# Coproducts of modules over an algebraic monad

Coproducts of modules over an algebraic monad $\Sigma$ are described in Section 4.16.14/15 in Durov's thesis. It is claimed there that for $\Sigma$-modules $M,N$, the set $M \coprod N$ generates $M \oplus N$ and that this implies that every element of $M \oplus N$ may be written as $t(x_1,...,x_n,y_1,...,y_n)$ for some $m,n \geq 0$, $t \in \Sigma(n)$ (surely he means $t \in \Sigma(n+m)$), $x_i \in M$ and $y_j \in N$.

Is this correct? What about taking the algebraic monad $\Sigma$ of groups (or even just monoids) so that $\mathrm{Mod}(\Sigma)$ is the category of groups. Here $G \oplus H$ is the free coproduct, it consists of the elements of the form $g_1 h_1 g_2 h_2 \dotsc g_n h_n$ with $g_i \in G$ and $h_i \in H$. We cannot write it as $g h$.

It seems to me that we can only find an expression as above if $\Sigma$ is commutative (5.1.1. loc.cit), so that in particular the example $\mathbb{Z}_{\infty}$ should work, but in the general case Durov wanted to write $t(x_1,...,x_n)$ with $x_i \in M$ or $x_i \in N$. Is this true?

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Could he just mean that there is a surjection $\Sigma(M\coprod N)\rightarrow M \oplus N$? Here $\Sigma$ is the free $\Sigma$-module functor. That seems reasonable. – Justin Noel Jan 27 '12 at 14:54
Yes this is true and actually is used for the construction of coproducts of modules in loc.cit. It is also equivalent to the fact that $M \coprod N$ generates $M \oplus N$. – Martin Brandenburg Jan 27 '12 at 15:08

## 1 Answer

I think the point is that an operation in an algebraic theory (even a noncommutative one) need not preserve the order of its inputs. There is a binary operation in the theory of groups which takes the input $(g,h)$ to the product $h g$. More generally, the symmetric group on $n$ letters acts on the set of $n$-ary operations, although in general the action may not have many fixed points. Thus, you can always find an operation which permutes all the $M$-inputs to the front and all the $N$-inputs to the rear.

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So you say that Durov is right here? If $\Sigma$ is an algebraic monad, then by functoriality $S_n$ acts on $\Sigma(n)$ for every $n$ and by construction for every $\Sigma$-module $X$ and $t \in \Sigma(n)$ we have $[\sigma t]_X(x_1,...,x_n) = [t]_X(x_{\sigma(1)},...,x_{\sigma(n)})$. – Martin Brandenburg Feb 2 '12 at 21:41
In particular if we have some term $[t]_X(x_1,...,x_n)$, where $x_i$ lie in $M$ or $N$, we may choose a permutation $\sigma$ such that $x_{\sigma(1)},...,x_{\sigma(k)}$ lie in $M$ and $x_{\sigma(k+1)},...,x_{\sigma(n)}$ lie in $N$, and write $[t]_X(x_1,...,x_n)=[\sigma^{-1} t]_X(x_{\sigma(1)},...,x_{\sigma(n)})$ as desired. Have I understood your answer correctly? – Martin Brandenburg Feb 2 '12 at 21:45
Yes, that's what I'm saying. – Mike Shulman Feb 3 '12 at 1:28
Alright, thank you very much for your answer. – Martin Brandenburg Feb 3 '12 at 16:24