Is $\mathbb{Q}(\sqrt[5]{7}),i)$ the splitting field for some polynomial in $\mathbb Q[x]$?

Question

My textbook is proving that $\mathbb{Q}(\sqrt[5]{7}),i)$ as a vector space over $\mathbb{Q}$ has dimension $10$. I understand this. I am wondering about the significance of the field $\mathbb{Q}(\sqrt[5]{7}),i)$.

Is there an easy way to prove that it is (or is not) the splitting field for some polynomial in $\mathbb{Q}[x]$?

Answer 1

The splitting field of a $\Bbb Q$-polynomial is a normal extension of $\Bbb Q$, but $\Bbb Q(\sqrt[5]7,i)$ is not normal over the rationals.

EDIT: A normal extension? That’s a field extension $K\supset k$ with this property: if $f(X)\in k[X]$ has one root of $f$ in $K$, then all roots of $f$ are in $K$. One of the standard exercises is to show that a finite extension $K\supset k$ is normal if and only if it’s the splitting field of a $k$-polynomial. For your extension, $X^5-7$ has only one root in the big field. Therefore not normal.

Answer 2

As a variation on the arguments already given, note the following characterization of splitting fields.

Let $E/F$ be a finite-dimensional extension. The following are equivalent

1. $E$ is the splitting field over $F$ of a polynomial $f \in F[x]$.
2. For every irreducible polynomial $f \in F[x]$, if $f$ has a root in $E$, then $f$ splits completely in $E$.