Function: nfcompositum
Section: number_fields
C-Name: nfcompositum
Prototype: GGGD0,L,
Help: nfcompositum(nf,P,Q,{flag=0}): vector of all possible compositums
of the number fields defined by the polynomials P and Q; flag is
optional, whose binary digits mean 1: output for each compositum, not only
the compositum polynomial pol, but a vector [R,a,b,k] where a (resp. b) is a
root of P (resp. Q) expressed as a polynomial modulo R, and a small integer k
such that al2+k*al1 is the chosen root of R; 2: assume that the number
fields defined by P and Q are linearly disjoint.
Doc: Let \var{nf} be a number field structure attached to the field $K$
and let \sidx{compositum} $P$ and $Q$
be squarefree polynomials in $K[X]$ in the same variable. Outputs
the simple factors of the \'etale $K$-algebra $A = K[X, Y] / (P(X), Q(Y))$.
The factors are given by a list of polynomials $R$ in $K[X]$, attached to
the number field $K[X]/ (R)$, and sorted by increasing degree (with respect
to lexicographic ordering for factors of equal degrees). Returns an error if
one of the polynomials is not squarefree.
Note that it is more efficient to reduce to the case where $P$ and $Q$ are
irreducible first. The routine will not perform this for you, since it may be
expensive, and the inputs are irreducible in most applications anyway. In
this case, there will be a single factor $R$ if and only if the number
fields defined by $P$ and $Q$ are linearly disjoint (their intersection is
$K$).
The binary digits of $\fl$ mean
1: outputs a vector of 4-component vectors $[R,a,b,k]$, where $R$
ranges through the list of all possible compositums as above, and $a$
(resp. $b$) expresses the root of $P$ (resp. $Q$) as an element of
$K[X]/(R)$. Finally, $k$ is a small integer such that $b + ka = X$ modulo
$R$.
2: assume that $P$ and $Q$ define number fields that are linearly disjoint:
both polynomials are irreducible and the corresponding number fields
have no common subfield besides $K$. This allows to save a costly
factorization over $K$. In this case return the single simple factor
instead of a vector with one element.
A compositum is often defined by a complicated polynomial, which it is
advisable to reduce before further work. Here is an example involving
the field $K(\zeta_5, 5^{1/10})$, $K=\Q(\sqrt{5})$:
\bprog
? K = nfinit(y^2-5);
? L = nfcompositum(K, x^5 - y, polcyclo(5), 1); \\@com list of $[R,a,b,k]$
? [R, a] = L[1]; \\@com pick the single factor, extract $R,a$ (ignore $b,k$)
? lift(R) \\@com defines the compositum
%4 = x^10 + (-5/2*y + 5/2)*x^9 + (-5*y + 20)*x^8 + (-20*y + 30)*x^7 + \
(-45/2*y + 145/2)*x^6 + (-71/2*y + 121/2)*x^5 + (-20*y + 60)*x^4 + \
(-25*y + 5)*x^3 + 45*x^2 + (-5*y + 15)*x + (-2*y + 6)
? a^5 - y \\@com a fifth root of $y$
%5 = 0
? [T, X] = rnfpolredbest(K, R, 1);
? lift(T) \\@com simpler defining polynomial for $K[x]/(R)$
%7 = x^10 + (-11/2*y + 25/2)
? liftall(X) \\ @com root of $R$ in $K[x]/(T(x))$
%8 = (3/4*y + 7/4)*x^7 + (-1/2*y - 1)*x^5 + 1/2*x^2 + (1/4*y - 1/4)
? a = subst(a.pol, 'x, X); \\@com \kbd{a} in the new coordinates
? liftall(a)
%10 = (-3/4*y - 7/4)*x^7 - 1/2*x^2
? a^5 - y
%11 = 0
@eprog
The main variables of $P$ and $Q$ must be the same and have higher priority
than that of \var{nf} (see~\kbd{varhigher} and~\kbd{varlower}).