# Extension Degree of Maximal Real Subfield of Cyclotomic Field

## Problem 362

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.

Note that since $n>2$, the primitive $n$-th root $\zeta$ is not a real number.
Also, we have
\begin{align*}
\zeta+\zeta^{-1}=2\cos(2\pi /n),
\end{align*}
which is a real number.

Thus the field $\Q(\zeta+\zeta^{-1})$ is real.
Therefore the degree of the extension satisfies
$[\Q(\zeta):\Q(\zeta+\zeta^{-1})] \geq 2.$

We actually prove that the degree is $2$.
To see this, consider the polynomial
$f(x)=x^2-(\zeta+\zeta^{-1})x+1$ in $\Q(\zeta+\zeta^{-1})[x]$.

The polynomial factos as
$f(x)=x^2-(\zeta+\zeta^{-1})x+1=(x-\zeta)(x-\zeta^{-1}).$ Hence $\zeta$ is a root of this polynomial.

It follows from $[\Q(\zeta):\Q(\zeta+\zeta^{-1})] \geq 2$ that $f(x)$ is the minimal polynomial of $\zeta$ over $\Q(\zeta+\zeta^{-1})$, and hence the extension degree is
$[\Q(\zeta):\Q(\zeta+\zeta^{-1})] =2.$

## Comment.

The subfield $\Q(\zeta+\zeta^{-1})$ is called the maximal real subfield.
The reason why it is called as such should be clear from the proof.

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Equation $x_1^2+\cdots +x_k^2=-1$ Doesn’t Have a Solution in Number Field $\Q(\sqrt[3]{2}e^{2\pi i/3})$

Let $\alpha= \sqrt[3]{2}e^{2\pi i/3}$. Prove that $x_1^2+\cdots +x_k^2=-1$ has no solutions with all $x_i\in \Q(\alpha)$ and $k\geq 1$.

Close