Exercise 7.5

Question

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

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

(continuation) Let $K$ be a field containing $F$ such that $[K:F] = n$. For all $\alpha \in F^*$, show that the equation $x^n = \alpha$ has $n$ solutions in $K$. [Hint: Show that $q^n-1$ is divisible by $n(q-1)$ and use the fact that $\alpha^{q-1} = 1$.]

Richard Ganaye · Accepted Answer

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

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
As $q\equiv 1 \ [n], \frac{q^n-1}{q-1} = 1+q+\cdots+q^{n-1} \equiv 0 \ [n]$, then $n \mid \frac{q^n-1}{q-1} : $
$$q^n-1 = k n (q-1), k \in \mathbb{N}.$$

Since $\alpha \in F^*$, $\alpha^{q-1} = 1$, thus $$\alpha^{(q^n-1)/n} = (\alpha^{q-1})^k = 1.$$

As $\vert K \vert = q^n$, Prop. 7.1.2 (or the final remark in Ex.7.3) show that there exists $x \in K^*$ such that $x^n = \alpha$. Then, from Ex.7.3, we know that there exist $n$ solutions in $K$.

Conclusion: if $[K:F] = n$, for all $\alpha \in F^*$, the equation $x^n = \alpha$ has $n$ solutions in $K$.

\end{proof}

Exercise 7.5

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Exercise 7.5

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