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It is well known that a localization $S^{-1}R$ of a commutative ring $R$ is flat as a $R$-module.
Rather, I am looking for extensions of rings which share certain properties of localizations, like flatness, while inverting nothing.

More precisely, I would like to find classes of examples, if such exist, of maps of monoids $f:M\to N$ verifying

  1. $N^{\times} = \{1\}$,

  2. $f$ is injective, but not surjective,

  3. "$M$ is cofinal in $N$" in the sense that the augmentation induces an isomorphism $\Bbb{Z}[N]\otimes_{\Bbb{Z}[M]}\Bbb{Z}[1]\cong\Bbb{Z}[1]$ of abelian groups (note that $\Bbb{Z}[1]\cong\Bbb{Z}$ as abelian groups),

for which $\Bbb{Z}[N]$ is flat as either a left or right $\Bbb{Z}[M]$-module.

The first two conditions listed above are simply to guarantee that $f$ does not invert any elements whatsoever, while not being an isomorphism.

For completeness, and in case it is easier, I am actually looking for examples of maps $f:M\to N$ for which $\text{Tor}^{\Bbb{Z}[M]}_\ast(\Bbb{Z}[N],\Bbb{Z}[N])$ is $\Bbb{Z}[N]$ concentrated in degree zero.
Remark: this condition is related to the forgetful functor from the positive derived category of $\Bbb{Z}[N]$-modules to the derived category of $\Bbb{Z}[M]$-modules being full and faithful.

Edit: I have removed the extraneous commutativity condition on the monoids.

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  • $\begingroup$ We're assuming the monoids are commutative? $\endgroup$
    – Todd Trimble
    Commented Feb 21, 2012 at 3:00
  • $\begingroup$ 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 $\endgroup$
    – Benjamin Steinberg
    Commented Feb 21, 2012 at 3:53
  • $\begingroup$ @BS: Thanks. I added a generation condition for the map $M\to N$. $\endgroup$
    – Ricardo Andrade
    Commented Feb 21, 2012 at 4:17
  • $\begingroup$ 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 $0$. $\endgroup$
    – Tom Goodwillie
    Commented Feb 21, 2012 at 5:46
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    $\begingroup$ Can you not do something silly like take a localization and then add an identity to each of the monoids? $\endgroup$
    – Benjamin Steinberg
    Commented Feb 21, 2012 at 14:57

1 Answer 1

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This is not really an answer, but it is too long for a comment. Suppose that $f\colon M\to N$ is a homomorphism of monoids (not necessarily commutative). Let's work with right modules.

First of all condition (3) can be phrased as follows: construct a graph with vertex set $N$ and an edge between $n$ and $nm$ for all $n\in N$ and $m\in M$. Then (3) happens iff this graph is connected.

Next there is a classical condition coming from category/topos theory that guarantees flatness of $\mathbb ZN$ over $\mathbb ZM$. In fact, it is equivalent to flatness of $N$ as a right $M$-set, but this is a stronger condition. Namely, consider the category of elements $\mathcal C$ of $N$ viewed as a presheaf over $M$. So the objects are elements of $N$ and there is an arrow $(n,m)\colon nm\to n$ associated to each $n\in N$ and $m\in M$. Composition is $(n,m)(nm,m')=(n,mm')$.

Fact. If the category $\mathcal C$ is filtered one has that $\mathbb ZN$ is flat over $\mathbb ZM$.

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