MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I attempted to get some information on this from MSE, but did not even receive a comment, so i'm trying my luck here:

Let p be an odd prime number and $p-1=m d$ a decomposition into positive factors. Then there is a unique cyclic extension $K_d/\mathbb Q$ of degree d ramified only at p. It is contained in the cyclotomic field $\mathbb Q(\zeta)$, where $\zeta$ is a primitive p-th root of unity. $K_d$ is generated over $\mathbb Q$ by the Gaussian period $\omega = \operatorname{Tr}_{~\mathbb Q(\zeta)/K} (\zeta)$ (see Gupta,Zagier).

Now there are results concerning the higher coefficients of the minimal polynomial of $\omega$ over the rationals, my question is: what can be done for the constant term of the minimal polynomial, could someone point out results in this direction.

share|cite|improve this question
up vote 5 down vote accepted

The constant term of the minimal polynomial for $\omega$ is (up to sign) the product of the conjugates of $\omega$ over the rationals. I studied it in my papers, A combinatorial problem in finite fields, I, Pacific J. Math. 82 (1979), no. 1, 179–187, MR0549842 (80i:05010) and A combinatorial problem in finite fields, II, Quart. J. Math. Oxford Ser. (2) 31 (1980), no. 122, 219–231, MR0576339 (81i:05014). I looked at (in your notation) fixing $m$ while $d$ varies, and fixing $d$ while $m$ varies.

In the case $m=2$, $\omega$ is a unit, and the product of the conjugates is the Legendre symbol, $2\overwithdelims() p$. The same formula holds, for the same reason, when $m=4$.

If $d=2$ and $m$ is even, the product is $(1-p)/4$; if $m$ is odd, $(p+1)/4$.

If $d=3$, the product is $((c+3)p-1)/27$, where $c$ is defined by $4p=c^2+27d^2$, $c\equiv1\pmod3$.

If $d=4$, and $m$ is even, the product is $((p-1)^2-4p(s-1)^2)/256$; if $m$ is odd, $((3p+1)^2-4p(s-1)^2)/256$, where $s$ is defined by $p=s^2+4t^2$, $s\equiv1\pmod4$.

A general estimate is that the constant term is bounded, in absolute value, by $((p-m)/d)^{d/2}$.

In the case $m=3$, I found no simple formula for the product, but I proved that it grows exponentially with $p$. The absolute value of the constant term, raised to the power $1/d$, approaches the limit $$\exp\left({2\over\pi}\int_0^{\pi/3}z\tan z\ dz\right)=1.38135644\dots$$ I gave a conjecture along these lines for other values of $m$.

Most of these results were stated for Gaussian periods associated to fields of $q$ elements, $q$ a prime power, but for this post I've stated them all for $p$ a prime.

I had another paper on the topic, Period polynomials and Gauss sums for finite fields, Acta Arith. 39 (1981), no. 3, 251–264, MR0640913 (83e:10058). Stan Gurak wrote several papers on the period polynomials. See also Duke, W., A combinatorial problem related to Mahler's measure, Bull. Lond. Math. Soc. 39 (2007), no. 5, 741–748, MR2365222 (2008m:11146).

I rehashed some of the results in my earlier papers (with, I think, better exposition) in A sampler of recent developments in the distribution of sequences, which appeared in the book, Number theory with an emphasis on the Markoff spectrum (Provo, UT, 1991), 163–190, Lecture Notes in Pure and Appl. Math., 147, Dekker, New York, 1993, MR1219333 (94a:11112).

share|cite|improve this answer
thank you for your comprehensive answer. could you provide a short argument for $m=4$, what about $m=2^k$? – Thomas Jun 22 '13 at 11:56
$\zeta^a+\zeta^{-a}+\zeta^b+\zeta^{-b}=(\zeta^{(a+b)/2)}+\zeta^{-(a+b)/2})(\zeta‌​^{(a-b)/2}+\zeta^{-(a-b)/2})$. No such trick for m=2k, k≥3. See the papers for a conjecture on all values of m. – Gerry Myerson Jun 23 '13 at 0:33

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.