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Unformatted text preview: Algebra , Assignment 12 Solutions I wasnt sure anymore and I will tell you, it is a strange process to feel ones mind changing, allowing ideas into your brain which it had once considered unthinkable. I cannot say its painful, or particularly pleasurable, but that it requires a certain relaxation of the hold one keeps over oneself, and is to that degree both a thrill and a horror. from The Chess Garden, by Brooks Hansen 1. These problems deal with the field K = Q [ x ] / ( x 3 2). (You can assume, without proof, that x 3 2 is irreducible over Q .) All elements should be written in the form a + bx + cx 2 with a,b,c Q . (There may be intermediate steps in other forms.) (a) Find ( x 2 + 3 x + 7)(2 x 2 + x + 3). Solution: 2 x 4 +7 x 3 +20 x 2 +16 x +21 but now we reduce by x 3 = 2 and x 4 = 2 x to get 4 x +14+20 x 2 +16 x +21 or 20 x 2 +20 x +35. (b) Find x 4 . Solution: As x 3 = 2, x 4 = 2 x . (c) Find x 30 . Solution: As x 3 = 2, x 30 = ( x 3 ) 10 = 2 10 = 1024. (d) Find the multiplicative inverse of x + 3. Solution: (Method 1) First we divide x + 3 into x 3 2: x 3 2 = ( x 2 3 x + 9)( x + 3) + ( 29) and we are done as 29 is a unit. Reversing 29 = ( x 3 2) + ( x 2 + 3 x 9)( x + 3) so that 1 = 1 29 ( x 3 2) + 1 29 ( x 2 + 3 x 9)( x + 3) and so the multiplictative inverse of x + 3 is 1 29 ( x 2 + 3 x 9) . (Method 2) ( + x + x 2 )( x + 3) = 3 + 3 x...
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This note was uploaded on 04/01/2011 for the course MATH 101 taught by Professor Josephs during the Fall '08 term at NYU.
 Fall '08
 JOSEPHS
 Algebra

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