From: elkies@ramanujan.harvard.edu (Noam Elkies)
Newsgroups: sci.math
Subject: Re: More strange properties of the number six
Date: 30 May 92 17:02:24 GMT
In article
ara@zurich.ai.mit.edu (Allan Adler) writes:
>In article <4149@daily-planet.concordia.ca>
>gsmith@concour.cs.concordia.ca (Gene Ward Smith) writes:
>[...]
> x^6 + 2x^5 + 3x^4 + 4x^3 + 5x^2 + 6x + 7
>
> has Galois group Pgl2(5). The question then asked was, is this
> "an accident", or is there some theory which explains this?
>
> A few months ago, I made the situation even worse. I noticed that
> if r_1, ..., r_6 are the roots of the above polynomial, then
> r_1^k + ... + r_6^k = -2 for various values of k, including
> k from 1 to 6. It then becomes natural to ask, what about the
> polynomial with r_1^k + ... + r_n^k = n?
>
> This is given by
>
> x^6 - B(n,1)x^5 + B(n,2)x^4 - B(n,3)x^3 + B(n,4)x^2 - B(n,5)x + B(n,6),
>
> where B(n,i) is a binomial coefficient (and allowing any value for n,
> of course.)
>
> We find if n = -1 that we get the cyclotomic polynomial for seventh roots
> of. unity, and that 6 - n gives a transformed version of n. We also find
> that n = -4 gives a *different* Galois extension, but that the group
> once again is Pgl2(5)! That is, the tetrahedronal number polynomial
>
> x^6 + 4x^5 + 10x^4 + 20x^3 + 35x^2 + 56x + 84
>
> is *also* a Pgl2(5) polynomial. Conway's comment on this was "the
> impossible has now happened twice." [...]
>
>I don't know, but the following occurs to me. In your polynomial with
>the binomial coefficients, replace n by the indeterminate t. You
>then get a polynomial whose coefficients are polynomials in t.
>The polynomial is irreducible over Q(t) and probably its
>Galois group is PGL(2,5).
Unfortunately the generic Galois group turns out to be the full symmetric
group S6. Still, it is an interesting polynomial. For t=0,1,2,3,4,5,6
it's just (x-1)^t*x^(6-t); these are the only t for which there are
repeated factors: the discriminant has degree at most 30 in t, and was
just found to vanish to order at least 5,4,4,4,4,4,5 at 0,1,2,3,4,5,6,
but does not vanish identically (this can be seen for instance by examining
the polynomial's behavior for large t). [In particular the discriminant
is a square for infinitely many rational t, but as it happens they are
all between 0 and 6 and so in particular we don't get to see an A6 extension
for integral t.]
So far much the same is true with 6 replaced by any integer k [or any even
integer for the bracketed comment]: we get a k-fold covering of P^1, with
ramification at t=0,1,2,...,k of type (t,k-t) and unramified at all other t;
as a curve in P^2 it is smooth of degree k. (For this we need again the lemma
that the polynomial 1+u+u^2/2+u^3/3!+...+u^k/k! has no repeated roots to handle
what happens at infinity.) When k=6 we can apply an outer automorphism to S6
to this covering to obtain another sixfold covering of P1 ramified above
0,1,2,3,4,5,6 with cycle structure (1,2,3) at 0 and 6, (1,5) at 1 and 5,
(2,4) at 2 and 4, and (1,1,1,3) at 3. Specialization at t produces in our
first curve Galois group (contained in) PGL_2(5) iff it produces in the
second curve Galois group (contained in) the image S5 of PGL_2(5) in S6
under the outer automorphism of S6; i.e. iff the second curve has a rational
point above t. But that curve has genus 7 by Riemann-Hurwitz; while this
is less than the genus 10 of our first curve it's still large enough that
it can have only finitely many integral points (indeed finitely many rational
ones by Faltings).
That still doesn't seem to help explain the unexpected rational points at
t=-2 and now t=-4. (And even the quotient by the t <--> 6-t symmetry
still leaves a curve of genus >1.) But it does indicate that there is no
infinite family of such points, and most likely no more rational points
at all (assuming that you have already searched other small t for further
examples).
--Noam D. Elkies (elkies@zariski.harvard.edu)
Department of Mathematics, Harvard University
P.S. G.W.Smith's observation that
> if r_1, ..., r_6 are the roots of [x^6+2x^5+3x^4+...+7], then
> r_1^k + ... + r_6^k = -2 for various values of k, including
> k from 1 to 6.
[specifically 1..6, 9..13, 17..20, 25..27, 33&34, and 41] can be explained
by writing that sextic as the quotient of the trinomial x^8-8*x+7 by (x-1)^2.
This too generlizes to other degrees k.