Timeline for Can different modules have the same symmetric algebra? (answered: no)
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2009 at 0:36 | comment | added | Kevin Buzzard | Your argument is close to working though. If you "remove the noise in degree 0" it's fine. | |
| Nov 24, 2009 at 0:22 | comment | added | pinaki | I see that I was wrong. Sorry! | |
| Nov 24, 2009 at 0:21 | history | edited | pinaki | CC BY-SA 2.5 |
added 37 characters in body
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| Nov 23, 2009 at 15:22 | comment | added | Mark Hovey | Here's an example. Consider the map f: Sym(k) = k[x] --> k[x] that takes x to x^2. The linear part of this map is 0. Consider also the map g: k[x] --> k[x] that takes x to x+1. The linear part of this map is the identity. So the composition of the linear parts is 0. However, the linear part of the composition gf, which takes x to 1 + 2x + x^2, is multiplication by 2. So this answer is not functorial, as buzzard points out, and is wrong. | |
| Nov 23, 2009 at 13:25 | comment | added | Kevin Buzzard | [and hence I don't see why your map M->N is definitely an isomorphism; that's the point of my comment.] | |
| Nov 23, 2009 at 13:24 | comment | added | Kevin Buzzard | The map from the A-alg homs to the A-module homs is just "M lives in Sym(M) and there's a natural A-module projection from Sym(N) onto N". I think you're better off not picking generators---just imagine that M is generated by M. What I am confused about is that I don't think your argument is sufficiently functorial. For example given maps Sym(M)->Sym(N)->Sym(P), does your argument produce a commuting triangle of maps M->N->P? I don't see immediately that this is the case. | |
| Nov 23, 2009 at 11:50 | history | answered | pinaki | CC BY-SA 2.5 |