Angle Trisection and Regular Polygons

Hubert Shutrick

The Classical Problems

The aim is to give a relatively elementary proof that angles cannot be trisected and that cube roots cannot be extracted using a straight edge and compass.

Given two points in the Euclidean plane, the classical problem is to find out which other points can be constructed using a straight edge and compass. A coordinate system can be set up by considering the line through the points as the x-axis, letting one of them be (0,0), the other (1,0) and letting the y-axis be the orthogonal line through (0,0). It is obvious then how any point (m,n) with integer coordinates can be constructed.

Real numbers are said to be constructible if they are one of the coordinates of a point that we can construct. If s and t are constructible reals, then (s+t,0) and (s-t,0) are the intersections of a circle of radius t with centre (s,0) with the x-axis, their product st is the x-coordinate of where the line through (0,t) parallel to the line joining (0,1) to (s,0) meets the x-axis and their quotient s/t is the x-coordinate of where the line through (0,1) parallel to the line joining (0,t) to (s,0) meets the x-axis.

The real numbers form what is called a field which means that one can do the arithmetic operations addition, subtraction, multiplication and division with pairs of reals and the usual rules apply. A subset of the set R of real numbers is itself a field, a subfield of R, if 0 and 1 are in it and the sum, difference, product and quotient with non-zero denominator, of its members are again in the subset. From this, it is clear that any intersection of subfields is itself a subfield. The subfield that is the intersection of all the subfields that include a given subset, the least subfield that contains it, is said to be generated by the subset.

The subfield generated by \{0, 1\} is the field Q of rational numbers. It is the smallest subfield of R and all rationals are constructible by the constructions described above.

Apart from getting new numbers by those constructions, there is one other possibility and that is to construct the square root of a number in the subfield. It is, in fact, the only possibility because any constructed points are:

1. the intersection of two lines, which just gives numbers in the subfield;
2. the intersection of a line with a circle, which gives the solution to a quadratic equation where the usual formula introduces a square root;
3. the intersection of two circles but subtracting the equations of the circles gives the equation of the line through their points of intersection returning to the second case.

An easy way to get the square root of s is to draw the circle whose diameter is the segment joining (-1,0) to (s,0) because it meets the y-axis at (0,±\sqrt{s}).

Suppose that r is not in a subfield F. The larger subfield F(r) generated by F \cup\{r\} is called the extension of F by r. In the case when r = \sqrt{f} with f in F, the extension will be the set of all reals of the form s\sqrt{f} + t where s and t are in F. It is obvious that the sum, difference and product of two reals s\sqrt{f} + t and s'\sqrt{f} + t' can be denoted in this way and, if s\sqrt{f} + t is the denominator of a quotient, one uses the usual method of multiplying both the numerator and denominator by -s\sqrt{f} + t.

This form is unique in the sense that, if s'\sqrt{f} + t' is the same number then s'=s and t'= t and, of course, if not (s'-s)\sqrt{f} + (t'-t) = 0 would imply that \sqrt{f} was in F. We can then say that the dimension of G over F is 2 as one does in linear algebra considering \sqrt{f},1 as a base. A usual notation is dim_{F}G = 2 and it can be called a quadratic extension.

If dim_{Q}F = n with a base a_{1},a_{2}, … ,a_{n-1},1, then, in the expression s\sqrt{f} + t ,

s = s_{1}a_{1} +s_{2}a_{2}+ … + s_{n-1}a_{n-1} + s_{n}
t = t_{1}a_{1} +t_{2}a_{2}+ … + t_{n-1}a_{n-1} + t_{n}

uniquely in both cases with the coefficients in Q. So a base for G over Q is

a_{1}\sqrt{f},a_{2}\sqrt{f}, … ,a_{n-1}\sqrt{f},\sqrt{f},a_{1},a_{2}, … ,a_{n-1},1.

It has dimension 2n over Q. The general rule is

dim_{Q}F\cdot dim_{F}G = dim_{Q}G.

To conclude, any subfield of constructible reals starting from the given two points will have dimension 2^{m} over Q for some integer m.

Theorem 1. If a polynomial of degree 3 with rational coefficients has no rational root, then its extension Q(r) by a real root r has dimension 3 over the rationals and therefore r is not constructible.

Corollary 1. Squaring the cube. The cube root of 2 is not constructible since x^{3} - 2 has no rational root.

Corollary 2. Trisecting the angle. The angle 60° cannot be trisected because cos(20°) satisfies 4x^{3} - 3x - 1/2 = 0, which has no rational root, as was verified elsewhere.

A rational polynomial is one whose coefficients are rational and two of their properties are useful in this context.

1.The division algorithm works for rational polynomials u(x) and v(x) just as it does for integers and gives

u(x) = q(x)v(x) + r(x)

where the remainder r(x) has lower degree than v(x) and this can be used to find the greatest common factor of u(x) and v(x) by using Euclid's algorithm which gives it in the form

h(x) = f(x)u(x) + g(x)v(x).

2. A primitive polynomial is one with coprime integer coefficients. It is clear that, if u(x) is a rational polynomial and n is the highest common factor of the numerators of its coefficients and m is the least common multiple of their denominators, then u(x) = m/np(x) where p(x) is primitive. The product of two primitive polynomials is also primitive so u(x) can be factorized if and only if p(x) is the product of two primitive polynomials.

From this second property, it can be shown that the primitive polynomials x^{3} – 2 and 4x^{3} - 3x – 1/2 can't have a rational root m/n since nx + m would be a factor with n dividing the coefficient of x^{3} and m dividing the constant term. There are not many possibilities to check.

Proof of the theorem. Suppose u(x) is the polynomial so u(r) = 0. The claim is that 1, r , r^{2} is a base for the extension by r so any real in the subfield can be expressed uniquely as v(r) = ar^{2} + br + c where a,b and c are rational.

It is obvious how to add and subtract such expressions.

The product of two of them v(r) and v'(r) gives a polynomial of degree (at most) 4 in r which can be expressed

v(x)v'(x) = w(x)u(x) + z(x)

by the division algorithm and the remainder z(x) of degree (at most) 2 evaluated at r is the product in the field since u(r) = 0.

To calculate the inverse of v(r), use Euclid's algorithm to obtain

f(x)u(x) + g(x)v(x) = q,

where q must be a non-zero rational since u(x) has no factors. Evaluating at r gives that g(r)/q is the inverse of v(r).

Uniqueness is also obtained from Euclid's algorithm. If v'(r) was the same number as v(r) then the polynomial v'(x) – v(x) of degree at most 2 would have a common root with u(x) and the algorithm would give a common factor that u(x) can't have.

Regular Polygons

Gauss' Formula. A necessary and sufficient condition that a regular polygon is constructible is that its number of vertices is of the form

p_{1}p_{2}...p_{m}2^{n},

where the p_{j} are distinct Fermat primes.

Fermat primes are of the form 2^{2^{k}}+1 but the only known examples are 3,5,17,257 and 65537.

Complex numbers are useful here. Consider points with coordinates (x,y) as z = x+iy in the usual way so that the vertices of the regular n-gon can be \epsilon_{n}^{k} for k = 0,1, … ,n-1, where \epsilon_{n}=e^{i\theta_{n}} and \theta_{n} = 2\pi /n . They are the roots of the polynomial x^{n} – 1 . In theorem 1, we used subfields of the reals R, but the method works also with subfields of the field of complex numbers C. The points with rational coordinates become complex numbers whose real and imaginary parts are rational and they form a subfield of C, the extension Q(i) of the rationals by i.

In Gauss' formula, we can concentrate on the prime factors because:

1. If the regular polygon with mn vertices is constructible, then the one with n is too since we can take every mth vertex.

2. If regular polygons with m and n vertices are constructible and m and n are coprime, then so is the one with mn vertices because Euclid's algorithm gives integers a and b such that am + bn = 1 which implies that a\theta_{n} + b\theta_{m} = \theta_{mn}.

3. It is obvious that, from \theta_{n} , one can get \theta_{2n} by bisecting.

The important property of the third degree rational polynomials in theorem 1 was that they were irreducible which means that they could not be expressed as a product of two non-constant rational polynomials. It was this property that ensured that division was possible in the extended subfield and gave the uniqueness of the linear combinations of the base elements that described its elements. Therefore the theorem can be generalized as follows.

Theorem 1\sharp. If a is a root of an irreducible rational polynomial of degree n, then the extension field Q(a) has dimension n with base 1,a,a^{2}, … ,a^{n-1}.

For constructibility of Q(a), it is necessary that the dimension is a power of 2 as shown above. The general proof that it is sufficient requires some results in Galois theory and the Cauchy-Sylow theorem in group theory and will therefore be omitted.

Eisenstein's criterion. If all the coefficients of a primitive polynomial except that of the highest power are divisible by the prime p and the constant term is not divisible by p^{2}, then it is irreducible.

Proof by induction. If it could be expressed as the product of two primitive polynomials

(a_{0} + a_{1}x + ...)(b_{0} + b_{1}x + …)

then a_{0}, say, is divisible by p but not b_{0}. Assume that all coefficients a_{k} with k < n are divisible by p. The coefficient of x^{n} is a_{n}b_{0} + a_{n-1}b_{1} + ... and a_{n} is therefore divisible by p. This contradicts the assumption that a_{0} + a_{1}x + ... is primitive.

Lemma 1. The dimension over Q of Q(\epsilon_{p}), where p is prime, is p-1 so it is constructible if and only if p-1 is a power of 2. Since this is satisfied by the Fermat primes, it will establish that regular polygons with that number of vertices are constructible

Proof. Consider the polynomial

((x + 1)^{p} – 1)/x = x^{p-1} + px^{p-2} + C^{p}_{2}x^{p-3} + … + p,

which is satisfied by x = \epsilon_{p} – 1. All its coefficients except that of the highest power are divisible by the prime p and the constant term is not divisible by p^{2}.

Lemma 2. Regular polygons with p^{2} vertices, where p is an odd prime, are not constructible, which is why the Fermat primes in the formula must be distinct.

Proof. The polynomial ((x + 1)^{p^{2}} – 1) is not irreducible since it is divisible by ((x + 1)^{p} – 1) but the quotient satisfies Eisenstein's criterion because calculating modulo p leaves only the leading term x^{p(p-1)} and the constant term is p^{2}/p. It has degree p(p-1) which can't be a power of 2.

Note that, for m odd and greater than 1,

2^{nm} +1 = ( 2^{n} +1)( 2^{n(m-1)} - 2^{n(m-2)} + … - 2^{n} +1)

which is not prime so the primes of constructible polygons must be Fermat.

Although the general proof of sufficiency in the theorem is omitted, it is possible to see, to some extent, how constructions can be done by examining the case of the regular 17-gon. Denote \epsilon_{17} by \epsilon so we know from the polynomial that

\epsilon^{16} + \epsilon^{15} + … + \epsilon + 1 = 0.

Let

x_{1} = \epsilon + \epsilon^{9} + \epsilon^{13} + \epsilon^{15} + \epsilon^{16} + \epsilon^{8} +\epsilon^{4} + \epsilon^{2}

and let x_{2} be the sum of the other roots. Now, x_{1} + x_{2} = -1 and, if you multiply x_{1} and x_{2}, remembering that \epsilon^{17} = 1, the product is 4(x_{1} + x_{2}) = -4 so they are roots of x^{2} + x – 4 and can be constructed. The work of multiplying them can be reduced by using Euler's formulae since

x_{1} = 2(\cos\theta + \cos2\theta +\cos4\theta +\cos8\theta)

with a similar formula for x_{2}.

The next step in the construction would be to split x_{1} above into the sum of

x'_{1} =\epsilon + \epsilon^{13} + \epsilon^{16} + \epsilon^{4}

and x'_{2}, the sum of the other terms.

The reason why this works depends on some field and group theory. The automorphisms of the extension field that leave the rationals pointwise fixed form a group called the Galois group. These automorphisms leave the set of roots of a rational polynomial unchanged so they must permute the roots. An automorphism is then uniquely defined by the image of \epsilon because, if it takes \epsilon to \epsilon^{k}, it must take \epsilon^{n} to \epsilon^{nk} to preserve multiplication. Consider, then, the automorphism that takes \epsilon to \epsilon^{3}. It will take each root to it's third power so repeating it 16 times gives

(\epsilon,\epsilon^{3},\epsilon^{9},\epsilon^{10},\epsilon^{13},\epsilon^{5},\epsilon^{15},\epsilon^{11},\epsilon^{16},\epsilon^{14},\epsilon^{8},\epsilon^{7},\epsilon^{4},\epsilon^{12},\epsilon^{2},\epsilon^{6}).

Therefore the group is cyclic. If the rationals are extended by a suitable square root, the automorphisms that leave this extension pointwise fixed should be the subgroup of order 8, whose elements form the cycle

(\epsilon,\epsilon^{9},\epsilon^{13},\epsilon^{15},\epsilon^{16},\epsilon^{8},\epsilon^{4},\epsilon^{2}).

Since the subgroup should permute the roots in this cycle, it will leave a symmetric function of them fixed which is why x_{1} was chosen.

The method works for any regular polygon with a Fermat prime number p of vertices because the corresponding Galois group is always cyclic. To see why, suppose that m is the least positive integer such that \epsilon_{p}^{nm} = 1 for all n. If m<p-1 then the polynomial x^{m}-1 would have more roots than its degree so m=p-1 and there must be an element of order p-1.

A more complete account can be found in, for instance, N. JACOBSON, Basic Algebra 1 or Conjecture and Proof.

References

1. N. Jacobson, Basic Algebra 1, Dover, 2009 (second edition)
2. M. Laczkovich, Conjecture and Proof, MAA, 2001