Quote:
Originally Posted by wildrabbitt
I'm lost.

Let's take q=3 as an example and see if it makes things clearer for you.
Let ζ be a primitive cube root of unity.
This means that \(\zeta^3=1\) but no smaller power of ζ equals 1.
So ζ is a root of the polynomial \(X^31\) but ζ is not equal to 1.
Let's factorise the polynomial \(X^31\).
As 1 is a root of this polynomial, X1 must be a factor.
Dividing \(X^31\) by X1, we get
\[X^31=(X^2+X+1)(X1)\]
Over the integers, we cannot factorize any further: for any integer x,
\(x^2+x+1\) is odd so it cannot be zero.
(It also follows that we cannot factorize any further over the rational numbers, either.)
Now \(\zeta^31=0\) so \((\zeta^2+\zeta+1)(\zeta1)=0\) but \(\zeta1\neq 0\)
and therefore \(\zeta^2+\zeta+1=0\).
Thus we can conclude that ζ is a root of the polynomial \(X^2+X+1\) but not a root of
any nonzero polynomial of smaller degree.
Let's take a polynomial expression in ζ, for example \(\zeta^3+2\zeta^2\zeta+3\).
As \(\zeta^2=\zeta1\) and \(\zeta^3=1\), we can simplify this:
\[ \zeta^3+2\zeta^2\zeta+3=1+2(\zeta1)\zeta+3=3\zeta+2\]
Moreover this expressions is unique:
take any integers (or rational numbers) r and s and suppose that
\(\zeta^3+2\zeta^2\zeta+3=r\zeta+s\) as well.
Then \(r\zeta+s=3\zeta+2\) so \((r+3)\zeta+(s2)=0\).
But ζ is not a root of any nonzero polynomial of degree 1 (with integer or rational coefficients)
so r+3=0 and s2=0 giving r=3 and s=2.
I hope this helps!