Unformatted text preview: x , x + 1 , x 2 + x + 1 , x 3 + x 2 + 1 , x 3 + x + 1 , x 4 + x 3 + 1 , x 4 + x + 1 , x 4 + x 3 + x 2 + x + 1 are irreducibles. Note that x 4 + x 2 + 1 = ( x 2 + x + 1) 2 is not irreducible. (7) Problem 47(iii) on page 305. (graded) Solution: x 7 + x 6 + x 4 + 1 = ( x + 1)( x 2 + x + 1)( x 4 + x 3 + 1). Note that these factors are irreducibles by problem 46 above. (8) Problem 2 on page 363. Solution: Units: 1 , 2 , x + 2 , 2 x + 1. Zero divisors: x , 2 x , x + 1 , 2 x + 2. (9) Problem 14(iii) on page 363. (graded) Solution: This can be done by noting that x 2 ≡ 2 mod x 2 + 2, and so x 9 ≡ ( x 2 ) 4 · x ≡ 16 x mod x 2 + 2 . So the question becomes to ﬁnd f ( x ) such that 16 x f ( x ) ≡ 1(mod x 2 + 2). Solving this congruence gives f ( x ) =1 32 x . 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell.
 Spring '08
 BILLERA
 Algebra, Prime Numbers

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