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 $\phi(n)$, 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
\[\Q(\zeta_8)=\Q(i, \sqrt{2}).\] Click here if solved 24

The post The Cyclotomic Field of 8-th Roots of Unity is $\Q(\zeta_8)=\Q(i, \sqrt{2})$ first appeared on Problems in Mathematics.

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.

The post Extension Degree of Maximal Real Subfield of Cyclotomic Field first appeared on Problems in Mathematics.

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 $\zeta$ is a primitive $p$-th root of unity.
(Explicitly, you may take $\zeta=e^{2\pi i/p}$.)

Thus $x^p-2$ is a separable polynomial over $\Q$. The Galois group of $x^p-2$ is the Galois group of the splitting field of $x^p-2$.
The splitting field of $x^p-2$ is $K:=\Q(\sqrt[p]{2}, \zeta)$ of extension degree $p(p-1)$.
(Check this.)

Let $G=\Gal(K/\Q)$ be the Galois group of $x^p-2$. The order of the Galois group $G$ is $p(p-1)$. Let $\sigma \in G$ be an automorphism.
Then $\sigma$ sends an element of $G$ to its conjugate (a root of the minimal polynomial of the element.)
The minimal polynomial of $\sqrt[p]{2}$ is $x^p-2$ since it is irreducible by Eisenstein’s criteria.
The minimal polynomial of $\zeta$ is the cyclotomic polynomial
\[\Phi(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.\]

Therefore $\sigma$ maps
\begin{align*}
\sqrt[p]{2} &\mapsto \sqrt[p]{2}\zeta^a \\
\zeta & \mapsto \zeta^b
\end{align*}
for some $a=0, 1, \dots, p-1$ and $b=1, 2, \dots, p-1$.

Thus there are $p(p-1)$ possible maps for $\sigma$.
Since the order of $G$ is $p(p-1)$, these are exactly the elements of the Galois group $G$ of the polynomial $x^p-2$.

The post Galois Group of the Polynomial $x^p-2$. first appeared on Problems in Mathematics.

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 unity.

Remark

The following proof proves more than enough. You might want to refine the proof for a simpler proof.

Proof.

We first determine the splitting field of $x^p-2$.
The roots of the polynomial $x^p-2$ are
\[\sqrt[p]{2}\zeta^i,\] where $\zeta$ is a primitive $p$-th root of unity and $i=0,1,\dots, p-1$.

Let $F$ be the splitting field of $x^p-2$. Since both $\sqrt[p]{2}$ and $\sqrt[p]{2}\zeta$ is in $F$, the quotient $\zeta=\sqrt[p]{2}\zeta/\sqrt[p]{2} \in F$.
Therefore we see that $\Q(\sqrt[p]{2}, \zeta)\subset F$. Since the field $\Q(\sqrt[p]{2}, \zeta)$ contains all the roots of $x^p-2$, we must have $F=\Q(\sqrt[p]{2}, \zeta)$.

Next, we find the degree of the extension $\Q(\sqrt[p]{2}, \zeta)$ over $\Q$.
The field $\Q(\sqrt[p]{2}, \zeta)$ contains the cyclotomic field $\Q(\zeta)$ as a subfield and we obtain $\Q(\sqrt[p]{2}, \zeta)$ by adjoining $\sqrt[p]{2}$ to $\Q(\zeta)$.
Since $\sqrt[p]{2}$ is a root of $x^p-2$, the degree of the extension $[\Q(\sqrt[p]{2}, \zeta) : \Q(\zeta)] \leq p$.

Thus we have
\[ [\Q(\sqrt[p]{2}, \zeta): \Q]=[\Q(\sqrt[p]{2}, \zeta): \Q(\zeta)] [\Q(\zeta): \Q]\leq p(p-1)\] since the degree of cyclotomic extension over $\Q$ is $\phi(p)=p-1$. (Here $\phi$ is the Euler phi function.)

Note that $\Q(\sqrt[p]{2})$ is also a subfield and $[\Q(\sqrt[p]{2}): \Q]=p$ since $x^p-2$ is irreducible over $\Q$ by Eisenstein’s criterion.
Hence both $p$ and $p-1$ divide $[\Q(\sqrt[p]{2}, \zeta): \Q] \leq p(p-1)$. Since $p$ is a prime, the numbers $p$ and $p-1$ are relatively prime. Thus we must have $[\Q(\sqrt[p]{2}, \zeta): \Q]=p(p-1)$.

In particular, the polynomial $x^p-2$ must be irreducible over $\Q(\zeta)$ otherwise the degree $[\Q(\sqrt[p]{2}, \zeta): \Q]$ is strictly less than $p(p-1)$.

The post The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity first appeared on Problems in Mathematics.