# Singular

 Page 1 of 1 [ 2 posts ]
 Print view | E-mail friend Previous topic | Next topic
Author Message
 Post subject: The largest "n" that GF(2^n) can support in Singular?Posted: Wed May 26, 2010 4:46 pm

Joined: Thu Jul 09, 2009 7:28 am
Posts: 24
What is the largest n that GF(2^n) can support in Singular?

From 6.1 Limitations of Singular http://www.singular.uni-kl.de/Manual/la ... htm#SEC386 ,
"the number of elements in GF(p,n) must be less than 65536 " which means the "n" can at most be 16, is this correct?

But in my computation, I found "n" could be much larger. I am confused here. Could you please explain this?

Thanks
Gepo

Top

 Post subject: Re: The largest "n" that GF(2^n) can support in Singular?Posted: Thu May 27, 2010 10:04 am

Joined: Wed May 25, 2005 4:16 pm
Posts: 215
That depends on the way the ring is defined.
For the form
Code:
[ing r1=(2^n,a),....

the maximum for n is 15 (not 16: less than 2^16).
For the form
Code:
ring r2=(2,a),...; minpoly=.....

the maximal n is the maximal degree of the minpoly, which is
at least 65536.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 2 posts ]

 You can post new topics in this forumYou can reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

 It is currently Thu Feb 21, 2019 11:42 pm