Galois Theory

What was the first mathematical mental exercise you learned? Probably counting: 1, 2, 3, … Or as computer scientists prefer: 0, 1, 2, 3, … Thus we have \(\mathbb{N}\), the set of natural numbers.

What next? Perhaps earlier than you can remember, you learned to add and subtract. Already, we encounter a complication: if you want to subtract any number from any other you’ll need negative numbers. Thus we have \(\mathbb{Z}\), the set of integers.

Then multiplication and division are introduced. Now fractions are needed to handle division properly. Thus we have \(\mathbb{Q}\), the set of rationals.

Some time later, one learns about radicals: square roots, cube roots, and their ilk. We’re taught about real numbers (\(\mathbb{R}\)) and complex numbers (\(\mathbb{C}\)) and we feel safe knowing we can compute an answer for any radical expression.

But in another sense, real numbers are unfulfilling. Often, we cannot write down real numbers: the most we can do is give rational approximations. (For some reason we tend to use a denominator that is a power of 10, e.g. 3.14159.) While indispensable, especially in real life, let’s backpedal away from the reals, and cling to numbers we can write down exactly.

We find ourselves back in the land of rationals, and want to regain the ability to compute roots. We need to conjure up some method to take the square root of 5, for example, without resorting to real numbers.

Let us lazily travel the path of minimum resistance. We invent a new number called \(\sqrt{5}\) and define it to satisfy \((\sqrt{5})(\sqrt{5}) = 5\). This seems too easy. Is there more to define?

No! After some work, we discover we can consistently carry out the four basic operations (\(+,-,\times,/\)) if we stick to numbers of the form \(a + b\sqrt{5}\) where both \(a, b \in \mathbb{Q}\). We’ve literally done the least we can do; we have constructed the smallest extension of \(\mathbb{Q}\) containing a square root of 5, and we call it \(\mathbb{Q}[\sqrt{5}]\). (It also goes by other names, such as \(\mathbb{Q}[x]_{x^2-5}\).)

In general, we construct \(\mathbb{Q}[\sqrt{n}]\) similarly for any square root.

Notice \(-\sqrt{5}\) also lies in \(\mathbb{Q}[\sqrt{5}]\) and also squares to give 5. But unlike the reals, we cannot relate either root to the rationals. Without the reals to light our way, we have no notion of the sign or magnitude of \(\sqrt{5}\) compared to any nonzero rational number. It is analgous to defining \(i\) as the square root of \(-1\). We never try to compare \(i\) against a nonzero real number, and it lives on an axis of its own; the same is true for \(\sqrt{5}\) and the rationals. In fact, consider the following game:

The polynomial \(x^2 - 5\) has two roots in \(\mathbb{Q}[\sqrt{5}]\): \(\sqrt{5}, -\sqrt{5}\). Alice picks one, and challenges Bob to discover it. Bob may only do the following. Let’s call Alice’s root \(a\), and the other one \(b\). Bob writes down any equation consisting of the four basic operations involving \(a, b\) and any rational numbers, and asks Alice if the equation holds.

What should Bob do? Give up! It’s impossible to discover which root Alice picked. Any equation involving rationals and the roots \(a\) and \(b\) still holds after swapping \(a\) and \(b\).

Now consider the quadratic polynomial \(x^2 - x - 1\), which has solutions \(\frac{1 + \sqrt{5}}{2} , \frac {1 - \sqrt{5}}{2}\). If we swap \(\sqrt{5}\) and \(-\sqrt{5}\), we find the solutions switch positions.

In short, switching one square root with its counterpart

We generalize these simple facts to show the quintic is insoluble.

This time we try for a cube root of 2. We define a new number \(\sqrt[3]{2}\), and with some work, we find that if focus on numbers of the form \(a (\sqrt[3]{2})^2 + b \sqrt[3]{2} + c\) where \(a,b,c \in \mathbb{Q}\), then we can compute with all four basic operations and also have a cube root of 2. We call this set \(\mathbb{Q}[\sqrt[3]{2}]\).

But unlike the square root of 5, it turns out we’re still missing the other two cube roots. We need the cube roots of unity to gain access to them. So define \(\omega\) to be a primitive cube root of unity, and after more work, we find the set of numbers of the form

is closed with respect to the four basic operations, and we have all cube roots of 2. We call this \(\mathbb{Q}[\sqrt[3]{2}, \omega]\).

It turns out for any prime \(p\), one can derive radical expressions for all \(p\)th roots of unity. For example, \(\omega = \frac{1 + \sqrt{-3}}{2}\), so \(\mathbb{Q}[\sqrt[3]{2}, \omega] = \mathbb{Q}[\sqrt[3]{2}, \sqrt{-3}]\).

By factorization, we can iteratively derive radical expressions for any \(n\)th root of unity. Hence we can inductively extend \(\mathbb{Q}\) to contain all solutions of any radical expression.

As before, it turns out we can permute the roots of \(x^3 - 2\) and leave any existing equation (involving basic operations on rationals and the roots) unharmed. But not all permutations work: in fact, once we’ve decided where to map one of the roots, the destinations of the other roots are determined, otherwise certain equations would break. On the other hand, there is always a permutation from one root to any chosen root that preserves the validity of all equations.

In general, for a \(p\)th root where \(p\) is prime, the group of permutations that leave all equations involving \(p\)th roots and rationals unharmed is precisely the cyclic group of size \(p\).

Cardano solved the general cubic by first substituting a variable to obtain \(x^3 + 3 p x + 2 q = 0 \) and then showing that

is a solution, where \(u = \sqrt[3]{-q + \sqrt{q^2 + p^3}}\). The other two solutions are obtained via the other cube roots:

and

Recall a primitive cube root involves \(\sqrt{-3}\) (which we must adjoin to \(\mathbb{Q}\) along with \(u\), and possibly \(\sqrt{q^2 + p^3}\)). If we swap \(\sqrt{-3}\) and \(-\sqrt{-3}\), then two of the cube roots exchange places while the other stays in place. This corresponds to one solution staying put while the other two trade positions.

Similarly, applying the permutation \(u \rightarrow u\omega \rightarrow u\omega^2 \rightarrow u\) rotates \(x,y,z\) in a similar fashion. Thus by permuting roots appropriately, we can achieve any permutation of \(x, y, z\).

A subtle but important observation: since the cube roots involve \(\sqrt{-3}\), permuting them happens at a level “higher” than the swapping of the square roots. From group theory this must be so because the only normal subgroup of \(S_3\) is \(\langle(123)\rangle\); if it were to work at all, the cubic extension must happen after a quadratic extension.

Suppose we have \(n\) explicit radical expressions for the roots of a general polynomial of degree \(n\). Suppose somewhere during the computation of the roots we take the \(p\)th root of some subexpression, where \(p\) is some prime.

Then as long as we permute the \(p\) roots cyclically, the solutions stay valid, but may shuffle around, since the cyclic permutations are precisely those that have no effect on equations involving roots and rationals (and in particular the polynomial we started with).

We saw above that we could permute the 3 solutions of a general cubic in any desired manner through suitable permutations of a certain square root and a certain cube root. Intuitively, this should be true for general equations of any degree, that is, for any given permutation of the solutions, there must be some way to achieve it by permuting various roots sprinkled throughout the expressions. The roots of a general polynomial have no distinguishing features so why should one permutation of the solutions be feasible when others are not?

We’ve transplanted the problem to the realm of permutations, and no longer need algebra to answer it. If there are explicit expressions for solving the quintic, then there must be a way to “build” \(S_5\), the group of permutations of five objects, using a series of cyclic groups of prime order, so that a permutation in \(S_5\) corresponds to permutations in each of the cyclic groups.

Through brute calculation, we can show this is impossible, hence the quintic has no general solution using radicals.

The converse turns out to be true too. Given an irreducible polynomial, examine the group of permutations of its roots that have no effect on equations involving its roots and rationals (its Galois group). Then if this group can be “built” from a series of cyclic groups of prime order, then there exist radical expressions for its roots. In fact, there are algorithms for finding them.