Moving Up the Ladder of Subfields of a Splitting Field

25 Apr

Given a number field K with a chain of subfields K_1, K_2, …, K, each of which is contained in the previous one, one can try to pass from the rational numbers Q to K by successively breaking symmetries of present in a given subfield to move to the next one up. For this, the following theorem is important:

Theorem: Suppose that L is the splitting field of a polynomial with rational coefficients, and suppose that M is a subfield of L. Let G be the Galois group of K, and let H be a subgroup of G such that M is symmetric with respect to H. Let ‘a’ be an element of L. Let

\large \sigma_1, \sigma_2, \ldots, \sigma_n

be the elements of H. Define a polynomial P(x) by

\large P(x) = (x - \sigma_{1}(a))(x - \sigma_{2}(a))\cdots(x - \sigma_{n}(x))

Then P(x) is a polynomial with coefficients in M and with ‘a’ as a root.

Proof: H contains the trivial permutation, so ‘a’ is a root of the polynomial. The coefficients of P(x) are invariant with respect to H, because if

\large \sigma \in H


\sigma: H \to H

defined by


permutes the elements of H, so the coefficients of P(x) are symmetric with respect to H, so by the Fundamental Theorem of Galois Theory the coefficients are in M.

Thus, we can pass from the rational numbers Q to K by (successively, for each i) expressing an element of K_i in terms of roots of polynomials with coefficients in K_(i – 1).


Gauss’ Strategy for Constructing the 17-gon

22 Apr

In the previous post, we described how to compute the Galois group of the splitting field of a cyclotomic polynomial, and use it to characterize the subfields of the splitting field.

Gauss used this theorem to solve the ancient problem of constructing the regular 17-gon with compass and straightedge. The Greeks had been able to construct equilateral triangle, the square, and the regular pentagon with compass and straightedge, but didn’t know how to go further. In fact, it’s not possible to construct the 7-gon, the 11-gon, or the 13-gon with straightedge and compass, but it is possible to construct the 17-gon, as Gauss did.

To this end, consider the regular 17-gon with vertices on the unit circle, and with one vertex at (1, 0). The vertices cut the unit circle into 17 arcs of equal length.

If one can construct a segment of length

\large \cos(\frac{2 \pi}{17})

then one can place it along the x-axis of the coordinate plane starting at the origin. If one draws a vertical perpendicular, the point of intersection of this perpendicular with the circle will be the next vertex of the 17-gon (if one orients the vertices counterclockwise), which one can then use to construct the entire regular 17-gon.  

The tools of straightedge and compass allow one to add, subtract, multiply, divide, and take square roots of lengths of segments. Thus, the problem of constructing the 17-gon is the same as that of expressing

\large \cos(\frac{2 \pi}{17})

by successively performing such operations.

 The connection with the 17-th cyclotomic polynomial is that the roots of this polynomial are positioned at the vertices of the 17-gon. This follows from the fact that multiplying two complex numbers on the unit circle yields the point on the unit circle obtained by rotating (1, 0) through the sum of the angles that the points make with the positive x-axis. The fact implies that the roots of the 17-th cyclotomic polynomial are the points [other than (1,0)] that end up (1, 0) when rotated 17 times, and these are the vertices of the 17-gon.

Let ‘z’ the complex root of the 17-th cyclotomic polynomial that’s immediately counterclockwise of (1, 0) on the unit circle. Then the 16-th power of z is the complex root of the 17-th cyclotomic polynomial immediately clockwise of (1, 0), and we have

\large z + z^{16} = \cos(\frac{2 \pi}{17})

 So it suffices to express the left-hand side by successively performing the operations mentioned two paragraphs above.

Galois theory provides a systematic framework for thinking about using the operations mentioned above to break the symmetry present in the coefficients of the 17-th cyclotomic polynomial, in order to find expressions for its roots. We’ll get into the technical details next time.

Cyclotomic Polynomials and their Galois Groups

22 Apr

In our last post, we defined the Galois group of a polynomial, and remarked that while it usually consists of all permutations of the roots of the polynomial, there are special polynomials, for which the Galois group is a proper subset of the permutations of the roots.

Here we’ll discuss the Galois group of an important special family of polynomials known as the cyclotomic polynomials, and the use of the subgroups of the Galois group to characterize subfields of their splitting fields.

Cyclotomic polynomials

Let p be a prime. The p-th cyclotomic polynomial is defined by

\large \Phi(x) = x^{p -1} + x^{p - 2} \cdots + x^2 + x + 1

By the formula for a geometric series, we have

\large \Phi(x) = \frac{x^p - 1}{x - 1}

so that the roots of of the p-th cyclotomic polynomial are precisely the solutions to

\large x^p = 1

aside from 1. Let z be such a number. Then each power of z satisfies the equation, and the numbers

\large z, z^2, z^3, z^4, \ldots z^{p - 1}

are distinct, and consist of all of the roots of the p-th cyclotomic polynomial. [This last fact needs to be checked, but is straightforward.]

Galois groups of cyclotomic polynomials

Let G be the Galois group of the p’th cyclotomic polynomial. By the above description of the polynomial’s roots, an element of G has the following effect on z:

\large \sigma(z) = z^a

for some ‘a’ between 1 and p – 1 . Knowing where this single root goes determines where all roots of the polynomial go, since the definition of automorphism implies that

\large \sigma(z^k) = z^{ak}

This contrasts sharply with a generic polynomial, where the Galois group can permute the roots arbitrarily. The point here is that the condition that the roots of a polynomial are perfect powers of one another is a very special condition.

Because the p-th power of ‘z’ is 1, the automorphism associated with ‘a’ actually depends only on the remainder of ‘a’ upon division by ‘p’. The phrase “upon division upon p” is abbreviated “(mod p).”  Thus, elements of G correspond to remainders (mod p). Every remainder (mod p) gives rise to an element of G, except for the remainder 0, which sends all roots of the polynomial to 1. [One has to check that every remainder (mod p)  does in fact give rise to an element of G, but this is straightforward.]


\large \sigma_{a}(z) = z^a         \large \sigma_{b}(z) = z^b


\large \sigma_{b}(\sigma_{a}(z)) = z^{ab}

So successive application of elements of G correspond to multiplication of the corresponding exponents, again, with the understanding that two exponents are equivalent if their remainders (mod p) are equal. In what follows, we refer to an element of G by the corresponding remainder (mod p).

The Fundamental Theorem of Galois Theory says that there’s a perfect correspondence between subgroups of G and subfields of K, defined by associating a subfield K of G with the subgroup of G consisting of permutations that leave every element of K invariant. We have the following theorem:

Theorem: For each ‘k’ that’s a factor of p – 1, the remainders of k-th powers (mod p) form a subgroup of G consisting of (p – 1)/k elements. There are no other subgroups of G.

The theorem follows immediately from existence of an integer ‘a’ such that every remainder (mod p) is the remainder of some power of ‘a’ (mod p). This result goes under the name of the existence of primitive roots mod p, and is proved in every elementary number theory textbook, for example, in section 2.5 of William Stein’s Elementary Number Theory.

Next, we use the classification of subgroups of G to describe the subfields of the splitting field K of the 17-th cyclotomic polynomial.

Subfields of K

Because the divisors of 17 – 1 are precisely the powers of 2 that are less than or equal to 16, the above theorem says that the subfields of K are those that symmetric with respect to each of the following subgroups of G:

  • G: The nonzero remainders (mod 17).
  • H1: The nonzero squares (mod 17). These are 1, 2, 4, 8, 9, 13, 15 and 16.
  • H2: The nonzero 4-th powers (mod 17). These are 1, 4, 13 and 16.
  • H3: The nonzero 8-th powers (mod 17). These are 1 and 16.
  • H4: The nonzero 16-th powers (mod 17), consisting of 1 alone.

Call the corresponding subfields K0, K1, K2, K3 and K. The simplest elements of K that are symmetric with respect to the above subgroups are (respectively):

\large 1

\large \mu_{1} = z + z^2 + z^4 + z^8 + z^9 + z^{13} + z^{15} + z^{16}

\large \mu_{2} = z^1 + z^4 + z^{13} + z^{16}

\large \mu_{3} = z + z^{16}

\large \mu_{4} = z

Coming up next:

In the subsequent post, we will begin to describe how Gauss used the characterization of the subfields of the 17-th cyclotomic polynomial to solve the ancient problem of constructing the 17-gon by compass and straightedge.

The Fundamental Theorem of Galois Theory

21 Apr

The problem of expressing the roots of a polynomial (in one variable) in terms of its coefficients is an old and fundamental mathematical problem. Work on this problem includes the quadratic formula, the cubic formula, the quartic formula. In ~1830, Evariste Galois created a conceptual framework for understanding the problem in full generality, and as a consequence deduced that the solutions to the general quintic polynomial equation can’t be expressed via nested radicals. In this post we describe Galois’ conceptual framework.

In a previous post, we explained how solving for the roots of a polynomial in terms of its coefficients entails breaking the symmetry present in the coefficients of the polynomial, and we discussed how to break this symmetry in the special case of a quadratic polynomial. In a subsequent post, we characterized the polynomial expressions of the coefficients as precisely the polynomial expressions in the roots that are invariant under all permutations of the roots.

In the case of a quadratic polynomial, the only symmetry that needs to be broken is invariance under swapping of the two roots. In the case of a generic polynomial of degree ‘n,’ it’s necessary to pass from expressions invariant under all permutations of the n roots to an expression that’s not invariant under any nontrivial permutation of the n roots. 

The strategy for doing this is to break the symmetry a little bit at a time: starting from expressions that are invariant under all permutations and successively passing to expressions that are invariant under fewer and fewer permutations, finally ending up with an expression that isn’t invariant under any nontrivial permutations of the roots.

To codify this perspective, we introduce the Galois group of a polynomial.

Let P(x) be a polynomial with rational coefficients. Let K be the smallest number system containing the roots of P(x) that is closed under addition, subtraction, multiplication and division. Then K is called the splitting field of P(x). The splitting field of P(x) can also be defined as the collection of multivariable polynomial expressions in the roots of P(x), with rational coefficients.

The idea then is to classify elements of K according to how symmetric they are with respect to permutations of the roots, with a view toward developing a hierarchy of subsets of K from “least symmetric” to “most symmetric” and then think about the different levels on the hierarchy, and their relations. To this end, we consider a collection of permutations of the roots of P(x) defined by Evariste Galois. Giving the definition will take a few paragraphs, and its utility is not immediately apparent, but it will be vindicated by the end of this post.

An automorphism of K is a bijection from K to K such that

\large \phi(a + b) = \phi(a) + \phi(b), \hspace{0.3 cm} \phi(ab) = \phi(a)\phi(b)

The reader can check that these properties imply that an automorphism of K leaves the rational numbers invariant. The action of an automorphism on the other elements of K can be described as follows. An element of K can be written as

\large R(r_1, r_2, ... r_n)

where R is a multivariable polynomial with rational coefficients. The reader can check that the definition of automorphism implies that

\large \phi(Q(r_1, r_2, \ldots r_n)) = Q(\phi(r_1), \phi(r_2), \ldots, \phi(r_n))

Thus, an automorphism of K is determined by where it sends the roots of P(x). Where a root of P(x) goes is heavily restricted: if

\large P(r) = 0

 then the properties of an automorphism imply that

\large \phi(P(r)) = P(\phi(r)) = \phi(0) = 0

so that an automorphism of K sends roots of P(x) to roots of P(x). Since automorphisms are bijective (by definition), they must induce permutations of the set of roots of P(x). Thus, an automorphism of K can represented by a permutation of the roots of P(x).

None of this says anything about which permutations of the roots of P(x) induce automorphisms of K, something which is dependent on the particular polynomial. In a later post, we’ll discuss this subject in detail. For most choices of P(x), every permutation induces an automorphism. But there are exceptional polynomials P(x) for which there are permutations that don’t induce automorphisms.

A collection of permutations is called a group if it’s closed under composition of permutations. Successive application of automorphisms yields another automorphism, so the set of [permutations corresponding to automorphisms of K] form a group, called the Galois Group of P(x) (or of K). This is denoted Gal(K/Q) (where Q is the rational numbers), but for simplicity, we’ll write it as G.

We now return to the idea of classifying elements of K by the permutations that leave them invariant:

Given an element ‘y’ of K, the set of elements of G that leave ‘y’ invariant forms a subgroup H of G. We want to classify elements of K by these subgroups of G. The collection of elements of K that are invariant under H is closed under addition, subtraction, multiplication and division, and so is called a subfield of K, which is referred to as a fixed field of H.

The Fundamental Theorem of Galois Theory: Every subfield of K is a fixed field of a unique subgroup of G.

A special case of the theorem, which is the key to its proof, is the case where the subgroup H consists of all of G:

Lemma: The elements of K that are invariant with respect to G are precisely the rational numbers.

The fact that the rational numbers are invariant with respect to G was mentioned earlier in the post: the substantive content of the lemma is that if an element ‘y’ of K is invariant with respect to G, then it’s a rational number.

Because P(x) has rational coefficients, Newton’s theorem on symmetric polynomials implies that if ‘y’ is invariant with respect to all permutations of the roots, then it’s a rational number. The lemma characterizes the smallest subgroup of H permutations of the roots such that [if ‘y’ is invariant under H, ‘y’ is a rational number] as being the Galois group of K.

Thus, we see that the Galois group holds the key to classifying all elements of K according to the permutations under which they’re invariant. This allows one to study the problem of breaking symmetry to find expressions for the roots of a polynomial in terms of its coefficients in a systematic way.

In future posts, we’ll pursue this line of thought further, as well as proving the lemma, and giving concrete examples of splitting fields, Galois groups, subgroups of Galois groups, the fixed fields of these subgroups, etc.

Symmetric Polynomials in Multiple Variables

20 Apr

In the previous blog post on the quadratic formula, I discussed how the coefficients of a polynomial in one variable are symmetric expressions in the roots, and that solving for the roots in terms of the coefficients requires that one break the symmetry. A question that arises is: what kind of symmetric expressions involving the roots can you form with the coefficients of a polynomial? After all, you need to know which symmetric expressions involving the roots you’re allowed to form and try to render asymmetric.

Consider the case of a cubic polynomial

\large P(x) = x^3 + bx^2 + cx + d

Letting the roots be ‘r’, ‘s’ and ‘t’, we can write the polynomial as

\large (x - r)(x - s)(x - t)

Multiplying out, we get

\large x^3 - (r + s + t)x^2 + (rs + rt + st)x - rst

 So that

\large -b = r + s + t

 \large c = rs + st + rt

 \large -d = rst

 The theorem is that if

\large Q(r, s, t)

is invariant under permutations of the roots, then it can be written as a polynomial in the coefficients ‘b’, ‘c’ and ‘d’ of P(x). For example, consider the case

\large Q(r, s, t) = r^3 + s^3 + t^3

We can express this in terms of ‘b’, ‘c’ and ‘d’ as follows. In order to involve the cubes of ‘r’, ‘s’ and ‘t’, we have to cube -b:

\large (-b)^3 = r^3 + s^3 + t^3 + \cdots

Here the ellipses denote the sum of

\large 3 \left[r^2 s + r^2 t + s^2 r + s^2 t + t^2 r + t^2 s \right]


\large 6rst

The last of these is nothing but -6d. In order to involve the second to last expression, we need to take the product of ‘3’, ‘-b’, and ‘c’. Doing so gives the second to last expression, in addition to

\large 3rst

which is nothing but -3d. Putting this all together, we get

 \large Q(r, s, t) =-b^3 -3bc - 9d

Still more generally, if P(x) is a degree n polynomial with leading coefficient 1, then any symmetric polynomial in the roots of P(x)  can be written as a polynomial in the coefficients of P(x). This theorem was essentially known to Isaac Newton, who seems to have been the first to discover it. The proof is left to the reader (-:

The Quadratic Formula

20 Apr

Suppose that a, b and c are constants with ‘a’ nonzero. Consider the equation

\large ax^2 + bx + c = 0

It is well known that the values of ‘x’ satisfying this equation are given by the formula

\large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

There is a subtlety in the case when the quantity under the square root is negative, because in this case the solutions are not real numbers. However, if the reader wishes to, he or she may assume the quantity under the square root to be real.

There are many ways to see why the quadratic formula given above is true, but the most conceptual one, which generalizes, is as follows.

For simplicity, assume that ‘a’ is equal to 1. (If a is not equal to 1, we can divide the equation through by ‘a’ to get a new equation with ‘a’ equal to 1 without changing the values of x that satisfy the equation, though the values of ‘b’ and ‘c’ change). So consider the equation

\large x^2 + bx + c = 0

Let’s call the numbers that satisfy the equation ‘r’ and ‘s.’ Then the expression on the left can be written as

 \large (x - r)(x - s)

Multiplying this expression out, we get

 \large x^2 -(r + s)x + rs = 0

Comparing this with the form of the expression involving ‘b’ and ‘c’, we can conclude that

 \large -b = r + s \hspace{0.5 cm} c = rs

This gives expressions for ‘b’ and ‘c’ in terms of ‘r’ and ‘s.’ We want to express ‘r’ and ‘s’ in terms of ‘b’ and ‘c.’ The difficulty is that r + s and rs are symmetric, in the sense that they’re invariant when ‘r’ and ‘s’ are flipped, whereas ‘r’ and ‘s’ alone are not symmetric in this sense. So we somehow want to break the symmetry present in ‘b’ and ‘c.’ Well, if we square ‘b,’ we get

 \large b^2 = (r + s)^2 = r^2 + 2rs + s^2

 and subtracting ‘c’ from this four times gives the expression

 \large b^2 - 4c = r^2 - 2rs + s^2 = (r - s)^2

which is symmetric, but which is the square of something that’s not symmetric. So if we take its square root, then we get something that’s not symmetric. Taking the square root, we get

 \large \sqrt{b^2 - 4c} = r - s


 \large - \sqrt{b^2 - 4c} = s - r

each of which is permuted when r and s are permuted. (Note that we’ve just made an arbitrary choice as to which root is called ‘r’ and which root is called ‘s’.)

Having broken the symmetry, we can solve for r and s by adding

 \large -b = r + s

to each of the two preceding equations and dividing through by 2. This gives

 \large r = \frac{- b + \sqrt{b^2 - 4c}}{2}


 \large s = \frac{- b - \sqrt{b^2 - 4c}}{2}

which is the well known quadratic formula.