Claim:$\;$If $K$ is a field with an element $\alpha$ and a subfield $F$ such that
- $F$ is a finite field.$\\[4pt]$
- $K=F(\alpha)$.$\\[4pt]$
- $\alpha^{15}=1$.
then $[K:F]\le 4$.
Proof:
For the subfield $F(\alpha^3,\alpha^5)$ of $K$, we have
\begin{align*}
&
\frac{\alpha^5}{\alpha^3}\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\alpha^2\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\frac{\alpha^3}{\alpha^2}\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\alpha\in F(\alpha^3,\alpha^5)
\\[4pt]
\end{align*}
hence $F(\alpha^3,\alpha^5)=F(\alpha)=K$.
Next consider $4$ cases . . .
Case $(1)$:$\;\alpha^3,\alpha^5\in F$.
Then $K=F$, hence $[K:F]=1$.
Case $(2)$:$\;\alpha^3\in F,\alpha^5\not\in F$.
\begin{align*}
\text{Then}\;\;&
\alpha^{15}-1=0
\\[4pt]
\implies\;&
(\alpha^5-1)(\alpha^{10}+\alpha^5+1)=0
\\[4pt]
\implies\;&
\alpha^{10}+\alpha^5+1=0
\\[4pt]
\end{align*}
so $\alpha^5$ is a root of $x^2+x+1$, hence $[K:F]=[F(\alpha^5):F]\le 2$.
Case $(3)$:$\;\alpha^5\in F,\alpha^3\not\in F$.
\begin{align*}
\text{Then}\;\;&
\alpha^{15}-1=0
\\[4pt]
\implies\;&
(\alpha^3-1)(\alpha^{12}+\alpha^9+\alpha^6+\alpha^3+1)=0
\\[4pt]
\implies\;&
\alpha^{12}+\alpha^9+\alpha^6+\alpha^3+1=0
\\[4pt]
\end{align*}
so $\alpha^3$ is a root of $x^4+x^3+x^2+1$, hence $[K:F]=[F(\alpha^3):F]\le 4$.
Case $(4)$:$\;\alpha^3,\alpha^5\not\in F$.
Then the order, $o(\alpha)$, of $\alpha$ in $K^*$ is $15$.
Hence $1,\alpha,\alpha^2,\alpha^3,...\alpha^{14}$ are all distinct.
As was shown in case $(2)$, $\alpha^5$ is a root of $x^2+x+1$, hence $[F(\alpha^5):F]\le 2$.
As was shown in case $(3)$, $\alpha^3$ is a root of $x^4+x^3+x^2+x+1$.
Moreover, $\alpha^6,\alpha^9,\alpha^{12}$ are also roots of $x^4+x^3+x^2+x+1$.
Since $\alpha^3\not\in F$, it follows that
- $\alpha^{6},\alpha^{9}\not\in F$, else $\alpha^{18}\in F$, contradiction, since $\alpha^{18}=\alpha^{3}$.$\\[4pt]$
- $\alpha^{12}\not\in F$ since $\alpha^{12}=\alpha^{-3}$.
Thus $\alpha^3,\alpha^6,\alpha^9,\alpha^{12}$ are distinct roots of $x^4+x^3+x^2+x+1$, and none of them are in $F$.
It follows that $x^4+x^3+x^2+x+1$ is either irreducible in $F[x]$, or else factors in $F[x]$ as a product of two irreducible quadratics.
Hence $[F(\alpha^3):F]$ equals $2$ or $4$.
Either way, since $[F(\alpha^5):F]\le 2$, it follows that $F(\alpha^5)\subseteq F(\alpha^3)$.
The above line is the only place where we use the hypothesis that $F$ is a finite field.
Hence $K=F(\alpha^3,\alpha^5)=F(\alpha^3)$, so $[K:F]=[F(\alpha^3):F]\le 4$.
Thus in all $4$ cases, we have $[K:F]\le 4$, as was to be shown.
Note:
With some minor adjustments to the above argument (a future edit), we can obtain the slightly stronger result: $[K:F]\in\{1,2,4\}$.
Best Answer
Let $F$ be a field, $f(x) \in F[x]$ irreducible of prime degree $p$. Let $E/F$ be a finite extension. Suppose $f(x)$ is reducible in $E[x]$. Let $\alpha$ be a root of $f(x)$. Consider $F(\alpha)$. It follows that $[F(\alpha) : F] = p = \deg \alpha$ since $f(x)$ is irreducible in $F$ of degree $p$. Now observe that $[E(\alpha) : E] < p$ since $f(x)$ is reducible in $E$ and $f(x)$ is not the minimal polynomial for $\alpha$ over $E$, but $m_{\alpha, E}(x) \mid f(x)$ so $[E(\alpha) : E] = \deg m_{\alpha, E}(x) < \deg f(x) = p$. Now \begin{align*} [E(\alpha) : F] &= [E(\alpha) : F(\alpha)][F(\alpha) : F] = p[E(\alpha) : F(\alpha)]\\ [E(\alpha) : F] &= [E(\alpha) : E][E : F] \end{align*} So $ [E(\alpha) : E][E : F] = p[E(\alpha) : F(\alpha)]$. It follows that $p \mid [E(\alpha) : E][E : F]$. It immediately follows that $p \mid [E(\alpha) : E]$ or $p \mid [E : F] $. Conclude that $p \mid [E:F]$.