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.

  • $\begingroup$ Are you using $\Bbb Z_p$ for the cyclic group of order $p$ and the phrase "$p$-adic" in the same span of text? $\endgroup$ – anon Dec 7 '13 at 23:43
  • $\begingroup$ @anon No, a $\mathbb{Z}_p$-extension is a Galois extension with Galois group the $p$-adic integers. $\endgroup$ – Alex Youcis Dec 7 '13 at 23:44
  • $\begingroup$ Oh wow. ${}{}{}$ $\endgroup$ – anon Dec 7 '13 at 23:46
  • $\begingroup$ Amine, doesn't this just follow from the fact that you are now looking above the inertial field of each? $\endgroup$ – Alex Youcis Dec 7 '13 at 23:46
  • $\begingroup$ @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. $\endgroup$ – Alex Youcis Dec 7 '13 at 23:48

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 $1$).

  • $\begingroup$ 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 $K_\infty$ ?? $\endgroup$ – Med Dec 8 '13 at 22:29
  • 1
    $\begingroup$ 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 $\mathfrak{P}$. $\endgroup$ – Keenan Kidwell Dec 8 '13 at 22:38
  • $\begingroup$ Ehm, of course. Thank you very very much :) $\endgroup$ – Med Dec 8 '13 at 22:40
  • $\begingroup$ @KeenanKidwell you can also answer this question if you are interested math.stackexchange.com/questions/2777979/… $\endgroup$ – Blacksmith May 12 '18 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.