Prove that there exists a non-principal ideal in $Q(\zeta_{23})$. So I've found this extract in 'intro to cyclotomic fields" from Washington on page 7. 
He wants to show that $Z[\zeta_{23}]($the ring of integers) is not a U.F.D. 
So I understand it is sufficient to prove that $Q(\zeta_{23})$ contains a non-principal ideal.
We have that the prime 2 splits in $Q(\sqrt{-23})$ as follows, $(2)_R=\mu \tilde{\mu}$ where $\mu=(2,\frac{1+\sqrt{-23}}{2})$. 
He then lets P be a prime ideal of $Q(\zeta_{23})$ which lies above $\mu$ and he claims that it is non-principal. He proves this claim as follows:
The norm from $Q(\zeta_{23})$ to $Q(\sqrt{-23})$ is $\mu^f$ where $f$ is the degree of the residue class field extension. I don't understand what this means, is this something to do with intertial degree? 
Then he says that $f$ divides the [$Q(\zeta_{23})$:$Q(\sqrt{-23})$]=11.  Again I don't understand where this restriction of $f$ arises from. 
He then goes on to say that $\mu^{11}$ and $\mu$ are non principal which seems fine to me. 
I've considered that the norm is $\mu^f$ where $f$ is the relative inertial degree but P is not the only ideal above $\mu$ so why does $f$ need to divide 11? I haven't got a good understanding of ramification theory so I apologise if I am missing something simple. 
 A: Let $A$ be a Dedekind domain with field of fractions $K$ and $B$ its integral closure (again a Dedekind domain) in a Galois extension $L$ of $K$.
A prime ideal $p$ of $A$ will factor in $B$ as $pB = P_1^{e_1}...P_g^{e_g}$.
$e(P/p)$ is called the ramification index and $f(P/p)$ is the degree of the field extension $(B/P) / (A/p)$, called the residue class degree.
Now let $[L:K] =m$. Then $m = efg$. A sketch proof goes as follows:
$\sigma \in \operatorname{Gal}(L/K)$ maps a prime ideal $P$ of $B$ to another prime ideal $\sigma P$. If $P$ lies above $p$ then $\sigma P$ will also lie above $p$. So $f(\sigma P/p) = f(P/p)$.
It remains to show that the Galois group acts transitively on prime ideals of $B$ lying above $p$. 
Suppose $P$ is not conjugate to $Q$, both prime ideals lying over $p$. Then there is $\beta \in Q$ with $\beta \notin \sigma P$ for all $\sigma$. Let $b = \operatorname{Norm}(\beta) = \prod \sigma \beta$. This lies in $A$ and since $\beta \in Q$, it also lies in $Q$ and therefore in $Q \cap A = p$. 
On the other hand, $\beta \notin \sigma^{-1}P$ for all $\sigma$, so $\sigma \beta \notin P$. But $\prod \sigma \beta \in p \subset P$. This contradicts the primality of $P$.
So $\operatorname{Gal}(L/K)$ acts transitively and thus $f$ divides $m$.
My proof is basically taken from J. Milne's ANT notes pp.58. They are available on his website. On p.67 he motivates very well why the norm is defined the way it is, too.
