j-invariant of Fp2

answered 2020-04-05 17:48:00 +0200

Sébastien
3153 ●5 ●39 ●85 http://www.slabbe.org/

updated 2020-04-06 09:40:56 +0200

The function supersingular_j returns one such supersingular j-invariant :

sage: supersingular_j(GF(19^2, 'i'))
18
sage: supersingular_j(GF(15073^2, 'i'))
4443*i + 13964

To get all of them you use the method supersingular_points:

sage: S = SupersingularModule(19)
sage: S
Module of supersingular points on X_0(1)/F_19 over Integer Ring
sage: L,d = S.supersingular_points()
sage: L
[18, 7]
sage: S = SupersingularModule(431)
sage: L,d = S.supersingular_points()
sage: len(L)
37

edit flag offensive delete link

Comments

Thanks a lot, @Sebastien.

niranjan ( 2020-04-06 07:43:06 +0200 )edit

Is there a method to get all or a list of j-variants? For example, GF(431^2, 'i') has 37 j-variants. Above function just returns one (e.g. 4). How can we get a subset of 37? Thanks.

niranjan ( 2020-04-06 09:25:34 +0200 )edit

The documentation I linked to contains the answer. I updated the answer to return the list. Note that I do not know the math involved, so please check that it does what you want, I only know how to search/read the documentation.

Sébastien ( 2020-04-06 09:43:07 +0200 )edit

@Sebastien: thank you so much! This code helps. And, my apologies for not looking into the documentation carefully.

niranjan ( 2020-04-06 11:32:13 +0200 )edit

add a comment

1 Answer

Comments

Your Answer

Question Tools

Stats

j-invariant of Fp2 edit

1 Answer

Comments