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I would like prove that, under the conditions described below, no non-trivial variety exists.

Let $\mathcal{V}$ be a variety of algebras e.g. rings, semigroups, semilattices.

Further suppose that:

  1. The empty algebra exists i.e. $\mathcal{V}$ has no constants.
  2. The dual condition also holds. That is, let $\mathtt{1} \in \mathcal{V}$ be the one element terminal algebra. Then $id_\mathtt{1}$ is the only homomorphism whose domain is $\mathtt{1}$.

Any help much appreciated.

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2 Answers 2

up vote 6 down vote accepted

It's false. Take the theory generated by unary operations $\zeta$ and $\iota$ and a binary operation $\cdot$, subject to the equations $$ \zeta(x) = \zeta(y), \qquad \iota(x) = \iota(y), \qquad \zeta(x) \cdot y = \zeta(x), \qquad \iota(x) \cdot y = y. $$ An example of an algebra is any ring, with $\zeta(x) = 0$, $\iota(x) = 1$, and the usual $\cdot$. So the theory is not trivial.

Obviously $\emptyset$ is an algebra. Now suppose we have a homomorphism from $1$ to an algebra $A$, picking out an element $a \in A$. The equation $\iota(x) = \zeta(x)$ holds in $1$, so $\iota(a) = \zeta(a)$ too. Hence for all $b \in A$, $$ b = \iota(a) \cdot b = \zeta(a) \cdot b = \zeta(a), $$ giving $A = \{\zeta(a)\} \cong 1$.

The trick is that although the theory has no actual constants, it has two "pseudo-constants", namely, $\zeta$ and $\iota$.

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Thanks very much for this example. It happens to extend a previous example I had on MO, i.e. with just the first two equations one finds that coproducts don't preserve monos. Actually I am working under further conditions e.g. that the variety is commutative, but this has been very helpful. –  Rob Myers Jan 16 at 18:24
    
You're welcome! –  Tom Leinster Jan 16 at 20:55

For varieties that only have unary function symbols, it is necessary to use constant functions as in Tom Leinster's answer.

$\mathbf{Proposition}$ Suppose that $V$ is a variety such that all the function symbols are unary and for all terms $t$, $V$ does not satisfy the identity $t(x)=t(y)$. Then there is some algebra $X\in V$ and multiple homomorphisms $1\rightarrow X$.

$\mathbf{Proof}$ Suppose that $V$ does not satisfy the identity $t(x)=t(y)$ for each term $t$. Then there does not exist terms $s,t$ that satisfy the identity $s(x)=t(y)$. Therefore every identity in $V$ is an identity of the form $s(x)=t(x)$. Let $G$ be the monoid of all terms in $V$ in one variable up to logical equivalence. Then $G$ is a monoid defined by $s(x)t(x)=s(t(x))$. We may therefore think of the variety $V$ as the collection of all actions from the monoid $G$ to some set $X$. On the other hand, let $G$ be the trivial action on a set $X$. In other words, $gx=x$ for all $g\in G$. Then every function $\phi:1\rightarrow G$ is a homomorphism.

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I am sure I don't understand this. How do we guarantee that there is a big enough set $X$ with such a trivial $G$-action? Gerhard "Thinking Of Single Unary Function" Paseman, 2014.01.13 –  Gerhard Paseman Jan 13 at 20:17
    
Any set $X$ has a trivial $G$-action simply by letting $gx=x$. –  Joseph Van Name Jan 13 at 21:20
    
OK. I guess I'm having trouble with the statement "We may therefore think of the variety V as the collection of all actions from the monoid G to some set X." I am having trouble thinking of V as such. I prefer to take a free algebra A, build X from A, and then show I will have 1 isomorphic to a subalgebra of X. Gerhard "Maybe I Need More Coffee" Paseman, 2014.01.13 –  Gerhard Paseman Jan 13 at 21:34

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