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Timeline for Decomposition of a quotient module

Current License: CC BY-SA 3.0

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Jun 28, 2013 at 19:38 history edited Graham Leuschke CC BY-SA 3.0
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Jun 27, 2013 at 15:16 comment added Steven Landsburg Graham: Yes, the bit about the projection was nonsense. I've got an inclusion $k\rightarrow Coker(M)$, but it's not split. I think I'm done saying wrong things now!
Jun 27, 2013 at 12:15 vote accept TmobiusX
Jun 27, 2013 at 12:14 comment added TmobiusX @Graham Leuschke: Your explanation implies that my thought is false. I understand, thank you and Steven.
Jun 27, 2013 at 12:09 vote accept TmobiusX
Jun 27, 2013 at 12:15
Jun 27, 2013 at 10:20 comment added Graham Leuschke Is the projection onto $k$ an $R$-map? I don't see why.
Jun 27, 2013 at 3:12 comment added Steven Landsburg So let me go farther out on a limb here. I claim $Coker(M)=(R/J_{xv}\oplus R/J_{yv})/((v,x),(y,z))$, with the ideals $J_t$ as in my answer above. In particular, this implies $(xv,yv)=0$. And I claim further that the map $k\rightarrow coker(M)$ given by $1\mapsto (x,y)$ is an $R$-map. (Check that $x,y,z,v$ all kill $(x,y)$ in $coker(M)$.) This map appears to split the projection from $coker(M)$ onto $k$. Have I managed to make another mistake yet?
Jun 27, 2013 at 1:26 history edited Graham Leuschke CC BY-SA 3.0
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Jun 27, 2013 at 1:23 comment added Graham Leuschke Doh! Turnabout is fair play indeed. Quite right.
Jun 27, 2013 at 0:49 comment added Steven Landsburg Having already embarrassed myself with silly mistakes both above and in private email, I'm posting this comment with considerable trepidation, but --- aren't $(v,x)^T$ and $(y,z)^T$ both zero in $coker(M)$?
Jun 26, 2013 at 18:19 history answered Graham Leuschke CC BY-SA 3.0