I had the same problem and deleting the 'sage_notebook.sagenb' directory
from '~/.sage/' resolved it.
Thanks for the tip.
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What exactly do you mean by the dual of an ideal? Do you mean dual
with respect to the trace pairing, so the dual of the ideal (1) is the
inverse different?
David
On 4 September 2012 04:15, Cindy cindy425192...@gmail.com wrote:
Hi,
How can I calculate the dual of an ideal using sage?
Cindy,
Could you elaborate little more, what is precisely you need.
Regards,
Vijay
On Tue, Sep 4, 2012 at 12:42 PM, David Loeffler
d.a.loeff...@warwick.ac.ukwrote:
What exactly do you mean by the dual of an ideal? Do you mean dual
with respect to the trace pairing, so the dual of the ideal
k=GF(2^11);
K=GF(2^33)
How to see K as a vector space over filed k ? How to form its basis ?
How to construct tower of field extensions ?
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On Tuesday, September 4, 2012 2:52:29 AM UTC-4, mazkime wrote:
I had the same problem and deleting the 'sage_notebook.sagenb' directory
from '~/.sage/' resolved it.
Thanks for the tip.
But note that all of your notebook files are probably in there, so don't
forget the I then just copied
Le mardi 4 septembre 2012 10:40:24 UTC+2, sha2nk a écrit :
k=GF(2^11);
K=GF(2^33)
How to see K as a vector space over filed k ? How to form its basis ?
How to construct tower of field extensions ?
Hi, Sadly enough non of these three things are implemented in sage yet for
finite fields.
Some time ago I asked if Sage can solve something like
If P(x)=x³+ax²+bx+c with roots r_1, r_2 and r_3, how to express
((r_1-r_2)(r_1-r_3)(r_2-r_3))^2 as a function of a, b and c?
Because a, b and c are symmetric functions of roots, I guess I should read
Hi, David,
Yes, that's what I mean. Can I find it using sage?
Thanks.
Cindy
On Tuesday, September 4, 2012 3:12:25 PM UTC+8, David Loeffler wrote:
What exactly do you mean by the dual of an ideal? Do you mean dual
with respect to the trace pairing, so the dual of the ideal (1) is the
Hi, Vijay,
Let K be a number field and O_k be its ring of integers. Given an ideal J
of O_k, I want to find the dual of J, which is defined as the O_k-module:
J^*={x\in K| Tr(xJ)\subset Z}.
Thanks.
Cindy
On Tuesday, September 4, 2012 3:20:35 PM UTC+8, Vj wrote:
Cindy,
Could you elaborate
Hi,
BTW, the ideals I am dealing with are ideals of the ring of integers of a
number field.
Cindy
On Tuesday, September 4, 2012 3:12:25 PM UTC+8, David Loeffler wrote:
What exactly do you mean by the dual of an ideal? Do you mean dual
with respect to the trace pairing, so the dual of the
Hi,
Let K be a number field and O_k denote its ring of integers. For an ideal,
J of O_k, we can have an ideal lattice (I,b_\alpha), where
b_\alpha: J\times J \to Z, b_\alpha(x,y)=Tr(\alpha xy), \forall x,y \in J
and \alpha is a totally positive element of K\{0}.
Suppose now I know J and
Thanks Maarten for a quick reply.
I tried to hack this functionality but as you pointed out sage does not
treat L(below) as a field.
sage: n=2
sage: m=3
sage: q=2;
sage: k=GF(q);
sage: K=GF(q^n,'w');K
Finite Field in w of size 2^2
sage: P1.t = PolynomialRing(K,'t'); P1
Univariate Polynomial
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