# The cubic formula is a scam (and Galois can prove it)

Suppose I want to calculate $\cos 20$. I have available my friendly calculator, with square and cube root buttons. I remember the handy triple angle formula: $\cos 3 \theta=4 \cos^3 \theta -3 \cos \theta$, so I know that $4 \cos^3 \theta-3 \cos \theta=\cos 60 = 1/2$. “Aha!” I think. The cubic formula will allow me to solve this equation. After some grinding, we get
$\cos 20 = \frac{1}{2} \left[ \sqrt[3]{\cos 60 + \sqrt{\cos^2 60 - 1}}+\sqrt[3]{\cos 60 - \sqrt{\cos^2 60 - 1}} \right]$
Looks great! Time to haul out the calculator. The quantity under the square root is negative but that’s fine, I’ll just yank out an $i$ and compute $\sqrt{3/4}$. Now to compute the cube root of … hmmm … $1/2+i\sqrt{3}/2$. My calculator doesn’t do complex numbers but I remember this from precalculus: I write $1/2+i\sqrt{3}/3=r (\cos \phi+i \sin \phi)$ and then the cube root is $r^{1/3} (\cos \phi/3+i \sin \phi/3)$. Now $r$ is just $1$ and $\phi$ is $60$ so I need to compute $\cos 20$ — and I’m right back where I started! The cubic formula was useless!
Alright, the cubic formula isn’t really a scam. It is useful, if you know how to compute cube roots of complex numbers. But it is a bit of a nasty surprise that we have to go through complex numbers to get purely real answers. It’s nasty enough to make one wonder whether $\cos 20$ could be expressed in terms of rational numbers and square and cube roots without stepping into the complex world. The answer is no. And seeing this is a very nice exercise in Galois theory.

Imagine computing $\cos 20$ by taking square and cube roots of real numbers, starting from some rational numbers. At every step of the process, we can consider the field generated by all of the numbers we have written down so far. So we get a sequence of fields $\mathbb{Q}=K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_N \subset \mathbb{R}$ with $\cos 20 \in K_N$. Let $L_0=\mathbb{Q}[e^{2 \pi i/3}]$. Let $L_i$ be the composite of $K_i$ and $L_0$. One of the confusing things about this picture is that the $K_i$ are the fields which seem natural to concern ourselves with, as they involve the real numbers which we are actually computing with, but we need the $L_i$ to get the Galois theory to work out nicely.

Let $G$ be the Galois group of the normal closure of $L_N$. Let $G_i$ be the subgroup of $G$ fixing $K_i$ and let $H_i$ be the subgroup fixing $L_i$. Now, $L_{i+1}$ is obtained from $L_i$ by adjoining a cube or square root. We’ll deal with the cube root case first. In that case, $L_{i+1}$ is a $\mathbb{Z}/3$-extension of $L_i$, and an $S_3$ extension of $K_i$. In other words, $G_i/H_{i+1} \cong S_3$. The image of $H_i$ in $G_i/H_{i+1}$ is $\mathbb{Z}/3$. If we have a square root extension, then $G_{i}/H_{i+1} \cong \mathbb{Z}/2 \times \mathbb{Z}/2$, with the image of $H_i$ generating one factor.

We want to use the assumption that $\cos 20$ lies in $K_N$. So $L_N \supseteq \mathbb{Q}[\cos 20]$. The extension $\mathbb{Q}[\cos 20]/\mathbb{Q}$ is Galois, with Galois group $\mathbb{Z}/3$ generated by $\cos 20 \mapsto 1- 2 \cos^2 20=\cos 140$. Probably the easiest way to see this Galois group is to notice that $\mathbb{Q}[\cos 20]$ lies in $\mathbb{Q}[e^{2 \pi i/18}]$, which is Galois over $\mathbb{Q}$ with Galois group the unit group of $\mathbb{Z}/18$. In any case, the assumption that $\cos 20$ lies in $K_N$ means that there is a surjection $\pi:G \to\mathbb{Z}/3$, whose kernel contains $G_N$.

Consider the restriction of $\pi$ to $G_{N-1}$. Since $\pi$ is trivial on $H_N$, we get a surjection $\pi : G_{N-1}/H_N \to \mathbb{Z}/3$. But $G_{N-1}/H_N \cong S_3$ or $\mathbb{Z}/2 \times \mathbb{Z}/2$ and there are no nontrivial maps from $S_3$ or $\mathbb{Z}/2 \times \mathbb{Z}/2$ to $\mathbb{Z}/3$. So $\pi$ dies on $G_{N-1}$ and, in particular, on $H_{N-1}$. Now, we can repeat the argument, looking at $\pi: G_{N-2}/H_{N-1} \to \mathbb{Z}/3$. Continuing inductively, we deduce that $\pi$ is trivial on all of $G$. But $\pi$ is supposed to be a surjection, a contradiction. QED

One of these days I’ll get a chance to teach Galois theory. There are so many pretty little examples like this, and so few of them find their way into textbooks.

## 4 thoughts on “The cubic formula is a scam (and Galois can prove it)”

1. Dave says:

If I’m not mistaken your argument is a bit more complicated than it has to be (and in simplifying it, one proves something slightly stronger). The claim is that there’s no subfield K of the reals not containing cos(20) such that K(cos(20)) is obtained by adjoining a square or cube root of an element in K. (The strengthening is that it’s irrelevant how K was built up.) Suppose such K exists. The extension K(cos(20),zeta_3)/K is Galois, and the Galois group G maps onto Gal(Q(cos(20)/Q)) by restriction. Hence #G=6 (it’s at most 6 but divisible by both 3 and 2), our “square root or cube root” was a cube root, and K(cos(20),zeta_3)/K has the form K(cbrt{x},zeta_3)/K and is an extension of degree 6. But this means it’s an S_3 extension, and now you conclude as before.

2. Yeah, that’s a nice problem. (Actually this holds for any irreducible cubic with three real roots.) I think it’s a fairly common exercise to assign; at least I was assigned it in my algebra class (at Stanford) last quarter. My solution involved first adjoining the square root of the discriminant of $8x^3-6x-1$. Then you can break up the tower so that the degree of each $K_i/K_{i-1}$ is prime and is a root field of some $x^p-a$. Then there is a first $K_N$ in which $\cos 20$ is in your tower. One can show pretty easily that [K_N : K_{N-1} ] cannot be any prime other than three. However, three doesn’t work either, since $K_N/K_{N-1}$ is normal since it is a splitting field for $8x^3-6x-1$ (minimal polynomial for $\cos 20$); it is also a root of $x^3-a$ for some $a$. But then it also contains $e^{2\pi i/3}$, so it’s not a real field.

3. Casus irreducibilis says:

This is an instance of the famous Casus Irreducibilis that can be found in Wikipedia or Planet Math. See also D.A. Cox, Galois Theory, Wiley-Interscience, 2004 for a good reference

4. Omar Antolín says:

I think you’re wrong about few examples appearing in textbooks, but I’m too lazy to check the texts used at my school. (They were, varying by prof: Rotman’s Galois Theory, Stewart’s Galois Theory and Clark’s Elements of Abstract Algebra.)