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By a commutative variety $\mathcal{V}$ I mean a classical variety of algebras for some $(\Sigma,E)$, such that each pair of operations in $\Sigma$ commutes.

Equivalently (i) every interpretation of every operation defines an algebra homomorphism, or (ii) the hom-sets have pointwise algebraic structure, or (iii) $\mathcal{V}$ forms a symmetric closed monoidal category with its tensor product (the latter being definable as a bi-functor in any variety).

By a cogenerator I mean some $K \in \mathcal{V}$ such that any two distinct morphisms $\alpha,\beta : A \to B$ have a respective `predicate' $h : B \to K$ satisfying $h \circ \alpha \neq h \circ \beta$.

I have two questions:

  1. Do commutative varieties necessarily have a cogenerator?
  2. Do locally finite commutative varieties necessarily have a finite cogenerator?

I include some positive examples.

  1. If $(\Sigma,E)$ at least contains a binary operation and a respective unit, then $\mathcal{V}$ is essentially the modules for some commutative semiring. Examples are abelian groups with cogenerator $\mathbb{Q}/\mathbb{Z}$, vector spaces with cogenerator $k$, join-semilattices with zero with cogenerator $2$.

  2. Other examples with a two element cogenerator include sets, pointed sets, semilattices without zero, and the variety defined by a single operation and equation $u(x) = u(y)$ (pointed sets with an additional initial object).

  3. Actions of a commutative monoid have a cogenerator, since they form a topos $[M,\mathsf{Set}]$. If the monoid is finite, there is a finite cogenerator.


Edit

Let me add that:

  1. I do not know if commutative monoids have a cogenerator, or whether there is one which works for the finitely-generated = finitely-presentable $\mathbb{N}$-modules.

  2. The commutative inverse monoids define a commutative variety. They extend commutative monoids with a single involutive unary operation, which (i) preserves the monoid structure, (ii) satisfies $x = x \cdot u(x) \cdot x$. They have a cogenerator $2 \times \mathbb{Q}/\mathbb{Z}$ where $2$ is the two-chain.

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Consider a commutative variety C of algebras with three unary operations. You can choose a sufficiently large 3 generated commutative monoid M and require all these algebras in C to be M-algebras. What does a cogenerator of C look like? –  The Masked Avenger Feb 9 '14 at 3:44
1  
@TheMaskedAvenger In $M$-sets you can do without commutativity; you can take e. g. $\Omega^M$ where $\Omega$ is the subobject classifier. This $\Omega^M$ is (isomorphic to) the set of all $M$-subsets of $M\times M$ (with componentwise $M$-action); $M$ acts on these subsets via $$ mS:=\{(m_1,m_2)|(m_1m,m_2)\in S\}. $$ –  მამუკა ჯიბლაძე Feb 9 '14 at 9:25
    
A cogenerator in this context will be a lot like a dualising object: indeed, given an object $T$, the canonical homomorphism $A \to \mathrm{Hom}(\mathrm{Hom}(A, T), T)$ (obtained by transposing the evaluation $A \otimes \mathrm{Hom}(A, T) \to T$) is injective if and only if $T$ is a cogenerator. –  Zhen Lin Feb 9 '14 at 18:59
    
@Zhen Lin: Yes, that is in fact the intention. This also suggests that it's going to be a problem, because I am unaware of a dual description for commutative monoids, aside from the free ones. However it would be nice to have at least one counterexample, perhaps at the level of modules for a finite commutative semiring (hopefully easier). Thanks for your remark. –  Rob Myers Feb 9 '14 at 19:23

1 Answer 1

up vote 7 down vote accepted

The answer is no.

Let $A$ be the algebra with universe $\{0,1\}$ and fundamental operations $f(x,y,z)=x+y+z \pmod{2}$ and $g(x)=x+1\pmod{2}$. Then $f$ and $g$ commute with each other and with themselves, so the variety generated by $A$ is commutative. This variety has a weird property: on every $B\in \mathcal V(A)$ the operation $g$ interprets either as a fixed point free involution (type 1) or as the identity function (type 2). Moreover, every group of exponent 2 can be modified slightly to make it an algebra of type i in this variety for i = 1 OR 2.

Now, suppose that $K$ is a cogenerator for $\mathcal V(A)$. Necessarily $|K|>1$.

$K$ is not of type 1.

Assume otherwise. Let $B\in\mathcal V(A)$ be the 2-element type 2 algebra. There are maps $\alpha,\beta\colon B\to B$ where $\alpha = id$ and $\beta$ is a constant function. These maps cannot be separated by a map $h\colon B\to K$, since there is no homomorphism from a type 2 algebra to a type 1 algebra.

$K$ is not of type 2.

Assume otherwise. If $B\in\mathcal V(A)$ is of type 1, then $id, g\colon B\to B$ cannot be separated by any map $h\colon B\to K$ since you cannot separate elements of the same $g$-orbit of a type 1 algebra by a homomorphism into a type 2 algebra.

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