# Is it true that every quasi-variety exends to a variety? [closed]

I am wondering if the following is true:

If $$Q$$ is a quasivariety in an algebraic signature $$S$$, must there be a variety $$V$$ in an extended signature $$S′⊇S$$ such that $$Q$$ is the class of algebras isomorphic to $$S$$-reducts of algebras in $$V$$?

• What makes you think so? Did you find such a conjecture somewhere? Do you have a few examples in which that happens? Whether this is true or not, the question is not a good one without some context like the one I mentioned. Otherwise every one could come here and make arbitrary questions, some with merit, other not so much. If you have some good reason to think this might be true, please edit the post to include that, or it may happen that the question will be closed. Sep 28 at 8:34
• Thank you for your comment. The question arose in my research about relation algebra. I will try to explain how, but I am afraid it will not be easy. Sep 28 at 14:30

I will rephrase the question slightly, then give an affirmative answer.

Rephrased version: If $$\mathcal Q$$ is a quasivariety in an algebraic signature $$S$$, must there be a variety $$\mathcal V$$ in an extended signature $$S'\supseteq S$$ such that $$\mathcal Q$$ is the class of algebras isomorphic to $$S$$-reducts of algebras in $$\mathcal V$$?

Now for the affirmative answer to this version of the question.

The discriminator operation on a set $$X$$ is the function $$d\colon X^3\to X$$ defined by $$d(x,y,z) = z$$ if $$x=y$$ and $$d(x,y,z)=x$$ if $$x\neq y$$. If $$S$$ is an algebraic signature and $$\mathcal K$$ is a class of algebras in a signature $$S$$, write $$S'$$ for the signature obtained by adding one ternary operation $$d(x,y,z)$$ to $$S$$, and write $$\mathcal K^d$$ to be the class obtained from $$\mathcal K$$ by interpreting $$d$$ as the discriminator operation on each member of $$\mathcal K$$.

Let $$\mathcal Q$$ be any $$S$$-quasivariety. Take $$\mathcal K$$ to be $$\mathcal Q$$, and take $$\mathcal V$$ to be the $$S'$$-variety generated by $$\mathcal K^d$$. It turns out that $$\mathcal Q$$ is the class of algebras isomorphic to $$S$$-reducts of algebras in $$\mathcal V$$. (See Theorem 9.4 of the book by Burris and Sankappanavar.)

Edit. Let me add more detail to this answer. Let $$\mathcal K$$ be any universal class. The class $$\mathcal K^d$$ (union the singleton $$S$$-algebra, if this class does not already contain the singleton $$S$$-algebra) is the class of algebras isomorphic to directly indecomposables in the variety $${\mathcal V}:=\textrm{HSP}({\mathcal K}^d)$$, and every algebra in $$\mathcal V$$ is isomorphic to a Boolean product of members of $${\mathcal K}^d$$. (This is easy to read off of Theorem 9.4, part (d).) But now, if $$\mathcal K = \mathcal Q$$ is a quasivariety, then it is a universal class containing a singleton algebra, so the variety $$\mathcal V$$ generated by $${\mathcal K}^d={\mathcal Q}^d$$ is obtained from $${\mathcal Q}^d$$ by closing under isomorphisms and Boolean products, i.e. $${\mathcal V}=I\Gamma^a({\mathcal Q}^d)$$. The reducts of algebras in $$\mathcal V$$ to the signature $$S$$ will be the algebras in $$I\Gamma^a({\mathcal Q}) = {\mathcal Q}$$. Here I am using that quasivarieties are closed under isomorphisms and Boolean products. (Every Boolean product of some factors algebras is a subalgebra of a product of the factor algebras, and quasivarieties are closed under subalgebras and products.)

• Thank you very much, this answers my question. Your formulation is better, I will edit my post and use yours. Sep 28 at 14:28
• Theorem 9.4 of chapter IV. By the way, @A.D., given that you're happy with this answer, you might also want to accept it and perhaps upvote it (now that you can). See What should I do when someone answers my question?. Sep 28 at 14:42
• @Keith Kearnes, I checked Theorem 9.4 and none of the four items seem to correpond to your claim. Could you please be more precise about the item your are using? Sep 28 at 14:54