Useful theorem for complex numbers [PHI_p is irreducible]

Zuza · New Member Offline Joined: 23 Nov 2006 Posts: 16

Let $\omega = e^{\frac {2\pi}{p}}$ where $p$ is a prime number and let $P(x) = A_0 + A_1x + A_2x^2 + ... + A_{p - 1}x^{p - 1}$ where $A_0, A_1, ..., A_{p - 1}$ are rational.
If we know that $P(\omega) = 0$ , then prove that $A_0 = A_1 = ... = A_{p - 1}$ .

**Posted:** Sun Mar 08, 2009 10:06 am

bambaman

Consider $F(x) = \sum_{i = 0}^{p - 1} x^i$ .
Let $G(x)$ be the minimal polynomial of $\omega$ - the monic polynomial of least degree such that $\omega$ is its zero.
Such polynomial exists, because $F(\omega) = 0$ and $F$ has rational coefficients. It is unique, because if $G_{1,2}$ are 2 polynomials that satisfy this property, their difference is also a polynomial with rational coefficient, but its degree is smaller (and we can normalize it so that it will be monic).

Fact 1: $F$ is irreducible - it is a known fact about Cyclotomic Polynomials, proofs can be found anywhere. A easy proof using Eisenstein's criterion can be found here:
http://en.wikipedia.org/wiki/Eisenstein%27s_criterion

Fact 2: $G$ divides any polynomial with rational coefficients such that $\omega$ is its zero.
Proof: assume otherwise. Divide those 2 polynomials, and call the residue $R(x)$ . R's degree is less than G's degree and $\omega$ is its zero. Normalize it (possible because it is non-zero) and you get a contradiction to minimality.

Combine those 2 facts and get that $F$ is the minimal polynomial: $F = G$ .

If $A_0 = 0$ , $P$ must be the zero polynomial (its degree is smaller than F's degree). Thus $A_0 = A_1 = \cdots = A_{p - 1} = 0$ .
Otherwise, $F | \frac {P}{A_0}, deg(F) = deg(P) \implies P(x) = A_0F(x)$ and the conclusion follows.

**Posted:** Sun Mar 08, 2009 10:36 am

Zuza · New Member Offline Joined: 23 Nov 2006 Posts: 16

Very nice

Thanks

**Posted:** Sun Mar 08, 2009 11:13 am

t0rajir0u

This is an interesting theorem when stated geometrically: it states that an equiangular polygon with a prime number of sides, all of which have rational length, must in fact be equilateral, or stated perhaps more naturally, an equiangular polygon with a prime number of sides which is not equilateral has at least one irrational side. See, for example, the discussion here.

**Posted:** Sun Mar 08, 2009 11:50 am

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boxedexe

I would also like to mention that this is not only useful geometrically but it also has applications in combinatorics. See, for example, the book Complex Numbers from A to...Z (english version by Titu Andreescu and Dorin Andrica).

**Posted:** Sun Mar 08, 2009 12:14 pm

The.Puppet.Master

Thankyou. I love simple little things to like this.

**Posted:** Fri Apr 10, 2009 5:57 am

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