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\textbf{18.785 Number Theory\hspace{228pt}Fall~2015}\\\vspace{4pt}
\textbf{Problem Set \#11\hspace{215pt}Due: 12/10/2015}
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\subsection*{Description}

The problems on this final problem set are slightly longer than usual, so you are only required to solve one of them.
Your solutions are to be written up in latex and submitted as a pdf-file with a filename of the form \texttt{SurnamePset11.pdf} via e-mail to \texttt{drew@math.mit.edu}.
Collaboration is permitted/encouraged, but you must identify your collaborators, and any references consulted.
If there are none, write ``\textbf{Sources consulted:\ none}" at the top of your problem set.
As usual, the first persion to spot each non-trivial typo/error in any of the problem sets or lecture notes will receive 1-5 points of extra credit.
\medskip

\noindent
\textbf{Instructions:} Pick 1 of the first 3 problems to solve and write up your answers in latex, then complete the survey problem 4.
Each of these contains at least one part that is potentially tricky; at the end of each problem I have provided a reference that you may consult if you get stuck, but I would encourage you to first attempt to solve as much as you can on your own using only the material we have covered in class.

\subsection*{Problem 1. The conductor ideal (100 points)}

The goal of this problem is to analyze the conductor of an abelian extension of number fields (without assuming the results of class field theory).
Recall that the conductor of an abelian extension $L/K$ is a modulus for $K$ that is defined as a product of local conductors
\[
\c(L/K):=\prod_v \c(L_w/K_v)
\]
where $v$ ranges over the places of $K$ and $w$ is any place of $L$ that extends $v$.
Each local conductor $\c(L_w/K_v)$ determines the exponent of $\c(L/K)$ at $v$, equivalently, the integer $\c(L/K)(v)$ (recall that a modulus is a function $M_K\to \Z$ that we are viewing as a formal product over $M_K$).

The conductor of an extension of archimedean local fields has exponent $1$ if it is nontrivial (isomorphic to $\C/\R$) and $0$ otherwise.
The exponent of the conductor of a finite extension of nonarchimedean local fields $L/K$ is the least integer $n$ for which
\[
U_K^n \subseteq \N_{L/K}(\O_L^\times)
\]
where the groups $U_K^n\subseteq\O_K^\times$ are defined by $U_K^0:=\O_K^\times$ and $U_K^n:=1+\p^n$ for $n>0$; here~$\p$ is the maximal ideal of $\O_K$.

\begin{enumerate}[{\bf(a)}]
\item Verify that the conductor of an abelian extension of number fields is well defined by showing that (1) it does not depend on the choice of the $w$'s extending $v$, and (2) if $L/K$ is a finite extension of nonarchimedean local fields then there is an integer $n$ for which $U_K^n\subseteq \N_{L/K}(\O_L^\times)$.
To prove (2), show that the groups $U_K^n$ form a fundamental system of neighborhoods of the identity in the topological group $\O_K^\times$ (this means the cosets of the $U_K^n$ form a basis for the topology).
\item Show that if $L/K$ is an abelian extension of number fields then the local extensions $L_w/K_v$ are all abelian.
Does the converse hold?  That is, if $L/K$ is a Galois extension of number fields for which $L_w/K_v$ is abelian for every $v\in M_K$, is $L/K$ necessarily abelian? (if your answer is no, give an explicit counter example).
\item While not directly relevant to computing conductors, its worth noting that if you replace ``abelian" with "solvable" in (b) the converse does not hold.  Prove that in fact every Galois extension of local fields is solvable, even though many (most) extensions of number fields are not solvable.
\end{enumerate}

Now let $L/K$ be an abelian extension of local fields.
The archimedean case is clear, so we assume $L/K$ is nonarchimedean.  Let $\F_\p:=\O_K/\p$ denote the residue field.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{3}
\item Show that $K^\times\simeq \Z\times U_K^0$, and that the reduction map $\O_K\to \F_\p$ induces isomorphisms $U_K^0/U_K^1\simeq \F_\p^\times$ and $U_K^n/U_K^{n+1}\simeq \F_\p$ for $n\ge 1$.
\item Prove that if $L/K$ is unramified then $\N_{L/K}(\O_L^\times)=\O_K^\times$ but $\N_{L/K}(L^\times)\ne K^\times$ unless $L=K$.
Show that in fact $K^\times/\N_{L/K}(L^\times)\simeq\Gal(L/K)$ and both groups are cyclic.
\item Prove that if $L/K$ is totally ramified then $K^\times/\N_{L/K}(L^\times)\simeq \O_K^\times/\N_{L/K}(\O_L^\times)$.
\item Prove that if $L/K$ is totally tamely ramified then $\N_{L/K}(U_L^1)=U_K^1$ and the ramification index divides $\#\F_\p^\times$.
\item Prove that if $L/K$ is totally wildly ramified then $\N_{L/K}(U_L^1)\ne U_K^1$.
\item Show that the conductor of $L/K$ has exponent $0$ if and only if $L/K$ is unramified,  and otherwise has exponent $1$ if and only if $L/K$ is tamely ramified.
\end{enumerate}

Useful references for this problem if you get stuck are \cite[Ch.\ III]{FV} and \cite[Ch.\ V]{serre79}.

\subsection*{Problem 2. Profinite groups (100 points)}

Recall that a topological space is \emph{totally disconnected} if very pair of distinct points can be separated by open neighborhoods that partition the space.

\begin{enumerate}[{\bf(a)}]
\item Show that any product or inverse limit of totally disconnected compact Hausdorff spaces is a totally disconnected compact Hausdorff space.  Conclude that every profinite group is a totally disconnected compact Hausdorff space.
\end{enumerate}

Let $G$ be a topological group that is a totally disconnected compact Hausdorff space, let $\widehat G:= \varprojlim G/N$ be its profinite completion (so $N$ varies over finite index open normal subgroups of $G$ ordered by containment), and let $\phi\colon G\to \widehat G$ be the natural map that sends each $g\in G$ to its images in the finite quotients $G/N$.

\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{1}
\item Show that every open subgroup of $G$ has finite index and contains an open normal subgroup (which necessarily also has finite index).
\item Show that $\phi(G)$ is both dense in $\widehat G$ and closed, hence equal to $\widehat G$; thus $\phi$ is surjective.
\item Show that to prove that $\phi$ is injective it suffices to show that the intersection of all open subgroups of $G$ is trivial.
Then show that for every $g\in G-\{1\}$ there is a neighborhood $U$ of $1$ that is both open and closed and does not contain $g$, and it is enough show show that every such $U$ contains an open subgroup $H$.
\item Let $U$ be a neighborhood of $1$ that is both open and closed.
Show that $U$ contains an open neighborhood of $1$ and is closed under multiplication and inversion, hence a subgroup (this requires some work; you will need to use the fact that the multiplication map $G\times G\to G$ is continuous and that $U$ is compact).
\item Show that $\phi$ is a continuous open map, hence a homeomorphism.  Conclude that $G$ is isomorphic to it profinite completion, and in particular, a profinite group.
\item Show that for a profinite group $G$ the following are equivalent: (i) the topology of $G$ is induced by a metric, (ii) $G\simeq \varprojlim G_n$, with $n\in \Z_{\ge 1}$, the $G_n$ finite, and $G_{n+1}\to G_n$ surjective, (iii) the number of open subgroups of $G$ is countable.
\item Show that the 3 equivalent conditions in (g) imply that $G$ contains a countable dense set, and give an example showing that the converse does not hold.
\item Let $p$ be prime, let $\Z_p:=\varprojlim_n \Z/p^n\Z$, and let $G:=\prod_n \Z/p^n\Z$.
Show that $G$ is isomorphic to $\Z_p\times (G/\Z_p)$ as a group, but not as a topological group.
\item Show that $\widehat \Z^\times \simeq \widehat\Z\times \prod_n\Z/n\Z$ as topological groups.
\end{enumerate}

A useful reference for this problem if you get stuck is the textbook by Ribes and Zalesskii \cite{RZ}, which contains just about everything you might ever want to know about profinite groups.

\subsection*{Problem 3. Arithmetically equivalent number fields (100 points)}

Two number fields are said to be \emph{arithmetically equivalent} if their Dedekind zeta functions coincide.
Isomorphic number fields obviously have the same zeta functions, but as proved by Gassmann \cite{gassmann}, the converse need not hold.
A particularly simple example is given by the fields $\Q(\sqrt[8]{a})$ and $\Q(\sqrt[8]{16a})$; as shown by Perlis \cite{perlis}, for $a\in \Z$ not square and not twice a square, these fields are arithmetically equivalent but nonisomorphic (they have the same Galois closure, which is obtained by adjoining a primitive eight root of unity).

\begin{enumerate}[{\bf (a)}]
\item Show that arithmetically equivalent number fields must have the same Galois closure (up to isomorphism).
Conclude that if $K/\Q$ is Galois then $K'$ is arithmetically equivalent to $K$ if and only if $K'\simeq K$.
\end{enumerate}

In view of (a) we now consider two non-Galois extensions $K$ and $K'$ of $\Q$ with Galois closure $L/\Q$.
Let $G:=\Gal(L/\Q)$ and put $H:=\Gal(L/K)$ and $H':=\Gal(L/K')$.

\begin{enumerate}[{\bf (a)}]
\setcounter{enumi}{1}
\item Show that if $K$ and $K'$ are arithmetically equivalent number fields then for every prime $p$ there is a bijection between the primes of $K$ lying above $p$ and the prime of $K'$ lying above $p$ that preserves inertia degrees.
Conclude that $K$ and $K'$ must have the same degree.
\item Given a cyclic $C\subseteq G$, we can partition $G$ into double cosets $H\sigma_1C,\ldots,H\sigma_gC$.
When $H$ is not normal these cosets need not have the same size; each will have cardinality $f_1\cdot \#H$ for some integer $f_i\ge 1$.
Assume the $f_i$ are in increasing order and call the tuple $(f_1,\ldots,f_g)$ the \emph{coset type} of the pair $(H,C)$.  Prove that if $p$ is a prime that is unramified in $L$ and $C$ is any decomposition group of $p$ in $G$, then the coset type of $(H,C)$ is equal to the tuple of inertia degrees of the primes of $K$ lying above $p$ when arranged in increasing order.  Conclude that if $K$ and $K'$ are arithmetically equivalent then $(H,C)$ and $(H',C)$ have the same coset type for every cyclic subgroup $C$ of $G$.
\item Two subgroups $H$ and $H'$ of a finite group $G$ are said to be \emph{Gassmann equivalent} if for conjugacy class $c$ of elements in $G$ the sets $c\cap H$ and $c\cap H'$ have the same cardinality.
Show that this holds if and only if for every cyclic group $C$ the coset types of $(H,C)$ and $(H',C)$ coincide.
\item Suppose $H$ and $H'$ are Gassmann equivalent.  Show that $K$ and $K'$ have the same number of real and complex places (hint: consider the ``decomposition group" of the prime $p=\infty$ in $G$).
\end{enumerate}

\noindent
Like the Riemann zeta function, the Dedekind zeta function has a meromorphic continuation to $\C$ that satisfies a functional equation.
Let $\Gamma_\R(s):=\pi^{-s/2}\Gamma(s/2)$ and let $\Gamma_\C(s):=2(2\pi)^{-s}\Gamma(s)$, and define
\[
Z_K(s):=|\disc \O_K|^{s/2}\Gamma_\R(s)^{n_1}\Gamma_\C(s)^{n_2}\zeta_K(s),
\]
where $n_1$ and $n_2$ are the number of real and complex places of $K$, respectively.
Then
\[
Z_K(s)=Z_K(1-s)
\]
as meromorphic functions on $\C$ (you may assume this in what follows).

\begin{enumerate}[{\bf (a)}]
\setcounter{enumi}{5}
\item Let $f(s)=f_1(s)/f_2(s)$ be a ratio of Euler products over a finite set of primes that extends to a meromorphic function on $\C$ that satisfies a functional equation $f(s)=g(s)f(1-s)$ for some meromorphic function $g(s)$ whose zeros and poles do not coincide with any zero or pole of $f$.  Prove that $g(s)=1$, then use this and the functional equations for $\zeta_K(s)$ and $\zeta_{K'}(s)$ to prove that if $H$ and $H'$ are Gassmann equivalent then $K$ and $K'$ are arithmetically equivalent.
\item A \emph{Gassmann-Sunada triple} is a triple $(G,X,Y)$ in which $G$ is a group that acts faithfully on sets $X$ and $Y$ such that every element of $G$ fixes the same number of elements in $X$ and $Y$ but $X$ and $Y$ are not isomorphic as $G$-sets.
Show that $K$ and~$K'$ are arithmetically equivalent but not isomorphic if and only if $(G,G/H,G/H')$ is a Gassmann triple, where $G/H$ and $G/H'$ denote sets of cosets (for consistency with part (c), use right cosets and put the $G$-action on the right).
\item Show that if $K$ and $K'$ are arithmetically equivalent then they have the same normal core (largest subfield that is a normal extension of $\Q$).
Use this to show that~$K$ and $K'$ contain the same roots of unity and conclude that the unit groups $\O_K^\times$ and $\O_{K'}^\times$ are isomorphic as abelian groups.
\item Show that if $K$ and $K'$ are arithmetically equivalent then $|\disc\O_K|=|\disc \O_{K'}|$ and conclude that $h_KR_K=h_{K'}R_{K'}$, where $h_K:=\#\Cl\O_K$ is the class number and $R_K$ is the regulator of $K$ (and similarly for $K'$).
\end{enumerate}

\noindent
It is natural to ask whether the class numbers and regulators must also match.
This is not the case; the arithmetically equivalent fields $K:=\Q(\sqrt[8]{-15})$ and $K':=\Q(\sqrt[8]{-240})$ have class numbers 8 and 4, respectively (you do not need to prove this).
\begin{enumerate}[{\bf(a)}]
\setcounter{enumi}{9}
\item You showed in (b) that in arithmetically equivalent fields the multisets of inertia degrees above any prime $p$ must match (including at ramified primes); it is natural to ask whether the same is true of the ramification indices.
Show that this is not the case by showing that the polynomials $x^8-97$ and $x^8-16\cdot 97$ (which you may assume define arithmetically equivalent fields $K$ and $K'$), do not have the same factorization pattern in $\Q_2[x]$.
Conclude that the different ideals $\delta_{K/\Q}$ and $\delta_{K'/\Q}$ do not necessarily have the same factorization pattern, even though  the discriminant ideals $D_{K/\Q}:=\N_{K/\Q}(\delta_{K/\Q})$ and $D_{K'/\Q}:=\N_{K'/\Q}(\delta_{K'/\Q})$, do.
\end{enumerate}

A useful reference for this problem if you get stuck is Perlis' paper \cite{perlis} (spoiler alert: it completely answers many of the questions above).

\subsection*{Problem 4. Survey}
Complete the following survey by rating each problem you attempted on a scale of 1 to~10 according to how interesting you found it (1 = ``mind-numbing," 10 = ``mind-blowing"), and how difficult you found it (1 = ``trivial," 10 = ``brutal").  Also estimate the amount of time you spent on each problem to the nearest half hour.

\begin{center}
\begin{tabular}{l|r|r|r|}
& Interest & Difficulty & Time Spent\\\hline
Problem 1 & & & \\\hline
Problem 2 & & & \\\hline
Problem 3 & & & \\\hline
Problem 4 & & & \\\hline
\end{tabular}
\end{center}
\noindent
Please rate each of the following lectures that you attended, according to the quality of the material (1=``useless", 10=``fascinating"), the quality of the presentation (1=``epic fail", 10=``perfection"), the pace (1=``way too slow", 10=``way too fast", 5=``just right") and the novelty of the material to you (1=``old hat", 10=``all new").

\begin{center}
\begin{tabular}{l|l|r|r|r|r|r}
Date & Lecture Topic & Material & Presentation & Pace & Novelty\\\hline
12/1 & The ring of adeles, strong approx & & & & \\\hline 
12/3 & The idele group, profinite groups & & & & \\\hline 
\end{tabular}
\end{center}

\noindent
Please feel free to record any additional comments you have on the problem sets and the lectures, in particular, ways in which they might be improved.

\begin{thebibliography}{99}
\bibitem{FV}
I.B. Fesenko and S.V. Vostokov, \href{https://www.maths.nottingham.ac.uk/personal/ibf/book/book.html}{\textit{Local fields and their extensions}}, 2nd edition, AMS Translations of Mathematics Monographs \textbf{121}, 2002.
\bibitem{gassmann}
F. Gassmann, \textit{Bemerkungen zu der vorstehenden Arbeit von Hurwitz}, Math. Z. \textbf{25} (1926), 124-143. 
\bibitem{perlis}
R. Perlis, \href{http://www.sciencedirect.com/science/article/pii/0022314X77900701}{\textit{On the equation $\zeta_K(s)=\zeta_{K'(s)}$}}, J. Number Theory \textbf{9} (1977), 342--360.
\bibitem{RZ}
L. Ribes and P. Zalesskii, \href{http://link.springer.com/book/10.1007\%2F978-3-642-01642-4}{\textit{Profinite groups}}, 2nd edition, Springer, 2010.
\bibitem{serre79}
Jean-Pierre Serre, \href{http://link.springer.com/book/10.1007\%2F978-1-4757-5673-9}{\textit{Local fields}}, Springer, 1979.
\end{thebibliography}
\end{document}