# HW7_soln - Homework#7 Solutions#1 Let I = f(x g(x Since F[x...

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Homework #7 Solutions #1 Let I = h f ( x ) , g ( x ) i . Since F [ x ] is a PID there is an h ( x ) F [ x ] so that I = h h ( x ) i . But this implies that h ( x ) divides both f ( x ) and g ( x ). As f ( x ) and g ( x ) are relatively prime this can only happen if h ( x ) is a unit in F [ x ]. Hence h f ( x ) , g ( x ) i = I = h h ( x ) i = F [ x ]. Since 1 F [ x ] we conclude that there exist r ( x ) , s ( x ) F [ x ] so that r ( x ) f ( x ) + s ( x ) g ( x ) = 1. #2 According to Exercise 1, there are polynomials r 1 ( x ) , s 1 ( x ) F [ x ] so that r ( x ) f ( x ) + s ( x ) g ( x ) = 1. Multiplying both sides of this equation by c ( x ) we obtain r 2 ( x ) f ( x ) + s 2 ( x ) g ( x ) = c ( x ), where r 2 ( x ) = c ( x ) r 1 ( x ) and s 2 ( x ) = c ( x ) s 1 ( x ). Now apply the divi- sion algorithm to obtain r 2 ( x ) = a ( x ) g ( x ) + r ( x ) s 2 ( x ) = b ( x ) f ( x ) + s ( x ) where a ( x ) , b ( x ) , r ( x ) , s ( x ) F [ x ], r ( x ) = 0 or deg r ( x ) <
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