Suppose I want to calculate

. I have available my friendly calculator, with square and cube root buttons. I remember the handy triple angle formula:

, so I know that

. “Aha!” I think. The cubic formula will allow me to solve this equation. After some grinding, we get

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

and compute

. Now to compute the cube root of … hmmm …

. My calculator doesn’t do complex numbers but I remember this from precalculus: I write

and then the cube root is

. Now

is just

and

is

so I need to compute

— 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

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 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 with . Let . Let be the composite of and . One of the confusing things about this picture is that the 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 to get the Galois theory to work out nicely.

Let be the Galois group of the normal closure of . Let be the subgroup of fixing and let be the subgroup fixing . Now, is obtained from by adjoining a cube or square root. We’ll deal with the cube root case first. In that case, is a -extension of , and an extension of . In other words, . The image of in is . If we have a square root extension, then , with the image of generating one factor.

We want to use the assumption that lies in . So . The extension is Galois, with Galois group generated by . Probably the easiest way to see this Galois group is to notice that lies in , which is Galois over with Galois group the unit group of . In any case, the assumption that lies in means that there is a surjection , whose kernel contains .

Consider the restriction of to . Since is trivial on , we get a surjection . But or and there are no nontrivial maps from or to . So dies on and, in particular, on . Now, we can repeat the argument, looking at . Continuing inductively, we deduce that is trivial on all of . But 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.

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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.

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 . Then you can break up the tower so that the degree of each is prime and is a root field of some . Then there is a first in which 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 is normal since it is a splitting field for (minimal polynomial for ); it is also a root of for some . But then it also contains , so it’s not a real field.

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

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.)