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From: David Bernier on 2 Aug 2010 16:47
A program tells me that the polynomial p(x) = x^4 + x^3 + x^2 + x + 1 is
irreducible in Q[x]. If a = exp(2pi i/5) is as usual a 5th root
of unity, then I think the polynomial p(x) splits or factors in C[x] as:

x^4 + x^3 + x^2 + x + 1 = (x- a)(x - a^2) ( x - a^3) (x - a^4).

I'm trying to understand what algebraists mean by conjugate roots.

So I's like to know which pairs of roots among a, a^2, a^3 and a^4
are conjugate roots and why.

David Bernier
From: quasi on 2 Aug 2010 18:27
On Mon, 02 Aug 2010 16:47:14 -0400, David Bernier
<david250(a)videotron.ca> wrote:

>A program tells me that the polynomial p(x) = x^4 + x^3 + x^2 + x + 1 is
>irreducible in Q[x]. If a = exp(2pi i/5) is as usual a 5th root
>of unity, then I think the polynomial p(x) splits or factors in C[x] as:
>
> x^4 + x^3 + x^2 + x + 1 = (x- a)(x - a^2) ( x - a^3) (x - a^4).
>
>I'm trying to understand what algebraists mean by conjugate roots.
>
>So I's like to know which pairs of roots among a, a^2, a^3 and a^4
>are conjugate roots and why.

They are all conjugates of each other over Q since they are roots of
the same irreducible polynomial f in Q[x], namely

f(x) = x^4 + x^3 + x^2 + x + 1

On the other hand, over R, f splits into 2 irreducible quadratics
(irreducible over R). Thus, we have

f(x) = g(x) h(x)

where

g(x) = (x - a) (x - a^4)
h(x) = (x - a^2) (x - a^3)

Hence,

a and a^4 are conjugate over R

and

a^2 and a^3 are conjugate over R.

In R[x], all irreducible polynomials of degree greater than 1 are
quadratic, and the 2 roots are just ordinary complex conjugates.

quasi
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