# 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.]

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

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

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.