Exercise 7.3

Question

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Let $F$ a field with $q$ elements and suppose that $q\equiv 1 \pmod n$. Show that for $\alpha \in \F^*$, the equation $x^n = \alpha$ has either no solutions or $n$ solutions.

Richard Ganaye · Accepted Answer

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\begin{proof} This is a particular case of Prop. 7.1.2., where $d = n \wedge (q-1) = n$ : the equation $x^n =\alpha$ has solutions iff $\alpha^{(q-1)/n} = 1$. In this case, there are exactly $d = n$ solutions.

We give  here a direct proof.

Let $g$ be a generator of $F^*$.  Write $x = g^y, \alpha = g^a$. Then
$$x^n = \alpha \iff g^{ny} = g^a\iff q-1 \mid ny -a.$$

Suppose that there exists $x\in F$ such that $x^n = \alpha$. Then there exists $y\in \Z$ such that $q-1 \mid ny -a$. Since $n \mid q-1$, then  $n \mid a$.
$$q-1 \mid ny -a \iff \frac{q-1}{n} \mid y - \frac{a}{n}\iff y= \frac{a}{n} + k \frac{q-1}{n}, k\in \mathbb{Z}.$$

As  $\frac{a}{n} + (k+n) \frac{q-1}{n} = \frac{a}{n} + k \frac{q-1}{n}, k\in \mathbb{Z}$, the values $k=0,1\cdots,n-1 $ are sufficient :
$$x^n=\alpha \iff y = \frac{a}{n} + k \frac{q-1}{n}, k\in \{0,1,\cdots,n-1\}.$$

Moreover, these solutions are all distinct : if $k,l\in  \{0,1,\cdots,n-1\}$,
\begin{align*}
g^{\frac{a}{n} + k \frac{q-1}{n}} = g^{\frac{a}{n} + l \frac{q-1}{n}} & \Rightarrow g^{(k-l) \frac{q-1}{n}} = 1\
&\Rightarrow q-1 \mid (k-l) \frac{q-1}{n}\
& \Rightarrow n \mid k-l \
&\Rightarrow k\equiv l \ [n] \Rightarrow k=l.
\end{align*}

Conclusion: if  $F$ is a field with $q$ elements and $n \mid q-1$, the equation $x^n = \alpha$ has either no solutions or $n$ solutions in $F$.

Note: $$\exists x \in F^*, x^n = \alpha \iff n \mid a \iff \alpha^{(q-1)/n} = 1.$$

Indeed, if $x^n = \alpha$ has a solution, we have proved that $n\mid a$, thus $\alpha^{(q-1)/n} = (g^{a/n})^{q-1} = 1$.

Conversely, if $\alpha^{(q-1)/n} = 1$, $g^{a(q-1)/ n }=1$, thus $q-1 \mid a(q-1)/n$, so $ n \mid a$ :  $\alpha = x^n$, with $x = g^{a/n}$.
\end{proof}

Exercise 7.3

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Exercise 7.3

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