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wildrabbitt 2020-03-03 14:24

algebraic numbers
 
Can anyone help me understand how to work out the degree of an algebraic number?

Dr Sardonicus 2020-03-03 16:05

[QUOTE=wildrabbitt;538806]Can anyone help me understand how to work out the degree of an algebraic number?[/QUOTE]That depends on [i]how[/i] the given algebraic number is given.

wildrabbitt 2020-03-03 21:36

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 [B]so [/B]\[\zeta\] [B]can't be a root of a polynomial of lower degree with integral coefficients[/B].


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.

Dr Sardonicus 2020-03-04 03:03

[QUOTE=wildrabbitt;538835]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 [B]so [/B]\[\zeta\] [B]can't be a root of a polynomial of lower degree with integral coefficients[/B].


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.[/QUOTE]
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.

Nick 2020-03-04 09:11

[QUOTE=wildrabbitt;538835]I'm lost.[/QUOTE]
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!

wildrabbitt 2020-03-04 11:39

Thanks so much. It'll take me a while to absorb what the last two posts say.
I'm very grateful for the input.

Nick 2020-03-04 14:17

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.

Dr Sardonicus 2020-03-04 15:07

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 [i]irreducible[/i], 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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