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Unformatted text preview: 8.4. FINITELY GENERATED MODULES OVER A PID, PART I 375 s t = = D : It follows that A D U. s t = = I 1;i/A has .1;1/ entry equal to . The case that the nonzero entry is in the first row is handled similarly, with column operations rather than row opera- tions. If divides all the entries in the first row and column, then row and column operations of type 1 can be used to replace the nonzero entries by zeros. n Remark 8.4.9. The proof of this lemma is non-constructive, because in general there is no constructive way to find s and t satisfying s C t D . However, if R is a Euclidean domain, we have an alternative constructive proof. If divides , proceed as before. Otherwise, write D q C r where d.r/ < d./ . A row operation of the first type gives a matrix with r in the .i;1/ position. Then interchanging the first and i th rows yields a matrix with r in the .1;1/ position. Since d.r/ < d./ , we are done....
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- Fall '08