Dear group!
I've been trying to find a minimal polynomial for an algebraic number
with maxima. I follow the Landau paper 'Simplification of nested
radicals' suggested by Dr. Fateman (over a year and a half ago). I need
advice on the following topics.
1. Suppose I want to find an element t such that an algebraic extension
Q(sqrt(3), sqrt(2)) is tha same as Q(t). After the algorithm, I obtain
the minimal polynomials g_{i-1}(x)=x^2-3, f_{i}(x)=x^2-2. Then I have to
find s_{i}(x)=Resultant(g_{i-1}(t),f_i(t),t),
h_{i}(x)=s_{i}(x)/gcd(s_{i}(x), s'_{i}(x)) and finally find c_{i} such
that g_i(t)=resultant(g_{i-1}(y),f_i(t-c_{i}*y,y)) is square free.
Here's what I obtain from Maxima:
(%i8) s: resultant(t^2-3,x^2-2,t);
2 2
(%o8) (x - 2)
(%i9) h: s/gcd(s,diff(s,x)),ratsimp;
2
(%o9) x - 2
(%i11) resultant(y^2-3,subst(t-c*y,x,h),y);
4 2 2 4 2
(%o11) t + (- 6 c - 4) t + 9 c - 12 c + 4
(On page 18 resultant(g_{i-1}(y),h_{i}(x)(t-c_{i}*y,y)) is written
instead of resultant(g_{i-1}(y),f_i(t-c_{i}*y,y)). Happily in this
example h_{i}=f_{i}).
Now, how do I deduce which c is needed from %o11?
2. After I learned somehow that t is in fact sqrt(2)+sqrt(3) I have to
express sqrt(2) and sqrt(3) as elements in Q(t). I tried the following
approach: factor x^2-2 over Q(t) into (2x-a^3+9a)(2x+a^2-9a)/4. Now one
of the factors should be equal to zero at x=sqrt(2). To find that factor
I do tellrat(a^4-10*a^2+1); rat(2*sqrt(2)-a^3+9*a) ==> -a^3+9a+2sqrt(2).
(I know that in this example what this factor is.) It's not simplifyed
to zero :( The only working way is to backsubstitute t=sqrt(2)+sqrt(3)
and to ratexpand. But I don't see any guerarntee that the expanded
expression will be simplified to 0 by Maxima simplifier every time. So,
what can be done here?
3. Does Maxima lisp kernel have a code for q a queue data structure? I
have to use my own so far, but will switch to Maxima's if any.
Thank you for attension
--
Andrei Zorine