This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. 14C.47(iii) [graded out of 3 points] 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. 17A.2 [graded out of 4 points] Units: 1 , 2 ,x + 2 , 2 x + 1. Zero divisors: x, 2 x,x + 1 , 2 x + 2. 1 17A.14(iii) [graded out of 5 points] This can be done by noting that x 9 16 x (mod x 2 + 2). So the question becomes to nd f ( x ) such that 16 xf ( x ) 1(mod x 2 + 2). Solving this congruence gives f ( x ) =1 32 x . 2...
View
Full
Document
 Spring '08
 BILLERA
 Algebra, Remainder

Click to edit the document details