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(New) Added drafts for lectures 03/{22,29} 04/05.

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*Lecture 2019.03.22*
# LECTURE 5
We discussed three equivalent characterisations of what it means to be a Galois extension last lecture. Now we will introduce the **Fundamental Theorem of Galois Thoery**. This theorem is served in three parts. Naturally, we begin with
## Part 1 - Bijective Correspondence
As always, let $E/F$ be a Galois extension, and $G = \mathrm{Gal}(E/F)$. We have seen that when there is a subextension $E/K/F$, then $\mathrm{Aut}(E/K)$ is a subgroup of $G$. We want to formalise this idea. In fact we will show that there is a bijection between the set of all subextensions of a Galois extension, and the subgroups of the Galois group.
*Remark.* This idea gives rise to the alternative notation of an extension:
\begin{figure}[!h]
\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=0.25em, column sep=4em, minimum width=1em]
{
E & \mathrm{Gal}(E/E) = \{\mathrm{id}\} \\
& \rotatebox[origin=c]{-90}{$\subseteq$} \\
K & \mathrm{Gal}(E/K) = S \\
& \rotatebox[origin=c]{-90}{$\subseteq$} \\
F & \mathrm{Gal}(E/F) = G \\
};
\path[-stealth]
(m-1-1) edge [-] (m-3-1)
(m-3-1) edge [-] (m-5-1)
(m-1-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-1-2)
(m-3-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-3-2)
(m-5-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-5-2)
;
\end{tikzpicture}
\end{figure}
**Theorem.** *Let $E/F$ be a Galois extension, and $G$ be the associated Galois group. Then the set of all subgroups of $G$ is isomorphic to the set of subextensions.*
*Proof.* (Construction of an isomorphism) We will approach this proof by showing that if we begin with a subextension, then find the corresponding automorphism group, and then look at the fixed field of that automorphism group, we expect to end up with the initial subfield. We will then show that if we begin with a subgroup, then find the corresponding fixed field and then look at the automorphism group of that fixed field, we expect to end up with the inital subgroup.
Let $E/K/F$ be a subextension. We need to show that the fixed field of $\mathrm{Aut}(E/K)$, $H$ is $K$, as *a priori* we only know that $K \subseteq H$. By the first equivalence, this means that $E/K$ must be a Galois extension. Let us prove this general
**Theorem.** (Subextensions of Galois extensions are Galois) We will make use of the second equivalence: As $E/F$ is Galois, it is also normal. Let $p \in F[Z]$ with distinct roots such that $E$ is the splitting field. Clearly $p \in K[Z] \supseteq F[Z]$ and thus $E/K$ is Galois.
This concludes one direction of our proof. Now we proceed in the other direction. To that end, let $S \leq G$ be a subgroup in $G$, and $K$ the field fixed by $S$. Consider the subextension $E/K/F$, and look at the automorphism group $\mathrm{Aut}(E/K)$. We want to show that $S = \mathrm{Aut}(E/K)$. This is easily done, as by definition, $K$ is Galois, and thus $\mathrm{Aut}(E/K) = \mathrm{Gal}(E/K) = S$. $\blacksquare$
This is a handy application of Galois theory, and gives us a new way to view subextensions, namely as subgroups of the automorphism group.
## Part 2 - Numerics
**Thoerem.** If $E/K/F$ is a subextension of a Galois extension $E/F$, then $\mathrm{dim}(E/K) = |\mathrm{Aut}(E/K)|$.
*Proof.* This is true as a direct result of the thoerem proved above: we now know that a subextension of a Galois extension is also Galois, and by use of the second equivalence, we have the claim. However, we also have a new way to think about this extension, namely as a subgroup of the automorphism group. Here we know from group theory, that
\begin{align*}
|G:S| = \frac{|G|}{|S|} = \frac{|\mathrm{Aut}(E/F)|}{|\mathrm{Aut}(E/K)|} = \frac{\mathrm{dim}(E/F)}{\mathrm{dim}(E/K)} = \mathrm{dim}(K/F) = |K:F|
\end{align*}
where on the left, we are referring to the group index, and on the right the extension dimension.$\blacksquare$
It is a satisfying result to see that these two notations turn out to be compatible.
## Part 3 -
We have seen that every subextension $E/K/F$ of a Galois extension is Galois, but what about the extension $K/F$? We claim that
**Thoerem.** *For a Galois extension $E/K/F$ with $G = \mathrm{Gal}(E/F), S = \mathrm{Aut}(K/F)$, $K/F$ is a Galois extension if and only if $S \triangleleft G$, $S$ is normal in $G$.*
We begin with $K/F$ begin Galois. We have for any element $g \in G$ that $F$ is fixed,
\begin{figure}[!h]
\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=2em, column sep=4em, minimum width=2em]
{
E & & E \\
& F & \\
};
\path[-stealth]
(m-1-1) edge [<->] node [above] {$\overset{g}{\cong}$} (m-1-3)
(m-2-2) edge [-] (m-1-1)
(m-2-2) edge [-] (m-1-3)
;
\end{tikzpicture}
\end{figure}
and we can ask ourselves what happens to $K$ under $g$
\begin{figure}[!h]
\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=2em, column sep=4em, minimum width=2em]
{
E & & & & E \\
& K & & ? & \\
& & F & & \\
};
\path[-stealth]
(m-1-1) edge [<->] node [above] {$\overset{g}{\cong}$} (m-1-5)
(m-2-2) edge [->] node [above] {$g$} (m-2-4)
(m-3-3) edge [-] (m-2-2)
(m-2-2) edge [-] (m-1-1)
;
\end{tikzpicture}
\end{figure}
We claim that $g$ *preserves* $K$. This means that $g(K) = K$ but **not** that $g|_K = \mathrm{id}$. If the latter were true, then $g \in \mathrm{Aut}(E/K)$ which clearly needn't be the case.
*Proof.* ($g$ preserves $K$) As $K$ is a subextension of a Galois extension, it is also Galois. By the third equivalence, $K/F$ is also normal, and hence $K$ is the splitting field of some polynomial $p \in F[Z]$ with distinct roots $\{r_1, \ldots, r_k\}$ in $K$. These roots generate $K$, so it suffices to look at the image of these roots. As the coefficients of $p$ lie in $F$, $g$ fixes them and
\begin{align*}
0 = g(0) \overset{\mathrm{root}}{=} g(p(r_1)) \overset{\mathrm{hom}}{=} p(g(r_1))
\end{align*}
we see that $g(r_1)$ is again a root, meaning that $g$ is simply a permutation on the roots. A permutation of a generating set doesn't change the span, and thus we may conclude that $g(K) = K$.
We can now proceed to prove that $S$ is normal in $G$. To this, we must show that $\forall s \in S, g \in G: g^{-1} s g \in S$. As we now know that $g(K) = K$, we have
\begin{figure}[!h]
\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=2em, column sep=4em, minimum width=2em]
{
K & K & K & K \\
};
\path[-stealth]
(m-1-1) edge [->] node [above] {$g$} (m-1-2)
(m-1-2) edge [->] node [above] {$s|_K = \mathrm{id}$} (m-1-3)
(m-1-3) edge [->] node [above] {$g^{-1}$} (m-1-4)
;
\end{tikzpicture}
\end{figure}
In other words, $(g s g^{-1})|_K = \mathrm{id} \implies g s g^{-1} \in S \implies S \triangleleft G$, and we have the first direction of our proof.
Let us now proceed with showing the other direction. To that end, let $S \triangleleft G$ be normal. To show that the field fixed by $S$, $K$ is Galois over $F$, we will make a similar argument as before, and first show that $g(K) = K$.
*Proof.* ($g$ preserves $K$) Lets assume, indirectly, that $g$ does not preserve $K$. Then there exists some $x \in K$, such that $y := g(x) \notin K$. Now that we're no longer in $K$, $S$ need not fix $y$ and there is some $s \in S$, that moves $y$ to some $z \neq y$. With this we have $sg(x) = z \notin K$. However, owing to the normality of $S$, we know that for $sg$ there exists a $t \in S$ such that $gt = sg \iff g^{-1} s g \in S$. We now have a contradiction on our hands, as
\begin{align*}
y = g(x) \overset{t \in \mathrm{Aut}(E/K)}{=} g t (x) = s g (x) = s(y) = z
\end{align*}
but $y \neq z$.
We now want to use the second equivalence to show that $K/F$ is Galois. We can do this by looking at the homomorphism
\begin{align*}
\Phi{:}\, G \to S = \mathrm{Aut}(K/F); g \mapsto g|_K
\end{align*}
and by noting that $S = \mathrm{Ker}(\Phi)$, we see that
\begin{align*}
G/S \hookrightarrow \mathrm{Aut}(K/F); gS \mapsto \Phi(g)
\end{align*}
is injective and thus
\begin{align*}
|\mathrm{Aut}(K/F)| \geq |G/S| = \frac{|G|}{|S|}.
\end{align*}
On the other hand, by **Part 2**, we know that $\mathrm{dim}(K/F) = \frac{|G|}{|S|}$ and with Lecture 3, we have
\begin{align*}
\frac{|G|}{|S|} = \mathrm{dim}(K/F) \geq |\mathrm{Aut}(K/F)| \geq \frac{|G|}{|S|}.
\end{align*}
We may conclude that $\mathrm{dim}(K/F) = |\mathrm{Aut}(K/F)|$ and thus $K/F$ is Galois. $\blacksquare$
*Remark.* To know that $\Phi$ is well defined (i.e. $g|_K \in \mathrm{Aut}(F/K$)), we use that $g$ preserves $K$, and $g$ is an isomorphism over $E \supseteq K$.
**Corollary.** For $G$ the Galois group of some field extension $E/F$, every subfield $K$ that creates a subfield extension $E/K/F$ is preserved by $G$.
*Remark.* This again shows the compatibility between terms that we learned in group thoery and field extensions. A subgroup of the automorphism group is normal, and then the corresponding fixed field is also normal (i.e. Galois).
---
# SUMMARY
**Theorem.** Bijection
A the field of subextension is preserved by $G$.
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*Lecture 2019.03.29*
# LECTURE 6 - Solvability
This lecture we will investigate a very specific field extension, namely that of the rational functions in variables in $X_1, \ldots, X_n$ $\mathrm{Frac}(\mathbb{C}[X_1, \ldots, X_n])$ over the field of rational functions in the elementary symmetric functions $\mathrm{Frac}(\mathbb{C}[e_1, \ldots, e_n])$. We have seen elementary symmetric functions before in some examples, but now we will finally define them in general.
## Elementary Symmetric Functions
**Definition.** (Elementary Symmetric Functions) Let $X_1, \ldots, X_n$ be variables. The set of elementary functions $\{e_0, \ldots, e_n \}$ on these $n$ variables is defined as the coefficients of the polynomial
\begin{align*}
\prod _{k=1} ^n (Z - X_k) = \sum _{k=0} ^n e_{n-k} Z^k
\end{align*}
we can also write them in a more concrete closed form:
\begin{gather*}
\begin{cases}
e_0 = 1 \\
e_1 = \sum _{i=1} ^n X_i \\
e_2 = \sum _{1 \leq i<j \leq n} X_i X_j \\
e_3 = \sum _{1 \leq i<j<k \leq n} X_i X_j X_k \\
\cdots \\
e_n = \prod _{i=1} ^n X_i \\
\end{cases}
\end{gather*}
in general we have
\begin{align*}
e_k =
\begin{cases}
1 & k=0 \\
\sum _{1 \leq i_1 < \ldots < i_k \leq n} \left ( \prod _{l=1} ^{k} X_{i_l} \right ) & \text{else} \\
\end{cases}
\end{align*}
*Remark.* Note the strict inequalities between the inner indexes. The $x_i$'s are distinct.
Their name stems from the fact, that owing to the nature of the product form $\prod _{k=1} ^n (Z-X_k)$, the $e_k$'s are invariant under permutation of the variables. That is $e_k (X_1, \ldots, X_n) = e_k (X_\sigma(1), \ldots, X_\sigma(n))$ for $\sigma \in S_n$.
## The extension over Elementary Symmetric Functions
We can now go back to our original example, and ask ourselves is $\mathbb{C}(X_1, \ldots, X_n) / \mathbb{C}(e_1, \ldots, e_n)$ a finite extesnion? If so, is it Galois?
We will find that the answer to both of these questions is yes. Define $E = \mathbb{C}(X_1, \ldots, X_n), F = \mathbb{C}(e_1, \ldots, e_n)$, and lets prove this result.
*Proof.* (The extension is finite) We claim that $E$ is the splitting field over $F$ of the polynomial $p(Z) = \prod _{j=1} ^n (Z-X_j)$. It surely, by construction, contains all the roots of $p$. We must now that $p$ does not factorise into linear factors in any other subfield of $E$... @ror. In particular, as $E$ is the splitting field over $F$ of some polynomial with distinct roots, the extension is normal, and thus Galois.
Now that we know that the extension is finite, we can continue to find the exact degree. From Algebra I, we know that
\begin{align*}
\mathrm{dim}(E/F) \leq n!
\end{align*}
*Remark.* The proof for this statement lies in the fact that $E$ is a splitting field, meaning that it is no more than the base field with the roots adjoined. When we adjoin the first root $r_1$, the maximal degree of the extension is $n$, if $p$ is irreducible. Then we may factor $p/(Z-r_1) \in F(r_1)$ and continue adjoining the second root $r_2$. Again we may be dealing with an irreducible polynomial (in $F(r_1)$), thus the degree of the next extension is maxiamlly $n-1$. If we continue this process iteratively, we obtain $n!$ as an upper bound of the extension $E/F$.
If we now look at the action of $S_n$ on $\{X_1, \ldots, X_n \}$, with $\sigma \in S_n$ acting by $\sigma(X_i) = X_{\sigma (i)}$ we see that $\sigma(e_k) = e_k$, $S_n$ poses a good candidate for the automorphism group as a group action. In particular, we know that the $e_k$'s form a basis of $F$, thus we *know* that $S_n$ fixes $F$, and hence must be a subset of the automorphism group. Of course however $|S_n| = n!$ and thus $S_n = \mathrm{Aut}(E/F)$. Finally we have
\begin{gather*}
n! \leq |\mathrm{Aut}(E/F)| \overset{\text{Gal.}}{=} \mathrm{dim}(E/F) \leq n! \\
\implies \mathrm{dim}(E/F) = n!
\end{gather*}
## The subgroups of the extension
Now that we know that $E/F$ is Galois with dimension $n!$ we can apply theorems from the last lecture. Specifically, to study the subextensions of $E/F$, we can look at the subgroups of the Galois group (Part 1 of FTGT).
\begin{figure}[!h]
\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=0.25em, column sep=4em, minimum width=1em]
{
E =\mathbb{C}(X_1, \ldots, X_n) & \mathrm{Gal}(E/E) = \{\mathrm{id}\} \\
& \rotatebox[origin=c]{-90}{$\subseteq$} \\
K & \mathrm{Gal}(E/K) = G \\
& \rotatebox[origin=c]{-90}{$\subseteq$} \\
F = \mathbb{C}(e_1, \ldots, e_n) & \mathrm{Gal}(E/F) = S_n \\
};
\path[-stealth]
(m-1-1) edge [-] (m-3-1)
(m-3-1) edge [-] (m-5-1)
(m-1-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-1-2)
(m-3-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-3-2)
(m-5-1) edge [draw,-,decorate,decoration={snake}] node [below] {\tiny corresponds} (m-5-2)
;
\end{tikzpicture}
\end{figure}
Before we do this in general, we can look at following
**Example.** We know from group theory that
\begin{align*}
\{\mathrm{id}\} \triangleleft \mathbb{Z}_2 \triangleleft A_n \triangleleft K_4 \triangleleft S_n
\end{align*}
with
\begin{gather*}
\mathbb{Z}_2 = \{\mathrm{id}, (12)(34)\} \\
A_4 = \{\mathrm{id}, (12)(34), (13)(24), (14)(23)\}
K_4 = \{\mathrm{id}, (12)(34), (13)(24), (14)(23)\}
\end{gather*}
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\tikzset{
}
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# LECTURE 7 - Constructions with a Straightedge and Compass
*2019.04.05*
*Remark.* A **straightedge** refers to a ruler without markings.
## (I) Basic Operations and Constructions
### Operations
When given two points, legal *operations* with a straightedge and compass are
(1) Create straight line between them
(2) Create circle with one point at its centre that passes through the other point.
and also
(3) Create new points at intersections of previously drawn lines or circles.
### Constructions
(a) Transfer a length onto a line segment
(b) Transfer an angle onto a line segment
(c) Construct arbitrarily large lengths by repeating (a).
(d) Find the midpoint of two points.
(e) Construct the perpendicular bisector of a line segment.
(f) Construct a perpendicular line to a given line through a given point.
(g) Construct a parallel line.
Let $\mathbb{E}$ be the Euclidean Plane. Let $S \subseteq \mathbb{E}$ be a set of points, and denote by $K_S$ the set of points which can be constructed by the basic operations (1)-(3). From now on, we will assume $|S| > 1$, otherwise $K_S = S$.
**Definition.** (Constructable) For $A, B, C \in K_S$, the Euclidean distance $\mathrm{d}(A, B) \in R_{\geq 0}$ is called an *S-Constructable length*; the angle $\measuredangle ABC$ is called an *S-Constructable angle*.
This is all still a very geometric way of thinking about these problems, we will now translate them into algebraic terms. To do this, we will identify $\mathbb{E} \cong \mathbb{R}^2$ equipped with the Euclidean norm, such that $(0, 0)^{\mathrm{T}}, (1, 0)^{\mathrm{T}} \in S$. We can do this, as we have assumed $|S| > 1$, and we can choose the coordinates arbitrarily. We are now ready to make our first observation.
*Remark.* For ease of notation we will henceforth drop the transpose on coordinates.
**Fact 1.** $\forall n \in \mathbb{Z} : (0, n), (n, 0) \in K_S$
However, as we are seeking an algebraic descripbtion of these problems, we still don't have enough structure on $\mathbb{R}^2$. So, instead, let's identify $\mathbb{E} \cong \mathbb{R}^2 \cong \mathbb{C}$ with the usual mapping $(x,y) \in \mathbb{R}^2 \mapsto x+iy \in \mathbb{C}$. Now without loss of generality we have $0, 1 \in S \subseteq \mathbb{C}$, which by construction (a) implies $\mathbb{Z}, i\mathbb{Z} \in K_S$.
**Theoerm.** ($K_S$ a field) We claim that $K_S$ is the smallest subfield of $\mathbb{C}$ with the properties
(1) Contains $S$
(2) Closed under complex conjugation.
(3) Closed under square root extraction.
*Proof.* First we will prove that $K_S$ is a field.
### Additive group
#### Neutral element
By assumption $0 \in S$.
#### Closed under Addition
We can do a short case evaluation:
(i) *If* $zw = 0$ , then without loss of generality $w = 0$ and $w + z = z \in K_S$;
(ii) *if* $z, w$ are collinear, then we can draw a circle around the point with a greater norm, with radius of the smaller norm. Then the intersection of the line going through the origin and $w$ with the greatest norm is the sum of $w$ and $z$;
\begin{center}
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = .2pt},
]
% The circle
\def\a{1.5}
\def\b{2.5}
\def\g{30}
% The Axes
\node (origin) at (0,0) [point, label = {below right:$\mathcal{O}$}]{};
% Points
\node (n0) at (0,0) [point, label = {below right:$\mathcal{O}$}]{};
\node (n1) at +(\g:\a) [point, label = {below:$z$}]{};
\node (n2) at +(\g:\b) [point, label = {below:$w$}]{};
\node (n3) at +(\g:\a+\b) [point, label = {above right:$z+w$}]{};
\node (n4) at +(\g:\b-\a) [point, label = {left:$z-w$}]{};
% Circle
\draw (n2) circle (\a);
% Line connecting
\draw[-] (n0) -- (n3);
\end{tikzpicture}
\end{center}
(iii) *else* we can draw a parallelogram with one corner the origin, another $w$ and the opposite corner $z$. The last corner will be the sum of $z$ and $w$.
\begin{center}
\makeatletter
\tikzset{nomorepostaction/.code={\let\tikz@postactions\pgfutil@empty}}
\makeatother
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = .2pt},
]
\def\rad{10}
\def\a{1}
\def\b{2}
\def\c{2}
\def\d{0.50}
% The Axes
\draw[->] (-0.5,0) -- (3,0);
\draw[->] (0,-0.5) -- (0,3);
% Points
\node (n0) at (0,0) [point, label = {below right:$\mathcal{O}$}] {};
\node (n1) at (\a, \b) [point, label = {above:$z$}] {};
\node (n2) at (\c, \d) [point, label = {below:$w$}] {};
\node (n3) at (\a+\c, \b+\d) [point, label = {above:$z+w$}] {};
% Line connecting
\draw[dashed] (n0) -- (n1);
\draw[dashed] (n0) -- (n2);
\draw[dashed] (n1) -- (n3);
\draw[dashed] (n2) -- (n3);
\draw[-] (n0) -- (n3);
\end{tikzpicture}
\end{center}
#### Closed under inverses
Let $z \in K_S$. Then $-z \in K_S$ by following construction: We know that we can draw a circle of radius $|z|$, that has it's centre at the origin. Then we can construct the line from $z$ though the origin, and define a new point at the intersection of that line and the circle. This will be $-z$.
\begin{center}
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = .2pt},
]
% The circle
\def\rad{1.5}
\node (origin) at (0,0) [point, label = {below right:$\mathcal{O}$}]{};
\draw (origin) circle (\rad);
% The Axes
\draw[->] (-1.25*\rad,0) -- (1.25*\rad,0);
\draw[->] (0,-1.25*\rad) -- (0, 1.25*\rad);
% Points
\node (n1) at +(60:\rad) [point, label = above:$z$] {};
\node (n2) at +(-120:\rad) [point, label = below:$-z$] {};
% Line connecting
\draw[dashed] (n2) -- (n1);
\end{tikzpicture}
\end{center}
### Multiplicative group
*Remark.* Note that, as we can write $z \in \mathbb{C}$ as $z = r e^{i \alpha}, r, \alpha \in \mathbb{R}$, $z$ is constructable *if and only if* $r$ and $\alpha$ are respectively a constructable length and angle.
**Lemma.** (Multiplicative closure of constructable lengths) Let $a, b, c$ be constructable lengths. Then $\frac{bc}{a}$ is also a constructable length. In particular for any $a, b$ constructable lengths, $ab$ is also constructable, and $\frac{a}{b}$ is also constructable.
*Proof.* We can exploit similar triangles to prove this. The points $P_1, P_2, P_3$ are all constructable,
\begin{center}
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = .2pt},
]
% The circle
\def\a{3}
\def\b{4.5}
\def\c{1.5}
% The Axes
\node (origin) at (0,0) [point, label = {below right:$\mathcal{O}$}]{};
\draw[->] (-0.5,0) -- (\b + 1, 0);
\draw[->] (0,-0.5) -- (0,\b * \c / \a +1);
% Points
\node (n1) at (\a, 0) [point, label = {below:$a, P_1$}]{};
\node (n2) at (\b, 0) [point, label = {below:$b, P_2$}]{};
\node (n3) at (\a, \c) [point, label = {below right:$P_3$}]{};
\node (n4) at (\b, \b*\c / \a) [point] {};
\node (n5) at (0, \c) [label = {left:$c$}] {};
\node (n6) at (0, \b * \c / \a) [label = {left:$\frac{bc}{a}$}] {};
% Line connecting
\draw[-] (0,0) -- (n4);
\draw[-] (n1) -- (n3);
\draw[-] (n2) -- (n4);
\draw[dashed] (n5) -- (n3);
\draw[dashed] (n6) -- (n4);
\end{tikzpicture}
\end{center}
Setting either $a=1$ or $c=1$ we obtain the desired result.
**Lemma.** (Additive closure of constructable angles) Let $\alpha, \beta$ be constructable angles. Then $|\alpha \pm \beta|$ is also a constructable angle.
*Proof by picture.*
\begin{center}
\begin{tikzpicture}
[
scale=1,
>=stealth,
point/.style = {draw, circle, fill = black, inner sep = 1pt},
dot/.style = {draw, circle, fill = black, inner sep = .2pt},
]
\def\rad{4}
% Note \a must be greater than \b
\def\a{63}
\def\b{23}
\def\offset{20}
\node (n0) at (0,0) [point, label = {below right:$\mathcal{O}$}] {};
\node (n1) at +(\offset:\rad) {};
\node (n2) at +(\offset+\a:\rad) {};
\node (n3) at +(\offset+\a+\b:\rad) {};
\node (n4) at +(\offset+\a-\b:\rad) {};
\draw
(n1) -- (n0) -- (n2)
pic["$\alpha$", draw=orange, <->, angle eccentricity=1.2, angle radius=1cm]
{angle=n1--n0--n2};
\draw
(n2) -- (n0) -- (n3)
pic["$\beta$", draw=orange, <->, angle eccentricity=1.2, angle radius=1cm]
{angle=n2--n0--n3};
\draw
(n1) -- (n0) -- (n3)
pic["$\beta+\alpha$", draw=orange, <->, angle eccentricity=1.2, angle radius=2cm]
{angle=n1--n0--n3};
\draw
(n1) -- (n0) -- (n4)
pic["$\beta-\alpha$", draw=orange, <->, angle eccentricity=1.2, angle radius=3cm]
{angle=n1--n0--n4};
\end{tikzpicture}
\end{center}
#### Closed under multiplication
Let $z = r e^ {i\alpha} \in \mathbb{C}$ and $w = s e^ {i \beta} \in \mathbb{C}$ with $r, s, \alpha, \beta \in \mathbb{R}$, then $wz = rs e^{i (\alpha + \beta)}$. We have seen that constructable lengths are closed under multiplication, so $rs$ is also constructable and that constructable angles are closed under addition, so $\alpha + \beta \in S$. We may conclude that $wz \in S$.
### Closed under inverses
Again, by writing $z = r e ^{i\alpha} \neq 0$, we have with $r^{-1}$ a constructable length and $-\alpha$ a constructable angle, that $z^{-1} = r^{-1} e ^{-i\alpha} \in K_S$.
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