**However this statement is false in general**. As we show below
this can be reduced to the case when
*M* = . 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* denote the trace map. The
-linear symmetric
form
(,) = Trace() is non-degenerate. Let
denote the collection of all in *K* such that
(, *M*) ; similarly, let denote the
collection of all in *K* such that
(, *R*) . Let
*C*_{R} = ()^{-1} be the collection of all elements
in *K* such that
^{ . } *R*. We claim
that:

M^{ . } |
= | ||

C_{R}^{ . } |
= | R |

Together these conditions imply that

Let *v*_{1}, ..., *v*_{r} be elements of *K* such that
(*v*_{i}, *w*_{j}) = , where the latter is the Kronecker delta;
is the group generated by the *v*_{i}'s. To every element
of *K* we can associate a matrix *A*() by putting
*A*()_{ij} = (, *w*_{i}*v*_{j}) and then
*w*_{i} = *A*()_{ij}*w*_{j}. This gives a homomorphism from *K* to *r*×*r*
matrices over
such that *R* is precisely the collection of
elements whose images are matrices with entries in
. Moreover,
Trace() = Trace(*A*()), where the latter is the trace
*A*()_{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*),) = Trace(*S*^{ . }*A*()). It follows that
*r*(*A*()) = .

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'', i. e. 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*) = . 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 in *K* we have
(*E*_{kl}, *A*()) = (, *w*_{k}*v*_{l}). Thus
*r*(*E*_{kl}) = *w*_{k}*v*_{l} which
is in
*M*^{ . }. It follows that
*M*^{ . } is *r*(*D*)
which is , thus proving the first equality above.