dummit foote solutions

dummit foote solutions - Homework #01, due 1/20/10 = 9.1.2,...

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Homework #01, due 1/20/10 = 9.1.2 , 9.1.4 , 9.1.6 , 9.1.8 , 9.2.3 Additional problems for study: 9.1.1 , 9.1.3 , 9.1.5 , 9.1.13 , 9.2.1 , 9.2.2 , 9.2.4 , 9.2.5 , 9.2.6 , 9.3.2 , 9.3.3 9.1.1 (This problem was not assigned except for study, but it’s useful for the next problem.) Let p and q be polynomials in Z [ x,y,z ], where p = p ( x,y,z ) = 2 x 2 y - 3 xy 3 z + 4 y 2 z 5 q = q ( x,y,z ) = 7 x 2 + 5 x 2 y 3 z 4 - 3 x 2 z 3 (a) Write each of p and q as a polynomial in x with coefficients in Z [ y,z ]. p = (2 y ) x 2 - (3 y 3 z ) x + 4 y 2 z 5 q = (7 + 5 y 3 z 4 - 3 z 3 ) x 2 (b) Find the degree of each of p and q . deg( p ) = 7 deg( q ) = 9 (c) Find the degree of p and q in each of the three variables x , y , and z . deg x ( p ) = 2 deg x ( q ) = 2 deg y ( p ) = 3 deg y ( q ) = 3 deg z ( p ) = 5 deg z ( q ) = 4 (d) Compute pq and find the degree of pq in each of the three variables x , y , and z . pq = (2 x 2 y - 3 xy 3 z + 4 y 2 z 5 )(7 x 2 + 5 x 2 y 3 z 4 - 3 x 2 z 3 ) = 2 x 2 y (7 x 2 + 5 x 2 y 3 z 4 - 3 x 2 z 3 ) - 3 xy 3 z (7 x 2 + 5 x 2 y 3 z 4 - 3 x 2 z 3 ) + 4 y 2 z 5 (7 x 2 + 5 x 2 y 3 z 4 - 3 x 2 z 3 ) = 2 x 2 y 7 x 2 + 2 x 2 y 5 x 2 y 3 z 4 - 2 x 2 y 3 x 2 z 3 - 3 xy 3 z 7 x 2 - 3 xy 3 z 5 x 2 y 3 z 4 + 3 xy 3 z 3 x 2 z 3
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2 + 4 y 2 z 5 7 x 2 + 4 y 2 z 5 5 x 2 y 3 z 4 - 4 y 2 z 5 3 x 2 z 3 = 14 x 4 y + 10 x 4 y 4 z 4 - 6 x 4 yz 3 - 21 x 3 y 3 z - 15 x 3 y 6 z 5 + 9 x 3 y 3 z 4 + 28 x 2 y 2 z 5 + 20 x 2 y 5 z 9 - 12 x 2 y 2 z 8 deg x ( pq ) = 2 + 2 = 4 deg y ( pq ) = 3 + 3 = 6 deg z ( pq ) = 5 + 4 = 9 (e) Write pq as a polynomial in the variable z with coefficients in Z [ x,y ]. pq = 14 x 4 y - (21 x 3 y 3 ) z - (6 x 4 y ) z 3 + (10 x 4 y 4 + 9 x 3 y 3 ) z 4 + (28 x 2 y 2 - 15 x 3 y 6 ) z 5 - (12 x 2 y 2 ) z 8 + (20 x 2 y 5 ) z 9 9.1.2 Repeat the preceding exercise under the assumption that the coefficients of p and q are in Z / 3 Z . Let p and q be polynomials in Z / 3 Z [ x,y,z ], where p = p ( x,y,z ) = 2 x 2 y + y 2 z 5 q = q ( x,y,z ) = x 2 + 2 x 2 y 3 z 4 (a) Write each of p and q as a polynomial in x with coefficients in Z / 3 Z [ y,z ]. We need only reduce the coefficients modulo 3, which gives us p = ( 2 y ) x 2 + y 2 z 5 q = ( 1 + 2 y 3 z 4 ) x 2 (b) Find the degree of each of p and q . deg( p ) = 7 deg( q ) = 9 (c) Find the degree of p and q in each of the three variables x , y , and z . deg x ( p ) = 2 deg x ( q ) = 2 deg y ( p ) = 2 deg y ( q ) = 3 deg z ( p ) = 5 deg z ( q ) = 4
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3 (d) Compute pq and find the degree of pq in each of the three variables x , y , and z . pq = ( 2 x 2 y + y 2 z 5 )( x 2 + 2 x 2 y 3 z 4 ) = 2 x 2 y ( x 2 + 2 x 2 y 3 z 4 ) + y 2 z 5 ( x 2 + 2 x 2 y 3 z 4 ) = 2 x 2 y ( x 2 ) + 2 x 2 y ( 2 x 2 y 3 z 4 ) + y 2 z 5 x 2 + y 2 z 5 ( 2 x 2 y 3 z 4 ) = 2 x 4 y + x 4 y 4 z 4 + x 2 y 2 z 5 + 2 x 2 y 5 z 9 deg x ( pq ) = 2 + 2 = 4 deg y ( pq ) = 3 + 3 = 6 deg z ( pq ) = 5 + 4 = 9 (e) Write pq as a polynomial in the variable z with coefficients in Z / 3 Z [ x,y ]. pq = 2 x 4 y + x 4 y 4 z 4 + x 2 y 2 z 5 + 2 x 2 y 5 z 9 9.1.4 Prove that the ideals ( x ) and ( x,y ) are prime ideals in Q [ x,y ] but only the latter ideal is a maximal ideal. To show ( x ) is a prime ideal of Q [ x,y ], we must assume pq ( x ), where p,q Q [ x,y ], and show that either p ( x ) or q ( x ). Suppose p = a n ( y ) x n + ··· + a 1 ( y ) x + a 0 ( y ) and q = b m ( y ) x m + ··· + b 1 ( y ) x + b 0 ( y ), where a n ( y ) ,...,a 0 ( y ) ,b m ( y ) ,...,b 0 ( y ) Q [ y ]. Then pq has exactly one term with x -degree 0, namely the polynomial a 0 ( y ) b 0 ( y ) Q [ y ].
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This note was uploaded on 11/08/2011 for the course MATH 224 taught by Professor Gurel during the Spring '05 term at Boğaziçi University.

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dummit foote solutions - Homework #01, due 1/20/10 = 9.1.2,...

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