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 ($\alpha$,$\beta$) = Trace($\alpha$$\beta$) denote the non-degenerate symmetric form on K as above. Also let $\check{R}$ be the collection of elements $\alpha$ in K such that ($\alpha$, R) $\subset$ $\mathbb {Z}$. Now let $\alpha$ be any element of K such that $\alpha$$\check{R}$ $\subset$ $\check{R}$. Then we have ($\alpha$,$\check{R}$) $\subset$ $\mathbb {Z}$. 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 $\check{R}^{-1}_{}$ ^. $\check{R}$ 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 $\mathbb {Q}$. Let a be any non-zero integer and let R be the subring of K generated by w₁ = 1, w₂ = a$\beta$ and w₃ = a$\beta^{2}_{}$. 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 $\alpha$ = a₁w₁ + a₂w₂ + a₃w₃ is such that ($\alpha$, R) $\subset$ $\mathbb {Z}$ we obtain the conditions

3a₁ $$\in$$ $$\mathbb {Z}$$;3a²ba₃ $$\in$$ $$\mathbb {Z}$$;3a²ba₂ $$\in$$ $$\mathbb {Z}$$

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

a₁ $$\in$$ 3a²b ^. $$\mathbb {Z}$$;a₂ $$\in$$ 3ab ^. $$\mathbb {Z}$$;a₃ $$\in$$ 3a ^. $$\mathbb {Z}$$;

A basis for $\check{R}^{-1}_{}$ 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 $\check{R}^{-1}_{}$ ^. $\check{R}$ but w₁ is not. Hence the product is strictly smaller than R.