MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.

share|cite|improve this question
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
@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
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.

share|cite|improve this answer

Your Answer


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.