Exercise 7.11

Proof. Indeed, for all $k\in \mathbb{Z}$,

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Proof. (Ex. 7.11)

Let $\alpha \in F^{\text{∗}}$ with $|\alpha |=q-1$, and $g$ a generator of $K^{\text{∗}}$, so $|g|=q^2-1$. We know from exercise 7.10 that there exists an integer i such that $\alpha =g^{i(q+1)}$.

Let $h=g^{q+1}$. As $h^{q-1}=1$, then $h\in F^{\text{∗}}$, and since $|g|=q^2-1$, $|h|=q-1$, so $h$ is a generator of $F^{\text{∗}}$.

Note that for all $s\in \mathbb{Z}$, $\alpha =g^{(i+s(q-1))(q+1)}$, since $g^{q^2-1}=1$.

We will show that we can choose $s$ such that $j=i+s(q-1)$ is relatively prime with $q+1$. Then $j$ is such that $\alpha =g^{j(q+1)}=h^j$.

$i$ is odd : if not, $\alpha$ is an element of the subgroup of squares in $F^{\text{∗}}$, so its order divides $(q-1)∕2$, in contradiction with $|\alpha |=q-1$.

$(q-1)\wedge (q+1)\mid 2$. Since $i-1$ is even, there exist integers $s,t$ verifying the Bézout’s equation

Then $j=i+s(q-1)=1+t(q+1)$ is relatively prime with $q+1$ : $j\wedge (q+1)=1$.

Moreover, as $\alpha =h^j$, with $|\alpha |=|h|=q-1$, the lemma implies that

so $(q-1)\wedge j=1$. As $(q+1)\wedge j=1$ and $(q-1)\wedge j=1$, then $(q^2-1)\wedge j=1$.

Let $\beta =g^j$. Then $\alpha ={\beta }^{1+q}$, and using the lemma,

Conclusion: there exists a $\beta \in K^{\text{∗}}$ with order $q^2-1$ such that ${\beta }^{1+q}=\alpha$. □