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| turning an ideal into a ring: division https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=1703 |
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| Author: | swehner [ Sat Mar 14, 2009 5:36 am ] |
| Post subject: | turning an ideal into a ring: division |
Let I = < i_n > be an ideal which is contained in a proper ideal J = < j_n > of some (commutative) polynomial ring R. J is a ring without a unit. I is an ideal of the ring J as well. Just as one can meaningfully divide polynomials in R with respect to one of its ideals (using Groebner Bases), can one divide polynomials in the ring J with respect to its ideal I? In other words, is there, given a polynomial p in J, a notion of q = p (mod I) within the ring J? By this I mean p = Sum(k) alpha_k * i_k + q where all the alpha_k are polynomials in J. If yes, can Singular compute the alpha's and q? |
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| Author: | greuel [ Sun Mar 29, 2009 5:14 pm ] |
| Post subject: | |
I do not quite understand the question. What do you impose on q? In any case, the expression Sum(k) alpha_k * i_k is in I*J and you could apply the Singular command division(p,I*J); The result can then be used to get an expression as you propose, with q the usual remainder of the division by I*J. Is this what you mean? |
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