Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ for all $m\in M$. For example the zero element of a ring is absorbing for multiplication. It is well known that the trivial $\mathbb ZM$-module $\mathbb Z$ is projective in this case. Indeed, $z$ is a central idempotent and $\mathbb ZMz\cong \mathbb Z$. Therefore, $\mathbb ZM=\mathbb ZMz\oplus \mathbb ZM(1-z)$ and hence $\mathbb Z$ is projective. Moreover, the above direct sum is a ring direct product decomposition.

Question 1. Can the trivial module $\mathbb Z$ be stably free?

Recall that a finitely generated projective module $P$ is stably free if $P\oplus F\cong F'$ for some finitely generated free modules $F,F'$. Note that in our setting if $\mathbb Z\oplus F\cong F'$ and $r$ is the rank of $F$ and $r'$ is the rank of $F'$, then

$$\mathbb Z^{r'}\cong zF'\cong \mathbb Z\oplus zF\cong \mathbb Z^{r+1}$$ and so $r'=r+1$ if this happens.

Also, $R=\mathbb ZM(1-z)$ is a ring, known as the contracted monoid ring of $M$. It is obtained by identifying the zero of $M$ with the zero of $\mathbb ZM$ and it has the property that $R$-modules are precisely those $M$-modules annihilated by the zero of $M$.

Then note that

$$R^{r+1}\cong (1-z)F'\cong (1-z)F\cong R^r$$

and hence if $\mathbb Z$ is stably free, then $R$ does not have the Invariant Basis Number Property, which says that finite rank free modules "know" their rank.

Question 1'. Can a contracted monoid ring fail to have the Invariant Basis Number Property?

If either $M$ is finite or $M\setminus \{z\}$ is a submonoid, then the contracted monoid ring has the Invariant Basis Number Property.

It has been suggested to me by both George Bergman and Jason Bell (independent private communications) that the polycyclic inverse monoid (or Cuntz monoid)

$$P_2=\langle x,x^*,y,y^*,z\mid xx^*=1,yy^*=1, xy^*=z=yx^*, z=0\rangle,$$

where $z=0$ is shorthand for $z$ is the absorbing element, may be relevant here for Question 1'.

The point is that if $R$ is the contracted monoid ring of $P_2$, then $R^2$ is isomorphic to a direct summand in $R$. One notes that $x^*x,y^*y$ are orthogonal idempotents and so $e=x^*x+y^*y$ is an idempotent and $(a,b)\mapsto ax+by$ is a module isomorphism $R^2\to Re$ with inverse $c\mapsto (cx^*,cy^*)$. Thus $Re$ is stably isomorphic to $R$. Since $e\neq 1$, this doesn't contradict the invariant basis number property, but it hints that the $K$-theory of this ring may not be so nice.

Note that $R/R(1-e)R=R/R(1-(x^*x+y^*y))R$ is a Leavitt algebra which fails to have the invariant basis number property because the free modules of rank $1$ and $2$ are isomorphic. The extension of scalars of this algebra over any field is simple and this ring is known to contain interesting groups inside of its group of units like the simple group Thompson's group $V$.

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    $\begingroup$ The contracted monoid ring of $P_2$ doesn't have the rank condition, but it actually does have the IBN property. See Lam's "Lectures on Modules and Rings" page 11, and exercise 1.5, for a discussion of this example and proof (originally due to Cohn). I'll think about your general questions some more. $\endgroup$ – Pace Nielsen Aug 26 '16 at 14:46
  • $\begingroup$ @PaceNielsen, Thanks! I'll have to get a hold of Lam. This gives me hope the more general question might work. But what happens if I take $P_{\infty}$ meaning I take countably infinitely many unilateral shifts xx*=1 with orthogonal x*x's. Does it still have IBN? $\endgroup$ – Benjamin Steinberg Aug 26 '16 at 14:57
  • $\begingroup$ The case $P_{\infty}$ should be the direct limit of the finite cases, which are dealt with in Cohn's paper. See doi:10.1016/0040-9383(66)90006-1 and in particular, Theorem 2.3 and Theorem 6.1 (in connection with the material from Lam's book, if necessary). $\endgroup$ – Pace Nielsen Aug 26 '16 at 16:30

I've had a little more time to think about your questions, and I believe the following is true: Contracted monoid rings have IBN. This might appear implicitly in Cohn's paper, but I couldn't find it on a causual reading.

We will need the following lemma, which appears (with proof) as Exercise 1.5 in "Exercises in Modules and Rings":

Suppose a ring $R$ admits an additive group homomorphism $T$ into an abelian group $(A,+)$ such that $$ (\ast)\quad T(cd)=T(dc)\ \text{ for all }\ c,d\in R. $$ If $T(1)$ has infinite addivite order in $A$, then $R$ must have IBN.

Now, we will proceed to prove the main claim, and assume that $R$ is a contracted monoid ring. Since we will be concerned with constructing an additive group homomorphism from $R$, we are concerned with the additive group structure of $R$, which is given as a free $\mathbb{Z}$-module with basis $M':=M\setminus\{z\}$.

Define an equivalence relation on $M$ by first saying $a,b\in M$ are "basically related" if there exist $x,y\in M$ with $a=xy$ and $b=yx$, and then taking the transitive closure of this relation.

Define $T:R\to \mathbb{Z}$ by first defining it on $M'$ and then extending $\mathbb{Z}$-linearly. We must have $T$ send $1_M$, and anything related to it (by a finite sequence of basically related elements), to $1_{\mathbb{Z}}$. We send everything else in $M'$ to $0$.

This is a well-defined additive group homomorphism. All that remains is to check that it satisfies $(\ast)$. Let $c,d\in R$. We can then write $$ c=\sum_{m\in M'}a_m\cdot m,\qquad d=\sum_{n\in M'}b_n\cdot n $$ for some unique $a_m,b_n\in \mathbb{Z}$ (for each $m,n\in M'$), all but finitely many of which are zero. Since $R$ is a contracted monoid ring, we have either $mn\in M'$ or $mn=z$, and similarly for $nm$. Thus $$ T(cd)=\sum_{m,n\in M'}a_mb_nT(mn),\qquad T(dc)=\sum_{m,n\in M'}a_mb_nT(nm), $$ where we define $T(z)=0$. Thus, these sums will be equal unless $z\in M$ is in the equivalence class of $1_M\in M$.

Suppose, by way of contradiction, that this were the case. Then we would have $$ 1_M=x_1 y_1, y_1x_1=x_2y_2,\cdots, y_n x_n=z. $$ Hence $$ 1_M = 1_M^{n+1} = (x_1y_1)^{n+1} = x_1 (y_1x_1)^{n}y_1 = x_1(x_2y_2)^{n}y_1 = x_1 x_2(y_2x_2)^{n-1}y_2y_1 = \cdots $$ $$ = x_1 x_2 \cdots x_n (y_n x_n) y_n y_{n-1} \cdots y_1 = z $$ contradicting the fact that $M$ is a nontrivial monoid.

  • $\begingroup$ Many thanks. I didn't know this trace trick. Can I reproduce this argument credited to you and MO in something I am writing? $\endgroup$ – Benjamin Steinberg Aug 26 '16 at 20:17
  • $\begingroup$ @BenjaminSteinberg Yes, you are welcome to reproduce this argument. I thought it was quite nice! (Also, please send what you are writing to me by email, when it is ready for readers.) $\endgroup$ – Pace Nielsen Aug 28 '16 at 3:44
  • $\begingroup$ this is a beautiful argument. I had no idea the Hattori-Stallings trace was applicable here. In fact the trivial module is projective if the monoid has a right zero. Using that contracted rings have IBN I can prove if there are finitely many right zeroes the trivial module is not stably free. Now I breed a new argument for the case of infinitely many right zeroes. $\endgroup$ – Benjamin Steinberg Aug 28 '16 at 10:55

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