Finite Atol, The Trace of the Trace

Free Modules

When refering to a tower of field extensions, and an element u in the larger field extension, the norm of the norm is the norm, and the trace of the trace is the trace. The Proof is based on the images of u under various field isomorphisms, and does not generalize well, so let's start again.

Let S be an R algebra and a free R module, with basis c, and let T be an S algebra and a free S module, with basis b. Thus trace make sense in both S/R and T/S.

Cross multiply basis elements to find a basis for T as a free R module. Thus trace makes sense in T/R.

given an R endomorphism e on T, let M be the matrix over R that corresponds uniquely to e. Cut M up into blocks, like a checkerboard, where each block has size equal to the rank of S. Consider block_i,j. Multiply M by a vector v on the right, and this block is subjected to a slice of v, the slice that is built upon b_j crossed with the basis c, having coefficients in R. The result is a slice of the output vector, the slice that is built upon b_i crossed with the basis c, having coefficients in R. This defines, and is defined by, the i^th projection of e(zb_j), where z is the element of S determined by the slice of v. This block is really an R endomorphism on S.

At this point we need another assumption - e is an S module endomorphism. thus e respects addition and scaling by S. This is necessary if we are going to take the trace of e in T/S. Using the basis b, build a matrix P, with elements in S, that implements e. The entry P_i,j establishes the i^th projection of e(zb_j), where z is the coefficient (from S) on b_j. Set z = 1 and P_i,j is the coefficient on b_i of e(b_j). Thus the i,j block in M implements multiplication by P_i,j in S.

The trace of e in T/S is the sum over the main diagonal of P. This is an element w in S, and multiplication by w is realized by adding up the diagonal blocks in M. The trace of w in S/R is the sum of these diagonal elements, but this in turn is the sum of all the diagonal elements of M. Therefore the trace of the trace is the trace.

Norm

There is no comparable result for norm, however, a modest generalization is possible. Let T S and R be integral domains, and let e be multiplication by an element u in T. Clearly this is an S endomorphism on T. Build the matrices P and M as above. Embed everything in the fraction field of T, and the norm of u does not change. It is still the determinant of P. Call this norm w and note that the norm of w has not changed either, within the fraction field of S. Similarly, the norm of u in T/R does not change. Apply the earlier result from field extensions, and the norm of the norm is the norm.

Finite Flat

Let S be finite flat over R, and let T be finite flat over S. Cross the generators to show T is a finitely generated R module. Next, T is the summand of a free S module, and when this free module is produced, each instance of S becomes the summand of a free R module. Bring in the other half of this free R module, for each instance of S, and T becomes the summand of a free R module. Thus T is projective, and a finite flat R module, and trace is well defined.

Let P be a maximal ideal in R and localize about P. Since R_P is a local ring, S_P and T_P become free R_P modules. For notational convenience, I will refer to these rings as S T and R, as though localization had already taken place.

Let b be a basis for S over R. Let b have coefficients in P, rather than R, and find an R submodule H inside S. Multiply b_i by an arbitrary x in H. Each b_ib_j is a linear combination of basis elements, and this is multiplied by the coefficient on b_j, which lies in P. The result lies in H. Put this all together and H is an ideal in S.

Raise H up to a maximal ideal Q in S. This contains the maximal ideal P in R. Since everything in R-P is a unit, Q lies over P. Localize again, with respect to Q, and R and S are both local rings. Now T becomes a free S module.

Start with an S module endomorphism e on T, and suppose the trace of the trace is not the trace. Equality is a local property, so find P in R such that these two values still disagree in R_P. Localize about Q, and remember that trace and localization commute. Within our tower of free extensions, the trace of the trace is not the trace, and that is a contradiction. Therefore the trace of the trace is the trace, for any tower of finite flat extensions.