College Algebra Exam Review 361

College Algebra Exam Review 361 - 8.4 FINITELY GENERATED...

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8.4. FINITELY GENERATED MODULES OVER A PID, PART I 371 Proof. We prove this by induction on n . A submodule (ideal) of R is generated by a single element, since R is a PID. This verifies the base case n D 1 . Suppose that F has rank n > 1 and that the assertion holds for free modules of smaller rank. Let f f 1 ; : : : ; f n g be a basis of F , put F 0 D span . f f 1 ; : : : ; f n 1 g / . Let N be a submodule of F and let N 0 D N \ F 0 . By the induction hypothesis, N 0 has a generating set with no more than n 1 elements. Every element x 2 F has a unique expansion x D P n i D 1 ˛ i .x/f i . The map x 7! ˛ n .x/ is an R –module homomorphism from F to R . If ˛ n .N / D f 0 g , then N D N 0 , and N is generated by no more than n 1 elements. Otherwise, the image of N under this map is a nonzero ideal of R , so of the form dR for some nonzero d 2 R . Choose h 2 N such that ˛ n .h/ D d . If x 2 N , then ˛ n .x/ D rd for some r 2 R . Then y D x rh satisfies ˛ n .y/ D 0 , so y 2 N \ F 0 D N 0 . Thus x D y C rh 2 N 0 C Rh .
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