- Thanks for the background, Marcel.

In such a situation an author might perhaps have written:

"I thank X for demonstrating that some of my earlier

ideas were more useful than I had supposed."

About Primo 1.2: If you leave out backtrack, please,

please, keep path-only. Then I can use 1.1.0 for path-only,

with advanced-setup backtrack, and still benefit from

even better factorization of polys in 1.2.

Else, you would divide your talents into incompatible packages,

which would be unfortunate for folk like Hans, Bouk and me.

Thanks, as ever, for everything that you do...

David - Marcel Martin wrote:

> I just saw that F. Morain (with G. Hanrot) published the following

Thank you for telling us very interesting paper.

> paper

>

> This is a description of the method I am using to split the Hilbert

> polynomials with their Galois groups.

Since I'm now developing the computation of the maximal order,

I could not read the paper in detail.

I guessed that your polynomial factorization used subfields of

the Hilbert class fields. But I was not able to think a concrete way.

I would like to learn the method developed by you.

Thank you, Marcel

Satoshi Tomabechi - Marcel Martin conjectured that
> among the 300 subscribers of this list, Satoshi

The conjecture is hereby validated.

> is not the only one interested in ECPP or in the

> factorization of class polynomials over Z/P.

Thanks, Marcel!

David - Marcel Martin taught:
> Here, we see the advantage of the method. Instead of factoring a

What I understand slightly is as follows.

> degree-15 polynomial, we only have to factor a degree-3 and a

> degree-5 polynomials.

Class field theory tells that CL(D) = Gal(K(D)/Q),

where K(D) is the Hilbert class field.

We suppose that there is a sequence of groups

CL(G) = G1 > G2> ... > {e}.

( G>H denotes that H is a subgrouop of G)

Corresponding to the sequence we get a sequence of fields

Q =K(D)^G1 < K(D)^G2 < ... < K(D)^{e}=K(D),

where K(D)^Gi is the field which consists of elements

fixed by Gi. Also we get the polynomials Fi(X) with

coefficients in K(D)^G{i-1} which define the field

extension K(D)^Gi/K(D)^G{i-1}. By finding roots of Fi(X)

successively we get a root of the Hilbert polynomial H(D)(X).

By reduction modulo P similar procedure works over Z/P

as well as over Q.

Indeed CL(-971) is not cyclic, but there is a sequence of

groups:

CL(-971) = G1 =Z_5*Z_3 > G2=Z_5 > {e}

K(D)^G2/Q is an extension of degree 3.

K(D)/K(D)^G2 is an extension of degree 5.

Therefore we should factor a degree-3 and a degree-5

polynomials instead of factoring a degree-15 polynomial.

I look forward to your next post, however I could not understand

as you expect.

Thank you for an exciting explanation.

Satoshi Tomabechi