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From: JEMebius on 3 Aug 2010 06:15
quasi wrote:
> 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

In my previous reply I mistyped something and forgot something...

I said

"A more well-known example of conjugacy: x + iy and x - iy (x, y real) in the field of
complex numbers. The field extension is {Q:C}. The base field is R, the superfield is C.
C as an R-linear space is 2-dimensional, in formula: [C:R] = 2. Conjugates come in pairs."

I meant of course "The field extension is {C:R}."

Forgot to mention that in the situation of a field extension {G:F} each element of G not
in F is transformed into all its conjugates in turn by the automorphisms of G:F; each
element of F is transformed into itself.
So conjugate elements belong together by means of the Galois group of {G:F}.

Johan E. Mebius
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