# A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids

In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following.

Lemma 5.2(a) Suppose that $\{M_i\}_{i\in I}$ is a collection of monoids with a common submonoid $W$ such that the monoid ring $\mathbb ZM_i$ is flat as a left $\mathbb ZW$-module for each $i\in I$. Let $M$ be the amalgamated free product (or pushout) of the $M_i$ over $W$. Then $\mathbb ZM$ is flat as a left $\mathbb ZM_j$-module for each $j\in I$.

His proof is one line, but I cannot understand it. I would very much love to use this result. I'm a bit nervous about it because the proof he sketches would seem to work in the category of rings, or more generally $k$-algebras over a commutative ring with unit $k$, but papers I saw of Cohn on amalgamated free products of rings seem to involve much more complicated colimits diagrams than Fiedorwicz is using. Moreover, Warren Dicks pointed out to me an example of an amalgamated free product of $k$-algebras, with $k$ a field, where the factors are flat over the subring but the amalgamated free product is not flat over the factors. He derives it from the paper https://arxiv.org/abs/math/0205034.

I would be happy for a proof I can understand in the case of two factors. I did try to contact the author, but he is retired and didn't respond. I now reproduce his proof.

Proof. This is immediate because $\mathbb ZM$ is the direct limit of

$$\mathbb ZM_j\otimes_{\mathbb ZW} \mathbb ZM_{i_1}\otimes_{\mathbb ZW} \mathbb ZM_{i_2}\otimes_{\mathbb ZW}\cdots \otimes_{\mathbb ZW}\mathbb ZM_{i_k}$$ $i_1,\ldots, i_k\in I$ as a left $\mathbb ZM_j$-module and hence is flat over $\mathbb ZM_j$.

What I don't understand is exactly what is the directed (or filtered) diagram that this direct limit is indexed over (i.e., what are the maps). Each of the terms is flat, so if $\mathbb ZM$ is such a direct limit, then I am fine with the proof. There is a map from each of these iterated tensors to $\mathbb ZM$ induced by multiplication (and the inclusions'' into the pushout) but I don't see how to organize this into a direct limit (=filtered colimit).

• I notice one of the tensor symbols is missing a ZW. Is this indeed a faithful reproduction, and does the omission have significance? Gerhard "Something Does Not Look Right" Paseman, 2018.04.11. – Gerhard Paseman Apr 11 '18 at 15:22
• Also (not that I understand it that well) M is described as a kind of pushout and ZM is a monoid ring over M. Is that not enough to pin down what ZM is? Gerhard "Not Feeling Very Decorative Presently" Paseman, 2018.04.11. – Gerhard Paseman Apr 11 '18 at 15:27
• @GerhardPaseman, I fixed the typo of the missing ZW. ZM is the ring pushout. This is why I think if the argument is correct it should apply to rings. Explicitly describing ring pushouts seems complicated. For example Mark Sapir showed amalgamated free products of finite monoids can have undecidable word problem so there is no nice normal form in general. Of course flatness is a strong condition. – Benjamin Steinberg Apr 11 '18 at 16:00

This appears to be false to me (unless maybe there are commutativity hypotheses), unless I've made a mistake below. (This is based on an example that I saw Andrew Ranicki give, a number of years ago, about non-exactness of Cohn localization.)

Consider the diagram of monoids $$\Bbb Z \leftarrow \Bbb N \rightarrow \Bbb N \ast \Bbb N$$ where the right-hand arrow is the inclusion of the first factor into the free product. On taking monoid algebras, this is a diagram of rings $$\Bbb Z[x^{\pm 1}] \leftarrow \Bbb Z[x] \rightarrow \Bbb Z\langle x,y\rangle.$$ The left factor is clearly flat over $\Bbb Z[x]$ and the right factor is free on the basis $$\{1\} \cup\{y\cdot w \mid w \text{ is a monomial in }x,y\}.$$

Thus both factors are flat left modules. The monoid algebra on the pushout $M$ is the ring $\Bbb Z\langle x^{\pm 1}, y\rangle$ by a universal property argument.

However, consider the exact sequence of left $\Bbb Z\langle x,y\rangle$-modules $$0 \to \Bbb Z\langle x,y\rangle \oplus \Bbb Z\langle x,y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x,y\rangle \to \Bbb Z \to 0,$$ where the first map sends $(a,b)$ to $ax+by$ and the second sends $x$ and $y$ to zero. Tensoring on the left with $\Bbb Z\langle x^{\pm 1}, y\rangle$ gives the sequence $$0 \to \Bbb Z\langle x^{\pm 1},y\rangle \oplus \Bbb Z\langle x^{\pm 1},y\rangle \xrightarrow{[x,y]} \Bbb Z\langle x^{\pm 1}, y\rangle \to 0 \to 0.$$ This is not left exact, because the first factor of the direct sum is taken isomorphically to the middle term by invertibility of $x$.

• Are you making $\mathbb Z$ a module over $\mathbb Z\langle x,y\rangle$ by having $x,y$ annihilate it? – Benjamin Steinberg Apr 12 '18 at 2:42
• It looks ok to me. Are there stronger hypotheses that would yield flatness besides freeness assumptions? – Benjamin Steinberg Apr 12 '18 at 2:52
• Yes, $x$ and $y$ annihilate $\Bbb Z$. I'm not sure what additional assumptions to give: the example I give here is of a localization, and if one can show that it is an Øre localization instead then you should get flatness. Ideally there would be a hypothesis that covers both the free case and the Øre localization case. – Tyler Lawson Apr 12 '18 at 3:00
• It is really quite shocking that the failure is so extreme. I never tried the case where one inclusion is a free factor and the other is a group completion of a cancellative commutative monoid. I would have thought such a nice case would work. – Benjamin Steinberg Apr 12 '18 at 13:08
• @BenjaminSteinberg Indeed, I find this example shocking not only because it is so simple but because it is so severe. In fact, this argument shows that if $f\colon \Bbb Z\langle x,y\rangle \to R \neq 0$ is a ring homomorphism and $x$ is a unit in $R$, then $f$ is not flat. – Tyler Lawson Apr 12 '18 at 13:37