The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity

Field theory problems and solution in abstract algebra

Problem 89

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.

 
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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 $p(p-1)$.


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