Thread: algebraic numbers View Single Post
2020-03-04, 09:11   #5
Nick

Dec 2012
The Netherlands

3·19·29 Posts

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 $$X^3-1$$ 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
$X^3-1=(X^2+X+1)(X-1)$
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!