# Is the following a sufficient condition for being a primal algebra?

I have a question regarding universal algebra and, in particular, primal algebras:

Suppose that A is a finite simple algebra with no proper subalgebra, no automorphism except the identity map, with a constant c for each of its elements, and that moreover the variety generated by A is c-regular for every element c of A and has equationally definable principal congruences (EDPC). Under these assumptions, is A primal? (Maybe it can be useful to recall that EDPC implies congruence distributivity and point-regularity implies n-permutability for some natural n).

A finite algebra is primal if every finitary function on its universe of arity bigger or equal to 1 is represented by a term of the language. Foster and Pixley characterized primal algebras as those finite algebras that are simple, have no proper subalgebra and no automorphism except the identity map and, moreover, generate a congruence distributive and permutable variety. Therefore what I am asking in the above questions is if there is a way to transform the n-permutability into the usual 2-permutability under the assumption of the question.

Many thanks for any help or comments!

• You might instead see if you can use the conditions to build a generalized Sheffer stroke function, or review the Foster-Pixley argument to see how they get congruence permutability, and then try a variation on that. (They may do it using the stroke function, in which case you have to see what you can say about the clone of A first). – The Masked Avenger May 28 '14 at 18:12

The answer is no. Here is a counterexample that Agnes Szendrei and I found.

Consider the reflexive, symmetric, binary relation $T = \{0,m\}^2\cup \{m,1\}^2$ on the set $S=\{0,m,1\}$. The counterexample is the algebra $A$ whose universe is $S$ and whose operations are all operations that preserve $T$. (I.e., all operations $f$ on $S$ where $T$ is a subalgebra of $\langle S,f\rangle^2$.) This algebra is clearly not primal, since $T$ is a compatible binary relation of $A$ that is different from the total binary relation or the equality relation, and a primal algebra has no such compatible relation.

The lattice operations for the order $0<m<1$ preserve $T$, so $A$ is an expansion of the 3-element lattice. It is easy to check that $A$ is finite, simple, has no proper subalgebra, has no automorphism except the identity map, and has constant operations naming each of its elements. It follows that the variety $\mathcal V$ generated by $A$ is semisimple and congruence distributive. It then follows from Theorems 2.12 and 2.21 of

Ervin Fried and Emil W. Kiss, Connections between congruence-lattices and polynomial properties, Algebra Universalis, 17 (1983) 227-262.

that $\mathcal V$ has EDPC.

What remains is to prove that $A$ is $c$-regular for $c = 0, m, 1$. I argue by contradiction. I only explain $0$-regularity, but the other two cases are similar. Assume that $B\in \mathcal V$ has congruences $\alpha\neq\beta$ such that $0/\alpha=0/\beta$. Replacing one of the congruences with their intersection and renaming we may assume that $\alpha < \beta$. Factoring by $\alpha$ we may assume it is zero. Now we have a congruence $\beta\neq 0$ with $0/\beta = \{0\}$. The properties that hold for $B$ and $\beta$ can be realized in a finite subalgebra of $B$, so assume that $B$ is finite. Now we can assume that $|B|$ is minimal and that $\beta$ is minimal in $Con(B)$ among algebras in $\mathcal V$ realizing this data. To summarize and refine: we may assume that $B\leq A^n$ is an irredundant subdirect power of $A$ and that $\beta$ is an atom in $Con(B)$. The atoms are the restrictions of the kernels of the projections onto all but one of the coordinates, so we may assume that $\beta$ relates tuples that are equal in all but the first coordinate.

Now we use the fact that $A$ has a majority operation (from the lattice structure) and that $T$ is the only proper nonequality compatible relation of $A$ (check). This tells us what the possible finite, irredundant subdirect powers of $A$ are. The algebra $B\leq A^n$ is definable by a simple graph (say $G$) on the index set $\{0,1,\ldots,n-1\}$. Each edge $\{i,j\}$ of $G$ imposes a restriction on $ij$-th projection of $B$: the pair $(b_i,b_j)$ must lie in $T$. Thus, $B$ is the set of all tuples $(b_0,\ldots,b_{n-1})$ such that $(b_i,b_j)\in T$ whenever $\{i,j\}$ is an edge in $G$. (If what I have written sounds confusing, let me summarize the key point as follows: the Baker Pixley Theorem implies that any irredundant subdirect subalgebra of $A^n$ contains $\{0,m\}^n\cup \{m,1\}^n$.)

Now it is easy to see from the above description that the tuple $t=(m,0,0,\ldots,0)$ belongs to $B$ no matter what $B$ is. Moreover, this tuple differs from $0=(0,0,0,\ldots,0)$ in the first coordinate only. I.e., the $\beta$-class of $0$ is not $\{0\}$ after all, since it contains $t\neq 0$.