Timeline for Are plethories a theory of basis-free polynomials?
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6 events
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| Oct 28, 2011 at 14:51 | comment | added | Qiaochu Yuan | @Jacques: I don't know what you mean by "it is there in all choices of basis." Certainly all the examples you've written down implicitly give the same filtration on polynomials, but it is possible to write down bases where this is not the case (for example $x, x^3 - x^2, x^3 + x^2, x^4, x^5, ...$). @Wilberd: what I meant by "ignore the grading" is giving an abstract description of a polynomial ring that doesn't distinguish any of its elements (in the symmetric algebra description we distinguish the elements of degree $1$). | |
| Oct 28, 2011 at 12:53 | comment | added | Jacques Carette | I do not want to ignore the grading - it is there in all choices of basis, so I don't see a good a priori reason why it would disappear when we go basis-free. | |
| Oct 28, 2011 at 7:11 | comment | added | Wilberd van der Kallen | One may ignore the grading when describing a symmetric algebra through its universal property: a module map from the vector space to a commutative algebra extends uniquely to an algebra map. It is a happy and useful circumstance that this ungraded problem has a graded solution. | |
| Oct 28, 2011 at 4:20 | comment | added | S. Carnahan♦ | +1. The distinction seems to be whether you take the left adjoint of the "underlying set" functor or the "underlying module" functor from commutative rings. | |
| Oct 28, 2011 at 3:56 | comment | added | Alexander Woo | Let me add that many basic (and some not so basic) homological constructions have basis free definitions - for example one can easily define the Koszul resolution using the exterior algebra. | |
| Oct 27, 2011 at 23:14 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |