# Galois’ Unsolvable Polynomials

Galois’ process for finding out if a polynomial is solvable by radicals (i.e., if you can write its roots using a finite number of rational numbers and the symbols +, -, ×, /, and √) is actually surprisingly simple to describe. Here’s a quick 4-step summary of the process:

1. Take a polynomial p(x).
2. Define E  to be the smallest field that contains all rational numbers, as well as all the roots of p(x). This field is called the splitting field of p(x).
3. Define Gal(E/Q) to be the set of all isomorphisms from E to itself that hold fixed all rational numbers. This set is a group, and is called the Galois group of p(x).
4. p(x) is solvable by radicals if and only if Gal(E/Q) is a solvable group.

The proof of (4) is pretty complicated (much more involved than the proof I gave in the last post). Galois’ method is also a little weaker in that it only allows you to conclude unsolvability using the symbols +, -, ×, /, and √, whereas the last post also concluded unsolvability with √ as well as any univalent function (exp, log, sin, or whatever). However, it does allow you to not just generally state that some order-five polynomials are not solvable by radicals, but to determine for any given polynomial whether it is solvable. The process also gives you a good deal of insight into the nature of the roots of the polynomial you’re studying.

Now, the crucial step in this 4-step process is (4), which equates solvable polynomials with solvable groups. There are a few different but equivalent definitions of solvable groups:

Composition Series Definition

• A composition series of G is a series of subgroups of G, such that each subgroup is a maximal normal subgroup of the previous subgroup (maximal means that no normal subgroup contains it besides G itself).
• {1} = G0 G1 Gn = G.
• G is solvable if and only if the factors of its composition series are all cyclic of prime order (i.e. if for all k, Gk+1/Gk p for prime p).

Derived Series Definition

• The derived series of G is the series of subgroups where each subgroup is the commutator subgroup of the previous subgroup.
• [[G,G], [G,G]] [G, G] G.
• G is solvable if and only if this series eventually reaches the trivial group.

Subnormal Series Definition

• A subnormal series of G is a series of subgroups of G where each subgroup is a normal subgroup of the previous subgroup, and which terminates in the trivial group.
• {1} = G0 G1 Gn = G.
• G is solvable if and only if all the factors of a subnormal series of G are abelian.

Solvability is an interesting group-theoretic property aside from its application in analyzing polynomial roots. Here are some things we know about solvable groups, in order of generally decreasing power:

• Every group of odd order is solvable.
• Every abelian group is solvable.
• Every group of prime power order is solvable. (i.e. if |G| = pn for prime p and any n)
• Every group of product of prime powers order is solvable. (i.e. if |G| = pnqm for primes p, q and any n, m)
• Every group whose Sylow subgroups are cyclic is solvable.
• No finite non-abelian simple groups are solvable.
• If H and G/H are solvable, then G is solvable.
• Sn is not solvable for all n ≥ 5.
• Dihedral groups are all solvable.

A fun exercise is to see how many of these you can prove. (Some are much easier than others, and in fact the first one took 255 pages to prove!)

It turns out that most groups are solvable. In fact, the smallest non-solvable group has 60 elements (A5). Here’s the first few numbers in the sequence of sizes of non-solvable groups:

• 60, 120, 168, 180, 240, 300, 336, 360, 420, 480, 504, 540, 600, 660, 672, 720, 780, 840, 900, 960, …

## Determining the elements of Gal(E/ℚ)

Practically, the hardest part of the above 4-step process is determining what the Galois group is for a given polynomial. Usually one knows very little about the roots of the polynomial in question, and therefore also knows little about its splitting field. So how to determine the Galois group (the set of automorphisms of the splitting field that fix ℚ) without even knowing what the splitting field is? Well, it turns out that there are some really useful general tricks.

For one: Sometimes you can look closely at the polynomial and discover some simple algebraic relations that must hold among the roots. For instance, say p(x) = x4 + 2x2 – 1. Since there are only even powers of x in p(x), for any root r of p(x), -r must also be a root. This means that roots of p(x) can be written {A, -A, B, -B} for some reals A and B. And by the properties of homomorphisms, whichever root X that A is mapped to by a member of the Galois group, -A must also map to -X. Another way to say this is that the Galois group of any even polynomial is not the full group of permutations of the roots Sn (as the evenness imposes the above restriction on the allowed automorphisms).

A superb trick related to this is to look at the modular algebraic relations between roots. In general, you can take a complicated irreducible polynomial, and break it into smaller irreducible factors modulo a prime p. Dedekind’s theorem tells us that if there are no repeated factors, then the Galois group contains permutations of the roots with cycle type corresponding to the degrees of the factors.

Example 1

• Let p(x) = x2 – 1.
• The splitting field E of p(x) is clearly ℚ(√2) = {a + b√2 for all a, b in ℚ}.
• Gal(E/ℚ) = the set of automorphisms on E that hold fixed all rationals.
• If f is in Gal(E/ℚ), then f(√2)2 = f(√2 × √2) = f(2) = 2. So f(√2) = ±√2. So there are two automorphisms in Gal(E/ℚ): f(a + b√2) = a + b√2 and f(a + b√2) = a – b√2.
• So Gal(E/ℚ) 2
• Since 2 is solvable, so p(x) is solvable (as was obvious from the outset).

Example 2

Let p(x) = x5 + x4 + x + 3. Then we can write:

1. p(x) = (x + 1)(x2 + x + 1)(x3 + x + 1) mod 2
2. p(x) = x(x + 2)(x4 + x3 + 2x2 + 2x + 2) mod 3
3. p(x) = (x + 3)2 (x4 + 4x3 + 3x2 + x + 2) mod 5
4. p(x) = (x2 + 5x + 2)(x4 + 2x3 + 3x2 + 2x + 5) mod 7
5. p(x) = (x + 6)(x5 + 5x4 + 4x3 + 9x2 + x + 6) mod 11
6. p(x) = (x2 + 8x + 1)(x2 + 9x + 10)(x2 + 9x + 12) mod 13

From this and Dedekind’s theorem we can conclude:

1. Gal(E/ℚ) contains a permutation of cycle type (1,2,3)
2. Gal(E/ℚ) contains a permutation of cycle type (1,1,4): a 4-cycle
3. Repeated factor of x+3, so we don’t learn anything
4. Gal(E/ℚ) contains a permutation of cycle type (2,4)
5. Gal(E/ℚ) contains a permutation of cycle type (1,5): a 5-cycle
6. Gal(E/ℚ) contains a permutation of cycle type (2,2,2)

This automatically gives us a lot of insight into Gal(E/ℚ)! We have permutations of the form:

1. (a b)(c d e)
2. (a b c d)
3. N/A
4. (a b)(c d e f)
5. (a b c d e)
6. (a b)(c d)(e f)

We can go further and show that the only group of automorphisms that contains all of these types of elements is S5. (Every symmetric group can generated by a transposition and a cycle. We have a cycle by #5, and we can get a cycle by cubing #1.) And since S5 is not solvable (its commutator subgroup is A5, whose commutator subgroup is itself, and thus the derived series never terminates), the polynomial x5 + x4 + x + 3 is not solvable by radicals.

One final note: There is a fantastic computational algebra system designed to solve problems in high-level mathematics known as Magma. There’s an online Magma calculator here which is free to use. To use it to find the Galois group of the above polynomial (for example), you type:

P<x>:=PolynomialRing(Rationals());
GaloisGroup(x^5+x^4+x+3);

## The Inverse Galois Problem

Now we get to the most tantalizing part!

Instead of starting with a polynomial p(x) and finding its Galois group, one could equally well start with a group G and ask whether G is the Galois group of any polynomial with rational coefficients! If so, we say that G is realizable, and is realized by p(x).

It turns out that whether or not every finite group is realizable turns out to be one of the big unsolved questions in mathematics. We’d like to prove that you can find a polynomial with coefficients from ℚ with Galois group G, for every finite group G, but in general it’s not known if this is possible! This paper lists all the non-abelian simple groups of cardinality 100 million that are currently not known to be realizable.

We do know the answer for some general categories of groups. Here are some things that are known, in order of decreasing strength.

• All solvable groups are realizable.
• All finite abelian groups are realizable.
• All the symmetric and alternating groups are realizable.
• All cyclic groups are realizable (special case of finite abelian groups being realizable)
• 25 of the 26 sporadic groups are known to be realizable (the missing sporadic group is the Mathieu group M23, whose realizability remains an open question).
• Amazingly, this implies the existence of a polynomial with rational coefficients whose Galois group is the Monster group!