View Single Post
Old 2020-03-04, 09:11   #5
Nick's Avatar
Dec 2012
The Netherlands

25·53 Posts

Originally Posted by wildrabbitt View Post
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 \(X^3-1\) but ζ is not equal to 1.
Let's factorise the polynomial \(X^3-1\).
As 1 is a root of this polynomial, X-1 must be a factor.
Dividing \(X^3-1\) by X-1, we get
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 \(\zeta^3-1=0\) so \((\zeta^2+\zeta+1)(\zeta-1)=0\) but \(\zeta-1\neq 0\)
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 non-zero polynomial of smaller degree.

Let's take a polynomial expression in ζ, for example \(\zeta^3+2\zeta^2-\zeta+3\).
As \(\zeta^2=-\zeta-1\) and \(\zeta^3=1\), we can simplify this:
\[ \zeta^3+2\zeta^2-\zeta+3=1+2(-\zeta-1)-\zeta+3=-3\zeta+2\]
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 \((r+3)\zeta+(s-2)=0\).
But ζ is not a root of any non-zero polynomial of degree 1 (with integer or rational coefficients)
so r+3=0 and s-2=0 giving r=-3 and s=2.

I hope this helps!
Nick is offline   Reply With Quote