Timeline for Projective modules and tensor products
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2011 at 1:08 | vote | accept | Clinton Boys | ||
| Jul 6, 2011 at 20:33 | comment | added | Julian Kuelshammer | But this argument also shows that $W$ has a filtration $0\subset W_1\subset W_2\subset W_3'$, where $W_3'\oplus V_p\cong W_3$. Now we use the injection of $V_p$ into $W_1$ into $V_2\otimes P_{p-1}$. By the universal property of injectives (since any projective module is injective) we now have that $V_p$ is a direct summand of $W_3'$. So there is $\tilde{W}$ with $\tilde{W}\oplus V_p^2\cong V_2\otimes P_{p-1}$. | |
| Jul 6, 2011 at 20:27 | comment | added | Julian Kuelshammer | Sorry. Now in more detail: First you get a filtration of $V_2\otimes P_{p-1}$: $0\subset W_1\subset W_2\subset W_3$ with $W_1\cong V_2\otimes V_{p-1}$, $W_2/W_1\cong V_2\otimes V_2$ and $W_3/W_2\cong V_2\otimes V_{p-1}$. Now we know that $V_2\otimes V_{p-1}$ has a direct summand isomorphic to $V_p$, i.e. there is a surjection $V_2\otimes P_{p-1}\twoheadrightarrow V_p$. By the universal property, this splits, i.e. $V_p$ is in fact a direct summand of $V_2\otimes P_{p-1}$, i.e. there is a module $W$, s.t. $W\otimes V_p\cong V_2\otimes P_{p-1}$. | |
| Jul 5, 2011 at 23:39 | comment | added | Clinton Boys | Unfortunately this is precisely the fluency with the concept that Alperin is using, which I don't yet have (and which the book assumes very quickly!). Could you explicitly show me how to show that $P_{p-2}$ and $V_p$, twice, are direct summands of $V_2\otimes P_{p-1}$? The only way I know how to do this is to exhibit a surjective map $V_2\otimes P_{p-1}$ onto some projective module, and show that each of these modules arises as the kernel of such a map. However, if we do this twice for $V_p$, how do we know we aren't picking up the same summand each time? | |
| Jul 5, 2011 at 14:10 | comment | added | Julian Kuelshammer | No, you won't get a composition series of $V_2\otimes P_{p-1}$, and it won't be uniserial (the tensored are not simple anymore). But you will get a filtration because $\otimes$ is exact. | |
| Jul 5, 2011 at 11:57 | comment | added | Clinton Boys | Right, and the uniseriality and composition series are the same (i.e. the factors of the unique composition series for $V_2\otimes P_{p-1}$ are just the factors for $P{p−1}$, tensored with $V_2$) because $V_2$ is simple? I suspected this was true but he never proved it anywhere. | |
| Jul 5, 2011 at 11:20 | comment | added | Julian Kuelshammer | And this follows because we know the explicit structure of $P_{p-1}$. It is uniserial and has a composition series, where $V_{p-1}$ is appearing twice and $V_2$ is appearing once. Now tensor with $V_2$. | |
| Jul 5, 2011 at 11:16 | comment | added | Julian Kuelshammer | Because $V_2\otimes V_{p-1}$ appears twice in a series of submodules. And each of these has a direct summand $V_p$ according to $V_2\otimes V_{n}\cong V_{n-1}\oplus V_{n+1}$ for $n<p$. | |
| Jul 5, 2011 at 10:19 | comment | added | Clinton Boys | That makes sense. But why must $V_p$ appear as a direct summand twice? | |
| Jul 5, 2011 at 8:56 | history | answered | Julian Kuelshammer | CC BY-SA 3.0 |