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Old 2020-03-03, 14:24   #1
wildrabbitt
 
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Can anyone help me understand how to work out the degree of an algebraic number?
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Old 2020-03-03, 16:05   #2
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Quote:
Originally Posted by wildrabbitt View Post
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.
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Old 2020-03-03, 21:36   #3
wildrabbitt
 
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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
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Old 2020-03-04, 03:03   #4
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Quote:
Originally Posted by wildrabbitt View Post
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
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Old 2020-03-04, 09:11   #5
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Quote:
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
\[X^3-1=(X^2+X+1)(X-1)\]
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!
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Old 2020-03-04, 11:39   #6
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Thanks so much. It'll take me a while to absorb what the last two posts say.
I'm very grateful for the input.
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Old 2020-03-04, 14:17   #7
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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.
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Old 2020-03-04, 15:07   #8
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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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