A. A missing Lemma

Let K be a finite field extension of $\mathbb {Q}$. Let w₁, ..., w_r be a basis of K as a vector space over $\mathbb {Q}$. 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 $\alpha$ ^. M $\subset$ M. Let M^-1 be the collection of all $\alpha$ in K such that $\alpha$ ^. M $\subset$ R. We had claimed in an earlier version of these notes that M ^. M^-1 = R.

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$$\mathbb {Q}$ denote the trace map. The $\mathbb {Q}$-linear symmetric form ($\alpha$,$\beta$) = Trace($\alpha$$\beta$) is non-degenerate. Let $\check{M}$ denote the collection of all $\alpha$ in K such that ($\alpha$, M) $\subset$ $\mathbb {Z}$; similarly, let $\check{R}$ denote the collection of all $\alpha$ in K such that ($\alpha$, R) $\subset$ $\mathbb {Z}$. Let C_R = ($\check{R}$)^-1 be the collection of all elements $\alpha$ in K such that $\alpha$ ^. $\check{R}$ $\subset$ R. We claim that:

$$\check{M} \check{R}$$ =

$$\check{R}$$ = R

Together these conditions imply that M ^. $\check{M}$ ^. C_R = R. It follows that $\check{M}$ ^. C_R $\subset$ M^-1 and hence M ^. M^-1 = R.

Let v₁, ..., 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_iv_j) and then $\alpha$w_i = $\sum_{j}^{}$A($\alpha$)_ijw_j. This gives a homomorphism from K to r×r matrices over $\mathbb {Q}$ such that R is precisely the collection of elements whose images are matrices with entries in $\mathbb {Z}$. 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 ^. 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'', 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) = $\check{R}$. Now, D is the free group on the elementary matrices E<sub>kl</sub> 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_kv_l). Thus r(E_kl) = w_kv_l which is in M ^. $\check{M}$. It follows that M ^. $\check{M}$ is r(D) which is $\check{R}$, thus proving the first equality above.