20200303, 14:24  #1 
Jul 2014
3×149 Posts 
algebraic numbers
Can anyone help me understand how to work out the degree of an algebraic number?

20200303, 16:05  #2 
Feb 2017
Nowhere
2^{6}×59 Posts 

20200303, 21:36  #3 
Jul 2014
3·149 Posts 
The book I'm reading says that given a qth root of unity \[\zeta\], every polyonomial in \[\zeta\] can be expressed as
\[A_1\zeta^1 + A_2\zeta^2 + A_3\zeta^3 +A_4\zeta^4...A_{q1} + \zeta^{q1}\] and the expression is unique because the cyclotomic polynomial of degree q1 of which \[\zeta\] is a zero is irreducible over the rational field so \[\zeta\] can't be a root of a polynomial of lower degree with integral coefficients. I know the proof that cyclotomic polynomials are irreducible but I don't get why..... (don't even know what I'm unclear about). I'm lost. I asked the question I asked orignally because I thought it might help me understand. Last fiddled with by wildrabbitt on 20200303 at 21:39 
20200304, 03:03  #4  
Feb 2017
Nowhere
2^{6}×59 Posts 
Quote:
Of course the degree will always divide q1. In general, what you need is the minimum polynomial. In PariGP, you can ask for minpoly(Mod(f, polcyclo(q))). This is the minimum polynomial, and its degree is the degree you want. Last fiddled with by Dr Sardonicus on 20200304 at 03:05 

20200304, 09:11  #5 
Dec 2012
The Netherlands
2×3^{2}×83 Posts 
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! 
20200304, 11:39  #6 
Jul 2014
3×149 Posts 
Thanks so much. It'll take me a while to absorb what the last two posts say.
I'm very grateful for the input. 
20200304, 14:17  #7 
Dec 2012
The Netherlands
2×3^{2}×83 Posts 
If you're comfortable with linear algebra, that is probably the easiest way to understand it.
The complex numbers form a vector space over the rational numbers. Take any complex number z and consider the sequence: \[ 1,z,z^2,z^3,\ldots\] It can happen that the terms remain linearly independent however far we go. In that case we call z a transendental number. Otherwise there exists a nonnegative integer n such that \(1,z,z^2,\ldots,z^{n1}\) are linearly independent but \(1,z,z^2,\ldots,z^n\) are not. This is the case in which we call z an algebraic number. It then follows that \(z^n\) can be written as a linear combination of \(1,z,z^2,\ldots,z^{n1}\) and therefore z is the root of a monic polynomial of degree n. 
20200304, 15:07  #8 
Feb 2017
Nowhere
2^{6}·59 Posts 
A "reduced" polynomial expression may also be obtained using polynomial division with quotient and remainder.
If h(x) is a polynomial in Q[x] (which I will assume to be of degree at least 1 to avoid trivialities) , and f(x) any polynomial in Q[x], we have f(x) = h(x)*q(x) + r(x), with q(x) and r(x) in Q[x], and the degree of r(x) is strictly less than the degree of h(x). Clearly r(x) is unique. If h(z) = 0, we then have f(z) = r(z). If h(x) is irreducible, h(z) = 0, and r(x) is not the zero polynomial [that is, if h(x) does not divide f(x)], then r(z) is not 0. 
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