Degrees of certain class of extensions of a field

Question

Let $F$ be a field of characteristic $0$ and $\overline{F}$, its algebraic closure. Let $p$ be a prime number.

Take $\alpha\in F^*$ and for this $\alpha$, choose $\beta\in\overline{F}$ such that $\beta^p=\alpha$. I am studying the possible values of $[F(\beta):F]$?

Suppose $\alpha=1$. Now $(x^p-1)/(x-1)$ factors into irreducible polynomials of same degree, say $d$. So $d$ is a divisor of $p-1$ and it is the degree of $[F(\beta):F]$. If $F$ would have been $\mathbb{Q}$, then $d=p-1$, and for given divisor $d$ of $p-1$, we can show that there is suitable $F$ an extension of $\mathbb{Q}$ for which $[f(\beta):F]=d$.

Q.1 Suppose $\alpha\neq 1$ but it is some root of unity in $F$. What are possibilities for $[F(\beta):F]$?

Q.2 If $\alpha$ is non-zero but not a root of unity, is it always the case that $[F(\beta):F]$ is either $1$ or $p$?

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By the way, is there special name for extension of a field $F$ for polynomials of the form $x^p-\alpha$? Any standard reference (among abstract/basic algebra books or field theory books) is there for study of such extensions?

Answer 1

A lot of this depends on what exactly $F$ is. One interesting property of polynomials of this form are that they will either be irreducible over the ground field of have at least one root in the ground field. In particular, this means that if your ground field $F$ has a $p$th root of unity (aside from $1$), then $x^p-a$ is either irreducible over $F$ or splits completely over $F$, i.e., yes, either $[F(\beta):F]$ is $1$ or it's $p$, regardless of the value of $a$.

Suppose $x^p-a$ ($p=$ prime, $a\in F$) is reducible in $F$ so that $x^p-a=g(x)h(x)$ where $deg(g(x))<p$, $deg(h(x))<p$, and $g(x),h(x)\in F[X]$. WLOG we may assume that the leading coefficient of $g(x)$ is $1$. Let $c$ be the constant coefficient of $g(x)$ and denote $deg(g(x))$ by $m$. Clearly, every root of $g(x)$ must be a root of $x^p-a$. We note that if $r$ is any root of $x^p-a$ and $\omega$ is any $p$th root of unity, then $r,r\omega,r\omega^2,\dots,r\omega^{p-1}$ are all the roots of $x^p-a$.

This means that the roots of $g(x)$ (possibly with repetition for multiplicity) are $r\omega^{s_1},r\omega^{s_2},\dots,r\omega^{s_m}$ where each $s_i$ is an integer.

Since $g(x)$ can be factored as a product of terms of the form $x-d\omega^{s_i}$ and $c$ is the constant coefficient of $g(x)$, we have $c=r^m\omega^{s_1+s_2+\dots+s_m}$. But $g(x)\in F[X]$ by assumption, so $c\in F$.

We have then that $c^p=(r^m\omega^{s_1+s_2+\dots+s_m})^p=(r^p)^m(\omega^p)^{s_1+s_2+\dots+s_m}=a^m$ (since $r$ is a root of $x^p-a$).

Recall that $m=deg(g(x))<p$. Since $p$ is a prime, this means that $deg(g(x))$ is coprime to $p$ giving us by Bezout's identity that there exist $s$ and $t$ such that $sm+pt=1$.

This yields $$c^p=a^m\\ c^{sp}=a^{sm}\\ c^{sp}=a^{1-pt}\\ (c^sa^t)^p=a$$

So $c^sa^t$ is our root of $x^p-a$ in $F$.

But, all of this aside, you can easily find situations where $[F(\beta):F]$ is not just $1$ or $p$, like when, e.g. we have $x^3-8=0$.