Exercise 10.1.8

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}}

In the Mathematical notes we defined the field ${\mathscr P} \subset \R$ and what it means for a subfield $F \subset \R$ to be Pythagorean.
\be

\item[(a)] Let $\alpha$ be a real number. Prove that $\alpha \in {\mathscr P} $ if and only if there is a sequence of fields $\Q = F_0\subset \ldots \subset F_n \subset \R$ such that $\alpha \in F_n$, and for $i=1,\ldots,n$ there are $a_i,b_i \in F_{i-1}$ such that $F_i = F_{i-1}\left(\sqrt{a_i^2+b_i^2}ight)$.
\item[(b)] Prove that ${\mathscr P} $ is the smallest Pythagorean subfield of $\R$.
\ee

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}}

Write $\mathbb{P}$ the set of points of $\C$ (identified with the Euclidean plane) constructible by a sequence of straightedge-and-dividers constructions, and $\mathbb{D}$ the set of lines passing through two distinct such points (so constructible by straightedge-and-dividers) .

\bigskip

{\bf Lemma 1. } {\it If $A,B,C\in {\mathbb P}$, with $A
e B$, then the parallel $l$ to $(AB)$ passing through $C$ is in $\mathbb{D}$.}
\begin{proof} Let $C'$ the symmetric point of $C$ relative to $A$, and $C''$ the symmetric point of $C'$ relative to $B$, so $\overrightarrow{AC'} = -\overrightarrow{AC}, \overrightarrow{BC''} = - \overrightarrow{BC'}$. Since $AC = AC'$ and $BC' = BC''$, $C',C''$ lie in  $\mathbb{P}$, and $\overrightarrow{CC''} = 2 \overrightarrow{AB}$, so $C 
e C''$ and $(CC'')$ is parallel to $(AB)$, with $(CC'') \in \mathbb{D}$. So $l = (CC'') \in \mathbb{D}$.
\end{proof}

{\bf Lemma 2.}  ${\mathscr P} $ {\it is a subfield of $\R$.}
\begin{proof}
Using Lemma 1, we can mimic the proof of first part of Theorem 10.1.4.
\end{proof}

{\bf Lemma 3.} {\it Let $\alpha = a+ib \in \C$, where $a,b \in \R$. Then $\alpha \in \mathbb{ P}$ if and only if $a,b \in {\mathscr P} $.}

%{\it  Let $l = (AB),\  A,B \in \mathbb{P}$, a straightedge-and-dividers constructible line, and $C \in \mathbb{P}$ a straightedge-and-dividers constructible point. Then line $l'$ passing trough $C$ and orthogonal to $l$ is a straightedge-and-dividers constructible line: $l' \in \mathbb{D}$.}

\begin{proof}

\be

\item[$\bullet$] Suppose that $\alpha =a +i b \in \mathbb{P}$. By Lemma 1, we can construct by straightedge-and-dividers constructions the lines passing through $\alpha$ and parallel to the axis, so $a, ib$ are constructible, and using the divider, so are  the real numbers $a,b$, therefore $a,b \in {\mathscr P} $.
\item[$\bullet$] Suppose that $a,b \in {\mathscr P} $. Then $a, ib \in \mathbb{P}$, and the intersection $\alpha = a+ib$ of the lines passing through these two points and parallel to the $y$-axis and $x$-axis respectively lies in $\mathbb{P}$.
\ee

\end{proof}

\begin{proof} (Ex. 10.1.8)
\be
\item[(a)] Let $\alpha$ be a real number.
\be
\item[$\bullet$]  Suppose that there is a sequence of fields $\Q = F_0\subset \ldots \subset F_n \subset \R$ such that $\alpha \in F_n$, and for $k=1,\ldots,n$, there are $a_k,b_k \in F_{k-1}$ such that $F_k= F_{k-1}\left(\sqrt{a_k^2+b_k^2}ight)$.

We prove by induction that $F_k \subset {\mathscr P} $. Since ${\mathscr P} $ is a subfield of $\R$, $F_0 = \Q \subset \mathscr{P}$. Suppose that $F_{k-1} \subset {\mathscr P} $. As $a_k,b_k\in F_{k-1} \subset {\mathscr P} $, $\sqrt{a_k^2+b_k^2} \in {\mathscr P} $ (see the Mathematical Notes), so $F_k=F_{k-1}\left(\sqrt{a_k^2+b_k^2}ight) \subset {\mathscr P} $, and the induction is done.

Therefore  $F_n \subset {\mathscr P} $, and $\alpha \in F_n$, so $\alpha \in {\mathscr P} $.

\item[$\bullet$]  Conversely, suppose that $\alpha = a+ib \in {\mathbb P}$. We use induction on the number $N$ of steps in the construction of $\alpha$ to prove that there is a sequence of subfields of $\R$, $\Q = F_0\subset \ldots \subset F_n \subset \R$, such that $a,b \in F_n$, where $F_k= F_{k-1}\left(\sqrt{a_k^2+b_k^2}ight),\ k=1,\ldots,n$.

When $N = 0$, we must have $\alpha = 0,1$ or $i$, in which case $F_n = F_0 = \Q$ contains the real and imaginary part of $\alpha$.

Now suppose that $\alpha$ is constructed in $N\geq1$ steps, where the last step use the intersection of two lines $l_1,l_2$. As in the proof of the Theorem 10.1.6, the coordinates of $\alpha$ lie in the same field $F_n$.

If the last step uses the divider, then their exist four points $\beta,\gamma, \delta, \varepsilon \in \mathbb{P}$ such that $\delta,\varepsilon, \alpha$ are collinear and $|\alpha - \delta| = |\beta - \gamma|$. Moreover, by the induction hypothesis,  the coordinates $\beta_x, \beta_y,\gamma_x,\ldots$ of $\beta,\gamma,\delta,\varepsilon$ are in $F_n$.

Then $\alpha - \delta = \lambda(\varepsilon - \delta), \lambda \in \R$, with
$$\lambda = \pm \frac{|\alpha - \delta|}{|\varepsilon - \delta|}=\pm \frac{|\beta - \gamma|}{|\varepsilon - \delta|}.$$
Let 
\begin{align*}
F_{n+1} &= F_n(|\beta - \gamma|) = F_n(\sqrt{s^2+t^2}), \mathrm{where}\  s = \beta_x - \gamma_x , t = \beta_y - \gamma_y \in F_n,\
F_{n+2} &=F_{n+1}(|\varepsilon - \delta|) = F_{n+1}(\sqrt{u^2+v^2}),  \mathrm{where}\  u = \varepsilon_x - \delta_x, v = \varepsilon_y - \delta_y \in F_n.
\end{align*}
Then $\lambda \in F_{n+2}$. As $\alpha = a+ i b =  \delta + \lambda (\varepsilon - \delta)$, $a, b \in F_{n+2}$, and the induction is done.

In particular, if $\alpha = a+ib \in {\mathscr P}  = \mathbb{P} \cap \R$, then $b=0, \alpha = a \in F_n$, where 
$\Q = F_0\subset \ldots \subset F_n \subset \R$ and there are $a_k,b_k\in F_{k-1}$ such that $F_k = F_{k-1}\left(\sqrt{a_k^2+b_k^2}ight),\ k= 1,\ldots,n$.
\ee

\item[(b)] By the Mathematical notes, if $a,b \in {\mathscr P} $, then $\sqrt{a^2+b^2} \in {\mathscr P} $, so $\mathscr P$ is a Pythagorean subfield of $\R$.

Let $K$ any Pythagorean subfield of $\R$, and take any $ \alpha \in {\mathscr P} $.  By part (a), there exists a sequence of subfields of $\R$,
$$\Q = F_0\subset \ldots \subset F_n \subset \R,$$
such that $\alpha \in F_n$, and for $k=1,\ldots,n$ there are $a_k,b_k \in F_{k-1}$ such that $F_k= F_{k-1}\left(\sqrt{a_k^2+b_k^2}ight)$.
As any subfield of $\R$, $K$ contains $\Q$, so $F_0 = \Q \subset K$.

By induction, suppose that $F_{k-1} \subset K$ for some integer $k, 1 \leq k\leq n$. Then $F_k = F_{k-1}(\sqrt{a_k^2+b_k^2})$, where $a_k,b_k\in F_{k-1} \subset K$. As $K$ is Pythagorean, $\sqrt{a_k^2+b_k^2} \in K$, so $F_k \subset K$, and the induction is done.

So $F_n \subset K$, and $\alpha \in F_n$, so $\alpha \in K$. Hence ${\mathscr P}  \subset K$, so $\mathscr P$ is the smallest Pythagorean subfield of $\R$.
\ee
\end{proof}

Exercise 10.1.8

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Exercise 10.1.8

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