How to prove that $x^5-5$ is irreducible over $\mathbb Q(\sqrt 2, \sqrt[3] 3) $

Question

How to prove that $x^5-5$ is irreducible over $K = \mathbb Q(\sqrt 2, \sqrt[3] 3)? $

I came across this problem while solving another one and I dunno exactly how to proceed. I know by Eisenstein's criterion that $p_c(x)= x^5-5 $ is irreducible over $\mathbb Q$. This gives me for instance that any field extension containing $\sqrt[5]5$ must be at least of degree $5$, but since the degree of $K$ over $\mathbb Q$ is $6$, this won't help me to guarantee that $\sqrt[5] 5 \not \in K$.

Any further directions? thanks.

Answer 1

Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.

In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $K\subseteq\mathbb{R}$ and $\sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $\sqrt[5]{5}\in K$, which you know why this is false.

In the case where $p(x)$ has an irreducible quadratic factor $q(x)\in K[x]$. Since $K\subseteq \mathbb{R}$, we must have $$q(x)=(x-\omega\sqrt[5]{5})(x-\bar{\omega}\sqrt[5]{5})\,,$$ where $\omega$ is a primitive $5$-th root of unity. This shows that the constant term $\sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $\sqrt[5]{25}$ is not in $K$.

Here is a proof. If $\sqrt[5]{25}$ is in $K$, then $\sqrt[5]{5}=\dfrac{(\sqrt[5]{25})^3}{5} \in K$.