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%--------------------------------------------------------------------
\begin{document}
\title{Algebraic Number Theory  $\cdot$ Spring 2004 \\ Assignment 2 }
\author{John Kerl}
\date{\today}
\maketitle

%% ----------------------------------------------------------------
\begin{problem*}[1]
Prove or disprove and salvage:  Let $K=\Q(\sqrt{m})$ and let $p$ be a prime
integer.  Suppose $m$ is squarefree and $m \equiv 1 \pmod{4}$.  If $p\OK =
\fp^2$, then $p \mid m$.

\begin{proof} Suppose $p\OK = \fp^2$ for some non-zero prime ideal of $\OK$.
Let $\omega=\frac{1+\sqrt m}{2}$ and let $\pi=a+b\omega \in \fp \setminus
\fp^2$.  Then $\pi^2 \in \fp^2 = \fp\OK$ implies $p|\pi^2$.  Now,
\begin{eqnarray*}
	\pi^2 &=& \left(a+\frac{1+\sqrt m}{2} \; b \right)^2 \\
	&=& a^2 + ab + ab \sqrt m + \frac{(1 + 2 \sqrt m + m) b^2}{4} \\
	&=& a^2 + ab + ab \sqrt m + \frac{b^2}{4} + \frac{b^2m}{4}
		+ \frac{b^2 \sqrt m}{2} \\
	&=& \left(a^2 + ab + \frac{b^2}{4} + \frac{b^2m}{4}\right)
		+ \left(ab + \frac{b^2}{2}\right) \sqrt m \\
	&=& \left( \left(a+\frac{b}{2}\right)^2 + \frac{b^2m}{4}\right)
		+ \left(a + \frac{b}{2}\right) b \sqrt m \\
	&\implies& \; p \mid (a+b/2)^2 + b^2m/4
		\; \textrm{ and } p \mid (a + b/2) b \\
\end{eqnarray*}

The latter admits two cases since $p$ is prime in $\Z$.  In the case $p \mid
b$, $p \mid a^2 \implies p\mid a$ which is contradictory since $a+bi \not\in
p\OK$.  Therefore the case $p|a+b/2$ applies and $p \mid b^2m/4$ which,
since $p \nmid b$, implies $p|m/4$.  Since $m \equiv 1 \pmod  4$, $p \mid m$.
\end{proof} \end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[2]
Let $K/F$ be a Galois extension of number fields, and suppose $\fP$ is a
prime ideal of $\OK$ above the prime ideal $\fp$ of $\OF$, where $e(\fP/\fp)=1$.
Let $\sigma \in \Gal(K/F)$.  Show that
$$
\sigma\left( \frac{K/F}{\fP}\right)\sigma^{-1} =
\left( \frac{K/F}{\fP^\sigma}\right)
$$

\begin{proof} To save ink and keystrokes, let $\tau = \left( \frac{K/F}{\fP}
\right)$ and $\tau' = \left( \frac{K/F}{\fP^\sigma} \right)$.  These are both
already Frobenius automorphisms with respect to their respective residue class
fields.  It remains to be shown that $\sigma \tau \sigma^{-1} = \tau'$.

First, to see that $\sigma\tau\sigma^{-1} \in Z(\fP^\sigma)$, let $\alpha \in
\fP^\sigma$.  Since $\sigma$ sends $\fP$ to $\fP^\sigma$, $\sigma^{-1}$ does
the opposite and so $\sigma^{-1}(\alpha) \in \fP$.  Since $\tau \in Z(\fP)$,
$\tau(\sigma^{-1}(\alpha) \in \fP$.  Then $\sigma\tau\sigma^{-1}(\alpha) \in
\fP^\sigma$.

Second, since $\fP$ is unramified, $\Z(\fP) \cong \Gal(\F_\fP/\F_\fp)$ and
$\Z(\fP^\sigma) \cong \Gal(\F_{\fP^\sigma}/\F_\fp)$.  Since I have
$\sigma\tau\sigma^{-1}$ in $Z(\fP^\sigma)$, $\sigma\tau\sigma^{-1}$ corresponds
to one of the $f$ finite-field automorphisms in $\Gal(\F_{\fP^\sigma}/\F_\fp)$.
It suffices to show that $\sigma\tau\sigma^{-1}$ is the one which is the
$q$-power map on $\Gal(\F_{\fP^\sigma}/\F_\fp)$.  Recall that the induced map
from $Z(\fP^\sigma)$ to $\Gal(\F_{\fP^\sigma}/\F_\fp)$ is given by $\phi
\mapsto \overline\phi$ where $\overline{\phi}(\alpha) = \phi(\alpha) +
\fP^\sigma$.  Let $\alpha \in \OK$.  Then
\begin{eqnarray*}
	\tau(\alpha) &\equiv& \alpha^q \pmod{\fP} ;\;
		\tau(\alpha) - \alpha^q \in \fP \\
	\tau'(\alpha) &\equiv& \alpha^q \pmod{\fP^\sigma} ;\;
		\tau'(\alpha) - \alpha^q \in \fP^\sigma \\
	\overline{\sigma\tau\sigma^{-1}}(\alpha)
		&=& \sigma \tau \sigma^{-1}(\alpha) + \fP^\sigma \\
	&=& \sigma (\tau \sigma^{-1}(\alpha) + \fP) \\
	&=& \sigma( (\sigma^{-1}(\alpha))^q + \fP) \\
	&=& \sigma(\sigma^{-1}(\alpha))^q + \fP^\sigma \\
	&=& \alpha^q + \fP^\sigma \\
	&=& \tau'(\alpha) + \fP^\sigma \\
	&=& \overline{\tau}'(\alpha) \\
\end{eqnarray*}
\end{proof}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[3]
Let $K$ be a number field, and let $K_\fp$ be the completion of $K$ with
respect to a discrete non-Archimedean absolute value $|\;|_\fp$ for some
non-zero prime ideal $\fp$ of $\OK$.  Let
$$
\mcO_\fp = \{ \alpha \in K_\fp : |\alpha|_\fp \le 1\} \; \textrm{ and } \;
\mcP_\fp = \{ \alpha \in K_\fp : |\alpha|_\fp < 1\}
$$
and let $S=\OK\setminus \fp$.  Show that
$\mcO_\fp \cap K = S^{-1}\OK$ and
$\mcP_\fp \cap K = \fp S^{-1}\OK$.
\begin{proof}
\ldots
\end{proof}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[4]
Let $K$ be a number field, and let $K_\fp$ be the completion of $K$ with
respect to a discrete non-Archimedean absolute value $|\;|_\fp$ for some
non-zero prime ideal $\fp$ of $\OK$.  Show that an infinite series
$\sum_{n=1}^\infty a_n$ converges in $K_\fp$ if and only if
$|a_n|_\fp \to 0$ as $n \to\infty$.  (This means that for $K_\fp$ there is
nothing analogous to the harmonic series of real analysis.)

\begin{proof} For the forward implication, suppose $\sum_{n=1}^\infty a_n$
converges in $K_\fp$, say to the limit $a$.  Let $(s_n)$ be the sequence of
partial sums, where $s_n = \sum_{k=1}^n a_k$.  Since $\sum_{n=1}^\infty a_n \to
a$, by definition $s_n \to a$ and likewise by reindexing $s_{n-1} \to a$.

Now I claim that if $s_n \to a$ and $s_{n-1} \to a$, $s_n - s_{n-1} \to
0$.  It suffices to show the more general result that the limit of sums is the
sum of limits.  (This is familiar from real analysis, but the present affords
the opportunity for a much-needed refresher.)  Let $c_n \to c$ and $d_n \to
d$.  For all $\veps/2$ there are $N_1$, $N_2$ such that $|c_n-c|_\fp < \veps/2$
and $|d_n-d|_\fp < \veps/2$.  Take $N=\max\{N_1,N_2\}$.  Then for all $n>N$, $$
|(c_n+d_n)-(c+d)|_\fp = |(c_n-c)+(d_n-d)|_\fp \le |c_n-c|_\fp + |d_n-d|_\fp <
\veps$$  Hence $c_n+d_n \to c+d$.

Having $s_n-s_{n-1} \to 0$ and recalling $s_n - s_{n-1} = a_n$, I have that
$a_n \to 0$ and thus $|a_n|_\fp \to 0$.  This much is just as in the
Archimedean case, familiar from calculus.

For the reverse implication, suppose $|a_n|_\fp \to 0$ and let $s_n =
\sum_{k=1}^n a_k$.  Since $a_n = s_n - s_{n-1}$, $|s_n - s_{n-1}|_\fp \to 0$.
I now want to show that $(s_n)$ is Cauchy.  It suffices to show that for all
$\veps > 0$ there is an $N > 0$ such that for all $j,k \ge 0$, $$|s_{N+j+k} -
s_{N+j}|_\fp < \veps$$ But
$$
|s_{N+j+k} - s_{N+j}|_\fp \le
\max\{ |s_{N+j+k} - s_{N+j+k-1}|_\fp, \ldots, |s_{N+j+1} - s_{N+j}|_\fp \} = M
$$
so I want $M < \veps$.  Fortunately, for $i=1, \ldots, k$  I have
that, given $\veps>0$, there is an $N_i$ such that
$$|s_{N+j+i+1}-s_{N+j+i}|_\fp < \veps $$
so I merely pick $N = \max\{N_i\}$ to make $(s_n)$ Cauchy.
(The difference from the Archimedean case is that I would have
$$|s_{N+j+k} - s_{N+j}| \le k \max\{ |s_{N+j+k} - s_{N+j+k-1}|, \ldots,
|s_{N+j+1} - s_{N+j}| \}
$$
where the $k$ multiplier can be arbitrarily large.)

Since $(s_n)$ is Cauchy, $(s_n)$ converges which by definition means that
the series $\sum_{n=1}^\infty a_n$ converges.
\end{proof}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5a]
Let $K=\Q(\sqrt{-1})$ and let $p=5$.  Give explicitly the ring $\OK$ and
the factorization of the ideal $5\OK$ as a product of prime ideals in $\OK$.
(You may use without proof that $\OK$ is a PID, so that what is required to
describe a prime ideal is a generator which you have shown to be irreducible.)

\begin{answer}
Since $-1 \equiv 3 \pmod 4$, $\OK=\Z[\sqrt{-1}] = \Z[i]$.  These are called
the \emph{Gaussian integers}.

%To see that the Gaussian integers are a PID, it suffices to show that they are
%in fact a Euclidean domain.
%xxx func

By inspection, $5$ factors at least as far as $(2+i)(2-i)$.  To see that $2\pm
i$ are irreducible, I could certainly show that factors would need to lie
within the radius-3 circle in the complex plane, and simply test all such.
This wouldn't be too tedious since $2\pm i$ have small magnitude.  However,
less tedious and of much broader applicability is the following.  First define
the norm $N: \Z[i] \to \N$ by $N(a+bi) = a^2+b^2$.

I claim the norm is multiplicative:
\begin{eqnarray*}
	N((a+bi)(c+di)) &=& N((ac-bd)+(ad+bc)i) \\
	&=& (ac-bd)^2+(ad+bc)^2 \\
	&=& a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2 \\
	&=& a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2 \\
	&=& (a^2+b^2)(c^2+d^2) \\
	&=& N(a+bi)N(c+di)
\end{eqnarray*}

I also claim that $u\in \Z[i]$ is a unit iff $N(u)=1$.  First I must enumerate
the units of $\Z[i]$.  (We know them to be $\pm 1$ and $\pm i$, but I should
prove that rather than merely asserting it.) Suppose $u=a+bi$ is a unit of
$\Z[i]$.   Then $u$ is invertible in $\Z[i]$, i.e.
$\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2} \in \Z[i]$ where $a,b$ are not both 0.
Then $a^2+b^2|a$ and $a^2+b^2|b$.  In the case $a=0$, $b^2|b$ implies $b=\pm 1$
since $\Z$ is an integral domain.  In the case $b=0$, $a^2|a$ implies $a=\pm
1$.  In the case $a,b \ne 0$, $a^2 \ge a$ and $b^2 \ge b$ which implies
$a^2+b^2 > |a|,|b|$ so $a^2+b^2|a$ and $a^2+b^2|b$ are impossible.  Therefore
$u=\pm 1$ or $\pm i$.

For the forward implication, let $u$ be a unit of $\Z[i]$.  Then trivially
$N(u)=1$.  For the reverse implication, let $u=a+bi$ and suppose $N(u)=1$.
Then $a^2+b^2=1$ for integer $a,b$.  In the case $a=b=0$, $a^2+b^2 = 0 \ne 1$.
In the case $a,b \ne 0$, $a^2+b^2 > 1$.  In the case $a=0$, $b=\pm 1$.  In the
case $b=0$, $a=\pm 1$.  Thus in all cases, $u$ is a unit of $\Z[i]$.

Now suppose $2\pm i= w z$ for $w,z \in \Z[i]$.  Since the norm is
multiplicative, $N(2 \pm i) = 5 = N(wz) = N(w)N(z)$.  Since 5 is prime, either
$N(w)$ or $N(z)$ is 1, hence one factor is a unit and so $2 \pm i$ are
irreducible in $\Z[i]$.
\end{answer}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5b]
Choose a prime ideal $\fp$ of $\OK$ occurring in the factorization you gave in
part (a).  Find $\nu_\fp(5)$.
\begin{answer}
Let $\fp=(2+i)\OK$ and $\fp' = (2-i)\OK$.  Since $5\OK = \fp \fp'$,
$\nu_\fp(5) = 1$.  This means that $5$ splits and is unramified in $\OK$.
\end{answer}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5c]
For the ideal $\fp$ you used in part (b), consider the completion $K_\fp$.
Show that $K_\fp$ contains an isomorphic copy of $\Q_5$ and that $K_\fp \cong
\Q_5(\sqrt{-1})$.
\begin{proof}
\ldots
\end{proof}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5d]
Since I have already shown $K_\fp \cong \Q_5(\sqrt{-1})$, $\sqrt{-1}$ is
necessarily in $K_\fp$.  Thus it is possible to give the first 4 terms of the
5-adic expansion of $\sqrt{-1}$.

\begin{answer}
Let
\begin{eqnarray*}
	\sqrt{-1} &=& a_0 + 5 a_1 + 25 a_2 + 125 a_3 + O(625) \\
\end{eqnarray*}

For the $a_0$ term,
\begin{eqnarray*}
	a_0^2 &\equiv& -1 (5) \\
	a_0 &=& 2, 3 (5)
\end{eqnarray*}

For the $a_1$ term,
$$
\begin{array}{rlrlrlr}
\sqrt{-1} &\equiv& 2 + 5 a_1 (25) &\qquad& \sqrt{-1} &\equiv& 3 + 5 a_1 (25) \\
-1 &\equiv& 4 + 20 a_1 (25) &\qquad& -1 &\equiv& 9 + 30 a_1 (25) \\
-5 &\equiv& 20 a_1 (25) &\qquad& -10 &\equiv& 30 a_1 (25) \\
-1 &\equiv& 4 a_1 (5) &\qquad& -2 &\equiv& a_1 (5) \\
a_1 &=& 1 &\qquad& a_1 &=& 3
\end{array}
$$

For the $a_2$ term,
$$
\begin{array}{rlrlrlr}
\sqrt{-1}&\equiv&7 + 25 a_2 (125)&\qquad&\sqrt{-1} &\equiv& 18 + 25 a_2 (125)\\
-1&\equiv& 49 + 350 a_2 (125)&\qquad&-1 &\equiv& 324 + 900 a_2 (125)\\
-50 &\equiv& 350 a_2 (125)&\qquad&-325 &\equiv& 900 a_2 (125)\\
-1 &\equiv& 7 a_2 (5)&\qquad&-13 &\equiv& 36 a_2 (5)\\
4 &\equiv& 2 a_2 (5)&\qquad&2 &\equiv& a_2 (5)\\
a_2 &=& 2&\qquad& a_2 &=& 2
\end{array}
$$

For the $a_3$ term,
$$
\begin{array}{rlrlrlr}
\sqrt{-1}&\equiv&57+125 a_3 (625)&\qquad&\sqrt{-1} &\equiv& 68 +125 a_3 (625)\\
-1&\equiv& 3249 + 14250 a_3 (625)&\qquad&-1 &\equiv& 4624 + 17000 a_3 (625)\\
-3250&\equiv& 14250 a_3 (625)&\qquad&-4625 &\equiv& 17000 a_3 (625)\\
500&\equiv& 500 a_3 (625)&\qquad&375 &\equiv& 125 a_3 (625)\\
1&\equiv& a_3 (5)&\qquad&3 &\equiv& a_3 (625)\\
a_3 &=& 1&\qquad& a_3 &=& 3
\end{array}
$$

So,
$$
	\sqrt{-1} = 2 + 1 \cdot 5 + 2 \cdot 25 + 1 \cdot 125 + O(625), \qquad
	3 + 3 \cdot 5 + 2 \cdot 25 + 3 \cdot 125 + O(625) \\
$$

To check, I multiply out (on paper I computed to $O(3125)$, obtaining another
digit of precision):

\begin{verbatim}
  .2 1 2 1 3                      .3 3 2 3 1          
* .2 1 2 1 3                    * .3 3 2 3 1          
-------------                   -------------         
   4 2 4 2 1 1                     4 0 3 0 0 1        
     2 1 2 1 3                       4 0 3 0 0 1      
       4 2 4 2 1 1                     1 2 0 2 3      
         2 1 2 1 3                       4 0 3 0 0 1  
           1 4 1 4 4 1                     3 3 2 3 1  
----------------------          ----------------------
  .4 4 4 4 4 3 ...                .4 4 4 4 4 4 2 ...
\end{verbatim}

which are good to 5 places as expected.
\end{answer}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5e]
Show that $5\mcO_\fp = \mcP_\fp$, where $\mcO_\fp$ and $\mcP_\fp$ are as in
problem 3.
\begin{proof}
\ldots
\end{proof}
\end{problem*}

%% ----------------------------------------------------------------
\begin{problem*}[5f]
What happens if we instead consider $K=\Q(\sqrt{5})$ and $p=5$?  (In this case,
once again $\OK$ is a PID.)

\begin{answer}
Since $5 \equiv 1 \pmod 4$, $\OK = \Z[\frac{1+\sqrt 5}{2}]$, which contains
the element $\sqrt{5}$.  I think [xxx check] that $\sqrt{5}\OK$ is prime.
If so, then $5=(\sqrt{5})^2$ is the prime decomposition of 5 in $\OK$,
and 5 ramifies here.

Letting $\fp=\sqrt{5}\OK$, $\nu_\fp(5) = 2$.

xxx part c

xxx part d

xxx part e

\end{answer}
\end{problem*}

%% ----------------------------------------------------------------
\end{document}
