Arithmetic in cyclotomic field

Question

I know that the basis of $\mathbb{Z}(\zeta_9)$ is: $\{1,\zeta_9,\zeta_9^2,\zeta_9^3,\zeta_9^4,\zeta_9^5\}$ .

If I have for example: $3+104\zeta_9+\zeta_9^7$ .

Is it a cyclotomic integer in $\mathbb{Z}(\zeta_9)$ ?

Can I reduce $\zeta_9^7$ to a linear combination of $\zeta_9,\zeta_9^2,\zeta_9^3,\zeta_9^4,\zeta_9^5$ to be inside the basis of the ring?

Thank you

Answer 1

The fact that $\{1,\zeta_9,\zeta_9^2,\zeta_9^3,\zeta_9^4,\zeta_9^5\}$ is a basis for $\Bbb Z(\zeta_9)$ automatically means that any element of $\Bbb Z(\zeta_9)$, including $\zeta_9^7$, can be written as a $\Bbb Z$-linear combination of $1,\zeta_9,\zeta_9^2,\zeta_9^3,\zeta_9^4,\zeta_9^5$. (Note that the $1$ needs to be included.)

The fact that the minimal polynomial of $\zeta_9$ is $x^6+x^3+1$, so that $\zeta_9^6 = -\zeta_9^3-1$, helps us do this concretely: we have $$ \zeta_9^7 = \zeta_9 \zeta_9^6 = \zeta_9 (-\zeta_9^3-1) = -\zeta_9^4-\zeta_9. $$

Answer 2

Yes. You can take advantage of the fact that $9$ is composite to do so.

Specifically, $9$ is divisible by $3$ and so your field contains the cube roots of unity, which appear as $1,\zeta_9^3,\zeta_9^6$.

Then these cube roots add up to zero from the properties of $\mathbb{Q}[\zeta_3]$. Thus all the "missing" powers of $\zeta_9$ are easily rendered:

$\zeta_9^6=-1-\zeta_9^3$

$\zeta_9^7=-\zeta_9-\zeta_9^4$

$\zeta_9^8=-\zeta_9^2-\zeta_9^5$