On Sunday, August 20, 2017 at 6:02:32 PM UTC+1, John Cremona wrote: > > On 20 August 2017 at 17:47, Johan S. H. Rosenkilde <mail...@atuin.dk > <javascript:>> wrote: > > > > Vincent Delecroix writes: > > > >> If the basis of a "Finite dimensional module with basis" is always > assumed to be > >> ordered, then such method make sense. However, the terminology is quite > strange. > >> I see 1+1/2 ambiguities for matrices over polynomial ring such as > Mat(ZZ[X], 3). > >> > >> 1) leading_coefficient might be a termwise application of > leading_coefficient > >> > >> [1 X^2+X+1] > >> [X+3 2*X-3 ] > >> > >> would result in > >> > >> [1 1] > >> [1 2] > >> > >> 2) There is an additional trouble if Mat(ZZ[X], 3) is intended to be > equivalent > >> to polynomial ring over matrices Mat(ZZ, 3)[X]. Such matrix can > naturally be > >> written as > >> > >> M0 + M1 X + M2 X^2 + ... + Md X^d > >> > >> where M0, M1, ..., Md are matrices with coefficients in ZZ. With the > above > >> writing, the leading coefficient is Md. > > > > Indeed. And then, as a third option, the one we are introducing in > > #23619 as leading_matrix, where we take for each row v, the coefficients > > of terms of deg(v): > > > > [1 X^2+X+1] [0 1] > > leading_matrix( [X+3 2*X-3 ] ) = [1 1] > > > > Best, > > Johan > > > > Don't get too carried away with thinking up ways in which users might > conceivably get confused! The rings "matrices over polynomials over > F" and "polynomials over matrices over F" are isomorphic but not the > same. > > More to the point, a free module over a ring R will know what its base > ring and rank are, in computer science terms, even if in mathematical > terms it might have the structure of a free module over different > rings. (For example every QQ-vector space is a free module over QQ, > and also over ZZ.) > > My criticism was robustly countered. I'll go back to just saying that > if you had asked me what the "leading monomial" of an element of R^n > is, I might have guessed it (probably not, or wrongly), but I have > never heard of this expression in this context. And I have been > around for a while, teaching graduate courses in algebra etc. >
Hmm, D.Eisenbud, "Commutative algebra with ...", Chapter 15, talks about this stuff at length. (Of course this book is too thick for a UK graduate course ;-)) > > John > > > -- > > You received this message because you are subscribed to the Google > Groups "sage-devel" group. > > To unsubscribe from this group and stop receiving emails from it, send > an email to sage-devel+...@googlegroups.com <javascript:>. > > To post to this group, send email to sage-...@googlegroups.com > <javascript:>. > > Visit this group at https://groups.google.com/group/sage-devel. > > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "sage-devel" group. To unsubscribe from this group and stop receiving emails from it, send an email to sage-devel+unsubscr...@googlegroups.com. To post to this group, send email to sage-devel@googlegroups.com. Visit this group at https://groups.google.com/group/sage-devel. For more options, visit https://groups.google.com/d/optout.
