Exercise 3.2.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

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

ewcommand{\im}{\,\mathrm{Im}\,}

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

ewcommand{\Gal}{\mathrm{Gal}}

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

A field $F$ is an ordered field if there is a subset $P \subset F$ such that:
\begin{enumerate}
\item[(a)] $P$ is closed under addition and multiplication.
\item[(b)] For any $a\in F$, exactly one of the following is true: $a \in P, a = 0$, or $-a\in P$.
\end{enumerate}
One then defines $a>b$ to mean $a-b\in P$ (so that $P$ becomes the set of positive elements). From this, one can prove all the typical properties of $>$. Now let $F$ be an ordered field. Prove that $-1$ is not a square in $F$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

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

ewcommand{\im}{\,\mathrm{Im}\,}

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

ewcommand{\Gal}{\mathrm{Gal}}

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

\begin{proof}
Let $F$ an ordered field.

Since  $P$ is closed under multiplication by (a), if $a \in P$, then $a^2 \in P$.

If $-a \in P$, $a^2 = (-a)(-a) \in P$. By (b), every $a\in F$ verifies $a\in P$, or $a=0$, or $-a\in P$, so we can conclude that 
\begin{align}
\forall  a,\  a \in F^* \Rightarrow \ a^2 \in P. \label{eq3.2.4:1}
\end{align}
So $P$ contains all squares in $F$, $0$ excluded.
By definition of fields, we know that $1
eq 0$, so $1 = 1^2 \in P$.

By (b), the three cases $a\in P$,  $a=0$, $-a\in P$ are mutually exclusive, thus $-1 
ot \in P$.
Therefore  $-1$ is not a square in $F$, otherwise $-1 = a^2 \in P$ by \eqref{eq3.2.4:1}.

Conclusion: $-1$ is not a square in the ordered field $F$.
\end{proof}

Exercise 3.2.4

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Exercise 3.2.4

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