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Aug 1, 2010 at 17:02 vote accept babubba
Jul 26, 2010 at 21:02 answer added mark timeline score: 6
Jul 7, 2010 at 17:50 comment added Graham Leuschke To see that they're not the same, consider the toy example $R=k[t]/(t2)$. There are only two indecomposable f.g. modules, namely $R$ itself and the residue field $k$. If you're killing relations from exact sequences, you have $[R]=2[k]$ so $L$ is $\mathbb{Z}$. If you're only killing direct−sum relations then you get $\mathbb{Z}^2$.
Jul 7, 2010 at 14:35 comment added t3suji This is pretty standard, but amazingly the standard online refs do not seem to have clear answers in obvious places. Answers are "No" and "L(A)" respectively. For fun, one can try computing the groups when $X$ is the projective line.
Jul 7, 2010 at 12:36 history edited babubba
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Jul 7, 2010 at 12:14 history asked babubba CC BY-SA 2.5