This is really cool and seems to be exactly what I need. Thank you
very much!
Cheers,
Johan
On Apr 5, 3:19 pm, luisfe wrote:
> On Apr 5, 2:10 pm, "Johan S. R. Nielsen" wrote:
>
> > Oops, continuing:
>
> > more precisely, we wish to find a q in Q[Y1, Y2] such that q(f1, f2) =
> > g. In this case
On Apr 5, 2:10 pm, "Johan S. R. Nielsen" wrote:
> Oops, continuing:
>
> more precisely, we wish to find a q in Q[Y1, Y2] such that q(f1, f2) =
> g. In this case, we have
> q(Y1, Y2) = Y1^2 + Y1*Y2 - Y2
> as a solution, as
> f1^2 + f1*f2 - f2 = g
This is an elimination problem. Note that it is n
Oops, continuing:
more precisely, we wish to find a q in Q[Y1, Y2] such that q(f1, f2) =
g. In this case, we have
q(Y1, Y2) = Y1^2 + Y1*Y2 - Y2
as a solution, as
f1^2 + f1*f2 - f2 = g
As far as I can see, I can't easily use the lift function for this, as
the ideal's polynomials will always be lin
Thanks for the swift reply! That is a neat function, but I don't think
it is what I need. I was being too unclear, so here is an example:
Let R = Q[x], f1 = x^2 + 1 and f2 = x + 3 and g = x4+x3+4x2+x+3. We
wish to write g as a polynomial in f1 and f2 over Q; more precisely,
we wish to find a q
