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).