A VARIATION ON GEOMETRIC CONSTRUCTIONS
By
James Robertson
B.A. University Of Maine, 2007
A THESIS
Submitted in Partial Fulfillment of the
Requirements for the Degree of
Master of Arts
(in Mathematics)
The Graduate School
The University of Maine
December, 2010
Advisory Committee:
William Snyder, Professor of Mathematics, Advisor
Ali Özlük, Professor of Mathematics
David Bradley, Associate Professor of Mathematics
Andrew Knightly, Assistant Professor of Mathematics
A VARIATION ON GEOMETRIC CONSTRUCTIONS
BY
JAMES ROBERTSON
Thesis Advisor: Dr. William Snyder
An Abstract of the Thesis Presented
in Partial Fulfillment of the Requirements for the
Degree of the Master of Arts
(in Mathematics)
December, 2010
Since the time of the Greeks straightedge and compass constructions have been
studied extensively. These constructions represent geometry in its purist form and
have been taught to students of mathematics since. As elegant as these constructions are they have very real limitations. The Greeks suspected these limitations
and modern algebra fully exposed them. In spite of these limitations innovative
tools have found their way into mathematical literature and have enabled us to
construct cube roots and trisect any angle. Some of these tools have been fully
characterized while some have not. This thesis investigates marked straightedge
and compass constructions through an intriguing analog tool set that is coined as
q-construction. It is known that taking cube roots and finding an arbitrary angle
trisection is possible with marked straightedge and compass. At least one construction that is unknown is the taking of fifth roots. We show that it is possible to take
fifth roots using q-construction. This non-trivial result suggests that it is possible
to take fifth roots using a marked ruler and compass.
ACKNOWLEDGMENTS
I would like to sincerely thank my advisor Dr. William Snyder, without his
patience and guidance I would not have written a thesis or been able to offer any
contribution to mathematics. I also would like to thank my family and friends who
have supported and harassed me the whole way.
TABLE OF CONTENTS
ACKNOWLEDGEMENTS
ni
LIST OF FIGURES
v
PART
1.
INTRODUCTION
1
2.
q-CONSTRUCTION
4
3.
CHARACERIZATION OF q-CONSTRUCTIBLE NUMBERS
6
4.
EXAMPLES OF q-CONSTRUCTIBLE NUMBERS
17
REFERENCES
25
BIOGRAPHY OF THE AUTHOR
26
iv
LIST OF FIGURES
1
Showing Claim 2
11
2
Showing Claim 3
11
3
Showing a + b
e
4
Showing -a
e
F
12
5
Showing ab
e
F
13
6
Showing Ma e F
14
7
Showing
F
12
j;' = V(l-((x-l)/U+l))2)
v
e F
15
Part 1. I N T R O D U C T I O N
In ancient Greece, there was considered a pure way to study constructive geometry. In this method we are given two points which we may assume to be one
unit apart and with one to be centered at the origin, a straightedge to make lines
between two points and a compass to make circles between two points. We will first
formally describe how to find the points constructible with these tools as given in
[3].
In the Cartesian plane, a point is considered straightedge and compass constructible if it is the last of a finite sequence of points Pi, Pi,..., Pn such that each
point is in
{(0,0),(1,0)}
or is obtained in one of the following three ways:
(i) as the intersection
of two lines, each of which passes through two points that
appear earlier in the sequence;
(ii) as the point(s)
of intersection
of two distinct circles, each of which passes
through an earlier point in the sequence and each of which has an earlier point as
center;
(Hi) as the point(s) of intersection
(ii),
of a line and a circle as described in (i) and
respectively.
These tools have been extensively used and analyzed throughout history and have
well documented limitations [2]. Using them it is impossible to trisect an arbitrary
angle and it is also impossible to find the cube root of an arbitrary number. The
second of these was posed to the Greeks by the Oracle at Delphi who decreed if they
could double the volume of the altar to Apollo (which was a perfect cube) using
only the straightedge and compass then a plague that had beset the Isle of Delos
would be defeated [3j. Both of these constructions were proved to be impossible,
but only with the help of modern algebra. It is interesting to see that in spite of the
impossibility of these constructions the ingenuity of the Greeks found ways around
these limitations. Pappus, Nicomedes, Menaechmus and Archimedes were able to
1
independently find solutions to these problems with the use of additional tools
[3, 4]. These tools have fallen under far less algebraic scrutiny than straightedge
and compass constructions, despite the rich fields they generate. We will briefly
go over a few additional tools that were discovered in the quest to construct the
impossible to familiarize ourselves with them and mention a few of their algebraic
properties before we delve into our variation on the tools set in place by the Greeks.
Let us now consider what the Greeks referred to as solid constructions
[1]. We
will specifically consider the conic section y = x2 in addition to our previous tools
the straightedge and compass. We will formally describe these constructions by
including our description of straightedge and compass constructions and elaborating
on it.
In the Cartesian plane, a point, P, is considered a solid construction
if it is a
straightedge and compass point or if using one or more given conic sections with
constructible focus A, constructible directix I, and constructible eccentricity e, P
can be found as the intersection of two distinct constructible lines, circles, or conic
sections [1].
With the use of conic sections Menaechmus was able to find ^/x [4] (Archimedes
was able to construct a regular heptagon [1]). It has been shown that solid constructions using just y = x2 produce 2-3 towers over Q [1] . We will formally define
2-3 towers in Part 2.
We will next consider what are referred to as marked straightedge
constructions
[3].
In the Cartesian plane, a point is considered marked straightedge constructible if
it is the last of a finite sequence of points P l l P 2 l . . . , P n such that each point is in
{(0,0),(1,0),(0,1)}
or is obtained in one of the two following ways:
(i) as the intersection of two lines, each of which passes through two points that
appear earlier in the sequence;
(ii) as either of two points that are one unit apart, that are collinear with a point
that appears earlier in the sequence, and that are such that each lies on a different
line through two points that appear earlier in the sequence.
It has been shown that solid constructions and marked straightedge constructions
produce the same set of constructible points [3, 1]. Martin in [3] poses the question
of marked straightedge and compass constructions and what constructions could
come of these. We now present a formal description of these constructions.
In the Cartesian plane, a point is considered marked straightedge and compass
constructible if it is a straightedge and compass point or is obtained in one of the
following ways:
(iv) as either of two points that are one unit apart, that are collinear with a point
that appears earlier in the sequence, and that are such that each lies on a different
line through two points that appear earlier in the sequence;
(v) as either of two points that are one unit apart, that are collinear with a
point that appears earlier in the sequence, and that are such that each lies on two
circles (not necessarily distinct) each of which passes through an earlier point in the
sequence and each of which has an earlier point as center;
(vi) as either of two points that are one unit apart, that are collinear with a
point that appears earlier in the sequence, and that are such that each lies on a
line through two points that appear earlier in the sequence and a circle which passes
through an earlier point in the sequence and which has an earlier point as center.
We exclude any single construction that produces infinitely many points.
With some work we can show that marked straightedge and compass constructions can produce solutions to certain fifth and sixth degree polynomials but it is
not known if we can in general find fifth roots using these tools. This question
motivates our analog to marked straightedge and compass constructions, which we
refer to as q-constructions. We show that in general you can find fifth roots of qconstructible numbers which strongly hints that we could do the same with marked
straightedge and compass numbers.
3
Part 2. q - C O N S T R U C T I O N
We will now describe the tools for our constructions and define how to obtain
constructible points and numbers.
Definition 1. In the Cartesian plane, a point is a q-constructible point if it is the
last of a finite sequence of points P\, P2,..., Pn such that each point is in
{(0,0), (1,0), (0,1)}
or is obtained in one of the following ways:
(i) as the intersection
of two lines, each of which passes through two points that
appear earlier in the sequence;
(ii) as an intersection
of a line passing through two earlier points and a circle
passing through (0,0) and centered at a point on the x-axis appearing earlier in the
sequence;
(Hi) as a point of intersection of the graph of y = x3 and a line described in (i);
(iv) as a point of intersection
of the graph of y = x3 and a circle described in
(ii).
A real number will be called q-constructible,
if it is the x-coordinate of a q-
constructible point lying on the x-axis. A line passing through two q-constructible
points will be called a q-constructible
line.
The motivation for using these tools as described can initially be attributed to
the behavior of using the conic y = x2 instead of y = x 3 . We find that using
y = x2 along with the straightedge and compass produces 2-3 towers over Q. This
means that a is constructible by this process if and only if there exists a sequence
of field extensions K0,K\,...
,Kn
such that Q = K0 C Kx
where [Kj : Kj.-{\ = 2 or 3, for j = 1 , . . . , n and a e Kn.
C . . . C Kn
C K
This set of numbers
is contained in the set of marked straightedge and compass numbers [1]. However
4
with some work we can see the marked straightedge and compass produces a richer
set of numbers than the addition of y = x 2 to the straightedge and compass. In
fact we find that the marked straightedge and compass numbers can satisfy certain
polynomials up to the sixth degree, while the use of y = x2 with the straightedge
and compass can only satisfy a fourth degree polynomial [1]. On the other hand
we can show that using y = x3 with straightedge and compass produces numbers
that can satisfy polynomials of degree at most six. However, the limitation placed
on circles limits us to fifth degree polynomials. Thus we coined "q-constructible"
as the q refers to quintic polynomials.
One of the results we obtain gives the following algebraic characterization of
q-constructible numbers.
A real number, a, is q-constructible if and only if there exists a sequence of field
extensions KQ, K\,..., Kn with a £ Kn such that
where [Kj : Kj-i]
G {1,2,3,5}, for j = l,...,n, and such that if [Kj : /fj_i] = 5,
then Kj — Kj^i(ffaj-i)
where dj~i £ Kj-\
a
the polynomial x + x — Oj_i.
5
and ^/a,-~i is the unique real root of
Part 3. C H A R A C T E R I Z A T I O N OF q - C O N S T R U C T I B L E N U M B E R S
In order to give a characterization of the q-constructible numbers we will first
lay down some notation and terminology for convenience.
• We will refer to the set of all q-constructible numbers as F.
• If a is a real number, then the unique real root of the polynomial x° + x — a is
called the ultra-radical of a and is denoted by y'a. If a e K, where K is a sub-field
of R, then K(\/a)
will be called an ultra-radical extension of K.
• Let Q = KQ C K\ C ... C Kn C R be an extension of fields with [K3 • Kj _ i ] =
1, 2, 3, or 5 (for j = 1,2,..., n) such that if the degree is 5, then if, is an ultra-radical
extension of Kj-\.
We will call such an extension a real q-tower of Q.
T h e o r e m 1. A real number a lies in F if and only if a G Kn for some real q-tower
K0C...CKn
ofQ.
To prove this we will need a few tools.
Proposition 1. Let Pj = (a3,bj) for j — l,2,...,ra, be a q-constructible
of points. Then there is a real q-tower K0....,Kn
of Q such thataj^bj
sequence
e Kn for all
j-
Proof. We
will
use
induction
on
m.
Pi = ( a i A ) e {(0,0), (0,1), (1,0)} and thus a^h
For
m
=
1
then
£Q=K0.
Now assume m > 1 and that the proposition holds for any sequence of less than
m points.
Let Pi,...,Prn
be a q-constructible sequence of points. Then for all
j < m there exists by the induction hypothesis some q-tower K0, ...Kn such that
a.j,bj € Kn for all j < m. Now consider the point Pm = ( a m , 6 m ) . We know this
must be an intersection of curves in one of four possible ways.
6
First consider the intersection of two lines, each passing through two points
earlier in the sequence.
Since we are in the Cartesian plane we know both lines must be of the form:
where a'it b'. cj € Kn. Since these lines intersect we know they must have the solution
set
and
Thus the point of intersection will be Pm = (x, y) and x, y £ Kn.
Next we consider the intersection of a line L passing through two previous qconstructible points and a circle C centered at a q-constructible point (a'2,Q) on
the x-axis and passing through (0,0). Line L and circle C have the equations
and
where a'i,Vi.,di £ Kn.
solutions:
and
The intersection of these two will result in the following
when b[ ^ 0 and
when b[ = 0. Or we can equivalent!}' say x =
'"},
and y =
2 J
^2
where
di,ei, and fi are just the respective part of their respective quadratic solutions and
di,ei,fi
& Kn with / ; ^ 0. When b[ / 0 then ei = e^. Since C and L intersect
we know that e t > 0. If e* = 0 then x,y € K„. If e^ 7^ 0 then either ^fel € Kn
thus x,j/ e Kn or ^/ej ^ if„ thus x,y
x,y e KJ^'a'l-C^-a'^)
e A'„( v / e7) = Kn+\.
When 6'x = 0 then
= Kn+1.
Third, we consider Pm as the point of intersection of the curve y = x3 and a
q-constructible line
where a^b^c^
€ Kn.
The x-coordinate of the point(s) of intersection Pm —
(am,bm) will satisfy the equation
Let x = o m ; this will be a real root of a cubic polynomial with coefficients in
Let Kn+\
— Kn(am).
Then Kn+i
C M such that | K n + i : Kn\
Kn.
< 3. Therefore
am 6 -ft"n+i by definition and since bm = a3n then bm £ -ft'n+iFinally we consider Pm as an intersection of the curve y = x 3 and the circle
defined by
where aj e Kn. Then a m will be a solution of the equation
or equivalently
8
If am —• 0 then bm = 0 and we are done. If not then am must be a real root of
the equation
Since this is an increasing polynomial we know it must have one unique real
root. This root is \/2ai = am.
[Kn+i
Let Kn+\
= Kn(am).
Then Kn^\
C M such that
: Kn) < 5. We now need to show that [A' n+1 : Kn] ^ 4.
Let us assume [A"n-ri : Kn\ = 4. Then we can say
where p(x) and q(x) are irreducible polynomials over Kn and p(x) and q(x) have
degrees of 4 and 1, respectively. By assumption </2a,i must be a real root of p(x).
However, since q(x) has a real root and the root of x° + x — 2at is unique, we have
the desired contradiction.
•
From this we have the following corollary.
C o r o l l a r y 1. If a £¥,
then a 6 Kn for some real q-tower, KQ C ... C Kn of Q.
To finish proving Theorem 1 we need to prove the converse of this corollary. To
do this we will show that F is a field and the next lemma establishes this. The
proof of Lemma 1 is inspired by and owes much to the ideas of Chapter 4 of [3].
L e m m a 1. The set F is a sub-field o/R.
Proof. We prove this after a series of claims.
Claim !• Every q-constructible line has at least three q-constructible points such
that one is the midpoint of the other two.
9
To prove this claim we will first show that
q-constructible.
(0,1), (1,0), (2,0), (0,2)
By definition (0,0), (0,1), and (1,0) are q-constructible.
are
If we
construct the line defined by (0,0) and (1,0) (the x-axis) and intersect it with the
circle centered at (1,0) passing through (0,0) we quickly obtain (2,0). If we consider
the points of intersection of y = x3 with the circle centered at (1,0) passing through
(0,0) we see that one is (1,1). If we consider the intersection of the y-axis (defined
by (0,0) and (0,1)) with the line defined by (1,1) and (2,0) we obtain the point
(0,2).
We have shown the points (0,1), (1,0), (0,2), (2,0) to be q-constructible points.
Thus the six lines determined by these points are q-constructible lines. The lines
have equations x — 0, y = 0, x + y = 1, x + y = 2,2x + y = 2, and x + 2y = 2, hence
(0,0) and (§, | ) are q-constructible points. Thus y = x is a q-constructible line.
So (1,1) and ( | , | ) are q-constructible points. Now we see x = 1 and y = 1 are
q-constructible lines. Now we can see (5,1) and (1,5) are q-constructible points
and consequently the lines x = | and y = i are also q-constructible lines. If we
consider any non-vertical q-constructible line, it will intersect x = 0, x = 5, and
x = 1 thus we see that it will have at least three q-constructible points such that
one is the midpoint of the other two. For any vertical q-constructible line it must
intersect y = 0, y = ^, and y = 1, so it will have at least three q-constructible
points such that one is the midpoint of the other two. Hence we have shown that
every q-constructible line has at least three q-constructible points such that one is
the midpoint of the other two. We now need a second claim.
Claim 2: Suppose we are given a q-constructible line L and q-constructible
points
A, B, M on the line L such that M is the midpoint of the line segment AB and qconstructible point P not on line L.
We know by our previous claim that there
exists at least a third q-constructible point R on the line AP and we claim that if
R is the midpoint of AP then there exists a fourth point on AP that we will call R'
such that li! is not the midpoint of AP.
10
FIGURE 1. Showing Claim 2
Construct PM and RB, we will call the point of intersection C. Construct AC
and RM and call the point of intersection D. Finally, construct the line BD and
consider where it intersects with AP and call this R! and we are done. So we can
say without loss of generality that given A, B, M, P as previously stated that there
exists a distinct point R on AP such that it is not the midpoint of AP.
Claim 3: Consider a q-constructible line L and q-constructible point P not on
line L. Then the line parallel to L passing through P is
q-constructible.
FIGURE 2. Showing Claim 3
There are three points on L that we will call A,B,M
of AB.
where M is the midpoint
Then by what have previously shown, there is a distinct q-constructible
point R on the q-constructible line AP such that R is the not the midpoint of AP
and thus MR ft BP.
We can now determine point S and point Q, in that order,
and by geometry we see that PQ is the desired parallel line.
Given Claim 3 we can now prove Lemma 1 and show that F is a field. Let
a, b 6 F, and without loss of generality o / O w e will show a + b, —a, ab, ^ 6 F .
F I G U R E 3. Showing a + b e¥
Since a,b € F we can construct A = (a, 0) and B = (b, 0). By the previous
claim, we can construct the line y = 1 and x = a by constructing the line parallel
to the i-axis passing through C = (0,1) and the line parallel to the y-axis passing
through A, respectively.
Thus D = (a, 1) is constructible.
Next construct the
line containing O = (0,0) and D. Construct the line passing through B that is
parallel to OD. The intersection of this parallel with y = 1 can be shown to be
the point E = (a + b, 1). Finally, we can construct the line parallel to the y-axis
passing through E. From the intersection of this line with the x-axis we see that
F = (a + b, 0) is q-constructible. Therefore a + b is a q-constructible number and
a + b 6 F and F is closed under addition.
F I G U R E 4. Showing - a 6 F
12
Consider the q-constructible point A = (a, 0). Construct the line y = 1. Construct the line containing A and B = (0,1). Construct the line parallel to this
line passing through C = (0,0). The intersection point of this line with y = 1 is
D = (—0,1). Construct the line parallel to the y-axis passing through D.
The
point where this line intersects the x-axis is E = (—a,0). Thus - a g F and F
closed under taking additive inverses.
F I G U R E 5. Showing ab G F
Consider the q-constructible points C = (1,0), D = (0,1), A = (a,0), and
B = (6,0). Construct the line containing C and D. Construct the line parallel to
this line passing through B. The intersection of this line with the y-axis gives us the
point E = (0,6). Now, construct the line containing D and A. Finally, construct
the line parallel to this line passing through E. The intersection of this new line
with the x-axis will produce the q-constructible point F = (a6,0). Thus ab € F is
closed under multiplication.
Consider the q-constructible points A = (a, 0), B = (0,1) and the line containing
these two points. This line will have slope —-.
Now construct the line parallel
passing through point C = (1,0). This new line will intersect the y-axis at D =
(0, - ) . Construct the line containing B and C. Finally, construct the line parallel
to this passing through D and consider the point of intersection with the x-axis.
This will be E = (-, 0), and i G F. Thus F is closed under multiplicative inverses.
Therefore F is a field and consequently is a sub-field of R.
13
FIGURE 6. Showing \ e F
D
Before proving the next proposition we will single out two definitions and a
theorem.
Definition 2. A sub-field F of R is a Euclidean field if x G F and x > 0 implies
y/x G F. In other words F is closed under taking square roots.
Definition 3. A Euclidean field F is a Vietian field if x G F implies \fx £ F and
if cos x G F then cos | G F. In other words F is Vietian if it is closed under taking
square roots, cube roots, trisection of given angles.
T h e o r e m 2. Let V be a Vietian field. / / i 6 l such that
ax'1 + bx3 + ex2 + dx + e = 0,
where a, b, c, d, eG V with e ^ 0, then x G V.
Proof: See Theorem 9.8 and its proof from [3].
14
Proposition 2. Let KQ, ...,Kn be a real q-tower of Q. Then Kn C F.
Proof. We proceed by induction on n. If n = 0 then K „ = Q C F since F is a field
by Lemma 1.
Let n > 0, we now check that given a real q-tower of Q, KQ, ..., Kn, that /C n C F,
where Kn._i C F by the induction hypothesis.
We now check the three cases
determined by the degree [Kn : i f n - i ] To start we will consider [KH : i<Tn_i] = 2 (and for this we can follow the
proof from [3] for the Poncelet-Steiner Theorem). Then Kn = Kn-.i(-Jx)
where
x 6 A ' n - i , x > 0. We wish to show that %fx G F.
Since x is q-constructible, and x > 0 then — 1 < ^ 4 < 1 so 0 < ^-r-f + 1 < 2.
By the induction hypothesis we know x G F and since F is a field, j-^I + 1 e F.
By Lemma 1 we can q-construct the line parallel to the y-axis passing through
(|^Y + 1,0), see the Figure 0.7 where x' — —^ + *• ^
we
intersect this line
with the circle of radius 1 centered at (1,0) we find the point of intersection to
15
be (§xy + l,±Jl
— ( f ^ j ) 2 ) - We can use Lemma 1 to find the parallel line to the
x-axis through this point. Then we can project this point to the x-axis using a line
— (fijjj) 2 £ IF- But we see
with a — 1 slope passing through the point. Thus J\
that
which is in F and we are done with the first part. By this argument we see that F
is a Euclidean field.
Now we consider [Kn : A"n_i] = 3. Then Kn = Kn-i(a)
irreducible cubic polynomial with coefficients in Kn-\
where a is the root of an
C F. We now claim that F is
Vietian. R o m this claim it will follow that a £ F by Theorem 2 and so Kn C F. We
first show that given a € F then ^/a € F. Consider some q-constructible number,
a, by our previous arguments we can construct the line y = a and intersect it with
the curve y = x3.
This produced the point {\/a,a).
WTe can then drop the line
to the x-axis so y/a 6 F. Now we consider a = 2cos 9 G F. We then can see that
x = 2cos(|) is a root of the polynomial x 3 — 3x — a, which is q-constructible by
intersecting the q-constructible line y — 3x + a with y = x 3 . Thus F is Vietian,
a e F, and Kn C F.
Finally we consider the case when [Kn : Kn-\]
a = <fa with a £ Kn~\,
= 5. Then Kn = Kn-i(a)
where
(and so recall t/a £ K, and y/a satisfies the polynomial
x° + x — a). The points of intersection of the curve y = x 3 with the circle centered
at ( f , 0 ) with radius | will be the points {a,a3)
and (0,0). Thus we can project
{a, a3) to the x-axis to see that a £ F so Kn C F. With this we have established
the proposition.
D
16
Part 4. E X A M P L E S OF q - C O N S T R U C T I B L E N U M B E R S
Now that we've established F to be a field we can begin to discuss the numbers
that are constructible within these confines. As shown in Proposition 2, F is Vietian
so we can q-construct real roots of any polynomials over F that have degree < 4 by
Theorem 2. But the most interesting property of F (and the point of this thesis)
is its closure under taking fifth roots, which we will formally state in the following
theorem.
T h e o r e m 3. If r € F, then the real fifth root f/r £ F.
This is the theorem that we've alluded to since the introduction, that the fifth
root of any number in F is q-constructible.
We only have to prove ^/r is q-constructible for r > 0 because F is a field and
— 1 6 F. Furthermore, we can restrict to r > 2 since we can multiply any positive
r by a sufficiently large integer that is a perfect fifth power.
This proof will be carried out in several steps, and when we discuss sfr we will
mean the real fifth root of r unless otherwise noted.
Proposition 3. Suppose K is a sub-field o / R and a finite extension o / Q .
Suppose,
too, that r € K with r > 2 such that the polynomial x° — r is irreducible in K[x\.
Then there exists an extension L in R of K of degree at most 3 and there are numbers
a,b 6 L with a > 0, such that if f) is the real root of the polynomial x° + ax + b,
then L{^/r) = L{(5).
Proof. Let a = f/r.
Assume that L has been determined for now. Then I,a,
a 3 , a 4 is a basis of L(a)/L,
2
a2.
since [L(a) : L] = 5 which we will show later. Let
3
fi = ao + a\a + 0,2a + 03a + a^a4 for some a7- € L and we will determine f) as
17
described in the proposition. Since x° - r = Ilj=i( x ~~ aCJ)> where C = (5 is a
primitive fifth root of unity in C, we have that the minimal polynomial p(x) of /?
over L is given as
where fy = J2i=0akCj'kak.
Also,
where ai are the elementary symmetric functions of 0i,...,
/J5, for i = 1,..., 5
Since the polynomial we wish to find the real root of must be of the form q(x) =
5
x + ax + (1 = 0, we can see that ax = <r2 = o"3 = 0 and a4 = a. Instead of working
directly with the <r's it is easier to use the following power sums of the roots. Let
s
i
= s
j ( A ' • • • i Po) = Sfc=i Pi • Relationships between the elementary symmetric
functions and the power sums are given by Newton's identities [6]. (There is a nice
expose on Newton's identities on Wikipedia.) We will only need the following four
of Newton's Identities:
18
Equating the polynomial we wish to find the roots of, q{x), to the polynomial
defined by the elementary symmetric functions, p(x) we obtain the relationship
Using this relationship with Newton's identities for o\ = 0 we obtain
Substituting si = 0 into 2<72 = a\S\ ~ s% then for a2 = 0,
then substituting the previous results into 3<T3 = <T2'*I — ffis2 + S3 then for <r3 = 0,
and finally for 4a 4 = cr3si — a2S2 + T1S3 — S4 then for 0-4 = a,
Keeping in mind that 1 + C1 + C2 + C3 + C4 = 0 and C5n = 1 we obtain the
following:
Since 0 = si, we have
or equivalently
Thus
or equivalently
19
(1)
Similarly since 0 = S2, we have
or equivalently
After expanding the inside of the summation we get
Using 1 this summation will reduce to
Since the roots of unity form a cyclic group and 1 + C1 + C2 + C3 + C4 = 0 after we
take the summation of each term we see that
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or equivalently
Therefore
(2)
Next, since 0 = S3 we have (we will skip some of the algebra here)
or equivalently
which after expansion and simplification will reduce to
We will now recall that a" = r, thus
or equivalently
(3)
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Finally since we know S4 = - 4 a we can then say
or equivalently after simplification
(4)
Let's review what we have so far:
is a root of q(x) = x° + ax + b if the aj's satisfy 2 and 3.
Now let Wj, = — 1 and 04 = 1; thus by 2, a\ — 02 and by 3,
(5)
Now we can see that 4 becomes
As we mentioned before this proposition we need to have a > 0 so we must have
«4 < 0 or equivalently that
(6)
Referring to 5 we can notice that if f(x)
= x3 — x2 + (x + l ) r then f{x)
increasing function of x in K. This is because the derivative f'(x)
is an
= 3x2 — 2x + r
and we notice the discriminant is 4 - 12r which by our restrictions on r is negative
so f'(x)
> 0. We also notice that / ( - l ) = - 2 and /(0) = r which implies that
f(x) has its real root ai such that — 1 < 01 < 0. Furthermore since — 1 < 01 < 0
when we minimize the constraints in 6 we notice
22
and when we recall r > 2 we see
thus c > 0 so a > 0 as desired.
Now let L = K(a{).
Then L c K and by 5 we see that [L : K\ < 3. We can
recall our real root of the polynomial q(x) = x^ + ax + b is 0 = aia + aia2 — a3 + a4.
We claim L(a) — L{(3). Notice since L = K{a{) is a field that (5 € L(o), and since
/ 3 ^ L w e have L c L(/?) C L(o). To show that L(a) C L(/3) we will need to call
upon some field theory. We recall that since a — {/r and x 5 — r is irreducible over
if the degree [K(a) : K] = 5 . We have
From this equality we can see 5 | [L(a) : K] thus 5 | [L(o) : L][L : A"]. But we've
already seen that [L : K] < 3 so it must be 5 | [L(a) : L\. However, since [L(a) :
L] < \K(a)
: K] = 5 we have [L{a) : L] = 5. When considering L(/?) we first
observe [L(/3) : L] ^ 1 because L c L(/?)- We now note that
We've already shown [L(a) : L) = 5 so we know 5 = [Ha) : L{0)\\L{(5) : L] but we
have already observed [L(/3) : L] / 1 so it must be that [L{a) : L{0)\ = 1. Thus
since we know L{0) C L(a) it must be that L{0) =
L(a).
a
With the proposition established we can now complete the proof of Theorem 3.
Proof. Let r e ¥ and assume, without loss of generality, that r > 2. We will
find a real q-tower containing f/r.
By Theorem I r e Kn for some real q-tower
K0 C • • • C Kn of Q. Consider the field Kn{^F).
If [Kn(^F)
: Kn) < 5 so that
\/r is the solution to at most a fourth degree polynomial in Kn and F is Vietian
23
then f/r 6 F. Hence, we may assume this field extension is of degree 5 and thus
x5 — r is irreducible in -KVja;]. But then by the previous proposition, there exists
an extension L C I
such that [L : Kn] < 3 and there is an element f3 which is
the real root of a polynomial k{x) = x° + ax + b € L[x] with a > 0 such that
L(\/r)
= L(/3). We will need to produce a change of variables such that a = 1
if we are to find an ultra-radical extension containing %/r. So we let M = L(y / a)
and N = M(^/^/a)
= L(tfa)
and let 7 = fij\fa where \fa denotes the positive
fourth root of a. Now with a quick substitution we can see that 7 is the real root
of polynomial j(x) = x° +x + b/ sfa € N[x] since j(-y) = -&=f = 0. We can see that
7 = , * / - T = . Now if we let R = N(-y) then R/N is an ultra-radical extension and
we see that K0 C • • • C Kn C L C M C TV C R is a real q-tower of Q such that
v^i" e R, as desired.
D
24
REFERENCES
[1] Baragar, Arthur. Using a Compass and Twice-Notched Straightedge. The MAA
Monthly.
February, 2002 151-164.
[6] Dummit, David and Foote, Richard. Abstract Algebra 2ed. Prentice-Hall. New Jersey, 1999.
[2] Hungerford, Thomas W. Abstract Algebra, An Introduction 2ed. Thomson Learning, Inc.
1997.
[3] Martin, George E. Geometric Constructions. Springer-Veriag New York, Inc. 1998.
[4] Videla, Carlos R. On Points Constructible from Conies. Math Intelligencer 19. No.2 1997
53-57.
25
BIOGRAPHY OF THE AUTHOR
.lames Robertson was born October 26. 1984 to Thyrele and Barbara Robertson.
lie grew up in Ladysmitli, Virginia until moving to Hodgdon, Maine in 2001. He
graduated from Hodgdon High School in Hodgdon Maine in 2003.
In 2007 he
graduated with a Bachelors of Arts in Mathematics from The University of Maine
in Orono, Maine. He is a candidate for the Masters of Arts degree in Mathematics
from The University of Maine in December, 2010.