Exercise 7.8

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}}

In a field with $2^n$ elements, what is the subgroup of squares.

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}}

\bigskip

Let $F$ be a field with $q = 2^n$ elements.

\bigskip

{\bf Proof 1}
\begin{proof}
$d = (q-1) \wedge 2 = (2^n - 1) \wedge 2 = 1$, thus each $\alpha \in F^*$ verifies $\alpha^{(q-1)/d} = \alpha^{q-1} = 1$. Theorem 7.1.2 show that $\alpha$ is a square in $F$, of exactly one root.

The subgroup of squares is $F^*$.
\end{proof}

{\bf Proof 2}
\begin{proof}
For all $x \in F$, $x = x^q = \left (x^{2^{n-1}}ight)^2$. So all elements in $F$ are squares. 
\end{proof}

Exercise 7.8

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Exercise 7.8

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