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Let $p \equiv 1 \bmod 4$ be a prime number. Define the polynomial $$ f(x) = \sum_{a=1}^{p-1} \Big(\frac{a}{p}\Big) x^a. $$ Then $f(x) = x(1-x)^2(1+x)g(x)$ for some polynomial $g \in {\mathbb Z}[x]$ (this follows from elementary properties of quadratic residues).

For $p = 5$, we have $g = 1$; for $p = 13$, we find $$ g(x) = x^8 + 2x^6 + 2x^5 + 3x^4 + 2x^3 + 2x^2 + 1. $$ pari tells me that the Galois group is "2^4 S(4)" and has order $384 = 16 \cdot 24$.

My questions:

  • Have these polynomials been studied anywhere? Since I did not just make them up, I am tempted to believe that they are natural enough to have shown up somewhere else.
  • Is $g$ always irreducible? pari says it is for all p < 400.
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3 Answers

up vote 10 down vote accepted

These are known as Fekete polynomials: http://en.wikipedia.org/wiki/Fekete_polynomial . I don't know of any results on their Galois groups.

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Thanks for the help and welcome to MO! There do not seem to exist an awful lot of results on Fekete polynomials, at least not in the direction I am looking. –  Franz Lemmermeyer Mar 10 '10 at 20:47
    
Hello Franz. At one stage I hoped that Fekete polynomials would help with my "evil determinant" problem (see my website) but I couldn't make them do so. –  Robin Chapman Mar 12 '10 at 16:16
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Hi,

In Remark 2 of "Zeros of Fekete Polynomials", (http://arxiv.org/PS_cache/math/pdf/9906/9906214v1.pdf), Conrey et. al. give $$\sup_{|z|=1}|f(z)| \ll p^{0.5} \log p.$$

But the Mahler measure of $f$, $M(f)$, is bounded from above by $\sup_{|z|=1}|f(z)|$. (I took this from (7.4) of "Experimental Number Theory" by F. R. Villegas.)

Since Mahler measure is multiplicative, then, letting $$ f = \prod_i f_i $$ where the product is over all irreducible factors, one has $$ M(f) = \prod_i M(f_i). $$

Using lower bounds for Mahler measures found in "The Mahler measure of Algebraic Numbers: A Survey" by C. Smyth,(http://www.maths.ed.ac.uk/~chris/Smyth240707.pdf), perhaps one can, comparing upper and lower bounds for Mahler measures, get a nontrivial upper bound on the number of irreducible factors of the Fekete polynomial? For example, (11) of the paper by Smyth gives, for the algebraic number $\alpha$ with minimal polynomial of degree $d \geq 2$, $$ M(\alpha) > 1 + \frac{1}{1200}\left(\frac{\log \log d}{\log d}\right)^3. $$

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Regarding the Galois group of the factor $g(x)=g(q,x)$of these Fekete polynomials that is conjectured to be irreducible, here is PARI code that counts, for each prime $q \equiv 1 \bmod 4$, $17 \leq q \leq 100$, the number of primes $p$ for which $g(q,x)$ modulo $p$, has $n$ linear factors, for primes $p$ up to half a million, stored in the matrix $c[q,n+1]$, done for $0\leq n\leq 14$.

f(m,x)=sum(i=1,m-1,kronecker(i,m)*x^i))

g(m,x)=f(m,x)/(x*(x-1)^2*(x+1))

for(n=0,14,forprime(q=17,100,if(q%4==1,forprime(p=2,500000,if(matsize(polrootsmod(g(q,x),p))==[n,1], c[q,n+1]++)))))

If I did not make any mistakes, then, for these nine primes $q=17,29,37,41,53,61,73,89,97$, here is the $9$ by $15$ matrix showing $n=0$ to $14$.

[25204 3 12634 0 3084 0 538 0 72 0 3 0 0 0 0]

[25360 2 12486 0 3101 0 530 0 50 0 9 0 0 0 0]

[25133 1 12637 0 3187 0 519 0 59 0 2 0 0 0 0]

[25320 2 12519 1 3081 0 543 0 65 0 6 0 1 0 0]

[25293 2 12449 0 3206 0 500 0 81 0 6 0 1 0 0]

[25176 2 12603 0 3157 0 527 0 69 0 4 0 0 0 0]

[25110 3 12564 0 3274 0 516 0 65 0 5 0 1 0 0]

[25286 3 12554 0 3116 0 502 0 71 0 6 0 0 0 0]

[25112 3 12711 0 3132 0 526 0 49 0 5 0 0 0 0]

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