Primes which can be Norms vs Primes which Split Completely in Galois Extensions Let $K/ \mathbb Q$ be a Galois extension. I read somewhere that the integer primes $p$ which can be norms of some integral ideal $\mathfrak a$ of $k$ are exactly the ones which split completely in $K$, however I have been unable to find a proof of the same. I have however managed to show the result for all but finitely many primes. More precisely, I could show that for all the primes $p$ which do not ramify in $K$, the equivalence 
$$p\text{ is a norm of some ideal }\mathfrak{a} \lhd \mathcal{O}_K \iff p \text{ splits completely in }\mathcal{O}_K$$
holds. Here is my proof:
If $p$ splits completely that is $p\mathcal{O}_K = \mathfrak{p_1} \cdots \mathfrak p_n$ (where $n = [K:\mathbb Q]$) for some distinct prime ideals $\mathfrak{p_1}, \cdots , \mathfrak p_n$ then all of these prime ideals must lie over $p$ whence since the residue degrees are all $1$, we have $N(\mathfrak{p_i}) = p^1 = p$ for every $i \in [k]$. 
Conversely, if $p$ is the norm of an integral ideal $\mathfrak{a}$, and if $p\mathcal{O}_K = \mathfrak{p_1} \cdots \mathfrak p_m$ for some $m \leq n$ (with $\mathfrak{p_1}, \cdots, \mathfrak p_m$ being the primes of $K$ lying over $p$), then factoring $\mathfrak{a} = \prod_{j=1}^t \mathfrak{q_j}^{\alpha_j}$ we see (by taking norms on both sides) that $\{\mathfrak{q_j} : 1 \leq j \leq t \} \subset \{\mathfrak{p_i} : 1 \leq i \leq m \} $. Therefore, we may let $\mathfrak{a} = \prod_{i=1}^m \mathfrak{p_i}^{\alpha_i}$ where the $\alpha_i$ are now allowed to be zero, and again taking norms shows that $(\sum_{i=1}^m \alpha_i)f = 1$ (with $f$ being the common residue degree of the primes $\mathfrak p_i$ over $p$) whereupon $f=1$ shows (by the result $n = \sum_{i=1}^m e_{\mathfrak{p}_i|p}f_{\mathfrak{p}_i|p} = fm$) that $m=n$, so that $p$ does split completely in $K$.
I would be really grateful if someone were to verify my proof and tell me how I could deal with the case when $p$ ramifies. 
 A: Generally, "splitting completely" is understood to imply lack of ramification, in which case your equivalence wouldn't work. For example, $ 2 $ is not totally split in $ \mathbf Z[i] $, and yet $ N((1+i)) = 2 $.
What you can say is that there is an integral ideal of norm $ p $ in $ \mathcal O_K $ if and only if the inertia degree of the primes over $ p $ is trivial. In this case, there are indeed $ n =  [K : \mathbf Q] $ primes lying over $ p $ when you count them with multiplicity, and the correct way to count them in this case is indeed to use multiplicity. If $ p $ is such a prime, then you have
$$ (p) = \prod_{k=1}^g (\mathfrak p_k)^e $$
or, upon taking norms and using the fact that the Galois group acts transitively on the prime ideals,
$$ p^n = N(\mathfrak p_1)^{ge} = N(\mathfrak p_1)^n $$
so that $ N(\mathfrak p_1) = p $. The other case is similar - the norm of a prime ideal $ \mathfrak p $ is the size of the quotient ring $ \mathcal O/\mathfrak p $, so a prime ideal lying over $ p $ having norm equal to $ p $ is the same as saying the field $ \mathcal O/p \cong \mathbb F_p $, i.e. the residue field extension is trivial and thus the inertia degree $ f_{\mathfrak p | p} = 1 $. By using the Galois extension properties, you deduce from this that all primes lying over $ p $ have trivial inertia degrees.
