Canonization Algorithm for Z

       ∗ ∗ · · · ∗ ∗ .. . ∗ ∗ · · · ∗ ∗ · · ·        Fig. 7.12. The column-echelon form. An ∗ indicates a nonzero values and the pivots are boxed. basis for Z k . In our example, we have ˜ M 1 =        cd bc ab z 1 z 2 d t c t 1 b t t a t        , 7.6 where z 1 = ad − cd − t · bc − t · ab and z 2 = ac − t 2 · bc − t 2 · ab form a homo- geneous basis for Z 1 . The procedure that arrives at the echelon form is Gaussian elimination on the columns, utilizing elementary column operations of types 1, 3 only. Starting with the left-most column, we eliminate nonzero entries occurring in pivot rows in order of increasing row. To eliminate an entry, we use an elementary column operation of type 3 that maintains the homogeneity of the basis and matrix elements. We continue until we either arrive at a zero column or we find a new pivot. If needed, we then perform a column exchange type 1 to reorder the columns appropriately. Theorem 7.4 echelon form The pivots in column-echelon form are the same as the diagonal elements in normal form. Moreover, the degree of the basis elements on pivot rows is the same in both forms. Proof Because of our sort, the degree of row basis elements ˆe i is monotonically decreasing from the top row down. Within each fixed column j, deg e j is a constant c. By Equation 7.4, deg M k i, j = c − deg ˆe i . Therefore, the degree of the elements in each column is monotonically increasing with row. We may eliminate nonzero elements below pivots using row operations that do not change the pivot elements or the degrees of the row basis elements. We then place the matrix in diagonal form with row and column swaps. The theorem states that if we are only interested in the degree of the basis M k+1 = 0 i j k M x m m k−1 k m m x k k+1 j i Fig. 7.13. As ∂ k ∂ k +1 = ∅, M k M k +1 = 0, and this is unchanged by elementary opera- tions. When M k is reduced to echelon form ˜ M k by column operations, the correspond- ing row operations zero out rows in M k +1 that correspond to pivot columns in ˜ M k . elements, we may read them off from the echelon form directly. That is, we may use the following corollary of the standard structure theorem to obtain the description. Corollary 7.2 Let ˜ M k be the column-echelon form for ∂ k relative to bases {e j } and { ˆe i } for C k and Z k − 1 , respectively. If row i has pivot ˜ M k i, j = t n , it contributes Σ deg ˆe i F [t]t n to the description of H k − 1 . Otherwise, it contributes Σ deg ˆe i F [t]. Equivalently, we get deg ˆe i , deg ˆe i + n and deg ˆe i , ∞, respec- tively, as P-intervals for H k − 1 . In our example, ˜ M 1 1, 1 = t in Equation 7.6. As deg d = 1, the element contributes Σ 1 R [t]t or the P-interval 1,2 to the description of H . We now wish to represent ∂ k +1 in terms of the basis we computed for Z k . We begin with the standard matrix representation M k +1 of ∂ k +1 . As ∂ k ∂ k +1 = ∅, M k M k +1 = 0, as shown in Figure 7.13. Furthermore, this relationship is un- changed by elementary operations. Since the domain of ∂ k is the codomain of ∂ k +1 , the elementary column operations we used to transform M k into echelon form ˜ M k give corresponding row operations on M k +1 . These row operations zero out rows in M k +1 that correspond to nonzero pivot columns in ˜ M k and give a representation of ∂ k +1 relative to the basis we just computed for Z k . This is precisely what we are after. We can get it, however, with hardly any work. Theorem 7.5 basis change To represent ∂ k +1 relative to the standard basis for C k +1 and the basis computed for Z k , simply delete rows in M k +1 that cor- respond to pivot columns in ˜ M k . Proof We only used elementary column operations of types 1,3 in our vari- ant of Gaussian elimination. Only the latter changes values in the matrix.