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 Post subject: The largest "n" that GF(2^n) can support in Singular?
PostPosted: 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


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

Joined: Wed May 25, 2005 4:16 pm
Posts: 210
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.


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