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Topic review - The largest "n" that GF(2^n) can support in Singular?
Author Message
  Post subject:  Re: The largest "n" that GF(2^n) can support in Singular?  Reply with quote
That depends on the way the ring is defined.
For the form
[ing r1=(2^n,a),....

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

the maximal n is the maximal degree of the minpoly, which is
at least 65536.
Post Posted: Thu May 27, 2010 10:04 am
  Post subject:  The largest "n" that GF(2^n) can support in Singular?  Reply with quote
What is the largest n that GF(2^n) can support in Singular?

From 6.1 Limitations of Singular ... 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?

Post Posted: Wed May 26, 2010 4:46 pm

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