Gerhard Niklasch on Tue, 29 Aug 2000 15:06:13 +0200 (MET DST)

 Re: polgalois

```In response to:
> Message-ID: <39AB8F4E.A6E62213@uibk.ac.at>
> Date: Tue, 29 Aug 2000 12:24:14 +0200
> From: Mathias Lederer <csab2185@uibk.ac.at>
>
> The output of the command "polgalois(x)" is a three-component-vector [n,
> s, k], where k is the number of the group corresponding to a given pair
> (n, s). In the user's guide there is a list that explains the meaning of
> k for all polynomials up to degree 7. In this case k=1 except in 2
> cases.
> But when polgalois(x) is applied to a polynomial  e.g. of degree 11, k
> can be a much higher number.
>
> My question is: Where can I find a list which tells me the meaning of k
> in the general case? In GTM 138 it says that the number of k refers to
> standard literature. So which literature is standard?

"The transitive groups of degree up to eleven" by G. Butler and J. McKay,
in Communications in Algebra, vol. 11, 1983, pages 863--911.

(quoting from the readme file of Michel Olivier's galp directory,
ftp://megrez.math.u-bordeaux.fr/pub/galois/ .  However, galp uses a
somewhat different convention for identifying the groups it finds in
degrees 8 through 11.  The actual PARI code is in src/modules/galois.c,
and the state machines in there together with the files in galdata.tgz
determine the mapping.  I should hope that the numbering used by the
code is the same as that in Butler&McKay, but I'm separated from my
copy of the paper by several miles at the minute, and therefore cannot
check this right now.)

> And second, can't I get any direct information about the action of the
> Galois group on the zeros of the polynomial out of the command
> "polgalois"?

No, you need nfgaloisconj for that.  polgalois cannot give this because
it does not use or imply any numbering or labeling of the roots, and

To give an example, detecting dihedral galois groups  (or their sub-
groups)  in degree 4 involves checking whether a cubic resolvent
polynomial like
(T - (x1+x2)(x3+x4)) (T - (x1+x3)(x2+x4)) (T - (x1+x4)(x2+x3))
is irreducible over Q or factors nontrivially, where x1,...,x4
are the roots of the given quartic, in some arbitrary ordering.
But when you find that it does factor, you don't know how the
root you see relates to any _other_ particular numbering of the
roots of the quartic in some splitting field.

(In higher degrees, resolvents alone don't suffice -- the coset
structure of the group must be used.  But the bottom line remains
the same;  you can determine the group without knowing much about
how any particular group element acts on any particular root.)

Enjoy, Gerhard
```