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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.

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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 1

up vote 1 down vote accepted

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 $1$).

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thank you so much –  Med Dec 8 '13 at 22:19
    
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
1  
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

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