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.
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.
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 […]
The Cyclotomic Field of 8-th Roots of Unity is $\Q(\zeta_8)=\Q(i, \sqrt{2})$
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 $\Q(i, \sqrt{2})$.
Proof.
Recall that the extension degree of the cyclotomic field of $n$-th roots of unity is given by […]
Degree of an Irreducible Factor of a Composition of Polynomials
Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$.
Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.
Hint.
Use the following fact.
Let $h(x)$ is an […]
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 $\zeta$ […]
$x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$
Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.
Hint.
Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.
Proof.
Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is […]
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 […]
Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension
Prove that the polynomial
\[f(x)=x^3+9x+6\]
is irreducible over the field of rational numbers $\Q$.
Let $\theta$ be a root of $f(x)$.
Then find the inverse of $1+\theta$ in the field $\Q(\theta)$.
Proof.
Note that $f(x)$ is a monic polynomial and the prime […]