Timeline for Second summand to make projective module free
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2020 at 3:19 | vote | accept | zjs | ||
| Jun 21, 2020 at 3:17 | comment | added | LSpice | The problem with a question like "how to carry $p \oplus m$ to an element of $F P$" is that, in the generality of a totally arbitrary (ring $R$ and) projective $P$, we just have no way of peeking into the structure of $P$ other than by using the universal property; and, in terms of the universal property, the answer is that we use a splitting $P \to F P$ of $M \hookrightarrow F P \twoheadrightarrow P$ to map $P \oplus M \to F P$, and I think that's all that can be said! Obviously @StevenLandsburg's answer is more interesting. | |
| Jun 21, 2020 at 2:09 | comment | added | zjs | Yes, I'd say that this solution falls into "in terms of $P$", thanks! To really get a sense of what direct-summing $M$ does, I was also hoping to be able to think about what the module $P\oplus M$ "looks like", separately from the exact sequence context (since that seems to be the most natural way to actually prove the freeness of $P\oplus M$) --- is there a nice way to think about the isomorphism between $P\oplus M$ and the free module $FP$ on $P$? I'm having trouble picturing exactly what that might be, i.e. how to carry $p\oplus\sum r_ip_i$ to a $FP$-element. | |
| Jun 21, 2020 at 2:08 | answer | added | Steven Landsburg | timeline score: 3 | |
| Jun 20, 2020 at 23:20 | comment | added | LSpice | I can imagine it's one of those "I'll know it when I see it", but do you have a rigorous definition of "in terms of $P$"? For example, one can take $M$ to be the kernel of the projection to the module $P$ from the free $R$-module on the set $P$. Does that count? | |
| Jun 20, 2020 at 23:15 | review | First posts | |||
| Jun 21, 2020 at 0:05 | |||||
| Jun 20, 2020 at 23:10 | history | asked | zjs | CC BY-SA 4.0 |