Function: bnfisprincipal
Section: number_fields
C-Name: bnfisprincipal0
Prototype: GGD1,L,
Help: bnfisprincipal(bnf,x,{flag=1}): bnf being output by bnfinit, gives
[e,t], where e is the vector of exponents on the class group generators and
t is the generator of the resulting principal ideal. In particular x is
principal if and only if e is the zero vector. flag is optional, whose
binary digits mean 1: output [e,t] (only e if unset); 2: increase precision
until alpha can be computed (do not insist if unset); 4: return alpha in
factored form (compact representation).
Doc: $\var{bnf}$ being the \sidx{principal ideal}
number field data output by \kbd{bnfinit}, and $x$ being an ideal, this
function tests whether the ideal is principal or not. The result is more
complete than a simple true/false answer and solves a general discrete
logarithm problem. Assume the class group is $\oplus (\Z/d_i\Z)g_i$
(where the generators $g_i$ and their orders $d_i$ are respectively given by
\kbd{bnf.gen} and \kbd{bnf.cyc}). The routine returns a row vector $[e,t]$,
where $e$ is a vector of exponents $0 \leq e_i < d_i$, and $t$ is a number
field element such that
$$ x = (t) \prod_i g_i^{e_i}.$$
For \emph{given} $g_i$ (i.e. for a given \kbd{bnf}), the $e_i$ are unique,
and $t$ is unique modulo units.
In particular, $x$ is principal if and only if $e$ is the zero vector. Note
that the empty vector, which is returned when the class number is $1$, is
considered to be a zero vector (of dimension $0$).
\bprog
? K = bnfinit(y^2+23);
? K.cyc
%2 = [3]
? K.gen
%3 = [[2, 0; 0, 1]] \\ a prime ideal above 2
? P = idealprimedec(K,3)[1]; \\ a prime ideal above 3
? v = bnfisprincipal(K, P)
%5 = [[2]~, [3/4, 1/4]~]
? idealmul(K, v[2], idealfactorback(K, K.gen, v[1]))
%6 =
[3 0]
[0 1]
? % == idealhnf(K, P)
%7 = 1
@eprog
\noindent The binary digits of \fl mean:
\item $1$: If set, outputs $[e,t]$ as explained above, otherwise returns
only $e$, which is much easier to compute. The following idiom only tests
whether an ideal is principal:
\bprog
is_principal(bnf, x) = !bnfisprincipal(bnf,x,0);
@eprog
\item $2$: It may not be possible to recover $t$, given the initial accuracy
to which the \kbd{bnf} structure was computed. In that case, a warning is
printed and $t$ is set equal to the empty vector \kbd{[]\til}. If this bit is
set, increase the precision and recompute needed quantities until $t$ can be
computed. Warning: setting this may induce \emph{lengthy} computations
and you should consider using flag $4$ instead.
\item $4$: Return $t$ in factored form (compact representation),
as a small product of $S$-units for a small set of finite places $S$,
possibly with huge exponents. This kind of result can be cheaply mapped to
$K^*/(K^*)^\ell$ or to $\C$ or $\Q_p$ to bounded accuracy and this is usually
enough for applications. Explicitly expanding such a compact representation
is possible using \kbd{nffactorback} but may be very costly. The algorithm
is guaranteed to succeed if the \kbd{bnf} was computed using
\kbd{bnfinit(,1)}. If not, the algorithm may fail to compute a huge
generator in this case (and replace it by \kbd{[]\til}). This is orders of
magnitude faster than flag $2$ when the generators are indeed large.
Variant: Instead of the above hardcoded numerical flags, one should
rather use an or-ed combination of the symbolic flags \tet{nf_GEN} (include
generators, possibly a place holder if too difficult), \tet{nf_GENMAT}
(include generators in compact form) and
\tet{nf_FORCE} (insist on finding the generators, a no-op if \tet{nf_GENMAT}
is included).