Exercise 7.6

Question

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

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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}}

Let $K \supset F$ be finite fields with $[K:F] = 3$. Show that if $\alpha \in F$ is not a square in $F$, it is not a square in $K$.

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}
Let $q = \vert F \vert$. Then $\vert K \vert = q^3.$

If the characteristic of $F$ is 2, then $q = 2^k$ for some integer $k\geq 1$, and for all $x \in F$, $x = x^q = \left (x^{2^{k-1}}ight)^2$. Therefore all elements in $F$ (or $K$) are squares.
We can now suppose that the characteristic of $F$ is not 2, so that $1 
eq -1$ in $F$.

As $\alpha$ is not a square in $F$, $\alpha^{(q-1)/2} 
e 1$ (Prop. 7.1.2). From  $0 = \alpha^{q-1} - 1 = (\alpha^{(q-1)/2}-1)(\alpha^{(q-1)/2}+1)$, we deduce that $\alpha^{(q-1)/2} = -1$. Then 
$$\alpha^{(q^3-1)/2} = (\alpha^{(q-1)/2})^{q^2+q+1} = (-1)^{q^2+q+1} = -1,$$
since $q^2+q+1$ is always odd.

$\alpha^{(q^3-1)/2} 
eq 1$ : this implies (Prop. 7.1.2) that $\alpha$ is not a square in $K$.
\end{proof}

Exercise 7.6

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Exercise 7.6

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