# Find the minimum polynomial of a sum of roots of unity.

Let $\omega$ be an 11-th primitive root of 1 over $\Bbb Q$

Let $\beta = \omega + \omega^9$

Find $[ \Bbb Q ( \beta) : \Bbb Q ) ]$ and Find the minimum polynomail of $\beta$.

I asked a question similar to this before posted here Minimal polynomial over Q and I have been able to use the hint given in this question to find the minimal polynomail of $\beta$ in this question which I got to be $\beta^5+\beta^4+\beta^3+\beta^2-21\beta-7$, however the next question supposes that

$\beta = \omega + \omega^3 + \omega^9$

Sqauring this out gets messy very quickly and I can't see a similiar trick as I used in the first part to solve it, do I just have to deal with the messy squares or is their a simplier way I'm not seeing?

Also in general for questions such as this, whats the best method for working out the minimum polynomial??

• You are not asking about a root of unity but about a sum of such. I am changing the title to match. – Marc van Leeuwen Apr 24 '15 at 8:50

## 1 Answer

Recall that the Galois group $G$ of $\mathbb Q(\omega)/\mathbb Q$ is cyclic, generated for example by the automorphism that sends $\omega\mapsto \omega^2$. The Galois group of $\mathbb Q(\omega)/\mathbb Q(\beta)$ is a subgroup thereof, namely the stabilizer of $\mathbb Q(\beta)$, and the Galois group of $\mathbb Q(\beta)/\mathbb Q$ is the quotient of these groups.

With $\beta=\omega+\omega^3+\omega^9$, the orbit of $\beta$ consists of elements of the same form $\omega^a+\omega^b+\omega^c$ and in fact must have length $10$, because any $\sigma\in G$ with $\sigma(\beta)=\beta$ would induce a permutation of $\{\omega,\omega^3,\omega^9\}$ (because otherwise we obtain a polynomial in $\omega$ involving less terms than its minimal polynomial). This permutation cannot be of order $3$ because $|G|=10$. Hence it has one of these three elements as fixedpoint, hence must itself be the identity (because $\mathbb Q(\omega^3)=\mathbb Q(\omega^9)=\mathbb Q(\omega)$). - The same argument works for $\beta=\omega+\omega^9$ because the permutation that sends $\omega\mapsto\omega^9$ sends $\omega^9\mapsto\omega^{9\cdot 9}=\omega^4\ne\omega$.

While this argument shows how to find the degree of the minimal polynomial, it does not produce the polynomial itself. You could try to write down the orbits of $\beta$ explicitly and compute the elementary symmetric polynomials in these. However, while we quickly find the higher terms $X^{10}+3X^9+\ldots$ (resp. $X^{10}+2X^9+\ldots$), even with the simple rules for multiplication of powers of $\omega$ this method also gets messy. The most generic method (especially if $\omega$ itself had a more complicated irreducible polynomial) is indeed simply to compute the powers of $\beta$ step by step and look for linear dependencies.