Next: B. Comparison with classical'' Up: A. A missing Lemma Previous: A. A missing Lemma

## A..1 Counterexample

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 w1 = 1, w2 = a and w3 = a. We have the identities

w12 = w1;w1w2 = w2;w1w3 = w3;w22 = aw3;w2w3 = a2b;w32 = abw1

It follows that Trace(w1) = 3 and Trace(w1) = Trace(w2) = 0. Thus if = a1w1 + a2w2 + a3w3 is such that (, R) we obtain the conditions

3a1 ;3a2ba3 ;3a2ba2

Thus a basis for is given by u1 = w1/3, u2 = w2/(3a2b) and u3 = w3/(3a2b). Now suppose that = a1w1 + a2w2 + a3w3 is such that R. We obtain the conditions

a1 3a2b . ;a2 3ab . ;a3 3a . ;

A basis for is thus given by v1 = 3a2bw1, v2 = 3abw2 and v3 = 3aw3. We then compute

u1v1 = a2bw1;u2v3 = aw1;u3v2 = abw1

It follows that aw1 is in the product . but w1 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