Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$ with $\alpha^{15}=1$ then $\mid K:F \mid \leq 4$.

Question

Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$. If $\alpha^{15}=1$ then show that $\mid K:F \mid \leq 4$.

Let $F=\mathbb F_q$, where $q$ is power of some prime $p$, say, and let $\mid K:F\mid=d$, so that $K=\mathbb F_{q^d}$. If $\min_{\alpha,F}(x)=f(x)$, then we have $\deg(f)=d$ and $f(x)\mid(x^{15}-1)$, hence $f(x)\mid (x^{16}-x)$. I wanted to use the fact that $g(x)$ monic irreducible in $\mathbb F_q[x]$. Then $g(x)\mid (x^{q^{n}}-x) \iff \deg(g)\mid n$. But how do I show that characteristic of the field $F$ is $2$. I need help. Thanks.

Answer 1

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: $[K:F]\in\{1,2,4\}$.