# 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$? thank you for your time and for your help. • Are you using$\Bbb Z_p$for the cyclic group of order$p$and the phrase "$p$-adic" in the same span of text? – anon Dec 7 '13 at 23:43 • @anon No, a$\mathbb{Z}_p$-extension is a Galois extension with Galois group the$p$-adic integers. – Alex Youcis Dec 7 '13 at 23:44 • Oh wow.${}{}{}$– anon Dec 7 '13 at 23:46 • Amine, doesn't this just follow from the fact that you are now looking above the inertial field of each? – Alex Youcis Dec 7 '13 at 23:46 • @anon It's not that crazy. The field$\mathbb{Q}(\mu_p)$(the field generated by all$p$-power roots of unity) is a$\mathbb{Z}_p$-extension of$\mathbb{Q}$. In some sense, much of the obstruction to the classic technique to solve FLT is contained in this$\mathbb{Z}_p$-extension. – Alex Youcis Dec 7 '13 at 23:48 ## 1 Answer It doesn't make sense to say that each$\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$1$). • Keenan just a question. why the cardinality of the set of primes above$\frak P$is equal to the index of the decomposition group of$\frak P$in$K_\infty$?? – Med Dec 8 '13 at 22:29 • Because the Galois group of$K_\infty$over$K_n$acts transitively on the set of primes lying over$\mathfrak{P}$, and the decomposition group is the stablizer of any prime of$K_\infty$above$\mathfrak{P}\$. – Keenan Kidwell Dec 8 '13 at 22:38
• Ehm, of course. Thank you very very much :) – Med Dec 8 '13 at 22:40
• @KeenanKidwell you can also answer this question if you are interested math.stackexchange.com/questions/2777979/… – Blacksmith May 12 '18 at 16:53