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 2020-03-03, 14:24 #1 wildrabbitt   Jul 2014 44710 Posts algebraic numbers Can anyone help me understand how to work out the degree of an algebraic number?
2020-03-03, 16:05   #2
Dr Sardonicus

Feb 2017
Nowhere

453610 Posts

Quote:
 Originally Posted by wildrabbitt Can anyone help me understand how to work out the degree of an algebraic number?
That depends on how the given algebraic number is given.

 2020-03-03, 21:36 #3 wildrabbitt   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 $A_1\zeta^1 + A_2\zeta^2 + A_3\zeta^3 +A_4\zeta^4...A_{q-1} + \zeta^{q-1}$ and the expression is unique because the cyclotomic polynomial of degree q-1 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 2020-03-03 at 21:39
2020-03-04, 03:03   #4
Dr Sardonicus

Feb 2017
Nowhere

106708 Posts

Quote:
 Originally Posted by wildrabbitt The book I'm reading says that given a qth root of unity $\zeta$, every polyonomial in $\zeta$ can be expressed as $A_1\zeta^1 + A_2\zeta^2 + A_3\zeta^3 +A_4\zeta^4...A_{q-1} + \zeta^{q-1}$ and the expression is unique because the cyclotomic polynomial of degree q-1 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.
OK, I'm assuming q is prime, and you want the degree of the number represented by some polynomial f (your polynomial, with zeta replaced by x, considered modulo the cyclotomic polynomial).

Of course the degree will always divide q-1. In general, what you need is the minimum polynomial.

In Pari-GP, 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 2020-03-04 at 03:05

2020-03-04, 09:11   #5
Nick

Dec 2012
The Netherlands

7·239 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!

 2020-03-04, 11:39 #6 wildrabbitt   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.
 2020-03-04, 14:17 #7 Nick     Dec 2012 The Netherlands 7×239 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 non-negative integer n such that $$1,z,z^2,\ldots,z^{n-1}$$ 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 $$1,z,z^2,\ldots,z^{n-1}$$ and therefore z is the root of a monic polynomial of degree n.
 2020-03-04, 15:07 #8 Dr Sardonicus     Feb 2017 Nowhere 10001101110002 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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