College Algebra Exam Review 378

College Algebra Exam Review 378 - i and y i 1 mod i : If...

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388 8. MODULES The result is u 1 .x/ D 1 500 .11x ± 2/ u 2 .x/ D 1 4000 .11x 3 ± 112x 2 C 405x ± 554/; and u 3 .x/ D ± x 2 ± 8x C 19 ² u 2 .x/: For any polynomial b.x/ , we have Œb.x/Ł D Œb.x/u 1 .x/r 1 .x/Ł C Œb.x/u 2 .x/r 2 .x/Ł C Œb.x/u 3 .x/r 3 .x/Ł is the explicit decomposition of Œb.x/Ł into primary components. For ex- ample, if we take b.x/ D x 5 C 4x 3 (and reduce polynomials mod a.x/ at every opportunity) we get Œb.x/Ł D 3 500 .2x C 11/Œr 1 .x/Ł ± 5 16 Œr 2 .x/Ł C 9 2000 .289x 2 ± 324x C 2271/Œr 3 .x/Ł: Using similar considerations, we can also obtain a procedure for solv- ing any number of simultaneous congruences. Theorem 8.5.13. (Chinese remainder theorem). Suppose ˛ 1 , ˛ 2 , . . . , ˛ s are pairwise relatively prime elements of a principal ideal domain R , and x 1 , x 2 , . . . , x s are arbitrary elements of R . There exists an x 2 R such that x ² x i mod ˛ i for 1 ³ i ³ s . Moreover, x is unique up to congruence mod a D ˛ 1 ˛ 2 ´´´ ˛ s . Proof. We wish to find elements y i for 1 ³ i ³ s such that y i ² 0 mod ˛ j for j ¤
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Unformatted text preview: i and y i 1 mod i : If this can be done, then x D x 1 y 1 C x 2 y 2 C x s y s is a solution to the simultaneous congruence problem. As a rst approx-imation to y i , take r i D a= i . Then r i mod j for j i . More-over, r i is relatively prime to i , so there exist elements u i ;v i such that 1 D u i r i C v i i . Set y i D u i r i . Then y i D u i r i 1 mod i and y i mod j for j i . The proof of the uniqueness statement is left to the reader. n Example 8.5.14. Consider 1 .x/ D x 2 C 1 , 2 .x/ D .x 1/ and 3 .x/ D .x 3/ 3 . Find polynomials y i .x/ in Q x for 1 i 3 such that y i .x/ 1 mod i .x/ and y i mod j .x/ for j i ....
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