Timeline for Commutator tensors and submodules
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 26, 2011 at 8:13 | comment | added | darij grinberg | Also, the "Note (for use in Step 2) that the conclusion can be restated as follows: every element of the kernel of the canonical map $Sym^{n-1}A\otimes B\to Sym^nA$ is in the image of $Sym^{n-2}A\otimes K^2(B)$." part took the longest time for me to understand. You are silently using that the kernel of the homomorphism $A^{\otimes\left(n-1\right)}\otimes B \to \mathrm{Sym}^{n-1}A \otimes B$ is $K^{n-1}\left(A\right) \otimes B$. This is because $A^{\otimes\left(n-1\right)}\to \mathrm{Sym}^{n-1}A$ is surjective and $\otimes$ is right-exact. Just leaving this here for future readers. | |
| May 25, 2011 at 22:55 | comment | added | darij grinberg | Looks very good! I have not checked Step 3 as I have yet to wrap my mind around the direct limits and fibered products involved, but the rest is already way better than I could do. One minor typo: $A^n$ should be $A^{\otimes n}$ in the beginning of Step 2? | |
| May 25, 2011 at 22:54 | vote | accept | darij grinberg | ||
| May 23, 2011 at 23:08 | comment | added | Tom Goodwillie | It's an enjoyable exercise: Prove first that every module is the colimit of "all" finitely generated free modules over it, then (this is the more interesting part) prove that if the module is flat then the indexing category for that colimit is filtered. | |
| May 23, 2011 at 14:58 | comment | added | Martin Brandenburg | (2) was proven by Lazard, but a good reference is Lam's book about module theory. | |
| May 23, 2011 at 13:05 | comment | added | darij grinberg | (1) Ah, thanks, that's the standard argument which still holds if $A/B$ is projective. I should have realized it. (2) Thanks, I found it in literature. | |
| May 23, 2011 at 11:25 | comment | added | Tom Goodwillie | (1) I just meant that if $B$ is a submodule of $A$ and $A/B$ is free then there exists a submodule $C\subset A$ such that $A$ is the direct sum of $B$ and $C$. I.e., the exact sequence $0\to B\to A\to A/B\to 0$ splits. (2) I don't know a reference, but the key point is that the category of all f.g. free modules over a given flat module is filtered. | |
| May 23, 2011 at 10:01 | comment | added | darij grinberg | OK, sorry, I don't have that much time at my disposal right now, so reading this will take a while. At the moment I have two questions: (1) why is $A=C\oplus B$ in Step 2? (2) Why is a flat module a direct limit of free modules? (This seems to be well-known, but I have no idea how to prove it.) | |
| May 22, 2011 at 15:48 | history | answered | Tom Goodwillie | CC BY-SA 3.0 |