Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Problem: Let $n$ be an integer and $p$ a prime dividing $5(n^2-n+\frac{3}{2})^2-\frac{1}{4}$. Prove that $p \equiv 1 \pmod{10}$.

The polynomial can be re-written as $(\sqrt{5}(n^2-n+\frac{3}{2})-\frac{1}{2})(\sqrt{5}(n^2-n+\frac{3}{2})+\frac{1}{2})$. If this vanishes mod $p$ then $5$ is a quadratic residue mod $p$, which shows that $p \equiv \pm 1 \pmod{5}$ (the primes 2 and 5 are easily ruled out). It feels like the problem should be solvable by understanding the splitting of primes in the splitting field of this polynomial, but I can't find an appropriate "reciprocity law".

The things I'm not sure about are:

  1. How does one rule out the primes congruent to $-1$ mod $5$?
  2. Under what circumstances is it the case that the set {rational primes that split in the ring of integers of some number field} is the union of arithmetic progressions? This a kind of generalized reciprocity law but I don't know in what generality they are known to hold.
share|cite|improve this question
up vote 6 down vote accepted

The only way that this situation could possibly arise is if the splitting field of $f(x)=5(x^2-x+3/2)^2-1/4$ is the cyclotomic field generated by the fifth roots of unity. To show this, it helps to get your hands dirty and actually solve the equation $f(x)=0$.

In general, number fields are determined up to isomorphism by their splitting laws. The splitting law will be determined by a congruence condition iff the extension is an Abelian extension of $\mathbb{Q}$ (by class field theory).

share|cite|improve this answer
I actually did solve the equation - it's just a matter of solving the two quadratic I indicated. I'm pretty sure it's indeed Abelian, but why are you saying it has to be the one generated by fifth roots of unity? The radical expressions contain things like sqrt(125-10 sqrt(5)), this sure doesn't look like Z[zeta] but I could be wrong... – Alon Amit Aug 11 '10 at 9:24
If you solve $x^4+x^3+x^2+x+1=0$ you'll find some expressions for $\zeta_5$ which will be reminiscent of the roots of $f$; the task is to express one lot in terms of the other lot :-) – Robin Chapman Aug 11 '10 at 9:31
I think i see what you're saying now - the primes that split in Z[zeta] where zeta is a 5th root of unity are the ones congruent to 1 mod 5, so if this question has the same answer it must be the same field. Neat. I'll see if I can prove my intuition wrong and show that this really is the 5 cyclotomic. – Alon Amit Aug 11 '10 at 9:34
If one root of $f$ is $a\zeta+b\zeta^2+c\zeta^3+d\zeta^4$ where $a,\ldots,d$ are rational, the others should be $a\zeta^2+b\zeta^4+c\zeta+d\zeta^3$ $a\zeta^3+b\zeta+c\zeta^4+d\zeta^2$ $a\zeta^4+b\zeta^3+c\zeta^2+d\zeta$ in some order. If you can compute the zeros of $f$ numerically and put them in the right order, then finding $a,\ldots,d$ is just linear algebra :-) – Robin Chapman Aug 11 '10 at 11:19
"In general, number fields are determined up to isomorphism by their splitting laws." I see that this is true for Galois extensions of $\mathbb{Q}$. For a non-Galois extension of $\mathbb{Q}$, the same primes split completely as split in the Galois closure, so by "splitting laws" you must mean more information than this. Could you say what, exactly? – Pete L. Clark Aug 11 '10 at 14:21

HINT Your polynomial $p(n)$ splits over ${\mathbb Q}(w), w = \zeta_5$, namely

$ 125 \; p(x) = 125 \; (5 x^4-10 x^3+20 x^2-15 x+11) $

$\quad\quad\quad\quad\quad\; = \;\; (5 x+3 w^3-4 w^2-w-3) (5 x+4 w^3+3 w^2+7 w+1)$

$\quad\quad\quad\quad\quad\quad\; * \; (5 x-3 w^3+4 w^2+w-2) (5 x-4 w^3-3 w^2-7 w-6) $

Regarding the other questions in the query and the comments: there has been much research on various ways of characterizing number fields by splitting behavior, norm sets, etc - going all the way back back to Kronecker. Searching on the terms "Kronecker equivalent" or "arithmetically equivalent" will find pertinent literature. E.g. below is one enlightening review

MR0485790 (58 #5595) 12A65 (12A75)
Gauthier, François
Ensembles de Kronecker et représentation des nombres premiers par une forme quadratique binaire.
Bull. Sci. Math. (2) 102 (1978), no. 2, 129--143.

L. Kronecker [Berlin Monatsber. 1880, 155--162; Jbuch 12, 65] first tried to characterize algebraic number fields by the decomposition behavior of primes. Recently, the Kronecker classes of algebraic number fields have been studied by W. Jehne [J. Number Theory 9 (1977), no. 2, 279--320; MR0447184 (56 #5499)] and others.

This article deals with the following types of questions:
(a) When does the set of primes having a given splitting type in an algebraic number field contain (up to a finite set) an arithmetic progression?
(b) When is this set a union of arithmetic progressions?

If $K$ is an algebraic number field, let $\text{spl}(K)$ denote the set of rational primes which split completely in $K$ and let $\text{spl}^1(K)$ denote the set of rational primes which have at least one linear factor in $K$. Moreover, if $K/Q$ is a Galois extension with Galosis group $G$, let ${\text Art}_{K/Q}$ denote the Artin map which assigns a conjugacy class of $G$ to almost all rational primes $p$. If $C$ is a conjugacy class of $G$ then $\text{Art}_{K/Q}^{-1}(C)$ is the set of primes having Artin symbol $C$. Finally a set $S$ of rational primes is said to contain an arithmetic progression or to be the union of arithmetic progressions if the set of primes in the arithmetic progression(s) differs from $S$ by at most a finite set.

Let $G'$ denote the commutator subgroup of the Galois group $G$. Two results proved in the article are:

Theorem A. The following statements are equivalent:
(a) $|C|=|G'|$;
(b) $\text{Art}_{K/Q}^{-1}(C)$ is the union of arithmetic progressions;
(c) $\text{Art}_{K/Q}^{-1}(C)$ contains an arithmetic progression.

Theorem B. The following statements are equivalent:
(a) $K/Q$ is abelian;
(b) $\text{spl}(K)$ contains an arithmetic progression;
(c) $\text{spl}(K)$ is the union of arithmetic progressions;
(d) there exist a modulus $m$ and a subgroup ${r_1,\cdots,r_t}$ of the multiplicative group modulo $m$ such that $\text{spl}(K)$ is the union of the arithmetic progressions $mx+r_i\ (i=1,\cdots,t)$.

When $K/Q$ is a non-Galois extension it is well known that $\text{spl}(K)=\text{spl}(\overline K)$ where $\overline K$ denotes the normal closure of $K$. It follows from Theorem B that $\text{spl}(K)$ cannot contain an arithmetic progression. However, the author gives two conditions, one necessary and the other sufficient, for $\text{spl}^1(K)$ to be the union of arithmetic progressions when $K/Q$ is non-Galois. As a final application of his result the author gives a necessary and sufficient condition for the set of primes represented by a quadratic form to be the union of arithmetic progressions.

The proofs use class field theory, properties of the Artin map and the Čebotarev density theorem.

Reviewed by Charles J. Parry

share|cite|improve this answer
Thanks so much - this is just what I was looking for in my second question! – Alon Amit Aug 11 '10 at 18:16

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.