Function: idealfactorback
Section: number_fields
C-Name: idealfactorback
Prototype: GGDGD0,L,
Help: idealfactorback(nf,f,{e},{flag = 0}): given a factorization f, gives the
ideal product back. If e is present, f has to be a
vector of the same length, and we return the product of the f[i]^e[i]. If
flag is non-zero, perform idealred along the way.
Doc: gives back the ideal corresponding to a factorization. The integer $1$
corresponds to the empty factorization.
If $e$ is present, $e$ and $f$ must be vectors of the same length ($e$ being
integral), and the corresponding factorization is the product of the
$f[i]^{e[i]}$.
If not, and $f$ is vector, it is understood as in the preceding case with $e$
a vector of 1s: we return the product of the $f[i]$. Finally, $f$ can be a
regular factorization, as produced by \kbd{idealfactor}.
\bprog
? nf = nfinit(y^2+1); idealfactor(nf, 4 + 2*y)
%1 =
[[2, [1, 1]~, 2, 1, [1, 1]~] 2]
[[5, [2, 1]~, 1, 1, [-2, 1]~] 1]
? idealfactorback(nf, %)
%2 =
[10 4]
[0 2]
? f = %1[,1]; e = %1[,2]; idealfactorback(nf, f, e)
%3 =
[10 4]
[0 2]
? % == idealhnf(nf, 4 + 2*y)
%4 = 1
@eprog
If \kbd{flag} is non-zero, perform ideal reductions (\tet{idealred}) along the
way. This is most useful if the ideals involved are all \emph{extended}
ideals (for instance with trivial principal part), so that the principal parts
extracted by \kbd{idealred} are not lost. Here is an example:
\bprog
? f = vector(#f, i, [f[i], [;]]); \\ transform to extended ideals
? idealfactorback(nf, f, e, 1)
%6 = [[1, 0; 0, 1], [2, 1; [2, 1]~, 1]]
? nffactorback(nf, %[2])
%7 = [4, 2]~
@eprog
The extended ideal returned in \kbd{\%6} is the trivial ideal $1$, extended
with a principal generator given in factored form. We use \tet{nffactorback}
to recover it in standard form.