Exercise 7.10

Question

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If $K \supset F$ be finite fields and $[K:F] = 2$. For $\beta \in K$, show that $\beta^{1+q} \in F$ and moreover that every element in $F$ is of the form $\beta^{1+q}$ for some $\beta \in K$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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\begin{proof}
If $\beta = 0$, $\beta^{1+q} = 0 \in F$, and if $\beta \in K^*$, $\beta^{q^2-1} = 1$, so $\left(\beta ^{1+q}ight)^{q-1} = 1$, thus $ \beta^{1+q} \in F$ (Prop. 7.1.1, Corollary 1).

Let  $g$ be a generator of $K^*$. Then $K^* = \{1,g,g^2,\cdots,g^{q^2-2}\}$.

For every integer $k \in \Z$, 
$$g^k  \in F^* \iff (g^k)^{q-1} = 1 \iff g^{k(q-1)}=1 \iff q^2-1 \mid k(q-1) \iff q+1 \mid k.$$
Thus $F^* = \{1,g^{q+1}, g^{2(q+1)},\cdots, g^{(q-2)(q+1)}\}$. If $\alpha \in F^*$, there exists $i, 0 \leq i \leq q-1,$ such that $\alpha = g^{i(q+1)}$.
If we write $\beta = g^i$, then $\alpha = \beta^{1+q}$ (and for $\alpha = 0$, we take $\beta = 0$).

Conclusion: if $K$ is a quadratic extension of $F$ ($F,K$ finite fields), every element in $F$ is of the form $\beta^{1+q}$ for some $\beta \in K$.
\end{proof}

Exercise 7.10

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Exercise 7.10

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