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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!

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    @Watson Thank you for your edits! – 2017-02-01

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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.

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    Thank you for your answer. I still have one question though: we defined that a prime ideal $p$ is completely decomposed (or totally split) if $r= [L:K]$ (which you used) and if $p$ is unramified (which would require that the ramification degree of all the $\mathfrak{p}$ in the decomposition is $1$). It probably is totally obvious why this is true here, but could maybe explain this to me? (I fail to see why this property holds.) – 2017-02-01
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    @SallyOwens : This is because $[L:K] = \sum_{i=1}^r e_i f_i$ always holds. – 2017-02-01
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    Thank you - I think I got it now :) (Just to be sure: in this case (Galois extension), I know that the $e_i$ and $f_i$ are the same for all $\mathfrak{p}_i$ and hence $[L:K]=rfe$ and as $[L:K]$ is already equal to $r$, I must have $e=f=1$.) – 2017-02-01
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    @Watson would you mind telling me what $r\ f\ e$ means? – 2017-09-22
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    @GuerlandoOCs : when we write $p \mathcal{O}_L= ({\mathfrak{p}_1} \cdot \dotsc \cdot \mathfrak{p}_r)^e$, the integer $e$ is the _ramification index_, $r$ is the number of prime ideals in $L$ above $p$, and $f = [\mathcal O_L / \mathfrak{p}_1 : \mathcal{O}_K / (p)]$ in the _inertia degree_. See [here](https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions) for some details. – 2017-09-22