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\begin{document}
\centerline{Math 361 lecture for Monday, Week 11}
\bigskip

\centerline{Lattices associated with number fields}
\bigskip

Let $K$ be a number field of degree~$n$.  An embedding~$\sigma_{i}\colon K\to
\C$ is \emph{real} if $\sigma_{i}(K)\subset \R$.  Otherwise~$\sigma_{i}$ is
\emph{complex}.  Say~$\sigma_{1},\ldots,\sigma_{s}$ are the real embeddings
of~$K$ and
$\sigma_{s+1},\overline{\sigma}_{s+1},\ldots,\sigma_{s+t},\overline{\sigma}_{s+t}$
are the complex embeddings. Here,
$\overline{\sigma}_{s+j}(\alpha):=\overline{\sigma}_{s+j(\alpha)}$ for
$\alpha\in K$.  Define 
\begin{align*}
  \mathbb{L}_{K}^{s,t}=\mathbb{L}^{s,t} :=\R^{s}\times \C^{t}&\simeq \R^{n}\\
 (x_{1},\ldots,x_{s},z_{1},\ldots,z_{t}) &\mapsto
 (x_{1},\ldots,x_{s},u_{1},v_{1},\ldots,u_{t},v_{t}) 
\end{align*}
where~$z_{j}=u_{j}+v_{j}i\in \C$ for $j=1,\ldots,t$. Then define
\begin{align*}
  \sigma_{K}=\sigma\colon K&\to\mathbb{L}^{s,t}\\
  \alpha&\mapsto
  (\sigma_{1}(\alpha),\ldots,\sigma_{s}(\alpha),\sigma_{s+1}(\alpha),\ldots,\sigma_{s+t}(\alpha)).
\end{align*}

\begin{ex}
  Check that~$\sigma$ is an injective ring homomorphism fixing~$\Q$.
\end{ex}

Define the \emph{norm} of $q=(x_{1},\ldots,x_{s},z_{1},\ldots,z_{t})\in
\mathbb{L}^{s,t}$ to be 
\[
  N(q) = x_{1}\cdots x_{s}z_{1}\overline{z_{1}}\cdots z_{t}\overline{z_{t}}
  =x_{1}\cdots x_{s}|z_{1}|^{2}\cdots|z_{t}|^{2}.
\]
Then~$N(q)\in \R$, and~$N(q q')=N(q)N(q')$ for all~$q,q'\in\mathbb{L}^{s,t}$.

This norm on~$\mathbb{L}^{s,t}$ is related to our old norm on $K$ (the product of the
conjugates) in an obvious way: for $\alpha\in K$
\[
  N(\alpha)=
\sigma_{1}(\alpha)\cdots\sigma_{s}(\alpha)\sigma_{s+1}(\alpha)\overline{\sigma}_{s+1}(\alpha)\cdots\sigma_{s+1}(\alpha)\overline{\sigma}_{s+t}(\alpha)
=N(\sigma(\alpha)).
\]
So if we identify~$\alpha\in K$ with its image in~$\mathbb{L}^{s,t}$, its norm is
well-defined.

\begin{eg}\ 
  \begin{enumerate}
    \item Let~$K=\Q(i)$.  Then the embeddings of~$K$ are $\sigma_{1}(x+yi)=x+yi$
      and~$\overline{\sigma}_{1}(x+yi)=x-yi$ where $x,y\in\Q$.  We have
      \[
	\sigma(x+yi)=x+yi\in \mathbb{L}^{0,1}\simeq \R^{2}.
      \]
      and $N(x+yi)=(x+yi)(x-yi)=x^2+y^2$.
    \item Let~$K=\Q(\sqrt{2})$.  Then the embeddings of~$K$ are
      $\sigma_{1}(x+y\sqrt{2})=x+y\sqrt{2}$
      and~$\sigma_{2}(x+y\sqrt{2})=x-y\sqrt{2}$ where $x,y\in\Q$.  We have
      \[
	\sigma(x+y\sqrt{2})=(x+y\sqrt{2},x-y\sqrt{2})\in \mathbb{L}^{2,0}= \R^{2}.
      \]
      and $N(x+y\sqrt{2})=(x+y\sqrt{2})(x-y\sqrt{2})=x^2+2y^2$.
    \item Let~$K=\Q(\theta)$ where~$\theta$ is the real cube root of~$2$.  The minimal
      polynomial for~$\theta$ is $x^{3}-2$, which factors as 
      \[
	x^3-2=(x-\theta)(x-\omega \theta)(x-\overline{\omega}\theta)
      \]
      where $\omega=e^{2\pi i/3}$. 
      The embeddings are determined by $\sigma_{1}(\theta)=\theta$,
      $\sigma_{2}(\theta)=\omega \theta$, and
      $\overline{\sigma}_{2}(\theta)=\overline{\omega}\theta=\omega^{2}\theta$.
      For~$\alpha=a+b\theta+c\theta^{2}\in K$ where $a,b,c\in\Q$, we have
      \[
	\sigma(\alpha)=(\sigma_{1}(\alpha),\sigma_{2}(\alpha))
	=(a+b\theta+c\theta^{2}, a+b\omega \theta+c\omega^{2}\theta^{2}),
      \]
      and
      \begin{align*}
        N(\alpha)
	&= \sigma_{1}(\alpha)\sigma_{2}(\alpha)\overline{\sigma}_{2}(\alpha)\\
	&= (a+b\theta+c\theta^{2})(a+b\omega
	\theta+c\omega^{2}\theta^{2})(a+b\omega^{2} \theta+c\omega\theta^{2})\\
	&=a^3 + 2b^3 + 4c^3  - 6abc.
      \end{align*}
  \end{enumerate}
\end{eg}

\begin{thm}
  Let $\alpha_{1},\ldots,\alpha_{n}$ be a $\Q$-basis for~$K$, and let
  \[
    \sigma_{k}(\alpha_{\ell}) =
    \begin{cases}
      x_{k,\ell}&\text{if~$1\leq k\leq s$}\\
      u_{k,\ell}+iv_{k,\ell}&\text{if~$s+1\leq k\leq s+t$}
    \end{cases}
  \]
  where the $x_{k,\ell},u_{k,\ell}$ and $v_{k,\ell}$ are real numbers.
  So
  \[
    \sigma(\alpha_{\ell})=(x_{1,\ell},\ldots,x_{s,\ell},u_{s+1,\ell}+iv_{s+1,\ell},\ldots,u_{s+t,\ell}+iv_{s+t,\ell})\in
    \mathbb{L}^{s,t}=\R^{s}\times \C^{t}.
  \]
  Identifying~$\mathbb{L}^{s,t}$ with $\R^{n}$, we have corresponding vectors
  \[
    \mathbf{w}_{\ell} =
(x_{1,\ell},\ldots,x_{s,\ell},u_{s+1,\ell},v_{s+1,\ell},\ldots,u_{s+t,\ell},v_{s+t,\ell})\in\R^{n}.
  \]
  Let
  \[
    F=\{\textstyle \sum_{i=\ell}^{n} c_i \mathbf{w}_{\ell}: 0\leq
    c_{\ell}<1\}\in\R^{n}.
  \]
  Then $\sigma(\alpha_{1}),\ldots,\sigma(\alpha_{n})$ is an~$\R$-basis
  for~$\mathbb{L}^{s,t}$ and
  \[
    \vol(F) =  2^{-t}\sqrt{|\Delta[\alpha_{1},\ldots,\alpha_{n}]|}.
  \]
\end{thm}
\begin{proof}
  Let 
  Think of~$\sigma(\alpha_{j})$ as an element of~$\R^{n}$ for all~$j$.  Then
  performing row operations,
  \begin{align*}
    &\det
    \left(\begin{array}{ccc}
	x_{1,1}&\ldots& x_{1,n}\\
	\vdots&\ddots&\vdots\\
	x_{s,1}&\ldots& x_{s,n}\\
	u_{s+1,1}&\ldots& u_{s+1,n}\\
	v_{s+1,1}&\ldots& v_{s+1,n}\\
	\vdots&\ddots&\vdots\\
	u_{s+t,1}&\ldots& u_{s+t,n}\\
	v_{s+t,1}&\ldots& v_{s+t,n}
    \end{array}\right)
    =
    \det
    \left(\begin{array}{ccc}
	\sigma_{1}(\alpha_{1})&\ldots& \sigma_{1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s}(\alpha_{1})&\ldots& \sigma_{s}(\alpha_{n})\\
	\frac{\sigma_{s+1}(\alpha_{1})+\overline{\sigma}_{s+1}(\alpha_{1})}{2}&\ldots& 
\frac{\sigma_{s+1}(\alpha_{n})+\overline{\sigma}_{s+1}(\alpha_{n})}{2}\\
	\frac{\sigma_{s+1}(\alpha_{1})-\overline{\sigma}_{s+1}(\alpha_{1})}{2i}&\ldots& 
\frac{\sigma_{s+1}(\alpha_{n})-\overline{\sigma}_{s+1}(\alpha_{n})}{2i}\\
	\vdots&\ddots&\vdots\\
	\frac{\sigma_{s+t}(\alpha_{1})+\overline{\sigma}_{s+t}(\alpha_{1})}{2}&\ldots& 
\frac{\sigma_{s+t}(\alpha_{n})+\overline{\sigma}_{s+1}(\alpha_{n})}{2}\\
	\frac{\sigma_{s+t}(\alpha_{1})-\overline{\sigma}_{s+t}(\alpha_{1})}{2i}&\ldots& 
\frac{\sigma_{s+t}(\alpha_{n})-\overline{\sigma}_{s+t}(\alpha_{n})}{2i}
    \end{array}\right)
    \\
    &=
    \left(\frac{1}{2}\right)^{2t}\left(\frac{1}{i}\right)^{t}
    \det
    \left(\begin{array}{ccc}
	\sigma_{1}(\alpha_{1})&\ldots& \sigma_{1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s}(\alpha_{1})&\ldots& \sigma_{s}(\alpha_{n})\\
	\sigma_{s+1}(\alpha_{1})+\overline{\sigma}_{s+1}(\alpha_{1})&\ldots& 
\sigma_{s+1}(\alpha_{n})+\overline{\sigma}_{s+1}(\alpha_{n})\\
	\sigma_{s+1}(\alpha_{1})-\overline{\sigma}_{s+1}(\alpha_{1})&\ldots& 
\sigma_{s+1}(\alpha_{n})-\overline{\sigma}_{s+1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s+t}(\alpha_{1})+\overline{\sigma}_{s+t}(\alpha_{1})&\ldots& 
\sigma_{s+t}(\alpha_{n})+\overline{\sigma}_{s+1}(\alpha_{n})\\
	\sigma_{s+t}(\alpha_{1})-\overline{\sigma}_{s+t}(\alpha_{1})&\ldots& 
\sigma_{s+t}(\alpha_{n})-\overline{\sigma}_{s+t}(\alpha_{n})
    \end{array}\right)
    \\[8pt]
    &=
    \left(\frac{1}{2}\right)^{2t}\left(\frac{1}{i}\right)^{t}
    \det
    \left(\begin{array}{ccc}
	\sigma_{1}(\alpha_{1})&\ldots& \sigma_{1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s}(\alpha_{1})&\ldots& \sigma_{s}(\alpha_{n})\\
	2\sigma_{s+1}(\alpha_{1})&\ldots& 
2\sigma_{s+1}(\alpha_{n})\\
	\sigma_{s+1}(\alpha_{1})-\overline{\sigma}_{s+1}(\alpha_{1})&\ldots& 
\sigma_{s+1}(\alpha_{n})-\overline{\sigma}_{s+1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	2\sigma_{s+t}(\alpha_{1})&\ldots& 
2\sigma_{s+t}(\alpha_{n})\\
	\sigma_{s+t}(\alpha_{1})-\overline{\sigma}_{s+t}(\alpha_{1})&\ldots& 
\sigma_{s+t}(\alpha_{n})-\overline{\sigma}_{s+t}(\alpha_{n})
    \end{array}\right)
    \\[8pt]
    &=
    \left(\frac{1}{2}\right)^{2t}\left(\frac{1}{i}\right)^{t}
    \det
    \left(\begin{array}{ccc}
	\sigma_{1}(\alpha_{1})&\ldots& \sigma_{1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s}(\alpha_{1})&\ldots& \sigma_{s}(\alpha_{n})\\
	2\sigma_{s+1}(\alpha_{1})&\ldots& 2\sigma_{s+1}(\alpha_{n})\\
	-\overline{\sigma}_{s+1}(\alpha_{1})&\ldots&-\overline{\sigma}_{s+1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	2\sigma_{s+t}(\alpha_{1})&\ldots& 2\sigma_{s+t}(\alpha_{n})\\
	-\overline{\sigma}_{s+t}(\alpha_{1})&\ldots&-\overline{\sigma}_{s+t}(\alpha_{n})
    \end{array}\right)
    \\[8pt]
    &=
    \left(\frac{1}{2}\right)^{t}\left(-\frac{1}{i}\right)^{t}
    \det
    \left(\begin{array}{ccc}
	\sigma_{1}(\alpha_{1})&\ldots& \sigma_{1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s}(\alpha_{1})&\ldots& \sigma_{s}(\alpha_{n})\\
	\sigma_{s+1}(\alpha_{1})&\ldots& \sigma_{s+1}(\alpha_{n})\\
	\overline{\sigma}_{s+1}(\alpha_{1})&\ldots&\overline{\sigma}_{s+1}(\alpha_{n})\\
	\vdots&\ddots&\vdots\\
	\sigma_{s+t}(\alpha_{1})&\ldots& \sigma_{s+t}(\alpha_{n})\\
	\overline{\sigma}_{s+t}(\alpha_{1})&\ldots&\overline{\sigma}_{s+t}(\alpha_{n})
    \end{array}\right)
    \\[8pt]
    &= \pm(-2i)^{-t}\sqrt{|\Delta[\alpha_{1},\ldots,\alpha_{n}]|}\neq0.
  \end{align*}
\end{proof}

\begin{cor}
  Let~$\mathfrak{a}$ be a nonzero ideal in~$\mathfrak{O}_{K}$.
  Identifying~$\mathbb{L}^{s,t}$ with~$\R^{n}$,
  regard~$\sigma(\mathfrak{a})\subset\R^{n}$.  Then~$\sigma(\mathfrak{a})$ is a
  lattice with fundamental domain of volume
  \[
    2^{-t}N(\mathfrak{a})\sqrt{|\Delta|}
  \]
  where~$\Delta$ is the discriminant of~$K$.
\end{cor}
\begin{proof}
 Let~$\alpha_{1},\ldots,\alpha_{n}$ be a~$\Z$-basis for~$\mathfrak{a}$.
 Then~$\alpha_{1},\ldots,\alpha_{n}$ is a $\Q$-basis for $K$.  (To see this,
 note that we can take a $\Q$-basis for $K$ consisting of integers.  Then
 $\alpha_{1},\ldots,\alpha_{n}$ are related to that basis by a matrix that is
 invertible over~$\Q$.)  Therefore, by the theorem we just proved,
 $\sigma(\alpha_{1}),\ldots,\sigma(\alpha_{n})$ are linearly independent
 over~$\R$.  Thus, their~$\Z$-span is a lattice in~$\R^{n}$.

 By the theorem, we have that a fundamental region for
 $\sigma(\mathfrak{a})\subset\R^{n}$ has volume
 \[
   2^{-t}\sqrt{\Delta[\alpha_{1},\ldots,\alpha_{n}]}.
 \]
 Our result follows from the fact, proved earlier, that
 \[
   N(\mathfrak{a})=\left|\frac{\Delta[\alpha_{1},\ldots,\alpha_{n}]}{\Delta}\right|^{\frac{1}{2}}.
 \]
\end{proof}

\begin{eg}
  Let $K=\Q(\sqrt{7})$, and let $\mathfrak{a}=(3,1+\sqrt{7})$ (this is one of the
  prime factors of~$(3)\subset\mathfrak{O}_{K}$).
  Let's check the formula for the area of the fundamental domain of the lattice~$\sigma(\mathfrak{a})$.

  The first task is to find a $\Z$-basis for $\mathfrak{a}$.
  An arbitrary element of~$\mathfrak{a}$ has the form
  \[
    (a+b\sqrt{7})\cdot 3+(c+d\sqrt{7})\sqrt{7}=(a+3c+7d)+(3b+c+d)\sqrt{7},
  \]
  for some $a,b,c,d\in\Z$. So this set is the $\Z$-image of the matrix
  \[
    \left(\begin{array}{cccc}
	3 & 0 & 1 & 7 \\
	0 & 3 & 1 & 1
    \end{array}\right).
  \]
  To find the $\Z$-column span, we may use invertible integer column operations.
  We find
  \[
    \left(\begin{array}{cccc}
	3 & 0 & 1 & 7 \\
	0 & 3 & 1 & 1
    \end{array}\right)
    \to
  \left(\begin{array}{cccc}
      1 & 0 & 0 & 0 \\
      1 & 3 & 0 & 0
  \end{array}\right).
  \]
  So a $\Z$-basis for $\mathfrak{a}$ is $\{1+\sqrt{7},3\sqrt{7}\}$.
  The lattice~$\sigma(\mathfrak{a})$ is spanned by
  $\sigma(1+\sqrt{7})=
  (1+\sqrt{7},1-\sqrt{7})$
  and
  $
  \sigma(3\sqrt{7})
  =(3\sqrt{7},-3\sqrt{7})
  =3\sqrt{7}(1,-1).
  $
  The area of a fundamental domain is 
  \[
    \left|  \det
	    \left(\begin{array}{cc}
	    1+\sqrt{7} & 1-\sqrt{7} \\
	  3\sqrt{7} & -3\sqrt{7}
	    \end{array}\right)\right|
  =
  6\sqrt{7}.
  \]
  We now check that this area equals $2^{-t}N(\mathfrak{a})\sqrt{|\Delta|}$.
  We have $\Z/3\Z\to\mathfrak{O}_{K}/\mathfrak{a}$,
  and $1\not\in\mathfrak{a}$.  
  Hence, $\Z/3\Z\simeq \mathfrak{O}_{K}/\mathfrak{a}$.
  So $N(\mathfrak{a})=3$.
  Since~$\mathfrak{O}_{K}=\Span_{Z}\{1,\sqrt{7}\}$, the discriminant of~$K$ is
  \[
    \Delta =
    \det
    \left(\begin{array}{ll}
	1 & \sqrt{7} \\
	1 & -\sqrt{7}
    \end{array}\right)^{2}
    =(-2\sqrt{7})^{2}.
  \]
  So $2^{-t}N(\mathfrak{a})\sqrt{|\Delta|} = 2^{0}\cdot3\cdot(2\sqrt{7})
    = 6\sqrt{7}$, which agrees with our earlier calculations.
  Here is a picture of the lattice and the fundamental domain we were
  considering:
  \begin{center}
    \begin{tikzpicture}[scale=0.2]
      \foreach \i in {-3,-2,-1,0,1,2,3} {
        \foreach \j in {-2,-1,0,1,2} {
	  \node[miniball] at ({\i*3.646-\j*4.292},{-\i*1.646+\j*6.292}) {};
	}
      }
      \draw[<->] (-10,0)--(10,0);
      \draw[<->] (0,-10)--(0,10);
      \filldraw[blue!60,opacity=0.5]
	(0,0)--(3.646,-1.646)--({3.646+7.937},{-1.646-7.937})--(7.937,-7.937)--(0,0);
      \node at (0,-20)  {The lattice $\sigma(\mathfrak{a})$ where
	$\mathfrak{a}=(3,1+\sqrt{7})\subset \mathfrak{O}_{\Q(\sqrt{7})}$.};
    \end{tikzpicture}
  \end{center}
  \end{eg}
\end{document}
