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. 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. 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. 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.)

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