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 $\Q(i, \sqrt{2})$.

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*}

Extension Degree of Maximal Real Subfield of Cyclotomic Field
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Proof. […]

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Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]

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Definition (Algebraic Element, Algebraic Extension).
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Let $p \in \Z$ be a prime number.
Then describe the elements of the Galois group of the polynomial $x^p-2$.
Solution.
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\[ \sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1\]
where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ […]

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Proof.
Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein's criterion (with prime $p=2$).
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Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
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Proof.
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Let $\alpha$ be any real root of $f(x)$.
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Proof.
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