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Sep 1, 2013 at 21:31 vote accept CommunityBot
Sep 1, 2013 at 20:36 comment added Noam D. Elkies Sorry this seems to be confusing you: for $r=3$, $M$ EQUALS ${\bf Z}^3$, just with a different choice of coordinates that makes it easy to see what $S$ is doing.
Sep 1, 2013 at 20:06 comment added user35375 For $r=3$, doesn't it hold that ${\bf Z}^3/M\simeq {\bf Z}_2 \oplus {\bf Z}_2$, which isn't free?
Sep 1, 2013 at 19:18 comment added Noam D. Elkies The construction assumes that ${\bf Z}^n/M$ is free. You could even take $n=r$ and then change coordinates to identify ${\bf Z}^r$ with $M$, e.g. by using $2e_1,\ldots,2e_{r-1}$ and $\sum_{i=1}^r e_i$ as the new basis vectors.
Sep 1, 2013 at 17:16 comment added user35375 But in this case, ${\bf Z}^n/M$ is not free?
Sep 1, 2013 at 16:59 vote accept CommunityBot
Sep 1, 2013 at 17:16
Sep 1, 2013 at 16:52 history answered Noam D. Elkies CC BY-SA 3.0