3
$\begingroup$

As is well known, the definition of an monoid can be generalised to the notion of a monoid $A$ in a monoidal category $C$ (see the n-lab entry here). What I would like to know is if the notion of generating subset of a monoid can be generalised to this context - precisely, by generating subset I mean a subset $S$ such that the smallest ideal of $A$ containing $S$ is $A$.

I would naively guess that one would need the existence of infinite sums in $C$, define a generating set to be a sub-object $X$ of $A$, such that there is an isomorphism $$ A \simeq \bigoplus_{i \in I} B_i \otimes S \otimes B'_i, $$ for some objects $B_i,B'_i \in C$.

Moreover, is this definition invariant under equivalence of categories, i.e. if $A$ is generated by a subject object $S$ in $C$, which is equivalent to $D$ by a functor $F$, then is $F(A)$ generated by $F(S)$?

Finally, what is a good reference for all this?

$\endgroup$
11
  • 3
    $\begingroup$ Sorry, I don't know what you have in mind in the first sentence. This is an algebra in the sense of rings and algebras? I know how to define ring and algebra objects in categories with finite cartesian products, but what is the well-known definition you have in mind here? $\endgroup$ – Todd Trimble Jun 14 '14 at 20:53
  • 1
    $\begingroup$ Perhaps by "algebra" the OP means monoids, but in that case I don't know the relevance of the finite coproducts. $\endgroup$ – Qiaochu Yuan Jun 15 '14 at 0:06
  • 4
    $\begingroup$ Anyway, a generating set should be a subobject $X$ such that the induced map $F(X) \to A$ is an epimorphism, or maybe a regular epimorphism, where $F(X)$ is the free algebra on $X$, whatever that happens to mean to you and provided that is well-defined. $\endgroup$ – Qiaochu Yuan Jun 15 '14 at 2:47
  • $\begingroup$ I guess the op means an algebra object also called monoid, which is an object $A$ and a morphism $m:A\otimes A \to A$, the multiplication, and a morphism $e:1\to A$ with $1$, the unit. E.g. take the category of $\mathbb Z_n$ graded vector spaces, then the $\bigoplus_{i=0}^{n-1} [i]$ has the structure of an algebra object and I would say it is "generated" by the subobject $[1]$. $\endgroup$ – Marcel Bischoff Jun 15 '14 at 16:22
  • $\begingroup$ I forgot: "$m$ and $e$ fulfilling the obvious identities" $\endgroup$ – Marcel Bischoff Jun 15 '14 at 16:29
5
$\begingroup$

Here is a proposal that avoids requiring a notion of free monoid. Let $M$ be a monoid in some monoidal category $C$ and let $s : S \to M$ be a morphism (there is really no reason to restrict our attention to subobjects / monomorphisms).

Definition #1: $S$ weakly generates $M$ if, for any parallel pair of morphisms of monoids $f, g : M \to N$, for $N$ a monoid, $f \circ s = g \circ s$ implies $f = g$.

That is, $s$ is right cancellable with respect to morphisms into other monoids. If you have a notion of free monoid, by which I mean a left adjoint $F$ to the forgetful functor from monoids in $C$ to $C$, then this is equivalent to the induced map $F(S) \to M$ being an epimorphism of monoids.

This definition has the substantial drawback that it already fails to reproduce the usual notion of generation for rings! The problem is that an epimorphism of rings need not be surjective. Here is a second proposal which at least reproduces the usual notion of generation for rings (more precisely, it reproduces the notion of an additive subgroup of a ring generating it under multiplication), but now I need to assume there is a left adjoint to the forgetful functor, and probably I should also assume that coequalizers exist or else the proposal will probably behave strangely.

Definition #2: $S$ generates $M$ if the induced morphism $F(S) \to M$ is a regular epimorphism; that is, if there is some other object $R$ and a pair of morphisms $f, g : R \to F(S)$ such that the induced morphism $F(S) \to M$ is their coequalizer.

This definition has the benefit that the object $R$ can be regarded as specifying relations that the generators satisfy. It also reproduces the usual notion of generation for rings. (In the case of rings it doesn't matter whether you require the morphism to be a regular epimorphism of rings or just a regular epimorphism of abelian groups; I am not sure whether the distinction matters in general.)

A more canonical definition that doesn't rely on assuming that some auxiliary object exists is given by the notion of effective epimorphism. Fortunately the distinction vanishes in a category with pullbacks and so is not too important in practice.

Both definitions are manifestly categorical and so manifestly invariant under monoidal equivalence of monoidal categories (which I presume is what you meant to ask).

Finally, it is probably worth remarking that if $C$ has countable coproducts and the monoidal structure distributes over countable coproducts in both variables then the free monoid functor $F$ can be constructed very explicitly as

$$\displaystyle S \mapsto F(S) = \bigsqcup_{n \ge 0} S^{\otimes n}.$$

$\endgroup$
1
  • 2
    $\begingroup$ Thanks; this is close to the answer I would have given myself. A really good situation to be in is where the underlying category of the monoidal category is regular, and the tensor product preserves colimits in each argument (or countable coproducts and coequalizers if you prefer). Then the category of monoids is also regular, and various notions of epimorphism of monoids (regular, strong, extremal) coincide. $\endgroup$ – Todd Trimble Jun 16 '14 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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