Thread: Irreducible polys

THis is really long but wanted to check my work

FInd all eighth-roots of 1. Use this to express f (x)=x^8-1 as a product of irreducible polynomials
a) in C [x]
b) in R [x]
c) in Q [x]

Originally Posted by chadlyter

THis is really long but wanted to check my work

FInd all eighth-roots of 1. Use this to express f (x)=x^8-1 as a product of irreducible polynomials
a) in C [x]
b) in R [x]
c) in Q [x]

As 1=exp(2 pi n i) n=0, +/-1, +/-2, .. , the 8th roots of unity are:

r(n) = exp(2 pi (n/8) i) n=0, +/-1, +/-2, .. ,

which give 8 distinct values for n=0, 1, 2, .., 7, with r(0)=1, and r(4)=-1
being the real roots, all the others are complex.

RonL

Hello, chadlyter!

Find all eighth-roots of 1.
Use this to express f (x) = x^8 -1 as a product of irreducible polynomials

a) in C[x] . . b) in R[x] . . c) in Q[x]

The problems are easier in reverse order.


(c) Factor the polynomial.

x^8 - 1 .= .(x^4 - 1)(x^4 + 1) .= .(x² - 1)(x² + 1)(x^4 + 1)

. . = .(x - 1)(x + 1)(x² + 1)(x^4 + 1)


(b) The quartic factor can be factored:
. . . . . . . . . . . . . . . . . . . . . . - . . - . ._ . . . . . . . . . _
x^8 - 1 .= .(x - 1)(x + 1)(x² + 1)(x² - √2x + 1)(x² + √2x + 1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ._
(a) The eight roots are: . ±1, ±i, (±1 ± i)/√2

To save me typing all those ugly radicals,
. . let: .x1 .= .(1 + i)/√2, .x2 .= .(1 - i)/√2, .x3 .= .(-1 + i)/√2, .x4 .= .(-1 - i)/√2

x^8 - 1 .= .(x - 1)(x + 1)(x - i)(x + i)(x - x1)(x - x2)(x - x3)(x - x4)