Galois

June 2024

When I was an undergrad, I was excited by the idea of Galois theory — but disappointed with the first couple books I read about it. You hear about some guy proving the unsolvability of the quintic right before dying in a duel (exciting!) and then you spend a bunch of time pushing around the definitions of “normal”, “separable”, etc. to prove a correspondence theorem that seems important, but is kind of abstract. It has the same feeling as general topology, where playing around “open”, “locally compact”, “hausdorff”, etc. doesn’t really give you a feel for the interesting ideas.

Lots of people have written lots of things about Galois theory, and nothing I’m going to say in this blog post is new. But, to me at least, it’s a more fun way to think about the theory this way, in terms of symmetries of polynomial roots. Most of this comes from Edwards.

Quadratics

To start things off, let’s look at quadratic polynomials. Let $\alpha$ and $\beta$ be solutions to $x^2 + bx + c = 0$ with $a,b \in \mathbb{Q}$. The numbers $\alpha$ and $\beta$ could be irrational, as in the solutions to $x^2-2=0$. They could be complex, as in the solutions to $x^2+1 = 0$.

There are some special relationships between these numbers though. They obey the identities $\alpha + \beta \in \mathbb{Q}$ and $\alpha\beta \in \mathbb{Q}$, something that’s obviously not true for two arbitrary complex numbers. You can see this by factoring:

$$x^2 + bx + c = (x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta,$$

so $\alpha + \beta = -b \in \mathbb{Q}$ and $\alpha\beta = c \in \mathbb{Q}$. This is all generic. If we suppose that $\alpha$ and $\beta$ are the two zeroes of a monic quadratic polynomial with rational coefficients, then $\alpha + \beta$ and $\alpha\beta$ must both be rational.

Let’s generalize this a little. Let’s call a polynomial in two variables $p(\alpha, \beta)$ “symmetric” if $p(\alpha,\beta) = p(\beta,\alpha)$ identically. The coefficients of $x^2 + bx + c = (x-\alpha)(x-\beta)$ are symmetric polynomials in $\alpha$ and $\beta$, if we treat $\alpha$ and $\beta$ as formal symbols. So are the expressions $\alpha^3\beta + \beta^3\alpha$ and $\alpha^2\beta^2$. None of these expression “can tell” when we swap $\alpha$ and $\beta$. That’s the point.

Now let’s go back to thinking of $\alpha$ and $\beta$ as roots. We know from primary school that $\alpha$ and $\beta$ satisfy the quadratic formula:

$$\alpha,\beta = \frac{-b \pm \sqrt{b^2-4c}}{2}.$$

Let’s write $D = b^2-4c$ to save money on typesetting. If we look at the set of all numbers $$\mathbb{Q}(\sqrt{D}) = \{r + s\sqrt{D} : r,s \in \mathbb{Q}\}$$

we see that swapping $\alpha$ and $\beta$ amounts to mapping $\sqrt{D}$ to $-\sqrt{D}$. In other words, the permutation $(\alpha,\beta) \mapsto (\beta,\alpha)$ corresponds exactly to a field automorphism $\sigma(r + s\sqrt{D}) = r-\sqrt{D}$. This field automorphism leaves the base field $\mathbb{Q}$ fixed. Indeed, $\sigma(r+s\sqrt{D}) = r + s\sqrt{D}$ exactly when $s = 0$. The “vibes” based interpretation here is that $\sigma$ leaves symmetric polynomials in $\alpha$ and $\beta$ fixed because it permutes them. Here’s the long version of this for $\alpha + \beta$:

$$\sigma(\alpha + \beta) = \sigma(\alpha) + \sigma(\beta) = \sigma(r + s\sqrt{D}) + \sigma(r - \sqrt{D}) = (r - \sqrt{D}) + (r + \sqrt{D})$$

$$=(r+\sqrt{D}) + (r-\sqrt{D}) = \alpha + \beta.$$

Similarly, $\sigma(\alpha\beta) = \alpha\beta$.

So there’s some kind of correspondence here. Permutations of the roots correspond to automorphisms of the field extension. This is all a little easier to see if we go from quadratic extensions to a more complex situation.

Higher Degrees

Let’s go up a degree and consider the monic polynomial with rational coefficients

$$x^3 + bx^2 + cx + d = (x-\alpha)(x-\beta)(x-\gamma)$$

We have that $-d = \alpha\beta\gamma$, that $c = \alpha\beta + \beta\gamma + \gamma\alpha$, and that $-b = \alpha + \beta + \gamma$. As before, any permutation of the roots $\alpha,\beta,\gamma$ leaves these coefficients unchanged.

How can we turn a permutation of the roots into a field automorphism? Unlike in the quadratic case, it’s not so obvious how to do this; we don’t have the quadratic formula to guide us any more, and I refuse to use the cubic formula on aesthetic grounds. Let’s consider

$$\mathbb{Q}[\alpha,\beta,\gamma] = \left\{\sum_i q_i\alpha^{k_i}\beta^{m_i}\gamma^{n_i} : q_i \in \mathbb{Q}, k_i,m_i,n_i \in\mathbb{N}\right\}.$$

This set has stuff like $\alpha^2 + (5/7)\beta\gamma$ in it. This is a field extension of $\mathbb{Q}$. The only thing that’s not obvious is closure under taking inverses; but $\alpha$ times $\beta\gamma$ is $d$, so $1/\alpha = \beta\gamma/d$, and you can extend from there.

Now it’s clear how to lift a permutation of $\alpha,\beta,\gamma$ to a function on this field. The permutation $(\alpha,\beta,\gamma) \to (\alpha, \gamma, \beta)$ maps $\alpha^k\beta^m\gamma^n$ to $\alpha^k\gamma^m\beta^n$, for example.

And similarly for higher degree polynomials.

Is this function a field automorphism? Things are tricker now. Let’s consider the specific polynomial

$$x^4 + x^3 + x^2 + x + 1 = (x - \omega)(x-\omega^2)(x-\omega^3)(x-\omega^4)$$

where $\omega = e^{2\pi i/5}$ is a fifth root of unity. Consider the permutation $(\omega, \omega^2, \omega^3, \omega^4) \to (\omega^2, \omega, \omega^3, \omega^4)$. This would induce a function $\sigma$ that maps $\sigma(\omega) = \omega^2$. But then $\sigma(\omega^2) = \sigma(\omega)^2 = \omega^4$. But the permutation also tells us that $\omega^2$ is supposed to map to $\omega$. Oops.

So not every permutation of the roots lifts to a field automorphism. The algebraic structure of the field extension constrains us.

Going in the opposite direction is easy. If you have a field extension $K/\mathbb{Q}$ and an automorphism $\sigma:K \to K$ that fixes $\mathbb{Q}$, then for any polynomial $p$ with solutions $\alpha_i$ in $K$ and coefficients $c_i$ in $\mathbb{Q}$:

$$\sigma(p(x)) = \sigma\left(\sum_i c_kx^k\right) = \sum_i c_k\sigma(x)^k = p(\sigma(x))$$

so that if $\alpha_i$ is a root, so is $\sigma(\alpha_i)$. So every automorphism gives us a permutation, but not every permutation gives us an automorphism. So which permutations give us automorphisms? This is exactly the problem of determining the Galois group of a field extension.

In the quadratic case, we already know that the Galois group is always either the trivial group (in which case $K/\mathbb{Q}$ is the trivial extension) or $S_2$. In the cubic case, we know that the Galois group is a subgroup of $S_3$. In fact, one can prove that the Galois group of an irreducible polynomial (the interesting case) has to act transitively on the roots. So for an irreducible degree 3 polynomial, the Galois group is either $S_3$ or the cyclic group $C_3$

The overall theme here is as follows. You start by treating the roots of $p(x)$ as abstract symbols $\alpha_i$. For an abstract $p(x)$, you’d expect these roots to be indistinguishable from each other, and the right group of symmetries is just $S_n$. But for a specific $p(x)$, there are additional restrictions that the $\alpha_i$ are going to have to obey, as we just saw for $x^4 + x^3 + x^2 + x + 1$. These additional restricitons constrain the Galois group, and might stop it from being the full $S_n$. You can also see these restrictions in the structure of the field extension $K/\mathbb{Q}$. That’s the right context for understanding the Galois correspondence.

I like the root permutation perspective more than starting with field extensions in the abstract. Check out Edwards for more detail.