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\begin{document}

\chapter*{October 5:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Does $\Z_F$ have a basis over $\Z$?
\item
Free abelian groups
\end{itemize}

\section{Does $\Z_F$ have a basis over $\Z$?}

Let $\F$~be a number field of degree $n$ over $\Q$.  Let $\alpha_{1},
\ldots , \alpha_{n}$ be a basis of $\F$~over $\Q$.  By 
multiplying each $\alpha_i$ by an integer, if necessary, we
can assume that each $\alpha_i$ is in $\Z_F$, i.e., that each
$\alpha_i$ is an algebraic integer.

The set
$$\spn_\Z(\alpha_1, \cdots , \alpha_n) = \left\{\sum_{i=1}^n
a_i \alpha_i : a_i \in \Z \mbox{ for all } i\right\}$$
of all integral linear combinations of the $\alpha_i$ is
a subgroup of (the additive group of)~$\Z_F$.
In fact we can bound the distance between $\spn_\Z(\alpha_i)$
and $\Z_F$ in terms of the discriminant of the basis.

\begin{theorem}
Let $\alpha_{1}, \ldots , \alpha_{n}$ be a
basis of $\F$~over $\Q$, where each $\alpha_i \in \Z_F$, and 
let $d = \disc(\alpha_1, \cdots , \alpha_n)$.  Then
for every $\alpha \in \Z_F$
$$d\alpha = \sum_{i=i}^{n}{a_{i}\alpha_{i}}, \quad a_i \in \Z,$$
i.e., when $\alpha$ is expressed as a linear combination of
the basis elements the rational numbers that occur as coefficients
have denominators that are divisors of~$d$.
\end{theorem}

\begin{proof}
Let $\alpha $ be an element of $ \ZF$.
Then we have
$$\alpha = \sum_{i=i}^{n}{b_{i}\alpha_{i}} \qquad {\rm where}~ b_{1},\ldots,b_{n} \in
\Q.$$  
For each $j$, $1 \le j \le n$, we have
$$\alpha^{(j)} = \sum_{i=i}^{n}{b_{i}\alpha_{i}^{(j)}} $$ 
where the superscripts denote the images under the various embeddings.
This system of equalities can be written in matrix form as
$v = Mb$, where
$$v= \left[ \begin{array}{c} \alpha^{(1)}\\ \alpha^{(2)} \\ \vdots \\
\alpha^{(n)} \end{array} \right],\quad M = \left[
\begin{array}{ccc} \multicolumn{3}{c}{\dotfill} \\ \alpha_{1}^{(i)} & \cdots &\alpha_{n}^{(i)}\\
\multicolumn{3}{c}{\dotfill}\\\multicolumn{3}{c}{\dotfill}
\end{array} \right], \quad {\rm and}\quad b = \left[ \begin{array}{c} b_{1}\\ b_{2} \\ \vdots \\
b_{n} \end{array} \right].$$ Now we have $M^{T}v = M^{T}Mb$, thus
$(M^{T}M)^{-1}M^{T}v = b$.  Now we see that $M^{T}M =
[\trr{F/\Q}(\alpha_{i}\alpha_{j})]$, and $M^{T}v = [\trr_{F/\Q}(\alpha \alpha_{i})]$.
However,,
$$(M^{T}M)^{-1} = \frac{1}{d}[(-1)^{i+j}(M^{T}M)_{ij}]$$
where $(M^{T}M)_{ij}$ is the $ij^{\rm th}$ cofactor, and $d =
\det(M^{T}M) = \disc(\alpha_{1}, \ldots , \alpha_{n}).$ Thus $db$
is a matrix of integers, so
$$\alpha = \sum_{i=i}^{n}{\frac{a_{i}}{d}\alpha_{i}} 
\qquad {\rm where}~ a_{1},\ldots,a_{n} \in
\Z.$$
\end{proof}

Thus we see that $\ZF \subset {\rm span}_{\Z} \left(
\alpha_{1}/d, \cdots,
\alpha_{n}/d \right).$

\section{Free abelian groups}

\begin{defin}
A finitely generated free abelian group, or \defn{fab} for short,
 is a finitely generated free $\Z$-module.
Equivalently, a fab is a finitely generated torsion-free
abelian group.  The \defn{rank} of a fab is its 
number of generators (from results below, it follows immediately 
that any two generating sets have the same number
of elements).
\end{defin}

\begin{example}
The lattice $\{(0,0)\}$ inside $\Z^2$ has rank 0, and $\Z^2$
itself has rank~2.  The set $\{nv: n \in \Z\}$ of all multiples of
a given nonzero vector $v \in \Z^2$ forms a line of lattice points,
and is a fab of rank~1.
\end{example}

\begin{example}
Any fab $G$ of rank~2 in $\Z^2$ is of finite index, where
the index is the number of cosets, i.e., the order of
the quotient group $\Z^2/G$.
\begin{itemize}
\item
The fab consisting of vectors $(x,y)$ where $x+y$ is even has index~2.
\item
The fab consisting of vectors $(x,y)$ where both $x$ and $y$ are even
has index~$4$, and is sometimes denoted~$2\Z^2$.
\item
The fab $G$ generated by $(2,1)$ and $(0,3)$ has coset representatives
$(i,j)$ where $0 \le i < 2$, $0 \le j < 3$, as can be checked by showing
that every element of $\Z^2$ is congruent to one of them modulo~$G$,
and no two of them are congruent modulo~$G$`
\end{itemize}
\end{example}

The key results that we will need about fab's are that any subgroup
$H$ of a fab $G$ of rank~$n$ is itself a fab of rank $m \le n$,
and if $m = n$ then $H$ has finite index in~$G$, and the index
can be obtained by taking the absolute value of the determinant of
the change of basis matrix.

In class these proofs were done in the case $n = 2$; the arguments in
the general case are similar.

\begin{theorem}
If $G$ is a fab with basis $e_1, \cdots , e_n$ and $H$ is a subgroup
of~$G$, then $H$ is a fab, and it has a basis $f_j$, $1 \le j \le m$
such that the change of basis matrix is upper triangular, i.e.,
$$ f_i = \sum_{j \ge i} a_{ij} e_j, \qquad a_{ij} \in \Z.$$
In particular, the rank of $H$ is less than or equal to the rank
of~$G$.
\end{theorem}

\begin{proof}
Without loss of generality we can take $G = \Z^n$, and the basis
$e_i$ to be the standard basis vectors.  (This is exactly the
picture we get by writing elements~$g$ of~$G$ as linear combinations
of the basis vectors, and mapping $g$ to the vector of coefficients.)

We prove the result by induction on~$n$.  The case~$n$ is trivial:
any subgroup of~$\Z$ is an ideal and is therefore principal.  So
either $H$ is~$\{0\}$ or else there is a vector $f_1$ such 
that $H = \Z f_1$ consists of all multiples of~$f_1$.

Let $n > 1$, assume that the result is true for all ranks smaller
than~$n$.  Consider the projection~$\pi$ of $\Z^n$ onto
its first $n-1$ components.  The kernel consists of all multiples
$\Z e_n$ of the last basis vector (i.e., all vectors whose first
$n-1$ coefficients are~$0$).  Let $H' = \pi(H)$ be the image
of $H$ under this map, and $K = \ker(\pi) \cap H$.  This
information is summarized by a diagram

By our induction assumption, $H'$ is a fab with rank at most $n-1$
and the change of basis matrix is upper triangular.  Let $f_1, \cdots , f_k$
be elements of~$H$ whose images under~$\pi$ are a basis of~$H'$.
If $K = \{0\}$ then $H = \spn_\Z(f_1, \cdots , f_k)$ and we are
done.  Otherwise, $K$ is an ideal in~$\Z$, and we let $f_{k+1}$ be
a generator.  One checks that $f_1, \cdots , f_{k+1}$ is the
desired basis of~$H$.
\end{proof}

\begin{remark}
The matrix interpretation of this result is that any matrix of
integers can be converted to upper triangular form by row operations
over the integers (i.e., division by an integer is not allowed).
\end{remark}

\begin{corollary}
The ring $\Z_F$ of algebraic integers in a number field~$F$ is
a fab of rank~$n$.
\end{corollary}

\begin{proof}
If $\alpha_i$ is an arbitrary basis of algebraic integers with
discriminant~$d$ then
$$\spn_\Z(\alpha_i) \subset \Z_F \subset \spn_\Z(\alpha_i/d)$$
and the corollary follows immediately, using the theorem.
\end{proof}

If the rank of the subgroup~$H$ is equal to the rank~$n$ of~$G$
then one checks that the set of vectors
$$\sum_{i=1}^n c_i e_i, \quad 0 \le c_i < |a_{ii}|$$
is a set of coset representatives for $H$ in~$G$.  Thus
$H$ is of finite index in~$G$, and the index is equal to
the product of the $|a_{ii}|$, i.e., the absolute value of the
determinant of the (upper triangular) change of basis matrix.

\begin{theorem}
If $e_1, \cdots , e_n$ and $e_1', \cdots , e_n'$ are two
different basis of the fab~$G$, then the change of basis
matrix~$A$ has the property that $\det(A) = \pm.$
\end{theorem}

\begin{proof}
Writing 
$$ e_i' = \sum_{j=1}^n a_{ij} e_j, \qquad 
 e_i = \sum_{j=1}^n b_{ij} e_j'$$
one finds by substitution of one set of equations into the other
that $AB = BA = I_n$.  Thus $\det(A)\det(B) = 1$, and since
the matrix entries, and the determinants of the matrices,
are integers it follows that both determinants are equal to~$\pm 1$.
\end{proof}

\begin{corollary}
If $G$ is a fab of rank~$n$, and $H$ is a subgroup of rank~$n$ then
the index of $H$ in~$G$ is the absolute value of the determinant
of a change of basis matrix.
\end{corollary}

\begin{proof}
The earlier theorem said that this was true for a specific
change of basis matrix~$A$.  Changing the basis for~$H$ changes
the change-of-basis matrix~$A$ my multiplying it on the left
by a matrix $M$ that is the change-of-basis matrix between
the two bases of~$H$.  By the preceding theorem, $\det(M) = \pm 1$
and therefore
$$|\det(MA) | = | \det(A)| = [G:H]$$
as claimed.
\end{proof}


\end{document}

