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# sol12 - Algebra Assignment 12 Solutions I wasnt sure...

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Algebra , Assignment 12 Solutions I wasn’t sure anymore and I will tell you, it is a strange process to feel one’s mind changing, allowing ideas into your brain which it had once considered unthinkable. I cannot say it’s 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) .

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(Method 2) ( α + + x 2 γ )( x + 3) = 3 α + 3 + 3 x 2 γ + + x 2 β + 2 γ as the x times x 2 γ term becomes x 3 γ which is 2
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sol12 - Algebra Assignment 12 Solutions I wasnt sure...

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