ASKSAGE: Sage Q&A Forum - Latest question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 24 Apr 2018 14:21:01 -0500Number field basis containing 1http://ask.sagemath.org/question/42150/number-field-basis-containing-1/ In Sage, the default basis for a maximal order $O_K$ often does not contain the element $1$:
sage: QuadraticField(-3).ring_of_integers().basis()
[1/2*a + 1/2, a]
However, $1$ is always a primitive lattice vector in $O_K$. Is there an elegant way to produce a basis containing $1$?eodorneyTue, 24 Apr 2018 14:21:01 -0500http://ask.sagemath.org/question/42150/Calculation of maximal order fails even when using "maximize_at_primes"http://ask.sagemath.org/question/40677/calculation-of-maximal-order-fails-even-when-using-maximize_at_primes/I would like to calculate the maximal order of a field for which I already know at which primes we should maximize.
This means that the discriminant does not have to be factored, which is normally the bottleneck of this algorithm.
I did
d = [2,3,5,7,11,13,17,19]
K.<y> = NumberField([x^2 - di for di in d], maximize_at_primes=[2])
RR = K.maximal_order()
The last command did not finish overnight, but gave the following warning:
"*** Warning: MPQS: number too big to be factored with MPQS,
giving up."
Which seems to indicate that the program is indeed trying to perform a large factorization despite the command "maximize_at_primes=[2]"
Meanwhile, Magma has no problem performing this computation in a few hours:
R<x> := PolynomialRing(Integers());
K := NumberField([x^2 - 2,x^2 - 3 , x^2-5,x^2-7,x^2 - 11,x^2 - 13 , x^2 - 17 , x^2 - 19]:Abs);
O := MaximalOrder(K: Ramification := [2]);
Am I doing something wrong ?
JF BiassebiasseThu, 18 Jan 2018 08:58:26 -0600http://ask.sagemath.org/question/40677/