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| The largest "n" that GF(2^n) can support in Singular? https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1832 |
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| Author: | gepo [ Wed May 26, 2010 4:46 pm ] |
| Post subject: | The largest "n" that GF(2^n) can support in Singular? |
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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| Author: | hannes [ Thu May 27, 2010 10:04 am ] |
| Post subject: | Re: The largest "n" that GF(2^n) can support in Singular? |
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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