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.

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 $p(p-1)$.

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 $\zeta$ […]

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 […]

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. […]

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 […]

$x^3-\sqrt{2}$ 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 […]

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 […]

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 […]