}$.
So we get
$$Z_{\mathfrak p} = L \iff G_{\mathfrak p} = \{1\}
\iff r=[G:G_{\mathfrak p}]=n \iff p \text{ is totally split in } L,$$
as desired.
