Download Field Theory and Galois Theory Part I: Ruler and

Field Theory and Galois Theory Part I:
Ruler and Compass Constructions
x
Mukund Thattai
\I must say I always think of the Greek constructions as being drawn in the
sand, and am reminded of Archimedes' death, as recounted by Plutarch in his Life
of Marcellus (the Roman general). Marcellus' soldiers had had a hard time dealing
with various devices that Archimedes had dreamed up for the defence of Syracuse,
but Marcellus himself had ordered that Archimedes be brought to him unharmed
(doubtless so that he might make use of his services).
However, a Roman soldier, who was rather too brusquely ordered by Archimedes
to stand away from his diagram, just slew him on the spot. Plutarch says Marcellus
later bumped him o in turn."
John Conway
The Problem
It's not hard to believe that ruler and compass constructions have had a
long and colorful history. What is hard to believe is, given two very simple
tools, and a set of basic rules, how much can be achieved. The ancient Greeks
placed great emphasis on these constructions precisely because they were so
powerful. Yet, in the course of their investigations, these ancient mathematicians came across certain problems that seemed intractable within the
allowed framework, the three famous `Impossible Classical Constructions':
1. To construct a cube twice the volume of a given cube.
2. To trisect an angle.
3. To construct a square equal in area to a given circle.
Intuitively, it seems that these problems should have solutions; in fact,
the Greeks did not know that they were impossible to carry out. (To be
sure, they did develop other methods of solving these problems. But it's
always nicer to be able to solve a problem without using heavy machinery,
and these were the solutions the Greeks were after.) It was only in the 19th
century, through the application of abstract algebra and eld theory, that
the `Impossibility' part was proved.
This is the quintessential mathematical problem: the question is simple;
the answer is beautiful; and you can do a lot of nice math on the way.
I have chosen to present this topic in exactly the opposite direction from
the treatment I have seen in most textbooks. They begin with basic eld
theory, with rulers and compasses being treated as an exercise in its application. In fact, the rst time I came across the proofs of the construction
impossibility theorems, I was pleasently surprised: I didn't even know the
topic was discussed in the book. In this paper, I present the constructions
rst, and use them as motivation to develop some basic eld theory.
1
The Rules
Before you know what can't be done with ruler and compass, it's a good
idea to state what you can do. So, here are the rules of the game.
What you have are a straightedge and a compass. 1 You're also given a
set of points Pn on the plane, and you can carry out the following operations:
1. Connecting two given points by a straight line
2. Drawing a circle with a given center, whose radius is equal to the
distance between two given points
A point r R2 that lies on the intersection of two distinct circles or lines
constructed in this manner is said to be constructible in one step from Pn .
You can add r to your given set of points to get Pn+1 = Pn [ frg, and do
more constructions. (Of course, you're only allowed to do nitely many of
them.)
The Algebra
Take P0 to be just two points, and dene their separation to be 1, the
unit distance. Using this as the scale, associate distances on the plane with
elements of R, and treat our constructions as happening in R2 .
Denition 1 A real number x is said to be constructible if it is the distance
between two constructible points.
It follows that a point (x; y) R2 is constructible if its coordinates x and
y are constructible.
Given two numbers a and b 6= 0, a b, ab, and a=b are all constructible
numbers (see Fig1). Thus, the set of constructible numbers E is a subeld
of R, the real numbers. Given the number 1, we can construct all the
integers simply by addition; then, by taking
inverses, we can construct the
p
rationals, Q. pAlso: given a number a, a is constructible as shown in Fig2.
In this way, 2, an irrational number, can be constructed, so it must be
that Q E R. I will show later that \taking square roots" is, in a sense,
the most complicated operation we can perform.
1
Yes, I must be picky. A straightedge is an unmarked ruler. The point is: if you are
allowed to slide your ruler around on your drawing board (or through the sand on a beach,
or whatever) then it turns out that there are more possibilities. In particular: There's a
simple construction for the trisection of the angle, demonstrated by Archimedes.
2
c = ab
b = c/a
1
a
Figure 0.1: Constructing a eld
Now that we're working with real numbers, we would like some way to
do our ruler and compass constructions without actually using a ruler and a
compass. Of course, what we should do is use coordinate geometry. In this
framework, circles and lines are dened by equations, with coecients in a
certain eld. For example, given P0 = f(0; 0); (0; 1)g, we construct the eld
of rationals by addition and multiplication. Our new set is then: P1 = Q,
and any new construction we perform on this set will be dened by some
equation with rational coecients.)
The straight line constructed from a given eld F is represented by the
equation
ax + by + c = 0
(0.1)
where a; b; c F . Similarly, a circle constructed from this eld, with center
(h; k) and radius r, is represented by
(x ? h)2 + (y ? k)2 = r2
(0.2)
h; k; r F . Intersection points can be found as solutions to simultaneous
equations. For example, the intersection of a line with a circle can by found
by substituting for y in terms of x, say, using eqn 0.1, into eqn 0.2. This gives
a
a
1
Figure 0.2: Extracting square roots
3
a quadratic for x (and y is linear in x). Solving for the intersection of two
distinct circles: cancelling the x2 and y2 terms between the two equations,
we are left with the equation for a straight line (the line joining the two
points of intersection of the original circle). The problem is then reduced to
nding the intersection of one of the circles with this straight line, so again
we have a quadratic for x. We have proved:
Theorem 1 Given a eld F , we can only construct roots of 2nd degree polynomials over that eld (i.e. solutions of quadratic equations with coecients
in that eld) using ruler and compass.
Field Theory
p
Given the eld of rationals, we can construct 2 simply as the diagonal
of the unit square. Now that our set of points P is larger, the eld we're
working
in is correspondingly larger: we have all numbers of the form p +
p
q 2, where p and q Q.
If F = Q is the eld we start o with,
p the base eld, and K is the larger
eld we have constructed by adjoining 2: we write
p
K = Q( 2):
p
The eld extension itself is written K : F or Q( 2) : Q, and read as `K
is an extension of F'.
This particular extension is a simple extension, because we've adjoined
just
p4 one element to the base eld. If we were, in this new eld K , to construct
2: thispwould not be an element of K , and we could work in a larger
p4 peld
Note
that
L
:
F
is
a
still
a
simple
extension:
L
=
F
(
L = K ( 4 2).
p4 2 2; 2);
p p4 2
but since 2 = ( 2) , L can be simply represented as F ( 2). For some
practice with the notation: notice that R(i) = C = fp + qi j p; q Rg.
Also notice that it is not always true that F () consists of elements of the
form p + q, p; q F . It certainly contains such elements, but they may
not form a eld. (For instance, it may not be possible to nd inverses for
There's a small theorem here, stating that F ()( ) = F (; ). The order in which
you add the new elements doesn't matter: check it. Also, a small matter of notation: if F
and K are elds, with F K , and Y is any subset of K , then F (Y ) is the eld generated
by F [Y . Equivalently: F (Y ) is the smallest subeld of K containing F [Y . Put simply:
if F is the eld you are working with, and Y is an extra set of numbers someone gives you
(they're feeling generous), then F (Y ) is the new set of points you can construct without
extracting square roots.
2
4
p
every element). Consider = 3 2. You can check that Q() is the set of all
elements of the form p + q + r2 , p; q; r Q.
In apshort while, we will
p see why there is a dierencep here between adjoining 2 and adjoining 3 2 to Q. For now, recall how 2 was constructed
in the coordinate geometry interpretation: it was a solution to the equation
x2 ? 2 = 0, or equivalently, a root of the polynomial t2 ? 2. 3 In fact, if
we don't have the real numbers
at our disposal 4 one of the ways to characp
terize the properties of 2 is by saying it is a root of this polynomial: that
statement in itself contains \almost all" the useful information. Let's try
and make this more rigorous.
Constructing Algebraic Extensions
p
We say that 2 is algebraic over the rationals since it is the root of some
polynomial over the rationals. In the same way, i, the square root of ?1,
is algebraic over Q, since it is a root of the polynomial t2 + 1. However,
is not the root of any polynomial over the rationals. 5 We say that is transcendental over Q. 6 Finally: An extension K : F is algebraic if
each element of K is algebraic over F . An extension that is not algebraic is
transcendental.
Given an algebraic extension K : F , pick some element K . Then,
there is some polynomial p K [t] that is a root of (since is algebraic).
There are, in fact, innitely many: any polynomial pq will have p()q() =
0. Sift through them, pick out the polynomial m of least degree, and scale
it so that it is monic. 7 It might be hard work, picking through these
polynomials. But the result is worth it, because we have:
Theorem 2 The polynomial m is unique and irreducible.
3
I like to write my polynomials in the `indeterminate' t rather than x, because the latter
comes with many connotations, eg, of working in cartesian coordinates. But a polynomial
is a polynomial irrespective of how we come across it.
The set of polynomials in the indeterminate t with coecients in the eld F is represented
as: F [t].
4
The Pythagoreans, for instance, believed that all numbers could be derived as the
ratio of two whole
p numbers. I.e. they believed that all numbers were rational. The
realization that 2 was irrational came as quite a shock to them, so much so that they
suppressed the discovery.
5
See Stewart, p68, for a proof of this.
6
Note that is algebraic over R: it is the root of t ? .
7
A polynomial p(t) is monic if the coecient of the highest power of t is 1.
5
What more could you ask for from any decent polynomial? The claim is
proved as a footnote. 8
Denition 2 Such a polynomial is known as `the minimal polynomial of over F '.
p
Examples: The minimal polynomial of 2 over Q is t2 ? 2; of i over R
is t2 + 1; and of i over C is t ? i. Suppose was a complex cube root of
unity. Its minimum polynomial over R would not be t3 ? 1. This polynomial
certainly has as a root, but it's reducible: it factors into (t ? 1)(t2 + t + 1).
is not a root of the rst factor, and the second factor is irreducible over
R, so m = t2 + t + 1.
I had said before thatp the equation x2 ?2 = 0 contained, in some sense, all
the information about 2 that we were interested in. So, here's a question:
given just some irreducible polynomial m F [t], is there a way to extend the
eld F to contain roots of that polynomial? As a particular example, given
the polynomial m(t) = t3 ? 2 over Q, what is the smallest eld in which m
has a root?
Here's the procedure. Someone gives you a polynomial m, irreducible
over a eld F. If you're lucky, you already know of a root of the polynomial,
perhaps in a larger eld. Thisp is the case with m = t2 ? 2 over Q: you
already know of the number 2 R. Otherwise: just pretend that you
know of some root of m. As an example: suppose you did not know about
complex numbers, but someone gave you m = t2 + 1 over the reals. Stay
calm, look at the challenger straight in the eye, and tell him: \I do have a
root; call it alpha." Glance around nonchalantly, and nish him o with:
\So I suppose R() is the eld you're looking for."
Of course, you've been doing some quick thinking in the meantime. How
to make up a eld? OK, you're pretending that 2 +1 = 0. Fine, that means
2 = ?1. Let's just work on addition and subtraction for now: under these
operations, R() will contain elements of the form r0 + r1 + r2 2 + r3 3 + :::.
But wait! All terms of order greater than 2 can be substituted for, if you
Uniqueness: suppose f and g are two lowest degree monic polynomials (of degree n,
say) such that f () = 0, and g() = 0. Then: (f ? g) has degree smaller than n. (Because
they're monic, the highest order terms cancel.) But (f ? g)() = f () ? g() = 0. We
have a lower degree polynomial that is a root of, and this contradicts our assumption
unless (f ? g) = 0, or f = g.
Irreducibility: suppose m was reducible, so m = pq for some polynomials p and q of
lower degree. Then m() = 0 = p()q(), so either p or q has as a root, contradicting
our assumption.
8
6
use your 2 = ?1 rule. Keep working at it, and you will end up with just
terms like r0 + r1 . Now, taking inverses is easy, because (r0 + r1 )?1 is
just rr002?+rr112 . Check! 9
This is, in fact, the procedure you would use in general.
Theorem 3 Given an irreducible polynomial m of degree n over a eld F :
pretend it has a root , and construct the set of all polynomials h(), where
the degree of h is less than the degree of m.
It will turn out that all these elements have inverses of that form, so
you have a bona de eld, containing , the root of the polynomial m. This
extended eld is, in fact, just K (). 10
Now, this is not the only method for nding a large eld that contains
a zero of the irreducible polynomial of interest. As I mentioned, you might
already know of some roots at the start. What if you adjoined one of these
roots to your base eld? Wouldn't it matter which one you adjoined, or if
you used some other method altogether? It turns out not to.
Theorem 4 Suppose F () : F and F ( ) : F are extensions such that and
beta have the same minimal polynomial m over F . Then the two extensions
are isomorphic.
Think of this result in the following way: The eld F you begin with is
coarse, in the sense that it doesn't \contain enough information": m doesn't
factor over it. Working in this coarse eld F, you don't have enough machinery to discriminate between dierent roots of an irreducible polynomial. As
far as you can tell, they all satisfy the same properties; I could switch one
for another, and you would never know. (Of course, if you had a larger eld,
you would be able to carry out more tests on these roots; your machinery
would be ner, and might detect dierences.) 11
If you're wondering why you've been so lucky, why every element has an inverse of the
same form: the answer lies in the fact that m is irreducible. Making a common analogy
between integers and polynomials: m is the equivalent of a prime number. You might
have heard of the result that the set of integers modulo p, written Z , is a eld i p is
prime. In the same way, the set of polynomials modulo m is a eld, so you can always take
inverses. The proofs of these two statements also proceed analagously. See for instance,
Stewart, p40.
10
Note that m is the minimal polynomial of over F, since m is irreducible over F .
11
See Stewart, p40-43, for proofs.
9
p
7
The Degree of an Extension
If you were watching carefully, you would have noticed that the eld R()
constructed above was just the eld of complex numbers, with substituted
for i. You already knew that every element of C could be written as p + qi,
with p; q R, and that the formula for inverses was as stated above.
Notice how i and 1 form a basis for C over R. I just showed that they
span C, with coecients in R. The statement that they're independent over
R is exactly the statement that i does not satisfy any 1-degree polynomial
over R. Of course it does not: its minimal polynomial is of degree two.
We shall say that C is an extension of degree two over R, or just write
[C : R] = 2, since C has a basis of two elements over R. 12
Denition 3 The degree of a eld extension K : F , written as [K : F ], is
the dimension of K considered as a vector space over F .
Given a simple algebraic extension F () : F , and m, the minimal polynomial, degree n, of over F ; we saw before that K = F () consisted of
elements of the form r0 + r1 + r2 2 + ::: + rn?1 n?1 . Therefore, the set of
elements f1; ; 2 ; :::; n?1 g spans L. Could it form a basis for K over F ?
We'll have to check independence.
Well: suppose our candidate basis vectors satised a relation of the form
s0 + s1 + s2 2 + ::: + sn?1 n?1 = 0, with coecients in F. If any of the
coecients were non-zero, we would have here a polynomial of degree less
than n, that was a root of. This contradicts our assumption about the
minimal polynomial; therefore, si = 0, for i = 1; :::; n ? 1. The vectors are
independent.
Note that, for an element transcendental over F , the innite set of
vectors f1; ; 2 ; :::; i ; :::g is independent. (Otherwise, would be the root
of some polynomial over F .)
I may have surprised some by suddenly bringing in the language of vector spaces in
a paper about elds. In fact, a eld can be thought of as a vector space over its subeld.
It already lets us add and multiply every element in the set, which is more than we need:
all we want is to be able to multiply by elements in the small eld. Thus: think of the
elements of the large eld as vectors, and those of the small eld as the allowed coecients.
12
8
This gives us:
Theorem 5 If F () : F is an algebraic extension,
the set f1; ; 2 ; :::; n?1 g forms a basis for F () over F , where n is the
degree of the minimal polynomial of over F . It follows that [F () : F ] = n.
If F () : F is a transcendental extension, then [F () : F ] = 1
p
Looking back at an old example: we saw that Q( 3 2)pconsisted of elements of the form p + q + r2 , p;pq; r Q. Therefore, [Q( 3 2) : Q] = 3. We
now see that this arises because 3 2 has a minimal polynomial of degree 3
over Q
So far, we have only been dealing with simple extensions. However,
these results can be extended. The following theorem allows us to calculate
the degree of a complicated extension based on the degree of some smaller
extension:
Theorem 6 If F ,K ,and L are elds such that F K L, then
[L : F ] = [L : K ][K : F ]
The method of proof is straightforward: we just nd a basis for the \big"
extension [L : F ]. Let m = [K : F ], and n = [L : K ]. Let xi ; i = 1; :::; m be
a basis for K over F , and yj ; j = 1; :::; n a basis for L over K . Then: the
set of mn elements fxi yj g forms a basis for L over F . The complete proof
is given as a footnote. 13
This gives us the simple
Corollary 1 If F ,K ,and L are elds such that F K L, then [K : F ]
divides [L : F ].
Independence: suppose the candidate basis vectors satised a relation
of the
form
P P
k x y = 0 where k F: Or, rearranging the summation,
k x y =0
P
(where k = k x K since the x span K). Now, the y are independent over K , so
k = 0 for each j. A similar argument (in K this time) shows that k = 0 for all i; j .
Therefore the vectors are independent.
P
Spanning: this is more or less obvious. For an arbitrary element z L, z = y ,
with P K since the y span L. But expanding
P as an element of K over F , we have:
x y with F: The vectors span.
= x , with F: Therefore, z =
13
P
i;j
ij
i
j
ij
j
i
ij
i
i
i
j
ij
j
i
j
j
ij
j
j
j
j
i
ij
i
j
ij
i;j
9
ij
i
j
ij
j
j
Bring out those Rulers and Compasses!
Recall, from Theorem 1, that \the best we can do" with a ruler and
compass, given a eld F , is to construct the roots of a 2nd degree polynomial
over F . Suppose you begin with the set of points P0 , and the corresponding
eld K0 . Each time you do a construction, and adjoin an element to your
given eld Ki , you have: [Ki ( ) : Ki ] = 1 or 2 (depending on whether the
2nd degree polynomial is reducible or not). At any rate: if is constructible,
then it must have been constructed in a nite number of steps from K0 . If
L = Kn is the nal eld that contains ,
[K : K0 ] = [Kn : Kn?1 ][Kn?1 : Kn?2 ]:::[K1 : K0 ] = 2m
for some m 0. In particular, K0 () is a subeld of K , so [K0 () : K0 ] is
also a power of 2 (since it must divide 2m , by Corollary 1).
Theorem 7 If is constructible, then [K0() : K0 ] is a power of 2.
The Impossibility Proofs
Theorem 8 The cube cannot be doubled using ruler and compass constructions.
Proof. Given a cube, with side of unit length (therefore, starting with pthe
eld of rationals Q): doubling its volume amounts to constructing = 3 2.
But has minimal polynomial t3 ? 2 over Q, so [Q() : Q] = 3, which is
not a power of 2. The construction is impossible.
2
Theorem 9 An arbitrary angle cannot be trisected using ruler and compass
constructions.
Proof. Of course, some angles can be trisected (for example, the angle
=2, by bisecting the angle of an equilateral triangle). I shall prove what
seems to be the standard impossible example, = =3. If an angle can be
constructed, its cosine and sine can be constructed as well, and vice versa.
14 Given the unit length, we are, then, trying to construct cos(=3).
If a line can be constructed,its x and y coordinates can be constructed; the trigonometric ratios are just the coordinates of the line of length 1, at the required angle to the
x-axis.
14
10
We have: cos() = 1=2. And, from the triple-angle formula,
1=2 = cos() = 4 cos3 (=3) ? 3 cos(=3):
Set = 2 cos(=3) to get: 3 ? 3 ? 1 = 0.
But the polynomial m(t) = t3 ? 3t ? 1 is irreducible over Q. 15 Therefore,
[Q( ) : Q] = 3. The construction is impossible.
2
Theorem 10 The circle cannot be squared using ruler and compass constructions.
Proof. Given apcircle with radius of unit length, we are trying to construct
a square of side . If this were possible, we would also be able to construct
, giving [Q() : Q] = 2k for some nite k. However, this would mean that
was algebraic over Q, a contradiction. The construction is impossible. 2
x
Bibliography
1. Stewart, Ian, Galois Theory, Chapman and Hall, London, 1973.
2. Dummit, D.S. and Foote, R.M., Abstract Algebra, Prentice Hall, N.J.,
1991.
3. Artin, Emil, Galois Theory, University of Notre Dame Press, 1971.
15
See Stewart, p63
11