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 $\Q(i, \sqrt{2})$.
Sponsored Links
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
Sponsored Links
More from my site
Extension Degree of Maximal Real Subfield of Cyclotomic Field
Let $n$ be an integer greater than $2$ and let $\zeta=e^{2\pi i/n}$ be a primitive $n$-th root of unity. Determine the degree of the extension of $\Q(\zeta)$ over $\Q(\zeta+\zeta^{-1})$.
The subfield $\Q(\zeta+\zeta^{-1})$ is called maximal real subfield.
Proof. […]- The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity
Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]
- Example of an Infinite Algebraic Extension
Find an example of an infinite algebraic extension over the field of rational numbers $\Q$ other than the algebraic closure $\bar{\Q}$ of $\Q$ in $\C$.
Definition (Algebraic Element, Algebraic Extension).
Let $F$ be a field and let $E$ be an extension of […]
Galois Group of the Polynomial $x^p-2$.
Let $p \in \Z$ be a prime number.
Then describe the elements of the Galois group of the polynomial $x^p-2$.
Solution.
The roots of the polynomial $x^p-2$ are
\[ \sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1\]
where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ […]- Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$
Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.
Proof.
Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein's criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the […]
- Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?
(b) […]
- Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group
Show that $\Q(\sqrt{2+\sqrt{2}})$ is a cyclic quartic field, that is, it is a Galois extension of degree $4$ with cyclic Galois group.
Proof.
Put $\alpha=\sqrt{2+\sqrt{2}}$. Then we have $\alpha^2=2+\sqrt{2}$. Taking square of $\alpha^2-2=\sqrt{2}$, we obtain […]
Application of Field Extension to Linear Combination
Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.
Proof.
We first prove that the polynomial […]
