{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

dummit foote solutions

# dummit foote solutions - Homework#01 due = 9.1.2 9.1.4...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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 ].

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 94

dummit foote solutions - Homework#01 due = 9.1.2 9.1.4...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online