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Consider the cubic polynomial $x^3-ax+b$ for $a,b\in\mathbb N$.

It has three real roots which, by Cardano's formula, can of course be written in closed form using thirds of angles or cube roots of complex numbers.

Now, in some cases, there is a much nicer way (at least in my opinion, how objective can this be??) of writing the roots: as a rational expression of trig functions of rational angles. Let's call such a polynomial nice in honour of Mathlinks, where this kind of roots are traditionally referred to as "nice roots", presumably starting with this one.

For example, we can write

  • the roots of $x^3-3x+1=0$ as $2\cos\dfrac{k\pi}9,\ k=2,4,8$,

  • the roots of $x^3-21x+7=0$ as $\dfrac3{2\cos\frac{k\pi}7}+2,\ k=2,4,8$,

  • the roots of $x^3-19x+19=0$ as $\dfrac7{2(\cos k\omega+\cos7k\omega+\cos49k\omega)}-2,$ where $\omega=\dfrac{\pi}{19}$ and $k=1,3,9$, or equivalently,

$ \dfrac{\sqrt{19}}{2(\sin k\omega+\sin 7k\omega+\sin 49k\omega)} $ or $ \dfrac{2\sqrt{19}}{\csc k\omega+\csc 7k\omega+\csc 49k\omega}. $

The last example (see also here) shows that it makes sense to also admit square roots. (Probably that doesn't change anything, by virtue of identities of type $ \tan\frac{{3\pi }}{{11}}+4\sin\frac{{2\pi }}{{11}}=\sqrt{11} $, but it can obviously allow to write roots in a more elegant way.)

Question:

Is it possible to characterize the pairs $(a,b)$ that yield nice polynomials?

Some ideas

  • For any angle $\frac{k\pi} n$ involved in a nice root, $\mathbb Z_3$ must be a subgroup of the multiplicative group $\mathbb Z_n^*$.
  • For a nice polynomial $x^3-ax+b$, its discriminant $\Delta=4a^3-27b^2$ is always a square. (See theorem 2 here).
  • It appears that if the involved fractions of $\pi$ have a common prime denominator, this prime divides both $a$ and $b$. (Why?)
  • The article Kurt Girstmair, On root polynomials of cyclic cubic equations, ARCHIV DER MATHEMATIK, Vol. 36, N° 1, 313-326, DOI: 10.1007/BF01223707 may possibly be helpful, but I can't access it.
  • Maybe nice polynomials have something in common with the ones mentioned in this MO thread, e.g. in the sense that in the definition of a nice cubic polynomial, if we replace "trig functions of rational angles" by "roots of unity", nothing essential changes?
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15  
What is going on is that the discriminant of the polynomial (which is $4a^3−27b^2$ in your notation) is a square in the "nice" examples -- then by Galois theory the roots generate an $A_3$, and hence abelian, extension of the rationals, and one can get explicit formulae for the roots just using roots of unity, by the Kronecker-Weber theorem. Finally, the roots of unity can be converted into sines and cosines via the usual methods. –  Kevin Buzzard Jan 22 '12 at 21:19
1  
As Kevin's answer shows, this follows by some fairly standard facts in algebraic number theory. I think the question is more appropriate for math.SE (if Kevin's comment hasn't already cleared it up for you). –  Qiaochu Yuan Jan 22 '12 at 21:36
    
I agree with Qiaochu and voted to close. –  Benjamin Steinberg Jan 23 '12 at 3:16
    
In fact every cyclic extension of odd degree has to be contained in $\mathbf{Q}(e^{2i\pi/N}) \cap \mathbf{R} = \mathbf{Q}(\cos 2\pi/N)$ so that the desired result is indeed true. Explicit expressions can be obtained by Gauss's method. See for example galois-group.net/theory/Galois_Kraus.pdf (p.24, "Annexe") –  François Brunault Jan 23 '12 at 10:53
    
@François: I am not really after that, that was just a little remark and BTW now the equivalence is very obvious to me. The real question is the one which $(a,b)$ are feasible. Is each such polynomial with a square determinant "nice"? I don't think so... –  Wolfgang Jan 23 '12 at 11:21

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