# Basis for the field extension $\mathbb{Q}(\zeta_{12})$

Consider the cyclotomic field $\mathbb{Q}(\zeta_{12})$ where $\zeta_{12}$ represents the $12$-th primitive root of unity. Since the minimal polynomial of $\zeta_{12}$ is given by $\Phi_{12}(x)$ which has degree $4$, we have that, $$[ \mathbb{Q}(\zeta_{12}) : \mathbb{Q} ] =4$$ And so I would think that $$a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$ would be a basis for $\mathbb{Q}(\zeta_{12})$.

Now consider the isomorphism that maps $5 \in (\mathbb{Z}_{12})^{\times}$ to $\rho_5 (\zeta_{12}) = \zeta_{12}^5$.

I compute the following, $$\rho_5 (a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3) = a + b \zeta_{12}^5 + \zeta_{12}^{10} + \zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$

I'm taking the exponents to be modulo $4$.

I don't think my calculation is correct since this isomorphism should not fix the entire field by Galois Theory. What am I doing wrong ?

## 1 Answer

What makes you think that $$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$ is true? The powers of $\zeta_{12}$ repeat modulo 12. In fact since the $12$th cyclotomic polynomial is $\Phi_{12}=x^4-x^2+1$, we have that \begin{align*} \zeta_{12}^5&=\zeta_{12}^5+0\\\\ &=\zeta_{12}^5+(-\zeta_{12})\underbrace{(\zeta_{12}^4-\zeta_{12}^2+1)}_{0}\\ &=\zeta_{12}^3-\zeta_{12} \end{align*} and by a similar computation, we can find that $\zeta_{12}^{10}=-\zeta_{12}^2+1$. Lastly, since the powers of $\zeta_{12}$ repeat modulo 12, we have $$\zeta_{12}^{15}=\zeta_{12}^{12}\cdot \zeta_{12}^3=1\cdot \zeta_{12}^3=\zeta_{12}^3$$ Thus the correct decomposition in the basis $\{1,\zeta_{12},\zeta_{12}^2,\zeta_{12}^3\}$ is $$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= (a+c) - b \zeta_{12} - c \zeta_{12}^2 + (b+d) \zeta_{12}^3$$