Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.