MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the written version of a talk Barry Mazur gave to Friends of the Harvard Mathematics Department on May 5, 2009, there is an interesting question in Footnote 5 (page 8).

He recalls how Gauss wrote $\sqrt p$ (where $p$ is an odd prime) as an explicit rational linear combination of roots of unity (using Gauss sums) and says that he doesn't know any such explicit expression for the roots $\alpha$ of an irreducible cubic polynomial $T^3+bT+c\in\mathbf{Q}[T]$ whose discriminant is a square (so that $\mathbf{Q}(\alpha)$ is a cyclic extension of $\mathbf{Q}$, and hence contained in $\mathbf{Q}(\zeta)$ for some root of unity $\zeta$).

Question. Does anyone know such an explicit expression for the roots of irreducible cubic polynomials whose discriminant is a square ?

share|cite|improve this question
Chapter 3 of H. Davenport, Multiplicative Number Theory, Third Edition, has some relevant material on this, I think. – Timothy Foo Jan 4 '12 at 9:54

I guess Mazur's remark just means that in the expression of the cubic polynomial whose roots generate the cubic subfield of the $p$-th roots of unity, there are numbers $L$ and $M$ with $L^2 + 27M^2 = p$. It reflects the fact from Kummer theory that for understanding cyclic cubic extensions you have to adjoin the cube roots of unity, whereas for quadratic extensions the necessary square roots of unity are already there.

The computation of the cubic subfield of a cyclotomic field with prime conductor using "Gaussian periods" is already in the Disquisitiones, but you certainly should be able to find a modern exposition in Cohen's books.

share|cite|improve this answer

For the simplest cubic $x^3-ax^2-(a+3)x-1$ which is cyclic and real with discriminant $p^2$ where $p=a^2+3a+9$, and when $p$ is prime, the roots $\theta_j,j=0,1,2$ are translates of the Gauss's cubic periods $\eta_j$. Explicitly $\theta_j=\eta_j+(L-1)/6$, where $4p=L^2+27$ and $\eta_j$ are the Gauss's periods. Since $\sum_{j=0}^2 \eta_j=-1$, one can certainly replace $(L-1)/6$ by linear combinations of $\eta_j$. This relation appeared (and also the quartic and sextic case) in E. Lehmer, "Connections between Gaussian periods and cyclotomic units", Maths Comp. Vol 50, No. 182 (1988) 535-541.

share|cite|improve this answer
That's a nice special case. Thanks for the reference. – Chandan Singh Dalawat Jan 4 '12 at 10:52
At least for this case, the polynomials are also "nice" in the sense I have just defined in the MO thread…. E.g., for $a=1$ we can write the roots as $2(\cos\frac{k\pi}{13}+\cos\frac{5k\pi}{13}),\ k=1,3,9$. for $a=2$ as $1-2 (\cos\frac{k\pi}{19}+\cos\frac{7k\pi}{19}+\cos\frac{11k\pi}{19}),\ k=1,3,9$, for $a=3$ as $1-\sqrt{12}\sin\frac{k\pi}{9},\ k=1,2,4$, etc. – Wolfgang Jan 22 '12 at 21:57
I wonder if the roots (i.e. real numbers) that can be written as an explicit rational linear combination of roots of unity are exactly those that can be written as an explicit rational linear combination of trig functions of rational angles (which I call "nice roots"). – Wolfgang Jan 22 '12 at 22:00
Yes, since the maxmimal real subfield of the field generated by $e^{2\pi i k/n)$ is the one generated by $\cos ( 2 \pi k/n)$. – Will Sawin Jan 22 '12 at 22:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.