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Feb 21, 2012 at 20:41 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 19:35 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 16:09 answer added Benjamin Steinberg timeline score: 2
Feb 21, 2012 at 14:57 comment added Benjamin Steinberg Can you not do something silly like take a localization and then add an identity to each of the monoids?
Feb 21, 2012 at 9:50 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 9:38 comment added Ricardo Andrade @Tom: I made a mistake. I hope I got it right now.
Feb 21, 2012 at 9:32 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 9:24 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 5:46 comment added Tom Goodwillie Either you did not write what you mean, or I am misunderstanding your notation. If $M$ is a proper submonoid of $N$ then $\mathbb Z[M]$ is a proper submodule of $\mathbb Z[N]$ and the quotient module is $0$.
Feb 21, 2012 at 4:17 comment added Ricardo Andrade @BS: Thanks. I added a generation condition for the map $M\to N$.
Feb 21, 2012 at 4:15 history edited Ricardo Andrade CC BY-SA 3.0
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Feb 21, 2012 at 3:53 comment added Benjamin Steinberg Let M be the trivial monoid and N be $\{0,1\}$ with multiplication. Then $\mathbb ZN\cong \mathbb Z\times \mathbb Z$ is free as a $\mathbb Z$-module. Actually $N$ can be any commutative monoid with trivial group of units. This is of course too trivial so you should clarify what you want
Feb 21, 2012 at 3:00 comment added Todd Trimble We're assuming the monoids are commutative?
Feb 21, 2012 at 2:34 history asked Ricardo Andrade CC BY-SA 3.0