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Let *R* be a subring of a number field *K* and suppose that *R* is
finitely generated as an abelian group. Let
(,) = Trace() denote the non-degenerate
symmetric form on *K* as above. Also let be the collection
of elements in *K* such that
(, *R*) . Now
let be any element of *K* such that
. Then we have
(,) . By the non-degeneracy of the pairing
we see that lies in *R*. Thus by putting
*M* = 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
^{ . } 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 of *b* to
. Let *a* be any
non-zero integer and let *R* be the subring of *K* generated by
*w*_{1} = 1,
*w*_{2} = *a* and
*w*_{3} = *a*. We have the identities

*w*_{1}^{2} = *w*_{1};*w*_{1}*w*_{2} = *w*_{2};*w*_{1}*w*_{3} = *w*_{3};*w*_{2}^{2} = *aw*_{3};*w*_{2}*w*_{3} = *a*^{2}*b*;*w*_{3}^{2} = *abw*_{1}

It follows that
Trace(*w*_{1}) = 3 and
Trace(*w*_{1}) = Trace(*w*_{2}) = 0.
Thus if
= *a*_{1}*w*_{1} + *a*_{2}*w*_{2} + *a*_{3}*w*_{3} is such that
(, *R*) we obtain the conditions
Thus a basis for is given by *u*_{1} = *w*_{1}/3,
*u*_{2} = *w*_{2}/(3*a*^{2}*b*) and
*u*_{3} = *w*_{3}/(3*a*^{2}*b*). Now suppose that
= *a*_{1}*w*_{1} + *a*_{2}*w*_{2} + *a*_{3}*w*_{3} is such that
*R*. We obtain the conditions
*a*_{1} 3

*a*^{2}*b*^{ . };

*a*_{2} 3

*ab*^{ . };

*a*_{3} 3

*a*^{ . };

A basis for
is thus given by
*v*_{1} = 3*a*^{2}*bw*_{1},
*v*_{2} = 3*abw*_{2} and *v*_{3} = 3*aw*_{3}. We then compute
*u*_{1}*v*_{1} = *a*^{2}*bw*_{1};*u*_{2}*v*_{3} = *aw*_{1};*u*_{3}*v*_{2} = *abw*_{1}

It follows that *aw*_{1} is in the product
^{ . } but *w*_{1} is not. Hence the product is
strictly smaller than *R*.

** Next:** B. Comparison with ``classical''
** Up:** A. A missing Lemma
** Previous:** A. A missing Lemma
Kapil Hari Paranjape
2002-10-20