\section{A missing Lemma}
Let $K$ be a finite field extension of $\bbQ$. Let $w_1$, \dots, $w_r$
be a basis of $K$ as a vector space over $\bbQ$. Let $M$ be the
subgroup of $K$ generated by the $w_i$'s. Let $R$ be the collection of
all elements $\alpha$ in $K$ such that $\alpha\cdot M\subset M$. Let
$M^{-1}$ be the collection of all $\alpha$ in $K$ such that
$\alpha\cdot M\subset R$. We had claimed in an earlier version of
these notes that $M\cdot M^{-1}=R$.

{\bf However this statement is false in general}. As we show below
this can be reduced to the case when $M=\check{R}$. In which case what
is being asserted is that $R$ is Gorenstein. But one can give an
example of a non-Gorenstein number ring $R$.

Let $\Trace:K\to\bbQ$ denote the trace map. The $\bbQ$-linear symmetric
form $(\alpha,\beta)=\Trace(\alpha\beta)$ is non-degenerate. Let
$\check{M}$ denote the collection of all $\alpha$ in $K$ such that
$(\alpha, M)\subset \bbZ$; similarly, let $\check{R}$ denote the
collection of all $\alpha$ in $K$ such that $(\alpha, R)\subset
\bbZ$. Let $C_R=(\check{R})^{-1}$ be the collection of all elements
$\alpha$ in $K$ such that $\alpha\cdot\check{R}\subset R$. We claim
that:
\begin{eqnarray*}
  M\cdot\check{M} & = & \check{R} \\
  C_R \cdot \check{R} & = & R
\end{eqnarray*}
Together these conditions imply that $M\cdot\check{M}\cdot C_R=R$. It
follows that $\check{M}\cdot C_R\subset M^{-1}$ and hence $M\cdot
M^{-1}=R$.

Let $v_1$, \dots, $v_r$ be elements of $K$ such that
$(v_i,w_j)=\delta_{ij}$, where the latter is the Kronecker delta;
$\check{M}$ is the group generated by the $v_i$'s. To every element
$\alpha$ of $K$ we can associate a matrix $A(\alpha)$ by putting
$A(\alpha)_{ij}=(\alpha, w_i v_j)$ and then $\alpha w_i = \sum_j
A(\alpha)_{ij} w_j$. This gives a homomorphism from $K$ to $r\times r$
matrices over $\bbQ$ such that $R$ is precisely the collection of
elements whose images are matrices with entries in $\bbZ$. Moreover,
$\Trace(\alpha)=\Trace(A(\alpha))$, where the latter is the trace
$\sum_i A(\alpha)_{ii}$ in the usual sense, of the matrix $A$. For any
matrix $S$ we can define an element $r(S)$ by the condition
$(r(S),\beta)=\Trace(S\cdot A(\beta))$. It follows that
$r(A(\alpha))=\alpha$.

Now, the group $D$ of integer matrices is self-dual under the pairing
$(T,S)=\Trace(TS)$. The image of $R$ in this group $D$ is
``saturated'', \ie if $T$ is a matrix such that $nT$ is in the image
of $R$ for some non-zero integer $n$, then $T$ is in the image of $R$.
It follows that $r(D)=\check{R}$. Now, $D$ is the free group on the
elementary matrices $E_{kl}$ whose only non-zero entry is a ``1'' in
the $(k,l)$-th place. For any matrix $S$ we have $(E_{kl},S)=S_{kl}$.
In particular, for any element $\alpha$ in $K$ we have
$(E_{kl},A(\alpha))=(\alpha,w_k v_l)$. Thus $r(E_{kl})=w_k v_l$ which
is in $M\cdot\check{M}$. It follows that $M\cdot\check{M}$ is $r(D)$
which is $\check{R}$, thus proving the first equality above.

\subsection{Counterexample}
Let $R$ be a subring of a number field $K$ and suppose that $R$ is
finitely generated as an abelian group. Let
$(\alpha,\beta)=\Trace(\alpha\beta)$ denote the non-degenerate
symmetric form on $K$ as above. Also let $\check{R}$ be the collection
of elements $\alpha$ in $K$ such that $(\alpha,R)\subset\bbZ$.  Now
let $\alpha$ be any element of $K$ such that
$\alpha\check{R}\subset\check{R}$. Then we have
$(\alpha,\check{R})\subset\bbZ$. By the non-degeneracy of the pairing
we see that $\alpha$ lies in $R$. Thus by putting $M=\check{R}$ we see
that $R$ is precisely the ring associated by $M$ in the leading
paragraph of this section. Thus to provide a counterexample to the
stated result it is enough to show that there is an $R$ such that
$\check{R}^{-1}\cdot\check{R}$ is strictly smaller than $R$.

Let $b$ be an integer which is not a cube. Let $K$ be the field obtain
by adjoining a cube root $\beta$ of $b$ to $\bbQ$. Let $a$ be any
non-zero integer and let $R$ be the subring of $K$ generated by
$w_1=1$, $w_2=a\beta$ and $w_3=a\beta^2$. We have the identities
\[ 
   w_1^2 = w_1; w_1 w_2 = w_2; w_1 w_3 = w_3;
   w_2^2 = a w_3; w_2 w_3 = a^2 b; w_3^2 = a b w_1
\]
It follows that $\Trace(w_1)=3$ and $\Trace(w_1)=\Trace(w_2)=0$. 
Thus if $\alpha=a_1w_1+a_2w_2+a_3w_3$ is such that
$(\alpha,R)\subset\bbZ$ we obtain the conditions
\[
   3 a_1 \in \bbZ; 3 a^2 b a_3 \in \bbZ; 3 a^2 b a_2 \in \bbZ
\]
Thus a basis for $\check{R}$ is given by $u_1=w_1/3$,
$u_2=w_2/(3a^2b)$ and $u_3=w_3/(3a^2b)$. Now suppose that
$\alpha=a_1w_1+a_2w_2+a_3w_3$  is such that $\alpha\check{R}\subset
R$. We obtain the conditions
\[
  a_1 \in 3a^2b\cdot\bbZ; a_2 \in 3ab\cdot\bbZ; a_3 \in 3a\cdot\bbZ; 
\]
A basis for $\check{R}^{-1}$ is thus given by $v_1=3a^2bw_1$,
$v_2=3abw_2$ and $v_3=3aw_3$. We then compute
\[
  u_1v_1 = a^2 b w_1; u_2 v_3 = a w_1; u_3 v_2 = a b w_1
\]
It follows that $a w_1$ is in the product
$\check{R}^{-1}\cdot\check{R}$ but $w_1$ is not. Hence the product is
strictly smaller than $R$.

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