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Topic review - turning an ideal into a ring: division |
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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?
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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Posted: Sun Mar 29, 2009 5:14 pm |
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turning an ideal into a ring: division |
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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?
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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Posted: Sat Mar 14, 2009 5:36 am |
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