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# A. A missing Lemma

Let K be a finite field extension of . Let w1, ..., wr be a basis of K as a vector space over . Let M be the subgroup of K generated by the wi's. Let R be the collection of all elements in K such that . M M. Let M-1 be the collection of all in K such that . M 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 = . 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 CR = ()-1 be the collection of all elements in K such that . R. We claim that:

 M . = CR . = R

Together these conditions imply that M . . CR = R. It follows that . CR M-1 and hence M . M-1 = R.

Let v1, ..., vr be elements of K such that (vi, wj) = , where the latter is the Kronecker delta; is the group generated by the vi's. To every element of K we can associate a matrix A() by putting A()ij = (, wivj) and then wi = A()ijwj. 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 Ekl whose only non-zero entry is a 1'' in the (k, l )-th place. For any matrix S we have (Ekl, S) = Skl. In particular, for any element in K we have (Ekl, A()) = (, wkvl). Thus r(Ekl) = wkvl which is in M . . It follows that M . is r(D) which is , thus proving the first equality above.

Subsections

Next: A..1 Counterexample Up: Some Lectures on Number Previous: 11.6 Index calculus
Kapil Hari Paranjape 2002-10-20