# The Cyclotomic Field of 8-th Roots of Unity is $\Q(\zeta_8)=\Q(i, \sqrt{2})$ ## Problem 491

Let $\zeta_8$ be a primitive $8$-th root of unity.
Prove that the cyclotomic field $\Q(\zeta_8)$ of the $8$-th root of unity is the field $\Q(i, \sqrt{2})$. Add to solve later

## Proof.

Recall that the extension degree of the cyclotomic field of $n$-th roots of unity is given by $\phi(n)$, the Euler totient function.
Thus we have
$[\Q(\zeta_8):\Q]=\phi(8)=4.$

Without loss of generality, we may assume that
$\zeta_8=e^{2 \pi i/8}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i.$

Then $i=\zeta_8^2 \in \Q(\zeta_8)$ and $\zeta_8+\zeta_8^7=\sqrt{2}\in \Q(\zeta_8)$.
Thus, we have
$\Q(i, \sqrt{2}) \subset \Q(\zeta_8).$

It suffices now to prove that $[\Q(i, \sqrt{2}):\Q]=4$.
Note that we have $[\Q(i):\Q]=[\Q(\sqrt{2}):\Q]=2$.
Since $\Q(\sqrt{2}) \subset \R$, we know that $i\not \in \Q(\sqrt{2})$.
Thus, we have
\begin{align*}
[\Q(i, \sqrt{2}):\Q]=[[\Q(\sqrt{2})(i):\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]=2\cdot 2=4.
\end{align*}

It follows that
$\Q(\zeta_8)=\Q(i, \sqrt{2}).$ Add to solve later

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