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\chapter{The Proofs of Impossibility}

It is known today that the proofs of impossibility of the three construction problems come from Pierre Wantzel and Ferdinand Lindemann, two nineteenth-century mathematicians whose publications brought the search for solutions to its end.  These proofs, however, did not solely come from two mathematical geniuses who happened to discover what two millennia of previous mathematicians did not have the right ideas to prove; rather, they build off previous advancements from both antiquity and the early modern period.  In fact, it was the translation of the three geometric problems into algebra that ultimately enabled Wantzel and Lindemann to demonstrate that their required constructions cannot be achieved by ruler and compass alone.   

The earliest argument for the impossibility of the cube duplication and angle trisection is attributed to Descartes in the seventeenth century, two hundred years before Wantzel’s proof was published.\footnote{For a thorough history of the proofs of the cube duplication and angle trisection, see Jesper Lutzen’s paper \textit{The Algebra of Geometric Impossibility: Descartes and Montucla on the Impossibility of the Duplication of the Cube and the Trisection of the Angle}.}  While it was believed among mathematicians since late antiquity that the two problems could not be solved by the ruler and compass constructions, no one formalized the notion into a proof.  While Pappus, for example, emphasized the impossibility of the problems by the Euclidean tools in his \textit{Collection}, he ``nowhere stated the necessity, desirability, or even possibility of giving a mathematical proof of this impossibility.''\footnote{Lutzen, p.6.}  The early modern geometers, despite providing their own solutions to the problems, also did not attempt to provide a proof of impossibility.  Why, then, was Descartes the first to address this issue? 

One possible explanation is that a proof of impossibility, by nature, requires a different sequence of logic than proving a theorem or proposition.  Since early antiquity, the mathematical enterprise could be viewed primarily as a problem-solving activity, and proving a statement is impossible is contrary to this end.  As Lutzen states, ``such an impossibility is not a proper mathematical result, but a sort of meta-statement regarding the problem solving activity.''\footnote{Lutzen, p.6,}  As both a mathematician and a philosopher, Pappus was the first of the Greek commentators to blur this distinction---he ``mixed solutions of mathematical problems with discussions about methodological issues.''\footnote{Lutzen, p.7.  He adds that it is therefore ``not so surprising that Pappus took up meta-mathematical questions such as the (un)solvability of problems with given means.''}  Thirteen centuries later, the philosopher-mathematician Descartes would also be interested in the solvability of the geometric problems.  Influenced by Pappus, Descartes classified problems based on the methods required for their solutions, suggesting it was prohibited to solve a planar problem---by Pappus's definition---by other means.  Thus, to clearly demonstrate that his construction using conic sections was the correct method for the solution, he needed to argue that a ruler and compass construction does not exist. 
	
Yet, to carry out this kind of proof, the geometric problems needed to be translated into algebra.  The method of using algebra to prove a theorem was not, however, a ``trivial matter of course'' or a ``natural line of argument.''\footnote{Lutzen, p.9, 10.}  Most early modern mathematicians were reluctant to use algebra in proofs.  Indeed, when algebraic methods were used in problems, they were used for finding the proper construction of the problem, but geometric methods were used for the proof of the correctness of the construction.  As Bos states, ``an algebra beset with such restrictions and uncertainties held little promise for geometry.''\footnote{Bos, p.133.}  For example, even in the early seventeenth century there was still an incomplete understanding of concepts like negative and complex numbers.  In the early modern period of mathematics, algebra problems usually contained a geometric proof, so the concept of using an algebraic proof for geometric problems was, in a sense, revolutionary.

The proofs Descartes provides do not, however, accurately give an algebraic synthesis of the impossibility of a ruler and compass construction to the cube duplication and angle trisection.\footnote{Thorough discussions of Descartes' method can be found in Lutzen and Bos.}  His first step was to assign letters to the known and unknown line segments and to formulate the relations of the problem into algebraic equations.  He would then reduce and rearrange the equations, leaving an equation in terms of just the unknowns, precisely the method that led Descartes to the correct equation corresponding to the two mean proportionals, as given by Hippocrates.  Descartes demonstrated that if the final equation is quadratic than the unknown segments can be constructed by the Euclidean tools, and thus the problem was planar by Pappus' definition.  ``If the equation is of degree higher than two,'' it was ``according to Descartes impossible to solve...by plane means.''\footnote{Lutzen, p.12.}  Later in his book, Descartes showed that the problems of the cube duplication and angle trisection lead to cubic equations.  Therefore, since problems constructible by planar means lead to quadratic equations and since the two geometric problems lead to cubic equations, Descartes asserted that the two problems cannot be solved by ruler and compass alone.  Yet, his logic for this argument is somewhat convoluted: he went on to prove the converse of the statement, that if ``the final equation is quadratic then the problem can be solved by'' the Euclidean tools.\footnote{Lutzen, p.14.}  

Much of the mathematics Descartes puts forth in his \textit{La G\`eom\`etrie} would become fundamental in the development of the theory of solutions of polynomial equations, which in turn helped Wantzel formulate his proof of the impossibility of the planar constructions.  He extensively describes the reducibility of polynomials and the factorization of equations.  Specifically, he considers the case where a given polynomial can be factored into two quadratic factors.  To do this, Descartes instructs to set up a resolvent of the form $p(x)=x^4+px^2+qx+r$ and to ``seek binomial factors of it considered as a third degree equation in $z=y^2$.''\footnote{Lutzen, p.16.}  If such binomial factors can be found, the roots of the two quadratic factors can be found by straightedge and compass constructions, and thus the problem is plane.  Lutzen summarizes Descartes’ method as follows: when the polynomial is determined for the ``principal unknown'' with coefficients in terms of the knowns, one carries out several ruler and compass constructions to determine ``new line segments from the givens,'' reducing the polynomial as much as possible.\footnote{Lutzen, p.19.}  If this produces a quadratic equation the problem is planar.  While Descartes' results are remarkable, he nevertheless ``did not develop an explicit algebraic formulation of the process of construction by ruler and compass.''\footnote{Lutzen, p.21.}  Moreover, Descartes lacked all the algebraic tools that would eventually be developed to finish the proofs. 

It is important to take stock of the many contributions Descartes gave to impossibility of geometric constructions, as well as some of his shortcomings.\footnote{Lutzen gives an extensive list of Descartes' contributions and failures, p.26-7.}  His contributions, among many, include formulating the impossibility of the cube duplication and angle trisection as theorems that should be proved, rather than just empirical statements.  Additionally, he attempted his own geometric proof, ``translated the geometric problems into algebraic ones concerning the solution of polynomial equations,'' and studied several methods to factorize polynomials.\footnote{Lutzen, p.26.}  For these advancements, the mathematical field is indebted to Descartes and his discoveries.  Yet, he did not provide the algebraic proof necessary to complete the study of the geometric problems.  He ``did not establish in a satisfactory way that...[the geometric]constructions could only solve quadratic equations,'' and he ``did not stress the relation between the algebraic process of factorization and geometric constructions but only focused on the construction of the final equation.''\footnote{Lutzen, p.27.}  It was these deficiencies, together with the lack of extensive knowledge of field theory, which led to an incomplete proof of impossibility.

Before turning to the use of this coordinate geometry in the proofs of the construction problems, it is important to first examine several algebraic definitions and theorems that will be used later in the proofs.  First, a field is an algebraic structure given by $(F, +, \cdot )$ that satisfies the following three conditions:\footnote{Gallian defines a field as ``simply an algebraic system that is closed under addition, subtraction, multiplication, and division (except by 0),'' where all the usual properties of these operations hold, p.176.} 
\begin{itemize}
\item $(F,+)$ is an abelian group 
\item $(F-{0}, \cdot )$ is an abelian group 
\item The distributive law $a\cdot (b+c)=a\cdot b +b\cdot c$ 
\end{itemize}

This leads to the following definitions.

\textit{Definition}: A field $E$ is an \textit{extension field} of a field $F$ if $F \subseteq E$ and the field operations on $E$ restrict to the field operations on $F$.  \\

\textit{Definition}: Let $E$ be a field extension of a field $F$.  Then $E$ has degree $n$ over $F$, written as [$E:F$]$=n$, if $E$ has dimension $n$ as a vector space over $F$.  \\

\textit{Definition}: A``tower'' of fields $F_1\subseteq F_2\subseteq ...\subseteq F_{k-1}\subseteq F_k$ is a finite sequence of field extensions $F_j\subseteq F_{j+1}$ where $j=1,...,k-1$.  \\

These lead to the following theorems.
\begin{theorem}
If $E\subseteq F\subseteq K$ is a ``tower'' of fields, then $[K:E]=[K:F][F:E]$.\footnote{As Gallian states, if $K$ is a finite extension field of the field $F$ and $F$ is a finite extension of the field $E$, then $K$ is a finite extension of field $E$, p.259.}
\end{theorem}

\textit{Proof}: Say $[F:E]=r$ and $\big\{\alpha_1...\alpha_r\big\}$ is a basis and $[K:F]=s$ with $\big\{\beta_1...\beta_s\big\}$ a basis. 
Let $\gamma \in K$, then $\gamma=C_1B_1+...+C_sB_s$ for some $C_i \in F$.  Then $C_i=b_{i1}\alpha_1+...b_{ir}\alpha_r$ for some $b_{ij} \in F$.  
So $\gamma = (b_{11}\alpha_1+...+b_{1r}\alpha_r)B_1 + (b_{s1}\alpha_1+...+b_{sr}\alpha_r)B_s$, which, when expanded, equals 

\begin{eqnarray*}
b_{i1}\alpha_1B_1+b_{i2}\alpha_2B_2+...+b_{ir}\alpha_rB_1+...+b_{sr}\alpha_rB_s
\end{eqnarray*}
So $\big\{\alpha_iB_j \vert 1\leq i \leq r, 1\leq j \leq s\big\}$ spans $K$ over $E$.
If 
\begin{eqnarray*}
b_{11}\alpha_1B_1+...+b_{1r}\alpha_rB_1+...+b_{sr}\alpha_rB_s=0
\end{eqnarray*}
then 
\begin{eqnarray*}
(b_{11}\alpha_1+...+b_{1r}\alpha_r)B_1+...+(b_{s1}\alpha_1+...+b_{sr}\alpha_r)B_s=0
\end{eqnarray*}
and hence $\big\{B_1...B_s\big\}$ is a basis and thus linearly independent.
So $b_{11}\alpha_1+...+b_{1r}\alpha_r=0$...$b_{s1}\alpha_1+...+b_sr\alpha_r=0$.  Therefore $\big\{\alpha_1...\alpha_r\big\}$ is linearly independent in $F$ over $E$, which is true for all $b_{ij}=0$.\\
\textit{Conclusion}: $\big\{\alpha_iB_j \vert 1\leq i \leq r, 1\leq j \leq s\big\}$ is a basis for $K$ over $E$.
Thus [$K:E$]$=r\cdot s =$ [$K:F$][$F:E$].  This proves the theorem.


\begin{theorem}
(Primitive Element Theorem) If $F_n$ is a finite extension of a field $F$, then there exists an element $a$ in $F_n$ such that $F_n=F_0(a)$.
\end{theorem}

The link between this field theory and the proofs of the geometric constructions comes in the form of irreducible polynomial equations.  \\  

\textit{Definition}: The irreducible polynomial of $a$ over $\mathbb{Q}$ is $p(x) \in \mathbb{Q}[x]$ of smallest degree with $p(a)=0$.  \\ 
 
This can be illustrated with the following example: 
Say $a=\sqrt{1+\sqrt{2}}$
So $a^2=1+\sqrt{2}$
$(a^2-1)=\sqrt{2}$, so $(a^2-1)^2=2$, which expanded gives $a^4-2a^2+1=2$.
Therefore $a^4-2a^2-1=0$. 
By the definition, $p(x)=x^4-2x^2-1 \in \mathbb{Q}[x]$ is the irreducible polynomial.  
It remains to show that $p(x)$ cannot be factored and thus is the irreducible polynomial.  First, try factoring $p(x)$ into the form $(x+a)(x^3+bx^2+cx+1)$.Expanding this gives
\begin{eqnarray*}
x^4+bx^3+cx^2+x+ax^3+abx^2+acx+a
\end{eqnarray*}
which is equivalent to
\begin{eqnarray*}
x^4+(a+b)x^3+(ab+c)x^2+(ac+1)x+a
\end{eqnarray*}  
and therefore $a+b=0$, $ab+c=-2$, $ac+1=0$, and $a=-1$.  Substituting backwards obtains $c=-1$ and $b=3$, but $-1+3 \neq 0$, and thus a factorization of this kind has no rational roots.  Now, try factoring $p(x)$ into the form $(x^2+ax+b)(x^2+cx+d)$.  Expanding this gives
\begin{eqnarray*}
x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd
\end{eqnarray*}
which is equivalent to
\begin{eqnarray*}
x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd
\end{eqnarray*}
and therefore $a+c=0$, $b+d+ac=-2$, $ad+bc=0$, $bd=-1$.  This implies $a=-c$, which substituted into the third equation gives $-cd+bc=0$, or $(b-d)c=0$.  This implies either $b-d=0$ or $c=0$.  If $b-d=0$, then $b=d$, which cannot be true since $bd=-1$.  If $c=0$, then $b+d=-2$, and since $bd=-1$, then $-2b+b^2=-1$.  Applying the quadratric formula gives 
\begin{eqnarray*}
b=\frac{-2 \pm\sqrt{4+4}}{2}
\end{eqnarray*}
which is not an element of $\mathbb{Q}$.  Therefore a factorization of this kind has no rational roots.  Thus $p(x)=x^4-2x^2-1 \in \mathbb{Q}[x]$ is the irreducible polynomial.      

With this algebraic background in hand, one can supply a modern account of the solutions provided by Wantzel and Lindemann.  First, it is important to discuss what points can be constructed by a straightedge and compass in the Cartesian plane.  Given a unit of distance, or a segment of length one, and the coordinate axes with the origin, what points can be constructed by a straightedge and compass?\footnote{A thorough examination of the abstract algebra and field theory that led to the proofs of impossibility can be found in Gallian's \textit{Contemporary Abstract Algebra} and Morandi's \textit{Field and Galois Theory}.}  It first must be shown that any point $(a, b)$ can be constructed with $a,b \in \mathbb{Q}$.  To show this, it is sufficient to construct $(a,0), (0,b)$, or even all $a \in \mathbb{Q}$ along one line.  Thus, suppose $a=\frac{m}{n}$ in lowest terms.  Since a unit length is assumed, a segment $AB$ of length one exists, therefore it remains to show that a segment of length $\frac{1}{n}$ exists.  Take segment $AB$.  From $A$, lay off points $C_1, C_2 ... C_n$ such that $AC_1=C_1C_2=...=C_{n-1}C_n$.  Join $BC_N$.  Construct parallels $C_1B_1, C_2B_2,...C_{n-1}B_{n-1}$ to $BC_n$, with $B_1, B_2...B_{n-1}$ on $AB$.  $AB_1$ then has length equal to $\frac{1}{n}$.  

Thus, it is possible to construct any $(a,b) \in \mathbb{Q}^2$.  Next, it must be shown that each step of a construction will involve new points generated by the intersections of either two lines, a line and a circle, or two circles.  Considering each of these cases in turn yield the following.  \\

\textit{Case 1: The Intersection of Two Lines} \\

If it is known that one can construct with the Euclidean tools any $(a,b) \in \mathbb{Q}^2$, assume $L_1$ and $L_2$ are two lines such that $L_1: (a_1,b_1)$ and $(c_1,d_1)$ and $L_2: (a_2,b_2)$ and $(c_2,d_2)$ with $a_i, b_i, c_i, d_i \in \mathbb{Q}$.  The lines can be expressed in point-slope form as

\begin{eqnarray*}
L_1: y-b_1=\left(\frac{d_1-b_1}{c_1-a_1}\right)(x-a_1)
\end{eqnarray*}
\begin{eqnarray*}
L_2: y-b_2=\left(\frac{d_2-b_2}{c_2-a_2}\right)(x-a_2)
\end{eqnarray*}
with each having the form 
\begin{eqnarray*}
A_1x+B_1y=C_1 
\end{eqnarray*}
\begin{eqnarray*}
A_2x+B_2y=C_2.
\end{eqnarray*}
Using determinants to solve for the point of intersection, $(x,y)$, gives

$$
x=\frac{\det
\begin{pmatrix}
C_1 & B_1 \\ 
C_2 & B_2
\end{pmatrix}}
{\det
\begin{pmatrix}
A_1 & B_1 \\ 
A_2 & B_2
\end{pmatrix}}
$$
and 
$$
y=\frac{\det
\begin{pmatrix}
A_1 & C_1 \\ 
A_2 & C_2
\end{pmatrix}}
{\det
\begin{pmatrix}
A_1 & B_1 \\ 
A_2 & B_2
\end{pmatrix}}
$$
and therefore both $x,y \in \mathbb{Q}$ and thus $(x,y) \in \mathbb{Q}^2$.

Thus, the intersection of two lines has rational coordinates in the same field as the two lines.  Or, as Gallian states, ``two lines in $F$ intersect in a point in the plane of $F$.''\footnote{Gallian, p.264.}\\

\textit{Case 2: The Intersection of a Line and a Circle} \\

Assume $L$ is a line such that $L: (a,b)$ and $(c,d)$ with $a, b, c, d \in F$ and $C$ is a circle with center $(p,q)$ and radius $r$ with $p,q,r \in F$

So $L$ can be expressed as 

\begin{eqnarray}
L: Ax+By-C
\end{eqnarray}
for some $A,B,C,D \in F$ and $C$ can be expressed as 
\begin{eqnarray}
C:(x-p)^2+(y-q)^2=r^2
\end{eqnarray}
In (1), assuming $B\neq 0$ and solving for $y$ gives
\begin{eqnarray}
y=-\frac{A}{B}x+\frac{C}{B}
\end{eqnarray}
which substituted into (2) gives 
\begin{eqnarray*}
(x-p)^2+\left(-\frac{A}{B}x+\frac{C}{B}-q^2\right)=r^2
\end{eqnarray*}  
Expanding this gives
\begin{eqnarray*}
\alpha x^2 + \beta x + \gamma =0
\end{eqnarray*} 
for some $\alpha,\beta,\gamma \in F$.  Applying the quadratic formula gives 
\begin{eqnarray*}
x=\frac{-\beta \pm\sqrt{\beta ^2 -4 \alpha \gamma}}{2\alpha}
\end{eqnarray*} 
Substituting this back into (3) gives 
\begin{eqnarray*}
y=-\frac{A}{B}\left(\frac{-\beta \pm\sqrt{\beta ^2 -4 \alpha \gamma}}{2\alpha}\right) +\frac{C}{B}
\end{eqnarray*} 

Define $\Delta$ as $\beta ^2 -4 \alpha \gamma$.

So if $\Delta$ is not a square in $F$, one needs to go to a bigger field to get the coordinates of the intersection. 
Given any field $F \subseteq \mathbb{R}$, $F(\sqrt{\Delta})$ is the smallest field containing $F$ and $\sqrt{\Delta}$.

\textit{Claim:} $F(\sqrt{\Delta})=\big\{f+g\sqrt{\Delta}  \vert  f,g \in F\big\}$

The proof of this claim follows the following argument.  First $\big\{f+g\sqrt{\Delta}  \vert  f,g \in F\big\}$ is contained in $F(\sqrt{\Delta})$ by the definition of a field.  So it suffices to show that $\big\{f+g\sqrt{\Delta}  \vert  f,g \in F\big\}$ is itself a field.  To demonstrate closure under multiplication, consider the product of
\begin{eqnarray*}
(f+g\sqrt{\Delta})(f'+g'\sqrt{\Delta})
\end{eqnarray*} 
which, when expanded, gives 
\begin{eqnarray*}
ff'+fg'\sqrt{\Delta}+f'g\sqrt{\Delta}+gg'\Delta
\end{eqnarray*}
which equals
\begin{eqnarray*}
ff'+gg'\Delta +(fg'+f'g)\sqrt{\Delta}
\end{eqnarray*}
which is in $F(\sqrt{\Delta})$.  Thus $F(\sqrt{\Delta})$ is closed under multiplication.
The additive inverse is given by $-f+g\sqrt{\Delta}$, since $f+g\sqrt{\Delta}+(-f+g\sqrt{\Delta})=0$. 
To demonstrate a multiplicative inverse, it might be tempting to use $(f-g\sqrt{\Delta})$.  However, 

\begin{eqnarray*}
(f+g\sqrt{\Delta})(f-g\sqrt{\Delta})=f^2-g^2\sqrt{\Delta}
\end{eqnarray*}
which, if equal to 0, would imply that $f^2=g^2\Delta$, or $\Delta=(\frac{f}{g})^2$, which contradicts $\Delta$ not being a square.  So the correct multiplicative inverse is given by 
\begin{eqnarray*}
\frac{1}{f^2-g^2\Delta}(f-g\sqrt{\Delta})
\end{eqnarray*}
since
\begin{eqnarray*}
(f+g\sqrt{\Delta})\left(\frac{1}{f^2-g^2\Delta}(f-g\sqrt{\Delta})\right)=1
\end{eqnarray*}
and thus the multiplicative inverse is in $F(\sqrt{\Delta})$.

Thus, for the intersection of a line and circle, if $\Delta$ is not a square in $F$ it is required to go to some field $F(\sqrt{\Delta})$. \\

\textit{Case 3: The Intersection of Two Circles} \\

The intersection of two circles will produce at most two points of intersection, unless the circles are tangent, in which case they intersect at one point.  Suppose the two circles, $C_1$ and $C_2$, intersect at two points.  Define the circles as 

\begin{eqnarray*}
C_1: (x-h_1)^2+(y-k_1)^2=r_1^2
\end{eqnarray*}
\begin{eqnarray*}
C_2: (x-h_2)^2+(y-k_2)^2=r_2^2
\end{eqnarray*}
Expanding these equations gives 
\begin{eqnarray}
C_1: x^2+y^2-2h_1x-2k_1y+h_1^2+k_1^2=r_1^2
\end{eqnarray}
\begin{eqnarray}
C_2: x^2+y^2-2h_2x-2k_2y+h_2^2+k_2^2=r_2^2
\end{eqnarray}
Subtracting (4) from (5) gives 
\begin{eqnarray}
2(h_1-h_2)x+2(k_1-k_2)y-h_1^2+h_2^2-k_1^2+k_2^2=r_2^2-r_1^2
\end{eqnarray}
which, together with (4), gives the intersection of a circle and a line.  Thus, the intersection of two circles yields the same results as the intersection of a line and a circle. \\

These three cases lead to the an important theorem and two corresponding corrolaries.  

\begin{theorem}
If $\alpha$ is constructible over $\mathbb{Q}$, then there exists a tower of fields $\mathbb{Q}=F_0\subseteq F_1\subseteq...\subseteq F_n$ where for each i either $F_{i+1}=F_i$ or $F_{i+1}=F_i(\sqrt{\Delta_i})$ where $\delta_i \in F_i$ not a square and $\alpha \in F_n$. 
\end{theorem}

\textit{Corollary}: If $F_1=Q(\sqrt{\Delta_0})$ with $\Delta_0$ not a square, then the dimension of $F_1$, given as dim$F_1$, is 2.

\textit{Proof}: $\big\{1,\sqrt{\Delta_0}\big\}$ spans $F_1$.  A typical element is $a+b\sqrt{\Delta_0}$, with $a,b \in \mathbb{Q}$.  If $a+b\sqrt{\Delta_0}=0$ with $b\neq 0$, then $\sqrt{\Delta_0}=\frac{-a}{b}$, or $\Delta_0=(\frac{-a}{b})^2$, which contradicts $\Delta_0$ not being a square.
Thus $\big\{1, \sqrt{\Delta_0}\big\}$ is linearly independent over $\mathbb{Q}$, and hence forms a basis.  Thus dim$F_1=2$, or [$F_1:\mathbb{Q}$]=2.

Similarly, for each $i$, $F_{i+1}$ is a vector space over $F_i$, so [$F_{i+1}:F_i$]=1 if $F_i$ or $2^l$ if $F_i(\sqrt{\Delta_i})$ with $l<i+1$.  \\ 

\textit{Corollary}: If $\alpha$ is constructible, [$\mathbb{Q(\alpha)}:\mathbb{Q}$]$=2^k$ for some nonnegative integer $k$.

\textit{Proof}: Say 
\begin{eqnarray*}
F_0=\mathbb{Q}\subseteq F_1\subseteq F_2 \subseteq ...\subseteq F_n 
\end{eqnarray*}
comes from a construction.  
It is known that [$F_{i+1}:F_1$]=1 or 2 for all $i$.  By Theorem 3, 
\begin{eqnarray*}
[F_n:\mathbb{Q}]=[F_n:F_{n-1}][F_{n-1}:\mathbb{Q}]
\end{eqnarray*}
which equals
\begin{eqnarray*}
[F_n:F_{n-1}][F_{n-1}:F_{n-2}][F_{n-2}:\mathbb{Q}]
\end{eqnarray*}
which equals
\begin{eqnarray*}
[F_n:F_{n-1}][F_{n-1}:F_{n-2}]...[F_1:\mathbb{Q}],
\end{eqnarray*}
with each dimension equalling 1 or 2.  So the dimension of the tower of fields is thus $2^k$ for some $k$.  

An application of this field theory can be used to demonstrate the impossibilty of the three construction problems.\footnote{An explanation of these proofs can be found in Gallian, p.265 and Morandi, p.145.}   First, consider the duplication of the cube.  Given a cube of volume 1, the problem compels the geometer to construct a cube of volume 2, thus having an edge of length equal to $\sqrt[3]{2}$.  The problem is thus reduced to determining if a length of $\sqrt[3]{2}$ is constructible with a straightedge and compass.  In this case, $x^3-2=0$ is the irreducible equation.  To show it is irreducible, assume it can be factored into the form $(x-a)(x^2+xa+a^2)$.  Expanding this gives $x^3+x^2a+xa^2-x^2a-xa^2-a^3$, or $x^3-a^3$.  So $a^3=2$, and thus $a=\sqrt[3]{2}$.  But $\sqrt[3]{2}$ is not an element of $\mathbb{Q}$, so the equation $x^3-2=0$ has not rational roots and is therefore irreducible.  Thus [$\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}$]=3, which is not equal to $2^s$ for some s.  Therefore $\sqrt[3]{2}$ is not constructible, and the duplication of the cube is impossible. 

The impossibility of the trisection of an angle can be proved with a particular example.  In order to trisect an angle of $60^{\circ}$, then $\cos20^{\circ}$ must be constructible.  The proof makes use of the trigonometric identity

\begin{eqnarray*}
\cos 3\theta=4\cos^3\theta-3\cos\theta 
\end{eqnarray*}    
with $\theta=20^{\circ}$.  Therefore 
\begin{eqnarray*}
\frac{1}{2}=4\cos^320^{\circ}-3\cos20^{\circ}
\end{eqnarray*} 

Substituting 
$\cos20^{\circ}=x$ gives $\frac{1}{2}=4x^3-3x$, or $1=8x^3-6x$.  Thus, $8x^3-6x-1=0$ is the irreducible equation over $\mathbb{Q}$.  To show this equation is irreducible involves the use of the Rational Roots Test, described as follows.

\textit{Claim}: If $a_nx^n+...+a_1x+a_0=0$ with each $a_i \in \mathbb{Z}$ has a rational root $x$, then $x=\frac{p}{q}$ in lowest terms such that $q$ divides $a_n$ and $p$ divides $a_0$. \\
\textit{Reason}: Substitute $x=\frac{p}{q}$ into the polynomial equation.  Then 
\begin{eqnarray*}
a_n\left(\frac{p}{q}\right)^n+...+a_1\left(\frac{p}{q}\right)+a_0=0.
\end{eqnarray*}
Multiplying both sides by $q^n$ gives 
\begin{eqnarray}
a_np^n+a_{n-1}p^{n-1}q+...+a_0q^n=0,
\end{eqnarray}
which is equivalent to 
\begin{eqnarray*}
a_np^n=-q\left(a_{n-1}p^{n-1}+...+a_0q^{n-1}\right).
\end{eqnarray*}  
Therefore, since $q$ and $p$ are relatively prime, $q$ must divide $a_n$.
Likewise, (7) is equivalent to 
\begin{eqnarray*}
a_0q^n=-p\left(a_np^{n-1}+...+a_1q^{n-1}\right).
\end{eqnarray*}
Again, since $q$ and $p$ are relatively prime, $p$ must divide $a_0$.

For the given polynomial, it follows from the Rational Roots Test that the only rational roots would be $x= \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$, or $\pm \frac{1}{8}$.  Substituting each of these into the polynomial yields: $x=1$ gives $1 \neq 0$, $x=-1$ gives $-3 \neq 0$, $x=\frac{1}{2}$ gives $-3 \neq 0$, $x=-\frac{1}{2}$ gives $-1 \neq 0$, $x=\frac{1}{4}$ gives $-frac{19}{8} \neq 0$, $x=-\frac{1}{4}$ gives $-\frac{3}{8} \neq 0$, $x=\frac{1}{8}$ gives $-\frac{111}{64} \neq 0$, and $x=-\frac{1}{8}$ gives $\frac{17}{64} \neq 0$.  Since none of the potential rational roots work, $8x^3-6x-1-0$ is the irreducible equation.  

Therefore, $[\mathbb{Q}(\cos 20^{\circ}):\mathbb{Q}]=3$, which is not equal to $2^s$ for some s.  Thus $\cos20^{\circ}$ is not constructible with the Euclidean tools, and the general trisection of an angle problem is impossible.

The proof of the squaring of the circle involves the use of theorem and a corresponding corollary: 

\begin{theorem}
(Lindemann-Weierstrass Theorem) Let $\alpha_1,...\alpha_m$ be distinct algebraic numbers.  Then the exponentials $e^{\alpha_1},...e^{\alpha_m}$ are linearly independent over $\mathbb{Q}$.\footnote{This is an alternative version of the theorem given by Morandi, since it is ``a little easier to prove than the original,'' (p.134).  He states the original, equivalent theorem on p.138 as follows: If $\alpha_1,...,\alpha_m$ are $\mathbb{Q}$-linearly-dependent algebraic numbers, then the exponentials $e^{\alpha_1},...,e^{\alpha_m}$ are algebraically independent, hence there is no non-zero polynomial $f(x_1,...,x_m) \in \mathbb{Q} [x_1,...,x_m]$ with $f(e^{\alpha_1},...,e^{\alpha_m})=0$.}
\end{theorem}

\textit{Corollary}: The numbers $\pi$ and $e$ are transcendental over $\mathbb{Q}$. 

\textit{Proof}: Suppose $e$ is algebraic over $\mathbb{Q}$.  Then there exists $r_i \in \mathbb{Q}$ such that $r_1e^{1}+...+r_ne^{n}=0$.  This implies $e^{0}, e^{1},...,e^{n}$ are linearly dependent over $\mathbb{Q}$.  But, by choosing $m=n+1$ and $\alpha_i=i-1$, this dependence contradicts the Lindemann-Weierstrass Theorem, and hence $e$ is transcendental over $\mathbb{Q}$.  If $\pi$ is algebraic over $\mathbb{Q}$, then so is $\pi i$.  This implies $e^{0}$ and $e^{\pi i}$ are linearly independent over $\mathbb{Q}$, which cannot be true since $e^{\pi i}=-1$.  Therefore, $\pi$ is transcendental over $\mathbb{Q}$. 

Hence, the impossibility of the squaring of the circle can also be proved with a specific example.  Consider a circle of radius 1.  The problem compels the geometer to construct a square with area $\pi$, thus having sides of length $\sqrt{\pi}$.  But, since $\pi$ is transcendental over $\mathbb{Q}$, a segment with length $\sqrt{\pi}$ cannot be constructed by the Euclidean tools since it is not algebraic of degree a power of 2.
 
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