Quadratic field extension of finite field $\mathbb{F}_{q}$.

Question

Question: Let $q = p^{n}$ be a prime power. For which $q$ is the quadratic extension $\mathbb{F}_{q^{2}}$ of $\mathbb{F}_{q}$ of the form $\mathbb{F}_{q}(\sqrt{x})$ for $x \in \mathbb{F}_{q}$? Furthermore, for what $q$ is the cubic extension $\mathbb{F}_{q^{3}}$ of $\mathbb{F}_{q}$ of the form $\mathbb{F}_{q}(\sqrt[3]{x})$ for $x \in \mathbb{F}_{q}$.

I am not sure how to approach this question. I have made a similar question to these where we look at quadratic extensions $\mathbb{Q}(\sqrt{d})$ of $\mathbb{Q}$. I tried to do a similar approach here by considering that we must adjoin a root of a minimal polynomial of degree 2, which we can find with the quadratic formule (since characteristic $\neq 2$). But I am not sure how I can find the conditions for $q$ such that the extension has this form. Same problem for the cubic extension.

Answer 1

The case of finite fields is easier because an extension of degree $n$ in this case is unique. So if you can find any element $\alpha\in\mathbb{F}_q$ such that $x^3-\alpha$ is irreducible over $\mathbb{F}_q$ then a root of this polynomial necessary generates $\mathbb{F}_{q^3}$ (unlike over $\mathbb{Q}$ where we can only conclude it generates some extension of degree $3$), and so the extension is indeed of the required form. On the other hand, if there is no such $\alpha$ then clearly the extension cannot be of this form.

So the question is now for which $q$ there is some $\alpha\in\mathbb{F}_q$ such that $x^3-\alpha$ is irreducible over $\mathbb{F}_q$. Since this is a polynomial of degree $3$, this is equivalent to asking for which $q$ there is some $\alpha$ such that $\alpha$ has no third root in $\mathbb{F}_q$. The answer follows from the following simple exercise in group theory:

Exercise: Let $G$ be a finite group. Then every element of $G$ has a third root if and only if $3\nmid |G|$.

Though in the case $G=\mathbb{F}_q^{\times}$ the statement is even easier if you remember this is a cyclic group.

Answer 2

In general, this question falls under the umbrella of Kummer theory. The basic result says: Let $n\geq 1$ be any integer. If $K$ is a field of characteristic coprime to $n$ (or $0$) that contains all $n$-th roots of unity, then an extension $L/K$ of degree $n$ is of the form $K(\sqrt[n]{a})$ if and only if it is galois with cyclic galois group. Also conversely, if $K$ admits such an extension that is simultaneously radical and cyclic of degree $n$, then $K$ has all $n$-th roots of unity and is of characteristic coprime to $n$ (or $0$).

For finite fields, all this is not difficult to translate into a concrete condition. Suppose we have the finite field $K$ of characteristic $p$ and want to know if the unique extension $L/K$ of degree $n$ is radical.

As every extensions of finite fields is cyclic, it will be radical if and only if $K$ has characteristic coprime to $n$ and contains all $n$-th roots of unity. The first is clearly the case if and only if $p\nmid n$, while the second happens if and only if the unit group of $K$ has a cyclic subgroup of order $n$. This in turn is the case if and only if $n$ divides the group order $|K|-1$.

Moreover, it is clear that the second condition already entails the first. Hence, the full characterization reads:

The unique degree $n$ extension $L$ of the finite field $K$ is of the form $K(\sqrt[n]{a})$ if and only if $n$ divides $|K|-1$.