Cyclotomic field of 9th root of unity

Question

What I know:

The set $\{1,\zeta,\,...\,,\zeta^{\varphi(n)-1}\}$ is an integral basis for the ring of algebraic integers of $\mathbb{Q}(\zeta_n)$ .

Then, if $n=9$ as $\varphi(9)=6$ the basis of $\mathbb{Z}(\zeta_9)$ will be $\{1,\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5\}$.

The questions:

1. Is $$1+\zeta+\zeta^2+\zeta^3+\zeta^4+\zeta^5+\zeta^6=0\ ?$$

2. Is it Galois group the multiplicative group $\mathbb{Z}/7\mathbb{Z}$ ?

And if the answers are yes:

Is this reasoning always the same for $\mathbb{Z}(\zeta_n)$ ?

Thank you

Answer 1

You can manually calculate $1+\zeta+\dots+\zeta^6$ by taking $\zeta=e^{2\pi i/9}$, and you will see that it is not zero.

For a cyclotomic field extension $\mathbb{Q}(\zeta_n)$, the Galois group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times$ because an automorphism is completely determined by where it sends a primitive $n$th root of unity, i.e. the roots of unity $\zeta^m_n$ where $\gcd(m,n)=1$.

Answer 2

The general rule is as follows:

1. For a given cyclotomic order $n$, select the residues from $1$ to $n-2$ inclusive that are prime to $n$.

2. Form the sum of powers of the primitive root $\zeta_n$ with exponents matching those above.

3. This sum is the Möbius function of $n$, which is

• $+1$ if $n$ is square-free with an even number of prime factors

• $-1$ if $n$ is square-free with an odd number of prime factors

• $0$ if $n$ is not square-free

For instance, with $n=6$ we have $1$ and $5$ as the only residues from $1$ to $5$ that are prime to $6$ (step 1), so the sum is $\zeta_6^1+\zeta_6^5$ (step 2), and with $6=2×3$, square-free with an even number of prime factors, the value of the sum $\zeta_6^1+\zeta_6^5$ will be $+1$ (step 3). You may easily check this with the familiar complex number representations for the sixth roots of unity.

With $n=9$ you should get

$\zeta_9^1+\zeta_9^2+\zeta_9^4+\zeta_9^5+\zeta_9^7+\zeta_9^8=0.$