Unformatted text preview: Q 2 Mat n. Z / such that PAQ is a diagonal 1 –by– n matrix, PAQ D .d;0;:::;0/ , with d ³ . P is just multiplication by ˙ 1 , so we can absorb it into Q , giving AQ D .d;0;:::;0/ . Let .t 1 ;:::;t n / denote the entries of the ﬁrst column of Q . Then we have d D t 1 a 1 C ±±± t n a n , and d is in the subgroup of Z generated by a 1 ;:::;a n . On the other hand, let .b 1 ;:::;b n / denote the entries of the ﬁrst row of Q ± 1 . Then A D .d;0;:::;0/Q ± 1 imples that a i D db i for 1 ² i ² n . Therefore, d is nonzero, and is a common divisor of a 1 ;:::;a n . It follows that d is the greatest common divisor of a 1 ;:::;a n . Lemma 3.5.12. Let .v 1 ;:::;v n / be a sequence of n elements of Z n . Let P D Œv 1 ;:::;v n Ł be the n –by– n matrix whose j th column is v j . The following conditions are equivalent: (a) f v 1 ;:::;v n g is an basis of Z n . (b) f v 1 ;:::;v n g generates Z n . (c) P is invertible in Mat n . Z / ....
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 Fall '08
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 Algebra, Greatest common divisor, Matn .Z/, Propsition 3.5.9

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