Total ramification in $\mathbb {Z}_p$-extension I've just started  reading a course on Iwasawa Theory in Washington's book Introduction to Cyclotomic fields, and have had some trouble with the proofs of theoremes. I would like someone  explain to me the proof of this proposition. 

Proposition : Let $K_\infty/K$ be a  $\mathbb{Z}_p$-extension. At least one prime ramifies in this 
  extension, , and there exists $n\geq 0$ such that every prime which ramifies in 
  $K_\infty/ K_n$ is totally ramified. 

Now, I understand that some prime ideal $\frak p$ of $K$ must ramify in $K_\infty$ because class field theory says the maximal unramified abelian extension of $K$ is finite. and I understand that only finitely many primes of $K$ ramify in $K_\infty/K$  ( the  $p$-adic  place). Call them $\frak{p}_1,...,\frak{p}$$_{r}$ and let $I_1, . . . , I_r$ be the corresponding inertia groups. Then  $$\bigcap I_i=p^n\mathbb {Z}_p.$$
for some $n$(because $\cap I_i$ is closed). The fixed field of $p^n\mathbb {Z}_p$ is $K_n$ and $Gal(K_\infty/ K_n)$ is contained in each $I_i.$
What bothers me is the following :  Washington immediately conclude that all primes above each $\frak {p}_{i}$ are totally ramified in $ K_\infty/K_n$   without giving the definition of total ramification in infinite Galois extension.
my question is : why  all primes above each $\frak {p}_{i}$ are totally ramified in $ K_\infty/K_n$ ?
thank you for your time and for your help.
 A: It doesn't make sense to say that each $\mathfrak{p}_i$ is totally ramified in $K_\infty/K_n$. What makes sense is total ramification of the primes of $K_n$ above each $\mathfrak{p}_i$ in $K_\infty/K_n$. So why is this true? Let $\mathfrak{P}$ be a prime of $K_n$ above some $\mathfrak{p}_i$. I want to show that the inertia group $I$ of $\mathfrak{P}$ in $\mathrm{Gal}(K_\infty/K_n)$ is equal to $\mathrm{Gal}(K_\infty/K_n)$. But it follows from the definition of inertia groups that $I=I_i\cap\mathrm{Gal}(K_\infty/K_n)=\mathrm{Gal}(K_\infty/K_n)$, the second equality holding because by construction $\mathrm{Gal}(K_\infty/K_n)\subseteq I_i$. So $\mathfrak{P}$ is totally ramified (incidentally this forces there to be a unique prime of $K_\infty$ above $\mathfrak{P}$ because the cardinality of the set of primes above $\mathfrak{P}$ is equal to the index of the decomposition group of $\mathfrak{P}$ in $K_\infty$, which we've just shown coincides with the inertia group and is equal to the entire Galois group, so the index is $1$). 
