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Here's a question that I bumped into a couple of times over the years when I was still working in maths, but which somehow never yielded to my efforts. Already when I was a student, I heard a professor remark that, although all cubic equations can be solved in terms of radicals, not all of them can be solved in terms of radicals where only roots of real numbers are extracted. Sadly, this piece of information diminished the appeal of the "cubic formula" for me.

  1. Is there a general characterization of algebraic numbers which can be expressed in terms of real-valued radicals? That is to say, all operations expressed by the formula should take place within the field of real algebraic numbers.

  2. As a more modest question, can it be shown that $\sin(2 \pi/7)$ can not be expressed in terms of real-valued radicals?

R.P.
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  • $\sin \frac {2\pi}{7}$ can be expressed in terms of radicals. $\cos \frac {2\pi}7$ is a solution of the polynomial equation $8x^3 + 4x^2 - 4x - 1 = 0$ and since that is a cubic, it can be expressed as radicals. And $\sin \frac {2\pi}7 = \sqrt {1-\cos^2 \frac {2\pi}{7}}$ I am not sure what the representation is, but it definitely is expressible as a radical. – Doug M Jan 03 '18 at 17:55
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    The fact that irreducible cubics with three real roots are not soluble via real radicals is known as the casus irreducibilis. – Angina Seng Jan 03 '18 at 18:00
  • @LordSharktheUnknown: thank you very much. Funnily enough, I remember coming across this term previously. How silly that I failed to realize that it covers exactly the remark of my professor! – R.P. Jan 03 '18 at 18:09
  • My knowledge of Galois theory is a little thin. But, a polynomial is solvable by radicals if the Galois group is solvable. $S_5$ is not solvable. To show that Galois group is at least $S_5$ showing that it has at least 5 distinct (possibly complex) roots is a good start. – Doug M Jan 03 '18 at 18:30
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    @DougM: the problem is the "real-valued" part. Even $\cos(2 \pi/7)$ can't be expressed in terms of real-valued radicals, since its (cubic) minimal polynomial falls in the casus irreducibilis. – R.P. Jan 03 '18 at 19:24
  • @DougM Try, say, Cardano's formula on the cubic minimal polynomial of $\cos(2\pi/7)$. You will see that the formula needs cubic roots of non-real numbers. That is the problem here. You cannot do it with real cubic roots. A usual formula for a complex cube root needs a real cube root of the modulus and the sine/cosine of one third of the argument. But, there is no formula for the latter (or rather it is equivalent to the problem at hand). – Jyrki Lahtonen Jan 04 '18 at 08:16
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    See Martin Isaacs, "Solution of polynomials by real radicals", Amer. Math. Monthly 92 (1985), 571-575. – i. m. soloveichik Jan 08 '18 at 15:49
  • Thank you very much, that looks like a very useful article! – R.P. Jan 08 '18 at 17:17

2 Answers2

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I will expand on i-m-soloveichik's comment, because it really answers the question. In Solution of Polynomials by Real Radicals (Amer. Math. Monthly 92 (1985), 571-575), Isaacs prove the following theorem.

Theorem. If $x \in \overline {\mathbb Q}$ is an algebraic number having the following properties:

  • $x$ is totally real: it is real, and so are all of its conjugates [in other words: its minimal polynomial splits over $\mathbb R$];
  • $x$ is a real radical element: it lies in a field $F_r \subset \mathbb R$ at the top of a tower of extensions $$F_0 = \mathbb Q \subset F_1 \subset F_2 \subset \ldots \subset F_r$$ such that forall $j$, $F_j/F_{j-1}$ is an extension generated by an element $\alpha_j$ which has a power in $F_{j-1}$. [In other words, $x$ is expressible in termes of real-valued radicals];

Then $x$ is constructible.

I find this theorem truly remarkable: if I allow you to take all the real $n$th roots that you want, you couldn't construct a single totally real number more than the kid down the block with just his square root.

Obviously, the situation is better for non totally real numbers, starting with $\sqrt[n]{2}$ and the like, but it makes you reconsider the definition of solvable by radicals, does it not?

For the second part of your question, it suffices to remark that $\kappa_n = \cos(2 \pi/n)$ is a totally real number. (Note that $\mathbb Q(\kappa_n)$ is a Galois extension of $\mathbb Q$, because it is a subextension of the $n$th cyclotomic field, whose Galois group is abelian, so every conjugate of $\kappa_n$ must lie in $\mathbb Q(\kappa_n) \subset \mathbb R$).

As such, one can apply your theorem to $\kappa_n$, which will be "expressible in terms of real-valued radicals" if and only if it is constructible, that is, if and only if the regular $n$-gon is straightedge-and-compass-constructible. It's famously not the case for $n = 7$.

(One easily transfers the result from $\kappa_n = \cos(i 2 \pi/n)$ to $\sigma_n = \sin(i 2 \pi/n)$, since the relation $\kappa_n^2 + \sigma_n^2 = 1$ means that one is constructible (resp. expressible in termes of real radicals) if and only if the other is).

PseudoNeo
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  • I may be missing something obvious, but what about real algebraic numbers that are not totally real? This was the same question that I had when reading the Isaacs paper, and I couldn't figure it out. It's probably just less easy to give a classification in that case (it's certainly not just the constructible numbers, for one). – R.P. Jan 19 '18 at 13:41
  • That's a good question, and I don't know the answer. It's clearly bigger than the constructible numbers, because it contains things like $\sqrt[n]{x}$, $x \in \mathbb Q$, but I don't know if a classification is possible. – PseudoNeo Jan 19 '18 at 17:21
  • Perhaps it would be better to first ask for a classification of quartic irrationals that can be given in terms of real radicals (the cubic case being dealt with). It would already be interesting to see what happens in that case, although I suspect it is doable. – R.P. Jan 20 '18 at 12:46
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Following up on the earlier answer given by @PseudoNeo and my comment on it, here are some more results, quoted from the 1998 paper Real fields and repeated radical extensions (on the arXiv here) by Isaacs and his student Moulton (numbering as in the paper, notation slightly altered):

Theorem B. Suppose that $F$ is a real field and that $F \subset L$ is a repeated radical extension with $[L:F]$ odd. If $F \subset K \subset L$ is an intermediate field extension, then $K$ is a repeated radical extension of $F$.

Here, a radical extension means an extension of the form $F[\alpha]$ with some power of $\alpha$ in $F$, and a repeated radical extension means what it appears to mean.

Theorem C. Suppose that $F$ is a real field and that $f \in F[X]$ is irreducible of odd degree. If $f$ has some root $\alpha$ in a real repeated radical extension of $F$, then $\alpha$ is the only real root of $f$.

In particular, this yields a large class of examples of non-"real-solvable" polynomials: all odd-degree irreducible polynomials with more than one real root!

Theorem D. Suppose that $F \subset L$, where $L$ is a real repeated radical extension of $F$. If $[L:F]$ is a power of two and $F \subset K \subset L$ is an intermediate field extension, then $K$ is a repeated radical extension of $F$.

I haven't looked at the particulars of the proofs too closely, but the ideas seem to be both accessible and interesting, so I will definitely keep that on my to-do list.

R.P.
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